PH 107 : Quantum Physics Solution Booklet - CSE-IITB

PH 107 : Quantum PhysicsSolution Booklet

Rwitaban Goswami

1

PH 107 : Quantum Physics Contents Rwitaban Goswami

Contents

1 Black Body Radiation 3

2 Compton Scattering 8

3 Specific Heat 15

4 Phase and Group Velocity 17

5 Wave packet, Fourier Theory, HUP 22

6 Wave function and Operators 36

7 Free Particle 42

8 Infinite Potential Box 46

9 Finite Potential Well 56

10 Step Potential 66

11 Harmonic Oscillator and Degenerate states 75

12 Statistical Mechanics and Density of States 8012.1 Statistical Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8012.2 Density of States and Fermi Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

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PH 107 : Quantum Physics Tutorial Sheet: 1 Rwitaban Goswami

1 Black Body Radiation

1.1. Rayleigh-Jean’s law gives the radiant energy density (energy per unit volume) as

uν(ν) dν =8πν2

c3kT dν

Show that in terms of wavelength interval, Rayleigh-Jean’s law can be expressed as

uλ(λ) dλ =8π

λ4kT dλ

Solution:Using physical intuition:Rayleigh-Jean’s law physically means that there is 8πν2

c3kT dν amount of energy density

between frequencies ν and ν+dν. Since ν = cλ, this means that there is 8πc2

c3λ2 kT dν = 8πcλ2kT dν

worth of energy density between frequencies ν and ν + dν.

But we want the amount of energy density between λ and λ+ dλ. To find that we need therelation between dλ and dν. We know that

ν =c

λ(1.1.1)

=⇒ dν = − c

λ2dλ (1.1.2)

So the amount of energy density between ν and ν + dν should be scaled by cλ2 to get the

amount between λ and λ + dλ (ignore the sign, it simply means that as λ decreases ν in-creases, so it’s actually the energy density between λ and λ− dλ, which makes no differencereally)

So the amount becomes cλ2

8πcλ2kT dλ = 8π

λ4 kT dλTherefore

uλ(λ) dλ =8π

λ4kT dλ (1.1.3)

A more mathematically rigorous method:Let ν1λ1 = ν2λ2 = c, where WLOG ν1 < ν2, λ2 < λ1

Since the range of frequencies (ν1, ν2) corresponds to the same modes as (λ2, λ1), they shouldcontain they same amount of total energy

=⇒ˆ ν2

ν1

uν(ν) dν =

ˆ λ1

λ2

uλ(λ) dλ (1.1.4)

(1.1.5)

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Notice the opposite limits for uλ and uν , this is because λ and ν are inversely related

Subsitute ν = cλ

in uν

=⇒ −ˆ λ2

λ1

uν(c

λ)c

λ2dλ =

ˆ λ1

λ2

uλ(λ) dλ (1.1.6)

=⇒ˆ λ1

λ2

uν(c

λ)c

λ2dλ =

ˆ λ1

λ2

uλ(λ) dλ (1.1.7)

Since the integrals are equal for all values of λ1 and λ2, this must mean the integrands mustalso be equal

=⇒ uλ(λ) =c

λ2uν(

c

λ) (1.1.8)

This is exactly the same result as before, we scale by cλ2 , and replace all ν by c

λ. Notice that

no negative sign appears here.

1.2. Use Planck’s equation and show that the expression for radiant intensity (in terms of λ) is givenby

I(λ) dλ =2πhc2

λ5

1

ehcλkT − 1

dλ

Solution: Planck’s equation states that

I(ν, T )dν =c

4

8πhν3

c3

1

ehνkT − 1

dν (1.2.1)

=2πhν3

c2

1

ehνkT − 1

dν (1.2.2)

(1.2.3)

Where I(ν, T ) is the radiant intensity in terms of νUsing similar logic as above, we scale this by c

λ2 to obtain

I(λ, T ) dλ =2πhc2

λ5

1

ehcλkT − 1

dλ (1.2.4)

1.3. According to Planck, the spectral energy density u(λ) of a black body maintained at temperatureT is given by

u(λ, T ) dλ =8πhc

λ5

1

ehc

λkBT − 1dλ

where λ denotes the wavelength of radiation emitted by the black body.

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(a) Find an expression for λmax at which u(λ, T ) attains its maximum value (at a fixed tem-perature T ). λmax should be in terms of T and fundamental constants h, c and kB.

Solution: We simply differentiate u(λ, T ) by λ and set to zero

∂u(λ, T )

∂λ

∣∣∣∣λ=λmax

= 0 (1.3.1)

=⇒ −58πhc

λ6max

1

ehc

λmaxkBT − 1+

8πhc

λ5max

hcehc

λmaxkBT

λ2maxkBT (e

hcλmaxkBT − 1)2

= 0 (1.3.2)

=⇒ 5λmaxT (ehc

λmaxkBT − 1) =hc

kBe

hcλmaxkBT (1.3.3)

We can further see that if we put x = hcλmaxkB

T

xex

ex − 1− 5 = 0 (1.3.4)

This equation can be solved numerically (you can use Desmos or Wolfram Alpha) toget x = 4.9651, so we get λmaxT = hc

4.9651kB= 0.0029 m K

This is the Wein’s law that you studied in JEE!Note that this constant is going to be different if you calculate the maximum of u(ν, T ),because the function u(ν, T ) 6= u(λ, T ), they represent different things physically (onerepresents energy density between ν and ν+dν, the other between λ and λ+dλ. Sincedλ and dν aren’t related proportionally, so the two functions are also different but havesimilar shape. Hence the constant will differ)

(b) Expressing λmax as αT

, obtain an expression for umax(T ) in terms of α, T and the funda-mental constants

Solution: We know from above that α = 0.0029 m KWe can simply put λmax = α

Tin u(λ, T ) to get umax(T )

umax(T ) = u(λmax, T ) (1.3.5)

= u(αT, T)

(1.3.6)

=8πhcT 5

α5

1

ehcαkB − 1

(1.3.7)

1.4. The earth rotates in a circular orbit about the sun. The radius of the orbit is 140×106 km. Theradius of the earth is 6000 km, and the radius of the sun is 700, 000 km. The surface temperatureof the sun is 6000 K. Assuming that the sun and the earth are perfect black bodies, calculatethe equilibrium temperature of the earth.

Solution: At equilibrium, energy absorbed by Earth (due to sun’s blackbody emission)should equal the energy emitted by Earth (due to its own blackbody emission)

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Now,

Eabsorbed = Eemitted by sunπr2

4πR2(1.4.1)

Where r = radius of Earth, R = radius of Earth’s orbitNow why do we divide by 4πR2? Because not all of Sun’s emission falls on Earth, so we wantto find the intensity of Sun’s emission falling on Earth, or Energy per unit area. The to-tal energy is spread out over a sphere surface of area 4πR2, so we divide by it to get intensity.

Why do we multiply by πr2? We need to multiply the intensity by the area seen by the sun.Note that this is πr2 (the area the disk of Earth that the Sun sees), not 2πr2 (the curvedarea of earth that receives the sunlight). This is made clear from the following diagram(notice that the sphere surface over which the energy is spread out is almost a plane at theEarth’s scale)

πr2

From Sun

Now, Eemitted by sun = σ4πr2sunT

4sun

=⇒ Eabsorbed = σπr2sunT

4sun

r2

R2

Now, Eemitted = σ4πr2T 4

But Eabsorbed = Eemitted, so σ4πr2T 4 = σπr2sunT

4sun

r2

R2

=⇒ T 4 = T 4sun

r2sun

4R2

Putting in values for Tsun, rsun, RT 4 = 60004 72×1010

4×1402×1012

=⇒ T = 300K = 27 ◦C, which does check out with the actual surface temperature

1.5. (a) Given Planck’s formula for the energy density, obtain an expression for the Rayleigh-Jean’sformula for U(ν, T )

Solution: Planck’s formula for energy density tells us that

UPlanck(ν, T ) =8πhν3

c3

1

ehνkBT − 1

(1.5.1)

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We know that limν→0

UPlanck(ν, T ) = URayleigh-Jean(ν, T )

URayleigh-Jean(ν, T ) = limν→0

UPlanck(ν, T ) (1.5.2)

=8πkBTν

2

c3limν→0

(hνkBT

ehνkBT − 1

)(1.5.3)

=8πkBTν

2

c3limx→0

(x

ex − 1

)(1.5.4)

=8πν2

c3kBT (1.5.5)

(b) For a black body at temperature T , U(ν, T ) was measured at ν = ν0. This value is foundto be one tenth of the value estimated using Rayleigh Jeans formula. Obtain an implicitequation in terms of hν0

kBT

Solution:

URayleigh-Jean(ν0, T ) = 10UPlanck(ν0, T ) (1.5.6)

=⇒ 8πν20

c3kBT = 10

8πhν30

c3

1

ehν0kBT − 1

(1.5.7)

=⇒ 10hν0

kBT= e

hν0kBT − 1 (1.5.8)

=⇒ 10x = ex − 1 (1.5.9)

where x = hν0

kBT

(c) Solve the above equation to obtain the value of hνkBT

, up to the first decimal place.

Solution: You can use Desmos or Wolfram Alpha to solve this numerically to getx = 3.615=⇒ νo

T= kB×3.615

h= 7.53× 1010 s−1K−1

1.6. Using appropriate approximations, derive Wiens’ displacement law from Planck’s formula forenergy density of black body radiation

Solution: Already solved in Q1.3(a)

1.7. Derive the Stefan-Boltzmann law from the expression from I(λ) given in Q1.2

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Solution:

I(λ, T ) dλ =2πhc2

λ5

1

ehcλkT − 1

dλ (1.7.1)

To obtain the total intensity of emission at a given temperature, we simply integrate thisfunction out in terms of λ

Itotal(T ) =

ˆ ∞λ=0

2πhc2

λ5

1

ehcλkT − 1

dλ (1.7.2)

= 2πhc2

ˆ ∞λ=0

1

λ5ehcλkT − 1

dλ (1.7.3)

First subsitute λ = cν

=2πh

c2

ˆ ∞ν=0

ν3

ehνkT − 1

dν (1.7.4)

Now subsitute x = hνkT

=2πk4T 4

h3c2

ˆ ∞x=0

x3

ex − 1dx (1.7.5)

ˆ ∞x=0

x3

ex − 1dx is a well known result =

π4

15(1.7.6)

=2πk4T 4

h3c2

π4

15(1.7.7)

=2π5k4

15h3c2T 4 (1.7.8)

= σT 4 (1.7.9)

Where σ = 2π5k4

15h3c2= 5.65× 10−8 W

m2K4 (1.7.10)

Now I(T )total = P (T )A

, where P (T ) = Power emitted by blackbody at temperature T, andA = Emission surface area of blackbody=⇒ P = σAT 4

2 Compton Scattering

2.1. A photon of energy hν is scattered through 90◦ by an electron initially at rest. The scatteredphoton has a wavelength twice that of the incident photon. Find the frequency of the incidentphoton and the recoil angle of the electron.

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Solution: First let us find the frequency of the incident photon

λ′ − λ = λc(1− cos θ) (2.1.1)

Where λc = hmec

, or the compton wavelength of an electron

=⇒ λ′ − λ = λc

(1− cos

(π2

))(2.1.2)

=⇒ 2λ− λ = λc (2.1.3)

=⇒ λ = λc (2.1.4)

=⇒ c

ν=

h

mec(2.1.5)

=⇒ ν =mec

2

h= 1.24× 1020 Hz (2.1.6)

Now to find the recoil angle ϕ of the electron

pλ

pλ′

pe′

ϕ

Through momentum conservation we have ~pe′ = ~pλ − ~p′λWhere ~pe′ is the electron momentum after scattering, ~pλ′ and ~pλ are the photon momentaafter and before scatteringConserving components || to incoming photon

pλ = pe′ cosϕ (2.1.7)

Conserving components ⊥ to incoming photon

p′λ = pe′ sinϕ (2.1.8)

Divide (2.7.12) by (2.7.11) to get

tanϕ =pλ′

pλ(2.1.9)

=λ

λ′=

1

2(2.1.10)

=⇒ ϕ = tan−1

(1

2

)= 26.56◦ (2.1.11)

2.2. Find the energy of the incident x-ray if the maximum kinetic energy of the Compton electronis mec2

2.5

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Solution: By Energy conservation

Ee′ − Ee = Eλ − Eλ′ (2.2.1)

=⇒ Ee′ = hc

(1

λ− 1

λ′

)+mec

2 (2.2.2)

= hc

(λ′ − λλλ′

)+mec

2 (2.2.3)

= mec2λ

2c

λ2

(1− cos θ

1 + λcλ

(1− cos θ)

)+mec

2 (2.2.4)

= mec2λ

2c

λ2

(1

11−cos θ

+ λcλ

)+mec

2 (2.2.5)

As we can easily see, for a given λ for the incident ray, the energy of the Compton electronis maximum at cos θ = −1, or θ = π

mec2

2.5= KEe′max

= Ee′max−mec

2 = mec2 λ2

c

λ(λ2

+ λc)(2.2.6)

=⇒ λ2 + 2λcλ− 5λ2c = 0 (2.2.7)

=⇒ λ =−2λc +

√24λ2

c

2(2.2.8)

= (√

6− 1)λc (2.2.9)

=⇒ Eλ =hc

λ=

1√6− 1

mec2 = 0.69 mec

2 (2.2.10)

2.3. Show that a free electron cannot absorb a photon so that a photoelectron requires boundelectron. However, the electron can be free in Compton Effect. Why?

Solution: Let us draw the scattering diagram for a free electron absorbing a photon

pλpe′

Apply Energy conservation

Eλ + Ee = Ee′ (2.3.1)

pλc+mec2 =

√p2e′c

2 +m2ec

4 (2.3.2)

=⇒ p2λ +m2

ec2 + 2mecpλ = p2

e′ +m2ec

2 (2.3.3)

But according to momentum conservation pλ = pe′

=⇒ 2mecpλ = 0 (2.3.4)

=⇒ pλ = 0 (2.3.5)

=⇒ ν = 0 (2.3.6)

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Which is not really possible. ∴ a free electron cannot absorb a photon.OTOH, drawing the scattering diagram and applying momentum and energy conservationfor Compton scattering yields a valid solution ∆λ = λc(1 − cos θ), ∴ Compton scatteringcan and does in fact involve a free electron

2.4. Two Compton scattering experiments were performed using x-rays (incident energies E1 andE2 = E1

2). In the first experiment, the increase in wavelength of the scattered x-ray, when

measured at an angle θ = 45◦ is 7× 10−14 m. In the second experiment, the wavelength of thescattered x-ray, when measured at an angle θ = 60◦ is 9.9× 10−12 m.

(a) Calculate the Compton wavelength and the mass (m) of the scatterer

Solution:

∆λ1 = λc(1− cos θ1) (2.4.1)

=⇒ λc =∆λ1

1− cos θ1

(2.4.2)

=7× 10−14

1− cos(π4

) (2.4.3)

=7√

2√2− 1

× 10−14 m (2.4.4)

= 2.39× 10−13 m (2.4.5)

Buth

mc= λc (2.4.6)

=⇒ m =h

λcc= 9.24× 10−30 kg (2.4.7)

(b) Find the wavelengths of the incident x-rays in the two experiments.

Solution:

λ′2 − λ2 = λc(1− cos θ2) (2.4.8)

=⇒ λ2 = λ′2 − λc(1− cos θ2) (2.4.9)

= 9.9× 10−12 − 2.39× 10−13(

1− cos(π

3

))(2.4.10)

= 9.9× 10−12 − 0.119× 10−12 (2.4.11)

= 9.781× 10−12 m (2.4.12)

Now, E1 = 2E2 (2.4.13)

=⇒ λ1 =λ2

2= 4.89× 10−12 m (2.4.14)

2.5. Find the smallest energy that a photon can have and still transfer 50% of its energy to anelectron initially at rest.

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Solution:

By Energy conservation,

Ee′ − Ee = Eλ − Eλ′ (2.5.1)

=⇒ 0.5Eλ = Eλ − Eλ′ (2.5.2)

=⇒ Eλ′ = 0.5Eλ (2.5.3)

=⇒ λ′ = 2λ (2.5.4)

Now, λ′ − λ = λc(1− cos θ) (2.5.5)

=⇒ λ = λc(1− cos θ) (2.5.6)

=⇒ Eλ =hc

λ=

hc

λc(1− cos θ)(2.5.7)

=mec

2

1− cos θ(2.5.8)

As we can see, Eλ is smallest for cos θ = −1, or θ = π

=⇒ Eλmin=mec

2

2=

0.511 MeV

2= 0.256 MeV (2.5.9)

2.6. In Compton Scattering, Show that if the angle of scattering θ increases beyond a certain valueθ0, the scattered photon will never have energy larger than 2m0c

2, irrespective of the energy ofthe incident photon. Find the value of θ0

Solution:

λ′ − λ = λc(1− cos θ) (2.6.1)

λ′ = λ+ λc(1− cos θ) (2.6.2)

=⇒ Eλ′ =hc

λ+ λc(1− cos θ)(2.6.3)

=m0c

2

λm0ch

+ 1− cos θ(2.6.4)

=m0c

2

moc2

Eλ+ 1− cos θ

(2.6.5)

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We want Eλ′ < 2m0c2 ∀λ > 0, θ > θ0

We know that

Eλ′|θ>θ0 =m0c

2

moc2

Eλ+ 1− cos θ

(2.6.6)

<m0c

2

moc2

Eλ+ 1− cos θ0

(2.6.7)

<m0c

2

1− cos θ0

(2.6.8)

Thus m0c2

1−cos θ0= 2m0c

2

=⇒ cos θ0 = 0.5 (2.6.9)

=⇒ θ0 = 60◦ (2.6.10)

2.7. In a Compton scattering experiment (see figure), X-rays scattered off a free electron initially atrest at an angle θ > π

4, gets re-scattered by another free electron, also initially at rest.

pλ1

pλ2

pλ3

θ

(a) If λ3 − λ1 = 1.538× 10−12 m, find the value of θ

Solution:

λ2 − λ1 = λc(1− cos θ) (2.7.1)

λ3 − λ2 = λc

(1− cos

(π2− θ))

= λc(1− sin θ) (2.7.2)

Adding (2.7.1) and (2.7.2),

=⇒ λ3 − λ1 = λc(2− sin θ − cos θ) (2.7.3)

=⇒ sin θ + cos θ = 2− ∆λ

λc(2.7.4)

=⇒ 1 + sin(2θ) =

(2− ∆λ

λc

)2

(2.7.5)

=⇒ sin(2θ) =

(2− ∆λ

λc

)2

− 1 (2.7.6)

= 0.867 (2.7.7)

=⇒ sin−1(sin(2θ)) = sin−1(0.867) ≈ π

3(2.7.8)

(2.7.9)

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The solutions to this are 2θ = · · · −5π3, π

3, 7π

3· · · and 2θ = · · · −4π

3, 2π

3, 8π

3· · ·

But we only care about π4< θ < π

2, =⇒ π

2< 2θ < π

=⇒ 2θ = 2π3

, =⇒ θ = π3

= 60◦

(b) If λ2 = 68× 10−12 m, find the angle at which the first electron recoils due to the collision.

Solution:

pλ1

pλ2

pe′

ϕ

θ

We know,

λ2 − λ1 = λc(1− cos θ) (2.7.10)

We can use momentum conservation || and ⊥ to ~pλ1

pe′ cosϕ = pλ1 − pλ2 cos θ (2.7.11)

pe′ sinϕ = pλ2 sin θ (2.7.12)

Dividing (2.7.12) by (2.7.11)

tanϕ =pλ2 sin θ

pλ1 − pλ2 cos θ(2.7.13)

=1

pλ1

sin θ1pλ2− 1

pλ1cos θ

(2.7.14)

=λ1 sin θ

λ2 − λ1 cos θ(2.7.15)

=λ1 sin θ

λ2 − λ1 + λ1(1− cos θ)(2.7.16)

=sin θ

1− cos θ

λ1

λc + λ1

(2.7.17)

=2 sin

(θ2

)cos(θ2

)

2 sin2(θ2

) λ1

λc + λ1

(2.7.18)

= cot

(θ

2

)1

λcλ1

+ 1(2.7.19)

Thus we need λ1 to find φUsing (2.7.10)

λ1 = λ2 − λc(1− cos θ) (2.7.20)

=⇒ λcλ1

=1

λ2

λc− (1− cos θ)

=1

68×10−12

λc− (1− cos

(π3

))

= 0.0364 (2.7.21)

Putting in (2.7.19)

tanϕ =cot(π6

)

1.0364= 1.67 (2.7.22)

=⇒ ϕ = tan−1 1.67 = 59.1◦ (2.7.23)

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3 Specific Heat

3.1. Show that the Einstein’s specific heat expression gives Cv = 3R as T → ∞ and Cv = 0 asT → 0

Solution:

U = (3N)

(hν

ehνkBT − 1

)(3.1.1)

=⇒ Cv =∂U

∂T=

3Nhν

kBT 2

hνehνkBT

(e

hνkBT − 1

)2 (3.1.2)

= 3NkB

(hν

kBT

)2e

hνkBT

(e

hνkBT − 1

)2 (3.1.3)

= 3R

(hν

kBT

)2e

hνkBT

(e

hνkBT − 1

)2 (3.1.4)

Now, we take limits

limT→∞

Cv = 3R limT→∞

(hν

kBT

)2e

hνkBT

(e

hνkBT − 1

)2 (3.1.5)

= 3R limx→0

x2

(ex − 1)2 limx→0

ex (3.1.6)

= 3R (3.1.7)

Similarly,

limT→0

Cv = 3R limT→0

(hν

kBT

)2e

hνkBT

(e

hνkBT − 1

)2 (3.1.8)

= 3R limx→∞

x2e−x

(1− e−x)2 (3.1.9)

= 0 (3.1.10)

3.2. (a) A diatomic molecule consists of two equal point masses (m) separated by a distance D(called the bond length). Consider a gas of such diatomic molecules. The molecularspecific heat at constant volume Cv of this gas changes from 1.5R to 2.5R at a temperatureT1. The quantum of rotational energy of the molecule is given by ~2

2I, where I is the moment

of inertia of the molecule. Obtain an expression for D in terms of m and T1

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Solution: First of all, Cv will have three contributions, one from translational, onefrom rotational, and one from vibrationalIn the translational dof, most of the molecules are in non ground state, hence contributeto 3

2R to Cv, as predicted classically

For the rotational dof, let E be the spacing between the ground and the first excited

state. As we know that there is a eE

kBT term in the relative number of molecules inthe ground and the first excited state, hence we know that if T � E

kB, then most of

the molecules are in ground state having no energy at all, and thus do not contributesignificantly to Cv. As we increase temperature to T � E

kBmore and more molecules

start to jump to first and higher excited states, and contribute 2× R2

to Cv.

Hence the Cv increases from 1.5R to 2.5R at T1 ≈ EkB

= ~2

2IkB= ~2

mD2kB

(b) Cv of the above diatomic gas changes from 2.5R to 3.5R at temperature T2. The quantumof vibration energy is given by ~ω, where ω is the angular frequency of vibrations. Obtainan expression for the bond strength (or the spring constant of the bond) in terms of m andT2

Solution: Using similar logic as above, T2 ≈ ~ωkB

= ~kB

√km

=⇒ k =mkBT

22

~2

3.3. Graphite structure consists of carbon layers arranged in the xy-plane. Each atom in the structurecan, in principle, perform simple harmonic motion in 3 mutually perpendicular directions. Therestoring forces in the xy-plane and hence the natural frequency of oscillations in the x andy directions are large such that ~ωx � 300kB, ~ωy � 300kB, the thermal energy at roomtemperature. On the other hand, the restoring force perpendicular to a layer is quite small andhence ~ωz � 300kB. On the basis of this information, find Cv of graphite at 300 K. You mayassume that the equipartition theorem is valid in this case.

Solution: As we can imagine, most of the atoms are in the excited state of the vibrationperpendicular to plane, because kBT � ~ωz. So that dof contributes R to Cv. On the otherhand, most of the atoms are in ground states of vibration in plane, so they do not contributeto CvHence Cv = R

3.4. CO2 is a linear tri-atomic molecule with the carbon atom in the middle connected to an oxygenatom on either side (O−C−O). This molecule has two rotational and four vibrational degrees offreedom. Assume that the excitation energy of the rotational degrees of freedom is very small.The quantum of energy required to excite the lowest vibrational mode is E, and and that forthe next mode is 4E. The next two modes have equal excitation energy of 6E. Given thatE = 12 × 10−3 eV, sketch the molar specific heat of CO2 (in units of R) from T = 100K toT = 1000K

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Solution:

Cv

T100K E

kB

4EkB

6EkB

1000K

2.5R

3.5R

4.5R

6.5R

4 Phase and Group Velocity

4.1. Two harmonic waves which travel simultaneously along a wire are represented by

y1 = 0.002 cos(8.0x− 400t) and y2 = 0.002 cos(7.6x− 380t)

where x, y are in meters and t is in sec.

(a) Find the resultant wave and its phase and group velocities

Solution:

ytotal = y1 + y2 (4.1.1)

= 0.002 (cos(8x− 400t) + cos(7.6x− 380t)) (4.1.2)

= 0.004 (cos(7.8x− 390t) cos(0.2x− 10t)) (Using trig identity) (4.1.3)

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−5 0 5−4

−2

0

2

4·10−3

−2.5π 2.5π

x

y

ytotal|t=0

±0.004 cos(0.2x)

Thus we see that the envelope is a sinusoid of frequency ω = ∆ω2

= 10 Hz and wavenum-ber k = ∆k

2= 0.2 m−1. Thus vg = ∆ω

∆k= 50 ms−1

The wave itself is a sinusoid of frequency ω = ωavg = 390 Hz and wavenumberk = kavg = 7.8 m−1. Thus vp = ωavg

kavg= 50 ms−1

(b) Calculate the range ∆x between the zeros of the group wave. Find the product of ∆x and∆k?

Solution:

∆x =1

2× 2× 2π

∆k= 5π m (4.1.4)

∆x∆k = 2π (4.1.5)

4.2. The dispersion relation for a lattice wave propagating in a 1-D chain of atoms of mass m bound

together by a force constant β is given by ω = ω0 sin(ka2

), where ω0 =

√4βm

and a is the distance

between the atoms.

(a) Show that the medium is non- dispersive in the long wavelength limits.

Solution: We have

vg = vp + kdvpdk

(4.2.1)

For non dispersive medium, we need

vp = vg (4.2.2)

=⇒ vp = vp + kdvpdk

(From (4.2.1)) (4.2.3)

=⇒ kdvpdk

= 0 (4.2.4)

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But,

vp =ω(k)

k= ω0

sin(ka2

)

k(4.2.5)

=⇒ kdvpdk

=aω0

2cos

(ka

2

)− aω0

2

sin(ka2

)ka2

(4.2.6)

=⇒ limk→0

kdvpdk

=aω0

2− aω0

2= 0 (4.2.7)

Hence medium is non dispersive in the long wavelength limit

(b) Find the group and phase velocities at k = πa.

Solution:

vp = ω0

sin(π2

)πa

=aω0

π(4.2.8)

kdvpdk

∣∣∣∣k=π

a

=aω0

2

(cos(π

2

)− 2

πsin(π

2

))(4.2.9)

= −aω0

π(4.2.10)

=⇒ vg = 0 (From (4.2.1)) (4.2.11)

4.3. Find the group and the phase velocities of the matter wave associated with a free particle underthe assumption that the frequency is defined using

(i) the kinetic energy

Solution:

vp =ω

k=E

p=

p

2m(4.3.1)

vg =dω

dk=dE

dp=

p

m(4.3.2)

(ii) total relativistic energy

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Solution:

vp =ω

k=E

p(4.3.3)

=

√p2c2 +m2

0c4

p(4.3.4)

vg =dω

dk=dE

dp(4.3.5)

=d

dp

√p2c2 +m2

0c4 (4.3.6)

=pc2

E=c2

vp(4.3.7)

4.4. The phase speed vp of light in a certain wavelength range in a dispersive medium is given bythe following expression:

vp = c

(A+

4π2B

λ2

)−1

where λ is the wavelength, c is the speed of light and A,B are constants.

(a) Find an expression for the group speed in terms of the wave vector ~k

Solution:

vp =c

A+ k2B(4.4.1)

=⇒ vg =c

A+ k2B− 2k2Bc

(A+ k2B)2(Using (4.2.1)) (4.4.2)

=c(A− ~k · ~kB)

(A+ ~k · ~kB)2(4.4.3)

(b) Taking A = 1.7 and B = 0.014π2 (µm)2, calculate the group and phase speed of light for a

wavelength of 400 nm. Which of the two speeds is physically important and why?

Solution:

vp =c

1.7 + 116

= 0.567c (4.4.4)

vg = c

(1.7− 1

16

)(1.7 + 1

16

)2 = 0.527c (4.4.5)

The reason for vg being physically important is deep. The wavepacket will be of theform Ψ(x, t) = E(x, t) · ei(x−vpt), where E is the envelope function. But the thingwhich is physical are the probability densities of measurement results, which is givenby ΨΨ = |E|2. Therefore the physical measurements depend only on the behaviour of

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the envelope, which is governed by vg.Even classically, |E|2 represents the intensity of light, which is physical, and the full Efield’s phase, ei(x−vpt) doesn’t really represent anything physical

4.5. Consider a square 2-D system with small balls (each of mass m) connected by springs. Thespring constants along the x- and y-directions are βx and βy, respectively. The dispersionrelation for this system is given by

−ω2m+ 2βx(1− cos(kxax)) + 2βy(1− cos(kyay)) = 0

where ~k = kx i + ky j is the wave vector and ax, ay are the natural distances between the twosuccessive masses along the x- and y-directions, respectively. Find the group velocity and theangle that it makes with the x-axis

Solution: We know that

ω(~k) =

√2βx(1− cos(kxax)) + 2βy(1− cos(kyay))

m(4.5.1)

=⇒ ~vg = ∇kω =dω

dkxi+

dω

dkyj (4.5.2)

=βxaxmω

sin(kxax) i+βyaymω

sin(kyay) j (4.5.3)

=⇒ ϕ = tan−1

(βyay sin(kyay)

βxax sin(kxax)

)(4.5.4)

Where ϕ is the angle made by vg with the x axis

4.6. Consider an electromagnetic (EM) wave of the form Aei(kx=wt). Its speed in free space is given

by c = ωk

=√

1ε0µ0

, where ε0, µ0 is the electric permittivity, magnetic permeability of free space,

respectively.

(a) Find an expression for the speed (v) of EM waves in a medium, in terms of its permittivityε and permeability µ

Solution: Here we want the speed of EM waves themselves, which is

vp = ωk

=√

1εµ

= c√εrµr

(b) Suppose the permittivity of the medium depends on the frequency, given by ε = ε0

(1− ω2

p

ω2

)

where ωp is a constant called the plasma frequency, find the dispersion relation for the EMwaves in a medium. (assume µ = µ0)

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Solution:

ω

k=

√1

εµ(4.6.1)

=c√

1− ω2p

ω2

(4.6.2)

=⇒ kc = ω

√1− ω2

p

ω2(4.6.3)

=⇒ k2c2 = ω2 + ω2p (4.6.4)

=⇒ ω(k) =√k2c2 − ω2

p (4.6.5)

(c) Consider waves with ω = 3ωp. Find the phase and group velocity of the waves. What isthe product of group and phase velocities?

Solution:

vp =ω

k=

3√

2

4c = 1.06c (4.6.6)

vg =dω

dk=

c2k

ω=c2

vp=

2√

2

3c = 0.94c (4.6.7)

5 Wave packet, Fourier Theory, HUP

Wave packet and Fourier Theory

5.1. A wave packet is of the form

f(x) = e−α|x|, −∞ < x <∞

where α is a +ve constant

(a) Plot f(x) versus x

Solution:

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−1.5 −1 −0.5 0 0.5 1 1.5

0.2

0.4

0.6

0.8

x

f(x)

(b) Find the values of x at which f(x) attains half of its maximum value

Solution:

fmax = f(0) = 1 (5.1.1)

=⇒ f(xhalf-width) = 0.5 (5.1.2)

=⇒ α|xhalf-width| = ln 2 (5.1.3)

=⇒ xhalf-width = ± ln 2

α(5.1.4)

(c) Calculate the Fourier transform of f(x) i.e. g(k) =´∞−∞ f(x)eikx dx

Solution:

g(k) =

ˆ ∞x=−∞

f(x)eikx dx (5.1.5)

=

ˆ ∞x=0

e(−α+ik)x dx+

ˆ 0

x=−∞e(α+ik)x dx (5.1.6)

=

(e(−α+ik)x

−α + ik+e(−α−ik)x

−α− ik

)∣∣∣∣∞

x=0

(5.1.7)

=1

α− ik +1

α + ik(5.1.8)

=2α

α2 + k2(5.1.9)

(d) Plot g(k) versus k

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Solution:

−20 −10 0 10 20

0.1

0.2

0.3

0.4

k

g(k)

(e) Find the values of k at which f(k) attains half of its maximum value

Solution:

gmax = g(0) =2

α(5.1.10)

=⇒ g(khalf-width) =1

α(5.1.11)

=⇒ 2α

α2 + k2=

1

α(5.1.12)

=⇒ khalf-width = ±α (5.1.13)

(f) From the values obtained in (b) and (e), find ∆x∆k

Solution: Taking ∆x and ∆k as half width at half maxima, we get∆x∆k = ln 2

α× α = ln 2

(We could have taken them as full width at half maxima and got four times this value)

5.2. A wave packet is of the form f(x) = x for −1 ≤ x ≤ 1 and f(x) = 0 elsewhere

(a) Plot f(x) vs x

Solution:

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−2 −1 0 1 2−1

−0.5

0

0.5

1

x

f(x)

(b) Calculate the Fourier transform of f(x)

Solution:

g(k) =

ˆ ∞x=−∞

f(x)e−ikx dx (5.2.1)

=

ˆ 1

x=−1

xe−ikx dx (5.2.2)

Integrating by parts

=i

k

(xe−ikx

∣∣1x=−1

−ˆ 1

x=−1

e−ikx)

(5.2.3)

=i

k

(xe−ikx +

1

ike−ikx

)∣∣∣∣1

x=−1

(5.2.4)

=2i

k

(cos k − sin k

k

)(5.2.5)

(5.2.6)

(c) Plot |g(k)| versus k

Solution:

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−8 −6 −4 −2 0 2 4 6 8

0.2

0.4

0.6

0.8

k

|g(k)|

(d) At what value of k, |g(k)| attains a minimum value?

Solution:

|g(k)|min = 0 (5.2.7)

=⇒ sin(kmin) = kmin cos(kmin) (5.2.8)

=⇒ tan(kmin) = kmin (5.2.9)

Thus, |g(k)| attains a minimum value at the solutions of tan k = k, one of them beingk = 0

5.3. A triangular pulse is represented by f(x) = 1− |x| for −1 ≤ x ≤ 1 and f(x) = 0 elsewhere

(a) Plot f(x) vs x

Solution:

−2 −1 0 1 20

0.2

0.4

0.6

0.8

1

x

f(x)

(b) Calculate the Fourier transform of f(x)

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Solution:

g(k) =

ˆ ∞x=−∞

f(x)e−ikx dx (5.3.1)

=

ˆ 1

x=0

(1− x)e−ikx dx+

ˆ 0

x=−1

(1 + x)e−ikx dx (5.3.2)

= 2

ˆ 1

x=0

(1− x) cos(kx) dx (5.3.3)


=2

k

((1− x) sin(kx)|1x=0 +

ˆ 1

x=0

sin(kx)

)(5.3.4)

=2

k

((1− x) sin(kx)− cos(kx)

k

)∣∣∣∣1

x=0

(5.3.5)

=2

k

(1− cos k

k

)(5.3.6)

(c) Plot g(k) versus k

Solution:

−8 −6 −4 −2 0 2 4 6 80

0.2

0.4

0.6

0.8

1

k

g(k)

(d) Find the value of k at which |g(k)| attains its maximum value?

Solution: At gmax = g(0) = 1

(e) Find the values of k, at which g(k) attains half of this maximum value

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Solution:

gmax = g(0) = 1 (5.3.7)

=⇒ g(khalf-width) =1

2(5.3.8)

=⇒√

2 sin

(khalf-width

2

)= ±khalf-width

2(5.3.9)

Solving this transcendental equation via a calculator,

=⇒ khalf-width = ±2.783 (5.3.10)

(f) From the values obtained in (e), find the spread ∆k. Calculate the spread ∆x using theuncertainty relation ∆x∆k = 1

2

Solution: Taking ∆k as half-width at half maxima,

∆k = 2.783 (5.3.11)

=⇒ ∆x =1

2× 2.783= 0.18 (5.3.12)

5.4. A wave packet is constructed by superposing waves, their wavelengths varying continuously as

y(x, t) =

ˆA(k) cos(kx− ωt) dk

where A(k) = A for(k0 − ∆k

2

)≤ k ≤

(k0 + ∆k

2

)and A(k) = 0 otherwise. Sketch y(x, t)

(approximately) and estimate ∆x by taking the difference between two values of x for whichthe central maximum and the nearest minimum is observed in the envelope. Verify uncertaintyprinciple from this.

Solution:

y(x, t) = A

ˆ k0+ ∆k2

k=k0−∆k2

cos(kx− ωt) dk (5.4.1)

= Asin(kx− ωt)

x

∣∣∣∣k0+ ∆k

2

k=k0−∆k2

(5.4.2)

=2A sin (∆k

2x)

xcos(k0x− ωt) (5.4.3)

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− 2π∆k

0 2π∆k

A∆k

x

y(x,0

)

We can see ∆x = 2π∆k

=⇒ ∆x∆k = 2π ≥ 1, hence HUP is satisfied

Uncertainty Principle

5.5. Position and momentum of an electron (E = 1 keV) are determined simultaneously. Its positionis known to an accuracy of only 1 A along the x-axis. Using uncertainty principle, find theminimum permissible uncertainty in its momentum along the x-axis? From the above data canyou determine the uncertainty along the y-axis?

Solution:

∆x∆p ≥ ~2

(5.5.1)

=⇒ ∆p ≥ ~2

A−1

(5.5.2)

=⇒ ∆p ≥ 5.27× 10−25 kg m s−1 (5.5.3)

We do not have any information about uncertainty of p or x along the y-axis.

5.6. An electron falls from a height of 10 m and passes through a hole of radius 1 cm. To study themotion of the electron afterwards, should we apply the wave aspect or the particle aspect?

29 of 95


Solution: When the electron is at the hole

〈p〉22m

= mgh (5.6.1)

=⇒ 〈p〉 = m√

2gh (5.6.2)

=⇒ λ ≈ h

〈p〉 =h

m√

2gh(5.6.3)

= 5.15× 10−5m (5.6.4)

=⇒ Fresnel Distance D =a2

λ≈ 7.8 m (5.6.5)

Hence, we should apply the wave aspect after 7.8 m distance from the hole, before it weshould apply the particle aspect (or ray optics)

5.7. A beam of electron of energy 0.025 eV moving along the x-direction passes through a slit ofvariable width w placed along the y-axis. Estimate the value of w for which the spot size on ascreen kept at a distance of 0.5 m from the slit would be a minimum.

Solution: We know that θmin = λw

for Fraunhofer diffraction, where θ is the angle madewith respect to the incoming beam direction (you can also get this using ∆py and ∆y). Thusfor small w,

Spot Size s = 2× θminD = 2× λD

w(5.7.1)

=2hD

w√

2mE(5.7.2)

But the spot size is just w when w is very large. Thus the minimum spot size is when thesetwo seem to agree

2hD

w√

2mE∼ w (5.7.3)

=⇒ w ∼ 4

√4h2D2

2mE≈ 88 µm (5.7.4)

5.8. If the average value of momentum 〈px〉 = 0, then the x-component of momentum ∆px is givenby ∆px =

√〈p2x〉. Using this relation, estimate the following (use ∆x∆px ≥ ~

2)

(a) Minimum kinetic energy that a proton and an electron would have if they were confined toa nucleus of approximate diameter 10−14 m. This is an argument used against the existenceof electron in nuclei.

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Solution: ∆x will be of the order 10−14 m.Thus ∆px will be of the order 10−20 kg m s−1

Therefore,

〈E2〉 = c2〈p2〉+m20c

4 (5.8.1)

∼ 1014 eV2 (5.8.2)

=⇒ 〈E〉 ∼ 10 MeV (5.8.3)

This back of the envelope estimation reveals that the energy is too high to remain stableinside the atom

(b) The ground state energy of a particle of mass m bound by a potential V = 12kx2

Solution:

〈E〉 =1

2m〈p2〉+

k

2〈x2〉 (5.8.4)

Since V is minimum at x = 0 and is symmetric about 0, we can safely assume 〈x〉 = 0

=∆p2

2m+k∆x2

2(5.8.5)

≥ ∆p2

2m+

k~2

8∆p2(5.8.6)

We can minimize the expected Energy by differentiating with ∆p and setting to 0

d〈E〉d∆p

= 0 (5.8.7)

=⇒ ∆p

m− k~2

4∆p3= 0 (5.8.8)

=⇒ ∆p =4

√k~2m

4(5.8.9)

Substituting this in (5.8.6),

Eground = ~

(1

4

√k

m+

1

4

√k

m

)=

1

2~√k

m(5.8.10)

Since the actual ground state energy is also 12~√

km

, we have done a pretty good job

with our back of the envelope estimation

(c) The radial distance r for which the sum of kinetic and potential energies is minimum in ahydrogen atom.

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Solution:

〈E〉 =1

2m〈p2〉 − q2

e

4πε0〈1r〉 (5.8.11)

∼ ∆p2

2m− q2

e

4πε0

1

∆r(5.8.12)

Using ∆r∆p ≥ ~2,

≥ ~2

8m∆r2− q2

e

4πε0

1

∆r(5.8.13)

Minimizing the Expected Energy with respect to ∆r

d〈E〉d∆r

= 0 (5.8.14)

=⇒ − ~2

4m∆r3+

q2e

4πε0

1

∆r2= 0 (5.8.15)

=⇒ ∆r =~2πε0mq2

e

= 0.13× 10−10 m (5.8.16)

This estimation is not bad considering that the real r is 4~2πε0mq2

e= 0.53× 10−10 m

5.9. Estimate the minimum possible energy (i.e., the ground state energy) consistent with the un-certainty principle for a particle of mass m bound by a potential V = kx4 (use ∆x∆px ≥ ~

2)

Solution:

〈E〉 =1

2m〈p2〉+ k〈x4〉 (5.9.1)

∼ ∆p2

2m+ k∆x4 (5.9.2)

≥ ∆p2

2m+ k

~4

16∆p4(5.9.3)

Minimizing the Expected Energy with respect to ∆p

d〈E〉d∆p

= 0 (5.9.4)

=⇒ ∆p

m− k~4

4∆p5= 0 (5.9.5)

=⇒ Eground ∼ 3

√k~4m

4

(1

2m+

1

4m

)(5.9.6)

=3

43

√k~4

4m2(5.9.7)

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5.10. A photon of energy E is emitted as a result of a particular transition.

(a) Find the value of the recoil energy, assuming that the atom recoils with a nonrelativisticspeed.

Solution: We begin by momentum as usual

pa′ = p′λ (5.10.1)

=⇒ pa′ =E

c(5.10.2)

=⇒ Ea′ =p2a′

2m=

E2

2mc2(5.10.3)

(b) Let the lifetime of the state be of the order of 10−8 s. Estimate the natural line width ofthe emitted line.

Solution: Natural line width refers to the spread in the energies of a line observed inemission/absorption spectrum, and this is nothing but ∆E.We know that ∆E∆t ≥ ~

2, where ∆t is the time it takes for the system to change it’s

observable value considerably. Thus here ∆t = 10−8 s.

=⇒ ∆E ∼ ~2∆t∼ 10−8 eV

(c) Find the value of E for which the recoil energy will be of the same order of magnitude asthe natural line width? For order of magnitude calculation take the mass number of theatom as 100.

Solution:

E2

2mc2∼ 10−8 eV (5.10.4)

=⇒ E ∼ 100 eV (5.10.5)

5.11. Take the hydrogen atoms in thermal equilibrium at a temperature T , for which kBT = 0.025eV. Let E1 be the difference in energy between the ground state and the first excited state whenthe atom is at rest. Let E2 be the energy of the photon (in the frame of container) required tomake this transition when the atom is moving towards the photon

(a) Find E1 − E2

Solution: We can assume that the translational modes are already activated at thesetemperatures and the average translational energy is 3

2kBT

We begin by conserving momentum and energy (taking note that we conserve theaverage over all atoms which move towards the incident photon)

pa −E2

c= p′a (Conserving Momentum)

(5.11.1)

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We know that 〈pa〉m

= vmean =√

8kBTπm

〈Etrans〉+ E2 = 〈E ′trans〉+ E1 (Conserving Energy)(5.11.2)

We know that 〈Etrans〉 = 32kBT

But,

〈E ′trans〉 =1

2m〈p′2a 〉 (5.11.3)

=1

2m〈p2

a〉 −E2

mc〈pa〉+

E22

2mc2(Using (5.11.1)) (5.11.4)

= 〈Etrans〉 −E2

mc〈pa〉+

E22

2mc2(5.11.5)

=3

2kBT − E2

√8kBT

πmc2+

E22

2mc2(5.11.6)

=⇒ E2 = E1 − E2

√8kBT

πmc2+

E22

2mc2(5.11.7)

=⇒(E2

E1

)2

− 2mc2

E1

(1 +

√8kBT

mc2

)E2

E1

+2mc2

E1

= 0 (5.11.8)

=⇒ E2 = 0.99999177E1 = 10.199916 eV (5.11.9)

=⇒ E1 − E2 = 0.000084 eV (5.11.10)

(b) After the absorption of the photon, find the final velocity of the hydrogen atom

Solution:

〈p′a〉m

=〈pa〉m− E2

mc(Using (5.11.1)) (5.11.11)

= 2466 m s−1 (5.11.12)

(c) If the lifetime of the first excited state is 10−8 s, will the photon with energy E2 be ableto cause a transition, had the atom been at rest? Discuss quantitatively. You are free tomake any assumption, provided you justify it

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Solution: Had the atom been at rest,

E2

c= p′a (5.11.13)

E2 = E ′trans + E1 (5.11.14)

=E2

2

2mc2+ E1 (5.11.15)

=⇒(E2

E1

)2

− 2mc2

E1

E2

E1

+2mc2

E1

= 0 (5.11.16)

=⇒ E2 = 10.200000055386 eV (5.11.17)

Thus the required photon energy is ∼ 10−5 eV more than the energy supplied. But aswe already know, the line width is ∼ 10−8 eV, thus the difference is more than that isacceptable, hence the photon will not be able to excite the atom at rest

5.12. For a non-relativistic electron, using the uncertainty relation ∆x∆px = ~2

(a) Derive an expression for the minimum kinetic energy when localized in a region of size a

Solution:

∆x ∼ a (5.12.1)

=⇒ ∆px ≥~2a

(5.12.2)

But, 〈E〉 =〈p2x〉

2m(5.12.3)

=1

2m

(∆p2

x + 〈px〉2)

(5.12.4)

≥ ~2

8ma2(5.12.5)

(b) Show that the uncertainty in the measurement of its velocity is the same as the particlevelocity if the uncertainty in the position of the particle is equal to its de Broglie wavelength.

Solution:

∆vx ≥~

2mλ=

v

4π∼ v (5.12.6)

(c) Using the expression in (b), calculate the uncertainty in the velocity of an electron havingenergy 0.2 keV

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Solution:

∆vx =1

4π

√2E

m= 6.7× 105 m s−1 (5.12.7)

(d) An electron of energy 0.2 keV is passed through a circular hole of radius 10−6 m. Find theuncertainty introduced in the angle of emergence in radians (given tan θ ∼= θ)

Solution:

∆y ∼ 10−6 m (5.12.8)

=⇒ ∆py =~

2∆y(5.12.9)

=⇒ ∆θ ∼= ∆py〈px〉

=~

2∆y√

2Em= 4.3× 10−5 rad (5.12.10)

6 Wave function and Operators

6.1. If φn(x) are the solutions of time independent Schrodinger equation with energies En, show that

ψ(x, t) =∑

n

Cnφn(x)e−iEn~ t

is a solution of time dependent Schrodinger equation (Cn are constants). However, show thatψ(x, 0) is not a solution of the time independent Schrodinger equation

Solution: According to TISE,

− ~2

2m

∂2φn(x)

∂2x+ V (x)φn(x) = Enφn(x) ∀ n (6.1.1)

We need to satisfy the TDSE,

− ~2

2m

∂2ψ(x, t)

∂2x+ V (x)ψ(x, t) = i~

∂ψ(x, t)

∂t(6.1.2)

=⇒ LHS =∑

n

− ~2

2m

∂2φn(x)

∂2xCne

−iEn~ t + Cn∑

n

V (x)φn(x)e−iEn~ t (6.1.3)

=∑

n

EnCnφn(x)e−iEn~ t (Using (6.1.1)) (6.1.4)

RHS = i~∑

n

Cnφn(x)∂ei

En~ t

∂t(6.1.5)

=∑

n

EnCnφn(x)e−iEn~ t = LHS (6.1.6)

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Hence ψ(x, t) satisfies TDSENow to check satisfiability of ψ(x, 0) on TISE

LHS =∑

n

− ~2

2mCn∂2φn(x)

∂2x+∑

n

CnV (x)φn(x) (6.1.7)

RHS = E∑

n

Cnφn(x) (6.1.8)

We can see that no single value of E satisfies TISE ∀ x, so ψ(x, 0) is not a solution of theTISE

6.2. Consider a wave function φk(x) = sin(kx). Is this an eigenfuction of the operators

p = −i~ ∂∂x

and K = − ~2

2m

∂2

∂x2

Find the eigenvalues

Solution:

pφk(x) = −i~∂ sin(kx)

∂x(6.2.1)

= −i~k cos(kx) (6.2.2)

Thus φk(x) is not an eigenfunction of p

Kφk(x) = − ~2

2m

∂2 sin(kx)

∂x2(6.2.3)

=(~k)2

2msin(kx) =

(~k)2

2mφk(x) (6.2.4)

Thus, φk(x) is an eigenfunction of K with eigenvalue (~k)2

2m

6.3. If ψ1(x, t) and ψ2(x, t) are solutions of time dependent Schrodinger equation, show thataψ1(x, t) + bψ2(x, t) is also a solution of the same, where a and b are constants.

Solution:

− ~2

2m

∂2ψ1(x, t)

∂2x+ V (x)ψ1(x, t) = i~

∂ψ1(x, t)

∂t(6.3.1)

− ~2

2m

∂2ψ2(x, t)

∂2x+ V (x)ψ2(x, t) = i~

∂ψ2(x, t)

∂t(6.3.2)

Adding a(6.3.1) + b(6.3.2),

− ~2

2m

∂2ψ(x, t)

∂2x+ V (x)ψ(x, t) = i~

∂ψ(x, t)

∂t(6.3.3)

Where ψ(x, t) = aψ1(x, t) + bψ2(x, t)

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6.4. Consider two operators O1 and O2, defined as:

O1 ≡ −i∂

∂x, O2 ≡ x− i ∂

∂x

Calculate O1ψ(x) and O2ψ(x), where ψ(x) is a function of x. Is ψ(x) an eigen function of thetwo operators if it has the form ψ(x) = eikx?

Solution:

O1ψ(x) = −i∂ψ(x)

∂x(6.4.1)

O2ψ(x) = xψ(x)− i∂ψ(x)

∂x(6.4.2)

=⇒ O1eikx = keikx (6.4.3)

=⇒ O2eikx = (x− k)eikx (6.4.4)

Thus, ψ(x) is an eigenfunction of O1 with eigenvalue k

6.5. An operator is given by

G ≡ i~∂

∂x+ Ax

where A is a constant. Find the eigenfunction φ(x). If this eigen function is subjected to aboundary condition φ(a) = φ(−a), find out the eigen values.

Solution: We need to solve the differential equation

Gφ(x) = λφ(x) (6.5.1)

=⇒ i~dφ(x)

dx+ Axφ(x) = λφ(x) (6.5.2)

=⇒ dφ(x)

dx=i

~(Ax− λ)φ(x) (6.5.3)

=⇒ˆ

1

φ(x)dφ(x) =

i

~

ˆAx− λ dx (6.5.4)

=⇒ lnφ(x) =i

~

(A

2x2 − λx

)+ ln(C) (6.5.5)

Where C is a constant

=⇒ φ(x) = Ce−iλ~ xei

A2~x

2

(6.5.6)

Applying the boundary condition φ(a) = φ(−a)

=⇒ Ce−iλ~ aei

A2~a

2

= Ceiλ~ aei

A2~a

2

(6.5.7)

=⇒ ei2aλ~ = 1 (6.5.8)

=⇒ 2aλ

~= 2nπ for n ∈ N (6.5.9)

=⇒ λn =nπ~a

(6.5.10)

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6.6. Consider a large number (N) of identical experimental set-ups. In each of these, a single particle

is described by a wave function φ(x) = Ae−x2

a2 at t = 0, where A is the normalization constantand a is another constant with the dimension of length. If a measurement of the position ofthe particle is carried out at time t = 0 in all these set-ups, it is found that in 100 of these, theparticle is found within an infinitesimal interval of x = 2a to 2a + dx. Find out, in how manyof the measurements, the particle would have been found in the infinitesimal interval of x = ato a+ dx

Solution:

N[2a,2a+dx]

N[a,a+dx]

=P (2a < x < 2a+ dx)

P (a < x < a+ dx)(6.6.1)

=|φ(2a)|2 dx|φ(a)|2 dx = e−

(2a)2−a2

a2 = e−6 (6.6.2)

=⇒ N[a,a+dx] ≈ 100e6 ≈ 40343 (6.6.3)

6.7. ψ1(x) and ψ2(x) are the normalized eigenfunctions of an operator P , with eigen values P1 andP2 respectively. If the wave function of a particle is 0.25ψ1(x) + 0.75ψ2(x) at t = 0, find theprobability of observing P1

Solution: Since P corresponds to an observable, therefore it is a Hermitian Operator. Thusit’s eigenfunctions with distinct eigenvalues must be orthogonal

=⇒ 〈ψ1|ψ2〉 = 〈ψ2|ψ1〉 = 0 (6.7.1)

And 〈ψ1|ψ1〉 = 〈ψ2|ψ2〉 = 1 (6.7.2)

According to the postulates of QM, the probability of observing P1 is |〈ψ1|ψ〉|2〈ψ|ψ〉

where ψ(x) = 0.25ψ1(x) + 0.75ψ2(x)

〈ψ1|ψ〉 = 0.25〈ψ1|ψ1〉+ 0.75〈ψ1|ψ2〉 = 0.25 (6.7.3)

〈ψ|ψ〉 = 0.252〈ψ1|ψ1〉+ 0.752〈ψ2|ψ2〉 (6.7.4)

+ 0.25 · 0.75〈ψ1|ψ2〉+ 0.25 · 0.75〈ψ2|ψ1〉 = 0.252 + 0.752 (6.7.5)

=⇒ Probability of P1 =〈|ψ1|ψ〉|2〈ψ|ψ〉 =

0.252

0.252 + 0.752=

1

10≈ 0.1 (6.7.6)

6.8. An observable A is represented by the operator A. Two of its normalized eigen functions aregiven as φ1(x) and φ2(x), corresponding to distinct eigenvalues a1 and a2, respectively. Anotherobservable B is represented by an operator B. Two normalized eigenfunctions of this operatorare given as u1(x) and u2(x) with distinct eigenvalues b1 and b2, respectively. The eigenfunctionsφ1(x) and φ2(x) are related to u1(x) and u2(x) as

φ1 = D(3u1 + 4u2); φ2 = F (4u1 − Pu2)

At time t = 0, a particle is in a state given by 23φ1 + 1

3φ2

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(a) Find the values of D, F and P

Solution: We know that φ1, φ2, u1, u2 are normalized, and φ1, φ2 are orthogonal andso are u1, u2.

We begin by using normalization of φ1

=⇒ 〈φ1|φ1〉 = 1 (6.8.1)

=⇒ |D|2 (9〈u1|u1〉+ 16〈u2|u2〉 (6.8.2)

+12〈u2|u1〉+ 12〈u1|u2〉) = 1 (6.8.3)

=⇒ 25|D|2 = 1 (6.8.4)

=⇒ D = 0.2eiθ (6.8.5)

Where θ is any value ∈ [0, 2π)

Now we can use orthogonality of φ1 and φ2

〈φ1|φ2〉 = 0 (6.8.6)

=⇒ DF (12〈u1|u1〉 − 4P 〈u2|u2〉 (6.8.7)

− 3P 〈u1|u2〉+ 16〈u2|u1〉) = 0 (6.8.8)

=⇒ P = 3 (6.8.9)

Finally we use the normalization of φ2

=⇒ 〈φ2|φ2〉 = 1 (6.8.10)

=⇒ |F |2 (16〈u1|u1〉+ 9〈u2|u2〉 (6.8.11)

−12〈u2|u1〉 − 12〈u1|u2〉) = 1 (6.8.12)

=⇒ 25|F |2 = 1 (6.8.13)

=⇒ F = 0.2eiφ (6.8.14)

(b) If a measurement of A is carried out at t = 0, what are the possible results and what aretheir probabilities?

Solution: According to the postulates of QM, the possible results of an A measurement

are the eigenvalues a1 and a2, and the probability is |〈φ|ψ〉|2

〈ψ|ψ〉 where φ is the corresponding

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eigenfunction and ψ = 23φ1 + 1

3φ2

〈φ1|ψ〉 =2

3(6.8.15)

〈φ2|ψ〉 =1

3(6.8.16)

〈ψ|ψ〉 =4

9+

1

9=

5

9(6.8.17)

=⇒ Probability(a1) =4959

= 0.8 (6.8.18)

=⇒ Probability(a2) =1959

= 0.2 (6.8.19)

(c) Assume that the measurement of A mentioned above yielded a value a1. If a measurementof B is carried out immediately after this, what would be the possible outcomes and whatwould be their probabilities?

Solution: According to the postulates of QM, once a1 is measured, the wavefunctionwill collapse to φ1. Thus the wavefunction at the moment after measurement is φ1(x).Again we repeat the same exercise as above for u1 and u2. The possible outcomes areof course b1 and b2.

〈u1|φ1〉 = 0.6eiθ (6.8.20)

〈u2|φ1〉 = 0.8eiθ (6.8.21)

〈φ1|φ1〉 = 1 (6.8.22)

=⇒ Probability(b1) = 0.36 (6.8.23)

=⇒ Probability(b2) = 0.64 (6.8.24)

(d) If instead of following the above path, a measurement of B was carried out initially att = 0, what would be the possible outcomes and what would be their probabilities?

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Solution: Following same steps as above, the possible outcomes are same, b1 and b2

ψ =2

3φ1 +

1

3φ2 (6.8.25)

=

(2

5eiθ +

4

15eiφ)u1 +

(8

15eiθ − 1

5eiφ)u2 (6.8.26)

〈u1|ψ〉 =2

5eiθ +

4

15eiφ (6.8.27)

〈u2|ψ〉 =8

15eiθ − 1

5eiφ (6.8.28)

〈ψ|ψ〉 =5

9(6.8.29)

=⇒ Probability(b1) =9

5

(4

25+

16

225+ 2

8

75cos(θ − φ)

)(6.8.30)

=468

1125+

144

375cos(θ − φ) (6.8.31)

=⇒ Probability(b2) =9

5

(64

225+

1

25− 2

8

75cos(θ − φ)

)(6.8.32)

=657

1125− 144

375cos(θ − φ) (6.8.33)

(e) Assume that after performing the measurements described in (c), the outcome was b2.What would be the possible outcomes, if A were measured immediately after this andwhat would be the probabilities?

Solution: So immediately after measurement the wavefunction collapses to u2. Nowonce again the possibilities are a1 and a2

〈φ1|u2〉 = 〈u2|φ1〉∗ = 0.8e−iθ (6.8.34)

〈φ2|u2〉 = 〈u2|φ2〉∗ = −0.6e−iφ (6.8.35)

〈u2|u2〉 = 1 (6.8.36)

=⇒ Probability(a1) = 0.64 (6.8.37)

=⇒ Probability(a2) = 0.36 (6.8.38)

7 Free Particle

7.1. Consider a particle with one-dimensional wave function

φ(x) = N(a2 + x2)−12 eip0x~

where a, p0, N are real constants.

(a) Find the normalization constant N

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Solution: ˆRφ(x)∗φ(x) dx = 1 (7.1.1)

=⇒ N2

ˆ ∞x=−∞

1

a2 + x2dx = 1 (7.1.2)

=⇒ N2

ˆ π2

θ=−π2

a sec2 θ

a2 sec2 θdθ = 1 (7.1.3)

=⇒ N2π

a= 1 (7.1.4)

=⇒ N =

√a

π(7.1.5)

(b) Determine the probability of finding the particle in the interval − a√5≤ x ≤ a√

5

Solution:

P (− a√5≤ x ≤ a√

5) =

ˆ a√5

x=− a√5

φ(x)∗φ(x) dx (7.1.6)

=a

π

ˆ a√5

x=− a√5

1

a2 + x2dx (7.1.7)

=a

π

ˆ tan−1 1√5

θ=− tan−1 1√5

1

adθ (7.1.8)

=2

πtan−1 1√

5≈ 0.27 (7.1.9)

(c) What is the expectation value of the momentum?

Solution:

〈p〉 = −i~ˆRφ(x)∗

∂φ(x)

∂xdx (7.1.10)

= −i~aπ

ˆ ∞x=−∞

1√a2 + x2

e−ip0x~

(−x

√a2 + x2

3 +ip0~√a2 + x2

)eip0x~ dx (7.1.11)

=−ia~π

ˆ ∞x=−∞

p0~1

a2 + x2i− x

(a2 + x2)2dx (7.1.12)

=−ia~π

(ip0~ˆ ∞x=−∞

1

a2 + x2dx−

ˆ ∞x=∞

x

(a2 + x2)2dx

)(7.1.13)

= p0~2 (7.1.14)

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7.2. Show that

ψ(x) = A sin(kx) +B cos(kx)

and

ψ(x) = Ceikx +De−ikx

are equivalent solutions of TISE of a free particle. A, B, C and D can be complex numbers.

Solution: We already know that eikx and e−ikx are solutions to TISE for a free particle.Hence ψ(x) = Ceikx +De−ikx is also a solution to the TISE. We can also see that by takingA = i(C −D) and B = (C +D), ψ(x) = A sin(kx) +B cos(kx)

7.3. A particle with mass m is described by the following wave function:

ψ(x) = A sin(kx) +B cos(kx)

where A, B, k are constants

(a) For a free particle, show that ψ(x) is a solution of the Schrodinger equation and find itsenergy

Solution: We already saw in Q7.2 that ψ(x) is a solution to TISEThis must be then an energy eigenstate, and we want to find the energy eigenvalue

Hψ(x) = −A ~2

2m

d2 sin(kx)

dx2+−B ~2

2m

d2 cos(kx)

dx2(7.3.1)

=~2k2

2m(A sin(kx) +B cos(kx)) (7.3.2)

=~2k2

2mψ(x) (7.3.3)

Therefore the energy eigenvalue is ~2k2

2m

(b) Check whether ψ(x) = A sin(kx) or ψ(x) = B cos(kx) is an eigenfunction of the momentumoperator.

Solution:

P (A sin(kx)) = −i~Ad sin(kx)

dx= −i~kA cos(kx) (7.3.4)

P (A cos(kx)) = −i~Ad cos(kx)

dx= i~kA sin(kx) (7.3.5)

Thus none of them are momentum eigenfunctions.

7.4. Show thatΨ(x, t) = A sin(kx− ωt) +B cos(kx− ωt)

does not obey the time-dependant Schrodinger’s equation for a free particle

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Solution: We have the TDSE,

− ~2

2m

∂2ψ(x, t)

∂x2= i~

∂ψ(x, t)

∂t(7.4.1)

LHS = − ~2

2m

∂2A sin(kx− ωt) +B cos(kx− ωt)∂x2

(7.4.2)

=~2k2

2mA sin(kx− ωt) +B cos(kx− ωt) (7.4.3)

RHS = i~∂A sin(kx− ωt) +B cos(kx− ωt)

∂t(7.4.4)

= −i~(A cos(kx− ωt)−B sin(kx− ωt)) 6= LHS (7.4.5)

7.5. Find the expectation value of the square of the momentum squared for the particle in the state:

Ψ(x, t) = Aei(kx−ωt)

What conclusion can you draw for this solution?

Solution:

〈P 2〉 =

´R Ψ(x, t)∗P PΨ(x, t) dx´

R Ψ(x, t)∗Ψ(x, t)dx (7.5.1)

= ~2k2

´R Ψ(x, t)∗Ψ(x, t) dx´R Ψ(x, t)∗Ψ(x, t) dx

(7.5.2)

= ~2k2 (7.5.3)

Conclusion is that this has expected energy equal to energy eigenvalue ~2k2

2m

7.6. A free proton has a wave function given by

Ψ(x, t) = Aei(5.02×1011x−8.00×1015t)

The coefficient of x is inverse meters, and the coefficient of t is inverse seconds. Find itsmomentum and energy.

Solution: We know that Ψ(x, t) is a momentum eigenfunction with momentum eigenvalue~k = 3.32× 10−22 kg ms−1, which is also the value of expected momentumWe also know that Ψ(x, t) is an energy eigenfunction with energy eigenvalue~ω = 5.3× 10−18 J, which is also the value of expected energy

7.7. The wave function for a particle is given by

φ(x) = Aeikx +Be−ikx

where A and B are real constants. Show that φ(x)∗φ(x) is always a positive quantity.

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Solution:

φ(x)∗φ(x) = (Ae−ikx +Beikx)(Aeikx +Be−ikx) (7.7.1)

= A2 +B2 + AB(e2ikx + e−2ikx) (7.7.2)

= A2 +B2 + 2AB cos(2kx) (7.7.3)

≥ A2 +B2 − 2AB (7.7.4)

= (A−B)2 ≥ 0 (7.7.5)

8 Infinite Potential Box

8.1. For a particle in a 1-D box of side L, show that the probability of finding the particle betweenx = a and x = a+ b approaches the classical value b

L, if the energy of the particle is very high.

Solution: Let us assume the particle to be in one of the energy eigenstates ψn(x)

Prob (a < x < a+ b) =

ˆ a+b

x=a

ψn(x)∗ψn(x) dx (8.1.1)

=2

L

ˆ a+b

x=a

sin2(nπ

Lx)dx (8.1.2)

=1

L

ˆ a+b

x=a

1− cos(

2nπ

Lx)dx (8.1.3)

=1

2nπ

ˆ 2nπL

(a+b)

x= 2nπLa

1− cosx dx (8.1.4)

=1

2nπ

(2nπ

Lb− sin

(2nπ

L(a+ b)

)+ sin

(2nπ

La

))(8.1.5)

→ b

Las n→∞

(8.1.6)

8.2. Consider a particle confined to a 1-D box. Find the probability that the particle in its groundstate will be in the central one-third region of the box.

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Solution:

Prob(L

3< x <

2L

3) =

ˆ 2L3

x=L3

ψ1(x)∗ψ1(x) dx (8.2.1)

=2

L

ˆ 2L3

x=L3

sin2(πLx)dx (8.2.2)

=1

L

ˆ 2L3

x=L3

1− cos(

2nπ

Lx)dx (8.2.3)

=1

2π

ˆ 4π3

x= 2π3

1− cosx dx (8.2.4)

=1

2π

(2π

3− sin

(4π

3

)+ sin

(2π

3

))(8.2.5)

=1

3+

√3

2π(8.2.6)

8.3. Consider a one dimensional infinite square well potential of length L. A particle is in n = 3state of this potential well. Find the probability that this particle will be observed betweenx = 0 and x = L/6. Can you guess the answer without solving the integral? Explain how.

Solution: As we can easily see,´ L

6

0sin2

(3 πLx)dx =

´ L3L6

sin2(3 πLx)dx = · · ·

=⇒ Prob(0 < x < L

6

)= 1

6

8.4. Solve the time independent Schrodinger equation for a particle in a 1-D box, taking the originat the centre of the box and the ends at ±L

2, where L is the length of the box.

Solution: We have

V (x) =

∞ x < −L2

0 −L2≤ x ≤ L

2

∞ x > L2

(8.4.1)

Applying TISE

=⇒ 1

ψ(x)

d2ψ(x)

dt2= −k2 − L

2≤ x ≤ L

2(8.4.2)

Where

−k2 = −2mE

~2(8.4.3)

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Solving the differential equation, we get

ψ(x) =

0 x < −L2

A cos(kx) +B sin(kx) −L2≤ x ≤ L

2

0 x > L2

(8.4.4)

Imposing the boundary conditions

Continuity at x = −L2

0 = A cos

(kL

2

)−B sin

(kL

2

)(8.4.5)

Continuity at x = L2

A cos

(kL

2

)+B sin

(kL

2

)= 0 (8.4.6)

Using (8.4.5) we get tan(kL2

)= A

B. Putting this in (8.4.6),

B sin(kL) = 0 (8.4.7)

=⇒ B = 0 or k =nπ

L(8.4.8)

=⇒ cot

(kL

2

)= 0 or k =

nπ

L(8.4.9)

=⇒ k =(2n′ + 1)π

Lor k =

nπ

L(8.4.10)

=⇒ k =nπ

L(8.4.11)

Because the second case covers the first case

tankL

2= 0 n even (8.4.12)

cotkL

2= 0 n odd (8.4.13)

=⇒ A = 0 n even (8.4.14)

=⇒ B = 0 n odd (8.4.15)

=⇒ ψn(x) =

{A cos

(n πLx)

n odd

B sin(n πLx)

n even(8.4.16)

Finally we can normalize ψ(x) to get A = B =√

2L

ψn(x) =

√2L

cos(n πLx)

n odd√2L

sin(n πLx)

n even(8.4.17)

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8.5. Consider a particle of mass m in an infinite potential well extending from x = 0 to x = L.Wavefunction of the particle is given by

ψ(x) = A

(sin(πLx)

+ sin

(2π

Lx

))

where A is the normalization constant

(a) Calculate A

Solution: ˆRψ(x)∗ψ(x) dx = 1 (8.5.1)

=⇒ |A|2ˆR

sin2(πLx)

+ sin2

(2π

Lx

)+��

��

��:0

2 sin(πLx)

sin

(2π

Lx

)= 1 (8.5.2)

=⇒ L|A|2 = 1 (8.5.3)

=⇒ A =eiθ√L

(8.5.4)

(b) Calculate the expectation values of x and x2 and hence the uncertainty ∆x

Solution:

〈x〉 =

ˆRψ(x)∗xψ(x) dx (8.5.5)

=1

L

ˆ L

x=0

x sin2(πLx)

+ x sin2

(2π

Lx

)+ 2x sin

(πLx)

sin

(2π

Lx

)(8.5.6)

=1

L

ˆ L

x=0

x− xcos(2 πLx)

2− xcos

(4 πLx)

2+ x cos

(πLx)− x cos

(3π

Lx)dx (8.5.7)

=1

L

ˆ L

x=0

x+sin(2 πLx)

4 πL

+sin(4 πLx)

8 πL

− sin(πLx)

πL

+sin(3 πLx)

3 πL

dx (8.5.8)

=L

2− 2L

π2+

2L

9π2(8.5.9)

= L

(1

2− 16

9π2

)(8.5.10)

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〈x2〉 =

ˆRψ(x)∗x2ψ(x) dx (8.5.11)

=1

L

ˆ L

x=0

x2 sin2(πLx)

+ x2 sin2

(2π

Lx

)+ 2x2 sin

(πLx)

sin

(2π

Lx

)(8.5.12)

=1

L

ˆ L

x=0

x2 − x2 cos(2 πLx)

2− x2 cos

(4 πLx)

2+ x2 cos

(πLx)− x2 cos

(3π

Lx)dx

(8.5.13)

=1

L

ˆ L

x=0

x2 + xsin(2 πLx)

2 πL

+ xsin(4 πLx)

4 πL

− 2xsin(πLx)

πL

+ 2xsin(3 πLx)

3 πL

dx (8.5.14)

= − 1

L

(x

cos(2 πLx)

4 π2

L2

+ xcos(4 πLx)

16 π2

L2

− 2xcos(πLx)

π2

L2

+ 2xcos(3 πLx)

9 π2

L2

)∣∣∣∣∣

L

x=0

(8.5.15)

+1

L

ˆ L

x=0

x2 +cos(2 πLx)

4 π2

L2

+cos(4 πLx)

16 π2

L2

− 2cos(πLx)

π2

L2

+ 2cos(3 πLx)

9 π2

L2

dx (8.5.16)

=L2

3− L2

4π2− L2

16π2− 2L2

π2+

2L2

9π2(8.5.17)

= L2

(1

3− 301

144π2

)(8.5.18)

∆x =√〈x2〉 − 〈x〉2 (8.5.19)

= L

√1

3− 301

144π2− 1

4− 256

81π4+

16

9π2(8.5.20)

= L

√1

12− 5

16π2− 256

81π4(8.5.21)

(c) Calculate the expectation values of p and p2 and hence the uncertainty ∆p

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Solution:

〈p〉 = −i~ˆRψ(x)∗

dψ(x)

dxdx (8.5.22)

= −i~ˆx∈R

ψ(x)dψ(x) (8.5.23)

=−i~

2ψ(x)2

∣∣∞x=−∞ (8.5.24)

= 0 (8.5.25)

〈p2〉 = −~2

ˆRψ(x)∗

d2ψ(x)

dx2dx (8.5.26)

= −~2

L

(−π

2

L2

ˆ L

x=0

sin2(πLx)dx− 4π2

L2

ˆ L

x=0

sin2

(2π

Lx

)dx (8.5.27)

−(π2

L2+

4π2

L2

) ˆ L

x=0

sin(πLx)

sin

(2π

Lx

)dx

)(8.5.28)

=~2π2

L3

(5L

2− 1

2

ˆ L

x=0

cos

(2π

Lx

)dx− 2

ˆ L

x=0

cos

(4π

Lx

)(8.5.29)

+5

2

ˆ L

x=0

cos(πLx)− cos

(3π

Lx

)dx

)(8.5.30)

=5~2π2

2L2(8.5.31)

∆p =√〈p2〉 − 〈p〉2 (8.5.32)

=

√5

2

~πL

(8.5.33)

(d) What is the probability of finding the particle in the first excited state, if an energy mea-surement is made?

Solution: As we know, the probability of finding the particle in first excited stateψ2(x) upon energy measurement is simply |〈ψ2(x)|ψ(x)〉|2

〈ψ2(x)|ψ(x)〉 = 〈ψ2(x)| 1√2

(ψ1(x) + ψ2(x))〉 =1√2

(8.5.34)

=⇒ Probability =1

2(8.5.35)

8.6. Suppose we have 10000 rigid boxes of same length L from x = 0 to x = L. Each box containsone particle of mass m. All these particles are in the ground state

(a) If a measurement of position of the particle is made in all the boxes at the same time, inhow many of them, the particle is expected to be found between x = 0 and L

4

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Solution: Before measurement the wavefunction is ψ(x) =√

2L

sin(πL

)x

=⇒ Probability(0 ≤ x ≤ L4) =´ L

4

x=0|ψ(x)|2 dx

=2

L

ˆ L4

x=0

sin2(πLx)dx (8.6.1)

=1

L

ˆ L4

x=0

1− cos

(2π

Lx

)dx (8.6.2)

=1

L

(L

4− L

2π

)(8.6.3)

= 0.091 (8.6.4)

=⇒ N(0 ≤ x ≤ L

4) = 910 (8.6.5)

(b) In a particular box, the particle was found to be between x = 0 and L4. Another measure-

ment of the position of the particle is carried out in this box immediately after the firstmeasurement. What is the probability that the particle is again found between x = 0 andL4

Solution:

0 L4

L

√2L

x

ψ(x)

ψ(x) before collapseψ(x) after collapse

As we can see, upon the first measurement the wavefunction must collapse to a functionwhich is non zero only inside x ∈ [0, L

4]. Thus if we again immediately carry out a

measurement, the only non zero values are inside x ∈ [0, L4]. Therefore the probability

for the second measurement to yield a value inside x ∈ [0, L4] is 1

8.7. An electron is bound in an infinite potential well extending from x = 0 to x = L. At time t = 0,its normalized wave function is given by

ψ(x, 0) =2√L

sin

(3πx

2L

)cos(πx

2L

)

(a) Calculate ψ(x, t) at a later time t.

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Solution: First we need to decompose ψ(x, 0) into energy eigenstates

ψ(x, 0) =2√L

sin

(3πx

2L

)cos(πx

2L

)(8.7.1)

=1√L

(sin(πLx)

+ sin(

2π

Lx))

(8.7.2)

=1√2

(ψ1(x) + ψ2(x)) (8.7.3)

=⇒ ψ(x, t) =1√2

(ψ1(x)e−i

E1~ t + ψ2(x)e−i

E2~ t)

(8.7.4)

=1√L

(sin(πLx)e−i

~π2

2mL2 t + sin(

2π

Lx)e−i

2~π2

mL2 t

)(8.7.5)

(b) Calculate the probability of finding the electron between x = L/4 and x = L/2 at time t.

Solution:

Prob

(L

4< x <

L

2

)=

ˆ L2

x=L4

ψ(x, t)∗ψ(x, t) dx (8.7.6)

=1

L

ˆ L2

x=L4

sin2(πLx)

+ sin2(

2π

Lx)dx (8.7.7)

+1

L

ˆ L2

x=L4

sin(πLx)

sin(

2π

Lx)(

ei3~π2

2mL2 t + e−i3~π2

2mL2 t

)dx (8.7.8)

=π + 2

8π+

1

8+

4−√

2

3πcos

(3~π2

2mL2t

)(8.7.9)

(8.7.10)

8.8. A speck of dust (m = 1 µg) is trapped to roll inside a tube of length L = 1.0 µm. The tubeis capped at both ends and the motion of the speck is considered to be along the length of thetube.

(a) Modeling this as a 1-D infinite square well, determine the value of the quantum number nif the speck has an energy of 1 µJ

Solution: Assuming the particle is in one of the energy eigenstates,

n2 ~2π2

2mL2= E (8.8.1)

=⇒ n =L

~π√

2mE ≈ 4.27× 1021 (8.8.2)

(b) What is the probability of finding this speck within 0.1 µm of the center of the tube(0.45 < x < 0.55)

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Solution: Using (9.6.4)

Prob (a < x < a+ b) =1

2nπ

(2nπ

Lb− sin

(2nπ

L(a+ b)

)+ sin

(2nπ

La

))(8.8.3)

≈ 1

10(8.8.4)

(c) How much energy is needed to excite this speck to an energy level next to 1 µJ? Comparethis excitation energy with the thermal energy at room temperature (T = 300K).

Solution:

∆E ≈ 2n~2π2

2mL2=

2E

n= 4.7× 10−28 J (8.8.5)

Thermal energy at room temperature

= 1.5kBT = 6.2× 10−21 J (8.8.6)

8.9. Consider a particle bound inside an infinite well whose floor is sloping (variation is small) asshown in the figure. Without solving the Schrodinger equation (provide proper justification foryour answers)

0

V0x0

E1

E2

(a) Sketch a plausible wave function when the energy is E1, assuming that it has no nodes.

Solution: We will use here what is commonly called the WKB approximation. Weknow that

V (x) =

∞ x < 0xLV0 0 ≤ x ≤ L

∞ x > L

(8.9.1)

We want the energy eigenfunction with energy eigenvalue E1 < V0. Let us say thatV (x0) = E1. Therefore, for 0 ≤ x ≤ x0, V (x) < E1, and for x0 < x ≤ L, V (x) > E1

Therefore we have the TISE as

1

ψ(x)

d2ψ(x)

dx2=

{−k(x)2 0 < x < x0

κ(x)2 x0 ≤ x < L(8.9.2)

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Where

k(x) =

√2m(E1 − V (x))

~2

κ(x) =

√2m(V (x)− E1)

~2

Thus, when 0 < x < x0 and x is not too close to x0, then k(x) is gradually varyingin its neighborhood, and we can treat it as such while solving TISE. Similarly whenx0 ≤ x < L, and x is not too close to x0, then κ(x) is gradually varying in itsneighborhood, and we can treat it as such while solving TISE.

=⇒ ψ(x) ≈

A sin(k(x)x) 0 < x < x0 − ε?? x0 − ε ≤ x ≤ x+ ε

Be−κ(x)x x0 + ε < x < L

(8.9.3)

Note that when x is close x0 we really cannot use these approximations, hence we cannotwrite the wavefunction so simply in a neighborhood of ε around x0. Thus the maintakeaway (and what you need to write in the examination, you really don’t need theabove explanation in that detail to be written in the exam), is that the wavefunction isapproximately sinusoidal for 0 < x < x0, and approximately exponentially decaying inx0 < x < L, with these solutions being patched together using appropriate boundaryconditions. The graph thus will be approximately

0 x0 Lx

ψ(x)

(b) Sketch the wave function with 5 nodes when the energy is E2

Solution: Using similar logic as above, we have

ψ(x) ≈ A sin(k(x)x) 0 < x < L (8.9.4)

Since V (x) is increasing as we increase x, therefore the wavelength 2πk(x)

will also increaseas we increase x. The graph is thus approximately

55 of 95


0 L

0

x

ψ(x)

9 Finite Potential Well

9.1. Consider an asymmetric finite potential well of width L, with a barrier V1 on one side and abarrier V2 on the other side (figure below). Obtain the energy quantization condition for thebound states in such a well. From this condition derive the energy quantization conditions fora semi-infinite potential well (when V1 →∞ and V2 is finite)

V (x) = V1 V (x) = 0 V (x) = V2

Solution: We can write the potential as

V (x) =

V1 x < 0

0 0 ≤ x ≤ L

V2 x > L

(9.1.1)

We are looking for a bound state, i.e. an energy eigenstate whose energy eigenvalue E <V (±∞) =⇒ E < V2 (WLOG we assume V2 < V1)Let us first solve the TISE differential equation regionwise which we will patch together by

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imposing boundary conditions

1

ψ(x)

d2ψ(x)

dx2=

κ21 x < 0

−k2 0 ≤ x ≤ L

κ2 x > L

(9.1.2)

ψ(x) =

0 x→ −∞Aeκ1x +Be−κ1x x < 0

C cos(kx) +D sin(kx) 0 ≤ x ≤ L

Feκ2x +Ge−κ2x x > L

0 x→∞

(9.1.3)

Where

κ21 =

2m(V1 − E)

~2

−k2 =−2mE

~2

κ22 =

2m(V2 − E)

~2

Since ψ(x) cannot blow up at ±∞, B = F = 0Since we want our solution to be fully real, we can safely take A, C, D, G to be realNow to impose boundary conditions

Continuity and Differentiability at x = 0

A = C (9.1.4)

Aκ1 = Dk (9.1.5)

Continuity and Differentiability at x = L

C cos(kL) +D sin(kL) = Ge−κ2L (9.1.6)

−k(C sin(kL)−D cos(kL)) = −κ2Ge−κ2L (9.1.7)

We put the values of C and D from (9.1.4) and (9.1.5) and write (9.1.7)(9.1.6)

sin(kL)− κ1

kcos(kL)

cos(kL) + κ1

ksin(kL)

=κ2

k(9.1.8)

=⇒ tan(kL)− κ1

k

1 + κ1

ktan(kL)

=κ2

k(9.1.9)

=⇒ tan(kL) =κ1

k+ κ2

k

1− κ1

kκ2

k

(9.1.10)

=⇒ tan(kL) = tan(

tan−1(κ1

k

)+ tan−1

(κ2

k

))(9.1.11)

=⇒ tan−1(κ1

k

)+ tan−1

(κ2

k

)= kL± nπ (9.1.12)

=⇒ tan−1

(√k2

1

k2− 1

)+ tan−1

(√k2

2

k2− 1

)= kL± nπ (9.1.13)

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Where

k1 =

√2mV1

~2(9.1.14)

k2 =

√2mV2

~2(9.1.15)

This was the quantization condition we were looking for. You can use any equation from(9.1.9)-(9.1.13) as the answer. We will solve this equation numerically to get the value of k,and hence E which are validTo find the condition for semi-infinite well, we will use V1 → ∞ =⇒ k1 → ∞ =⇒tan−1

(√k2

1

k2 − 1

)→ π

2

=⇒ tan−1

(√k2

2

k2− 1

)= kL± (2n′ + 1)

π

2(9.1.16)

=⇒√k2

2

k2− 1 = cot(kL) (9.1.17)

9.2. A particle of mass m is bound in a double well potential shown in the figure below. Its energyeigenstate ψ(x) has energy eigenvalue E = V0, where V0 is the energy of the plateau in themiddle of the potential well. It is known that ψ(x) = C, a constant in the plateau region.

V (x) = ∞ V (x) = 0 V (x) = V0 V (x) = 0 V (x) = ∞

(a) Obtain ψ(x) in the regions −2L ≤ x ≤ −L and L ≤ x ≤ 2L and the relation between thewave number k and L

Solution: Again, we write the potential as

V (x) =

∞ x < −2L

0 −2L ≤ x < −LV0 −L ≤ x ≤ L

0 L < x ≤ 2L

∞ x > 2L

(9.2.1)

Next we can solve the TISE differential equation regionwise and patch them usingboundary conditions, keeping in mind that we are looking for the stationary state with

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energy eigenvalue E = V0, if at all that exists

1

ψ(x)

d2ψ(x)

dx2=

−k2 −2L ≤ x < −L0 −L ≤ x ≤ L

−k2 L < x ≤ 2L

(9.2.2)

=⇒ ψ(x) =

0 x < −2L

A cos(kx) +B sin(kx) −2L ≤ x < −LDx+ C −L ≤ x ≤ L

F cos(kx) +G sin(kx) L < x ≤ 2L

0 x > 2L

(9.2.3)

Where

−k2 = −2mE

~2(9.2.4)

Imposing boundary conditions

Continuity at x = −2L

0 = A cos(2kL)−B sin(2kL) (9.2.5)

Continuity and Differentiability at x = −L

A cos(kL)−B sin(kL) = −DL+ C (9.2.6)

k(A sin(kL) +B cos(kL)) = D (9.2.7)


DL+ C = F cos(kL) +G sin(kL) (9.2.8)

D = k(−F sin(kL) +G cos(kL)) (9.2.9)

Continuity at x = 2L

F cos(2kL) +G sin(2kL) = 0 (9.2.10)

Thus now we have 6 equations and 6 variables, with an additional normalization con-straint. Thus we will get a set of k, and thus E which will satisfy these. If E = V0 is notone of them, then we don’t have solution with E = V0 at all. And we additionally needthe condition such that this wavefunction Dx+ C is constant in the middle region (asspecified in the question). If this condition is not satisfied for E = V0, then again we donot have any solution in which ψ(x) is constant in the middle region. Thus =⇒ D = 0

From (9.2.5) and (9.2.10) we get

tan(2kL) =A

B=−FG

(9.2.11)

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Putting in (9.2.6), (9.2.7), (9.2.8), (9.2.9),

B sin(kL) = C cos(2kL) (9.2.12)

B cos(kL) = 0 (9.2.13)

C cos(2kL) = −G sin(kL) (9.2.14)

0 = G cos(kL) (9.2.15)

Thus we get cos(kL) = 0 =⇒ kL =(2n+ 1)π

2. This was the condition we were

looking for. This also makes total sense if we look at the problem physicallyWe also get tan(2kL) = 0 =⇒ A = F = 0, and −B = G = (−1)nCThus our wavefunction becomes

ψ(x) =

0 x < −2L

(−1)n+1C sin(

(2n+1)2

π xL

)−2L ≤ x < −L

C −L ≤ x ≤ L

(−1)nC sin(

(2n+1)2

π xL

)L < x ≤ 2L

0 x > 2L

(9.2.16)

Since the boundary conditions are satisfied fully, the necessary and sufficient conditionfor such a solution to exist is

√2mV0

~ L = (2n+1)π2

(b) Determine C in terms of L

Solution: To find ψ(x) fully we must also impose the normalization condition

2C2

ˆ 2L

x=L

sin2

(2n+ 1

2πx

L

)dx+ 2C2L = 1 (9.2.17)

=⇒ 3C2L = 1 (9.2.18)

=⇒ C =

√1

3L(9.2.19)

=⇒ ψ(x) =

0 x < −2L

−(−1)n√

13L

sin(

(2n+1)2

π xL

)−2L ≤ x < −L√

13L

−L ≤ x ≤ L

(−1)n√

13L

sin(

(2n+1)2

π xL

)L < x ≤ 2L

0 x > 2L

(9.2.20)

(c) Assume the bound particle to be an electron and L = 1 A. Calculate the two lowest valuesof V0 (in eV) for which such a solution exists.

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Solution: As solved above, the necessary and sufficient condition for such a solutionto exist is

√2mV0

~ L = (2n+1)π2

=⇒ V0 =(2n+ 1)2π2~2

8mL2(9.2.21)

Thus the two lowest values of V0 are obtained by putting n = 0 and n = 1, giving usV0 = 9.34 eV, V0 = 84 eV

(d) For the smallest allowed k, calculate the expectation values for x, x2, p, p2 and show thatthe uncertainty principle is obeyed.

Solution: The smallest allowed value of k is for n = 0

=⇒ ψ(x) =

0 x < −2L

−√

13L

sin(π2xL

)−2L ≤ x < −L√

13L

−L ≤ x ≤ L√1

3Lsin(π2xL

)L < x ≤ 2L

0 x > 2L

(9.2.22)

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〈x〉 =

ˆ x=−L

x=−2L

1

3Lx sin2

(πx2L

)dx+

ˆ x=2L

x=L

1

3Lx sin2

(πx2L

)dx+

ˆ x=L

x=−L

1

3Lxdx

(9.2.23)

= 0 (9.2.24)

〈x2〉 =

ˆ x=−L

x=−2L

1

3Lx2 sin2

(πx2L

)dx+

ˆ x=2L

x=L

1

3Lx2 sin2

(πx2L

)dx+

ˆ x=L

x=−L

1

3Lx2 dx

(9.2.25)

=2

3L

(ˆ x=2L

x=L

x2 sin2(πx

2L

)dx+

ˆ x=L

x=0

x2 dx

)(9.2.26)

= 2L2

(1

2− 1

π2

)(9.2.27)

〈p〉 =

ˆ x=−L

x=−2L

−iπ~6L2

sin(πx

2L

)cos(πx

2L

)dx+

ˆ x=2L

x=L

−iπ~6L2

sin(πx

2L

)cos(πx

2L

)dx

(9.2.28)

= 0 (9.2.29)

〈p2〉 =

ˆ x=−L

x=−2L

π2~2

12L3sin2

(πx2L

)dx+

ˆ x=2L

x=L

π2~2

12L3sin2

(πx2L

)dx (9.2.30)

=π2~2

6L3

ˆ x=2L

x=L

sin2(πx

2L

)dx (9.2.31)

=π2~2

12L2(9.2.32)

=⇒ ∆x∆p =√〈x2〉〈p2〉 (9.2.33)

=

√π2

12− 1

6~ = 0.81~ > 0.5~ (9.2.34)

9.3. You are given an arbitrary potential V (x) and the corresponding orthogonal and normalizedbound-state solutions to the time independent Schrodinger’s equation ψn(x), with the cor-responding energy eigenvalues En. At time t = 0, the system is in the state, ψ(x, 0) =A(ψ1(x) + ψ2(x) + ψ4(x))

(a) Find the value of A

Solution: We start with normalization of ψ at t = 0

〈ψ|ψ〉 = 1 (9.3.1)

=⇒ 3|A|2 = 1 (9.3.2)

=⇒ A =1√3

(9.3.3)

(In general A = eiθ√3

but we will force our stationary states to be real, which we can

always do)

(b) What is the wave function at time t > 0

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Solution: We already have our decomposition of ψ into energy eigenstates, hence timeevolution is trivial

ψ(x, t) =1√3

(ψ1(x)e−iE1~ t + ψ2(x)e−i

E2~ t + ψ4(x)e−i

E4~ t) (9.3.4)

(c) What is the expectation value of the energy at time t > 0

Solution:

〈E〉 = 〈ψ(x, t)|Hψ(x, t)〉 (9.3.5)

=E1 + E2 + E4

3(9.3.6)

9.4. A finite square well (height = 30 eV, width = 2a, from −a to a), has six bound states 3, 7, 12,17, 21, and 24 eV. If instead, the potential is semi infinite, with an infnite wall at x = 0, howmany bound states will exist and what are the energies associated with it? Justify your answer

Solution: For the finite well the potential is

V1(x) =

V0 x < −a0 −a ≤ x ≤ a

V0 x > a

(9.4.1)

For the semi infinite well the potential is

V2(x) =

∞ x < 0

0 0 ≤ x ≤ a

V0 x > a

(9.4.2)

where V0 = 30 eVNow we can see that for x > 0, V1(x) = V2(x). We also know that ψ(2)(x) = 0 for x < 0,where ψ(2)(x) is any stationary state for V2(x). Therefore any stationary state ψ(1)(x) forthe finite well should also match a stationary state ψ(2)(x) for x ≥ 0, as long as ψ(1)(0) = 0,since the differential equation for both V1(x) and V2(x) are same for the region x > 0. Weknow that ψ(1)(x) = 0 for the odd states, which will thus have energies 7, 17, 24 eV. Thusthose are the energies for V2(x) as well, and thus the semi infinite well will have 3 boundstates

9.5. A particle with energy E is bound in a finite square well potential with height U and width 2L(from −L to L)

(a) If E < U, obtain the energy quantization condition for the symmetric wave functions in

terms of k and α, where k =√

2mE~ and α =

√2m(U−E)

~

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Solution: We can write the potential as

V (x) =

U x < −L0 −L ≤ x ≤ L

U x > L

(9.5.1)

We are looking for a bound state, i.e. an energy eigenstate whose energy eigenvalueE < V (±∞) =⇒ E < ULet us first solve the TISE differential equation regionwise which we will patch togetherby imposing boundary conditions

1

ψ(x)

d2ψ(x)

dx2=

α2 x < −L−k2 −L ≤ x ≤ L

α2 x > L

(9.5.2)

ψ(x) =

0 x→ −∞Aeαx +Be−αx x < −LC cos(kx) +D sin(kx) −L ≤ x ≤ L

Feαx +Ge−αx x > L

0 x→∞

(9.5.3)

Again, B = F = 0 to prevent ψ(x) from blowing up at ±∞.Since we are looking for symmetric solutions, we have A = G, D = 0Imposing boundary conditions, we have

Continuity and Differentiability at x = −L

Ae−αL = C cos(kL) (9.5.4)

αAe−αL = kC sin(kL) (9.5.5)


C cos(kL) = Ae−αL (9.5.6)

kC sin(kL) = αAe−αL (9.5.7)

We have two redundant equations, so the assumption of symmetric solution was con-sistentLet us write (9.5.5)

(9.5.4)

tan kL =α

k(9.5.8)

Thus this is the quantization condition we were looking for

(b) Apply this result to an electron trapped inside a defect site in a crystal. Modeling thisdefect as a finite square well potential with height 5 eV and width 200 pm, calculate theground state energy

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Solution: As we know, the ground state is a symmetric one, hence the ground stateenergy is the lowest k and thus lowest E satisfying tan(kL) = α

k

Writing k0 =√

2mU~ , we have α

k=

√k2

0L2

k2L2 − 1. We also have k0L = 2.298Thus we can

now solve this equation numerically to get kL = 1.081 =⇒ E = ~2(kL)2

2mL2 = 1.1 eV

(c) Calculate the total number of bound states with symmetric wave-function

Solution:

0 1 2 3 40

1

2

3

4

kL

√k0LkL

− 1tan(kL)

As is clear from the picture, as we increase k0, a new bound state is possible whenever

the x intercept of√

k0LkL− 1 crosses a multiple of π. Thus the number of bound states

is dk0Lπe = d

√2mUL~π e

9.6. A potential V (x) is defined over a region R, which consists of two sub regions R1 and R2

(R = R1

⋃R2). This potential has two normalized energy eigenfunctions Ψ1(x) and Ψ2(x)

with energy eigenvalues E1 and E2 (E1 6= E2), respectively. ψ1(x) = 0 outside region R1 andψ2(x) = 0 outside region R2

(a) Suppose the regions R1 and R2 do not overlap, show that the particle will stay there forever,if it is in the region R1

Solution: The wavefunction Ψ(x, 0) for the particle at t = 0 can be written as

Ψ(x, 0) = aΨ1(x) + bΨ2(x) (9.6.1)

Since the particle is in region R1 at t = 0, Ψ(x, 0) = 0 ∀x ∈ R2

Since R1

⋂R2 = φ =⇒ b = 0, a = 1

Thus the wavefunction evolved to some time t is

Ψ(x, t) = Ψ1(x)e−iE1~ t (9.6.2)

which is again = 0 in region R2

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(b) If the initial state is Ψ(x, 0) = 1√2(Ψ1(x) + Ψ2(x)), show that the probability density

|Ψ(x, t)|2 is independant of time

Solution:

Ψ(x, t) =1√2

(Ψ1(x)e−iE1~ t + Ψ2(x)e−i

E2~ t) (9.6.3)

=⇒ |Ψ(x, t)|2 = Ψ(x, t)∗Ψ(x, t) =1

2

(|Ψ1(x)|2 + |Ψ(x)|2

+ Ψ1(x)∗Ψ2(x)eiE1−E2

~ t + Ψ2(x)∗Ψ1(x)eiE2−E1

~ t)

(9.6.4)

=1

2

(|Ψ1(x)|2 + |Ψ(x)|2

)(9.6.5)

∵ R1

⋂R2 = φ, Ψ1(x)∗Ψ2(x) = Ψ1(x)Ψ2(x)∗ = 0

=⇒ d|Ψ(x, t)|2dt

= 0 (9.6.6)

(c) If the regions R1 and R2 overlap, show that the probability density |Ψ(x, t)|2 oscillate intime for the initial state given in (b)

Solution: From (9.6.4)

|ψ(x, t)|2 =1

2

(|Ψ1(x)|2 + |Ψ(x)|2 + Re

[Ψ1(x)∗Ψ2(x)ei

E1−E2~ t

])(9.6.7)

=1

2

(|Ψ1(x)|2 + |Ψ(x)|2

+ Re [Ψ1(x)∗Ψ2(x)] cos

(E1 − E2

~t

)− Im [Ψ1(x)∗Ψ2(x)] sin

(E1 − E2

~t

))

(9.6.8)

10 Step Potential

10.1. A potential barrier is defined by V = 0 for x < 0 and V = V0 for x > 0. Particles with energyE < V0 approaches the barrier from the left

(a) Find the value of x = x0 (x0 > 0), for which the probability density is 1e

times theprobability density at x = 0.

Solution: The step potential can be written like

V (x) =

{0 x < 0

V0 x ≥ 0(10.1.1)

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Thus we can write the TISE for energy eigenvalue E < V0 as

1

ψ(x)

d2ψ(x)

dx2=

{−k2 x < 0

κ2 x ≥ 0(10.1.2)

=⇒ ψ(x) =

{Aeikx +Be−ikx x < 0

Ceκx +De−κx x ≥ 0(10.1.3)

Where

−k2 = −2mE

~2(10.1.4)

κ2 =2m(V0 − E)

~2(10.1.5)

We must have C = 0 such that ψ(x) does not blow up at infinity.We could impose the boundary conditions at this point but that would not lead toanything useful or relevant to the question itself. We finally want the probabilitydensity at x ≥ 0

=⇒ ProbDens(x) = ψ(x)∗ψ(x) (10.1.6)

= |C|2e−2κx (10.1.7)

We wantProbDens(x0)

ProbDens(0)=

1

e(10.1.8)

=⇒ e−2κx0 =1

e(10.1.9)

=⇒ x0 =1

2κ=

~√8m(V0 − E)

(10.1.10)

(b) Take the maximum allowed uncertainty ∆x for the particle to be localized in the classicallyforbidden region as x0. Find the uncertainty this would cause in the energy of the particle.Can then one be sure that its energy E is less than V0?

Solution: We want to analyse what happens to the wavefunction, and hence theparticle if it was originally in the energy eigenstate with eigenvalue E < V0, and aposition observation leads us to find the position result as x > 0Of course our position detector will not be perfect, and won’t perform a perfect positionmeasurement.Hence after the measurement the wavefunction will collapse to a state close to theposition eigenstate (a delta function), i.e. it will collapse to a gaussian packet instead,one with σx = ∆x = x0. Thus is an assumption of the physical imperfection of ourdetectorThis collapsed wavefunction is no longer an energy eigenstate, hence it does nothave any energy eigenvalue, and the expectation of energy is certainly not E < V0

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Instead this collapsed wavefunction will be composed of a spread in energies, with

∆E =√〈E2〉 − 〈E〉2 (10.1.11)

∼ ∆p2

2m=

~2

8m2∆x2(10.1.12)

= V0 − E (10.1.13)

Therefore the energy has enough spread such that it takes up the expectation of energywell above V0

10.2. A potential barrier is defined by V = 0 eV for x < 0 and V = 7 eV for x > 0. . A beam ofelectrons with energy 3 eV collides with this barrier from left. Find the value of x for whichthe probability of detecting the electron will be half the probability of detecting it at x = 0.

Solution: This is the same as the setup of Q10.1, where V0 = 7 eV, E = 3 eVUsing the same formulation as above,

We want x0 such thatProbDens(x0)

ProbDens(0)=

1

2(10.2.1)

=⇒ e−2κx0 =1

2(10.2.2)

=⇒ x0 =log 2

2κ=

~ log 2√8m(V0 − E)

(10.2.3)

= 0.33 A (10.2.4)

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10.3. A beam of particles of energy E and de Broglie wavelength λ, traveling along the positive x-axis in a potential free region, encounters a one-dimensional potential barrier of height V = Eand width L.

(a) Obtain an expression for the transmission coefficient.

Solution: We have the potential function

V (x) =

0 x < 0

E 0 ≤ x < L

0 x ≥ L

(10.3.1)

Thus we can write the TISE for the energy eigenstate with eigenvalue E as

1

ψ(x)

d2ψ(x)

dx2=

−k2 x < 0

0 0 ≤ x < L

−k2 x ≥ L

(10.3.2)

=⇒ ψ(x) =

Ae−ikx +Beikx x < 0

Cx+D 0 ≤ x < L

Fe−ikx +Geikx x ≥ L

(10.3.3)

Where

−k2 = −2mE

~2(10.3.4)

We want to analyse the case where the particle is coming in from x < 0 and gettingreflected and transmitted. Hence F = 0.Imposing the boundary conditions,


A+B = D (10.3.5)

ik(−A+B) = C (10.3.6)


CL+D = GeikL (10.3.7)

C = ikGeikL (10.3.8)

From (10.3.7) and (10.3.8)

D = GeikL(1− ikL) (10.3.9)

Putting C and D from (10.3.8) and (10.3.9) into (10.3.5) and (10.3.6)

A+B = GeikL(1− ikL) (10.3.10)

−A+B = GeikL (10.3.11)

=⇒ B = G(1− ikL

2) (10.3.12)

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We know that the transmission coefficient T = |G|2|B|2

=⇒ T =1

1 + k2L2

4

(10.3.13)

=1

1 + mEL2

2~2

(10.3.14)

(b) Find the value of L (in terms of λ) for which the reflection coefficient will be half.

Solution: We know that if R = 12

=⇒ T = 12

Using the result (10.3.13),

=⇒ 1

1 + k2L2

4

=1

2(10.3.15)

=⇒ kL = 4 (10.3.16)

=⇒ L =2λ

π(10.3.17)

10.4. A beam of particles of energy E < V0 is incident on a barrier (see figure below) of height 2V0.It is claimed that the solution is ψI = Ae−k1x for region I (0 < x < L) and ψII = Be−k2x for

region II (x > L), where k1 =√

2m(2V0−E)~2 and k2 =

√2m(V0−E)

~2 . Is this claim correct? Justifyyour answer.

V (x) = 0 V (x) = 2V0 V (x) = V0

Solution: If we apply boundary conditions to the purported solution, we will get


Ae−ik1L = Be−ik2LAk1e−ik1L = Bk2e

−ik2L (10.4.1)

=⇒ k1 = k2 (10.4.2)

=⇒ V0 = 0 (10.4.3)

Thus the only possibility is if both the regions have same potential height, which is notwhat happens. Hence the claim is incorrect

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10.5. A beam of particles of mass m and energy 9V0 (V0 is a positive constant with the dimension ofenergy) is incident from left on a barrier, as shown in figure below. V = 0 for x < 0, V = 5V0

for x ≤ d and V = nV0 for x > d. Here n is a number, positive or negative and d = π~√8mV0

. Itis found that the transmission coefficient from x < 0 region to x > d region is 0.75.

V (x) = 0 V (x) = 5V0 V (x) = nV0

(a) Find n. Are there more than one possible values for n?

Solution: We have the potential function

V (x) =

0 x < 0

5V0 0 ≤ x < d

nV0 x ≥ d

(10.5.1)

Thus we can write the TISE for the energy eigenstate with eigenvalue 9V0 as

1

ψ(x)

d2ψ(x)

dx2=

−k21 x < 0

−k22 0 ≤ x < d

−k23 x ≥ d

(10.5.2)

=⇒ ψ(x) =

Ae−ik1x +Beik1x x < 0

Ce−ik2x +Deik2x 0 ≤ x < d

Fe−ik3x +Geik3x x ≥ d

(10.5.3)

Where

−k21 = −18mV0

~2(10.5.4)

−k22 = −8mV0

~2(10.5.5)

−k23 = −2mV0(9− n)

~2(10.5.6)

Now according to the ansatz of the question, F = 0First we take note that k2d = π =⇒ eik2d = −1.Let us impose boundary conditions


A+B = C +D (10.5.7)

k1(−A+B) = k2(−C +D) (10.5.8)

Continuity and Differentiability at x = d

−(C +D) = Geik3d (Using eik2d = −1) (10.5.9)

−k2(−C +D) = k3Geik3d (10.5.10)

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Due to the fact that k2d being a multiple of π has significantly reduced the calculationeffort for us. We now directly get

A+B = −Geik3d (10.5.11)

−A+B = −k3

k1

Geik3d (10.5.12)

=⇒ B = −k1 + k3

2k1

Geik3d (10.5.13)

We know that that T = k3|G|2k1|B|2 = 0.75

=⇒ T =4k3k1

(k1 + k3)2= 0.75 (10.5.14)

=⇒16k3

k1(1 + k3

k1

)2 = 3 (10.5.15)

=⇒ 3

(k3

k1

)2

− 10k3

k1

+ 3 = 0 (10.5.16)

=⇒ k3

k1

= 3 ork3

k1

=1

3(10.5.17)

Putting in the values of k1 and k3

=⇒ 9

9− n = 9 or9

9− n =1

9(10.5.18)

=⇒ n = 8 or n = −72 (10.5.19)

(b) Find the un-normalized wave function in all the regions in terms of the amplitude of theincident wave for each possible value of n.

Solution: Solving for A, B, C, D using the boundary condition equations,

B = −k1 + k3

2k1

Geik3d (10.5.20)

A = −k1 − k3

2k1

Geik3d (10.5.21)

D = −k2 + k3

2k2

Geik3d (10.5.22)

C = −k2 − k3

2k2

Geik3d (10.5.23)

For n = 8, k3d = π2

=⇒ eik3d = iFor n = −72, k3d = 9π

2=⇒ eik3d = i.

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Thus for both values of n, eik3d = i. Thus the above equations become

G = i2k1

k1 + k3

B (10.5.24)

A =k1 − k3

k1 + k3

B (10.5.25)

D =k1(k2 + k3)

k2(k1 + k3)B (10.5.26)

C =k1(k2 − k3)

k2(k1 + k3)B (10.5.27)

Since we cannot normalize our wavefunction, and it can be scaled by any constant,we can set B = 1 WLOGThus our wavefunction becomes

ψ(x) =

eik1x + k1−k3

k1+k3e−ik1x x < 0

k1(k2+k3)k2(k1+k3)

eik2x + k1(k2−k3)k2(k1+k3)

e−ik2x 0 ≤ x < d

− 2k1

k1+k3eik3(x−d) x ≥ d

(10.5.28)

(c) Is there a phase change between the incident and the reflected beam at x = 0? If yes,determine the phase change for each possible value of n. Give your answers by explainingall the steps and clearly writing the boundary conditions used

Solution: As we can see, for n = 8, k1 > k3, hence A has the same sign as B.Therefore no phase changeIf n = −72, k1 < k3, hence A has the opposite sign as B, therefore there is a phasechange of π rad.

10.6. A monoenergetic parallel beam of non-relativistic neutrons of energy E is incident on an infinitemetal surface. Within the metal, the neutrons experience a uniform negative potential V . Theincident beam makes an angle θ with respect to the surface normal. Find the fraction of theincident beam that is reflected.

Solution: Let us assume that the x-axis perpendicular to the metal surface, and y-axisalong the metal surface.Our potential looks like

V (x, y) =

{0 x < 0, ∀y ∈ R−V x ≥ 0, ∀y ∈ R

(10.6.1)

We can write the TISE for an energy eigenstate with eigenvalue E as

1

ψ(x, y)∇2ψ(x, y) =

{−2mE

~2 x < 0, ∀y ∈ R−2m(E+V )

~2 x ≥ 0, ∀y ∈ R(10.6.2)

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Assuming an ansatz where there is an incident and reflected part, and a transmitted part,we know that the solution to this is

ψ(x, y) =

{Aeik1xx+ik1yy +Be−ik1xx+ik1yy x < 0, ∀y ∈ RCeik2xx+ik2yy x ≥ 0, ∀y ∈ R

(10.6.3)

Where

k21x + k2

1y =2mE

~2(10.6.4)

k22x + k2

2y =2m(E + V )

~2(10.6.5)

According to the ansatz of our question, k1y

k1x= tan θ, as k1x ı + k1y is the direction the

incident wave travels in region x < 0, i.e. outside the metalImposing boundary conditions

Continuity at x = 0

(A+B)eik1yy = Ceik2yy ∀y ∈ R (10.6.6)

Continuity of ∇ψ(x, y) at x = 0

k1x(A−B)eik1yy ı+ k1y(A+B)eik1yy = k2xCeik2yy ı+ k2yCe

ik2yy ∀y ∈ R (10.6.7)

The only way for (10.6.6) to be true ∀y ∈ R is if k1y = k2y and A+B = CPutting this in (10.6.7) gives us

k1x(A−B) = k2xC (10.6.8)

=⇒ A =k1x + k2x

2k1x

C (10.6.9)

B =k1x − k2x

2k1x

C (10.6.10)

Thus the fraction that gets reflected is R = |B|2|A|2 .

Since we know that k1y

k1x= tan θ, we know that k1x =

√2mE~2 cos θ, k1y =

√2mE~2 sin θ

We also know that k2y = k1y =√

2mE~2 sin θ

Thus we can see that k2x =

√2m(E cos2 θ+V )

~Putting all of this in R, we get

R =|B|2|A|2 =

(k1x − k2x)2

(k1x + k2x)2(10.6.11)

=(√E cos θ −

√E cos2 θ + V )2

(√E cos θ +

√E cos2 θ + V )2

(10.6.12)

10.7. A scanning tunneling microscope (STM) can be approximated as an electron tunneling into

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a step potential [V (x) = 0 for x ≤ 0, V (x) = V0 for x > 0]. The tunneling current (orprobability) in an STM reduces exponentially as a function of the distance from the sample.Considering only a single electron-electron interaction, an applied voltage of 5 V and the samplework function of 7 eV, calculate the amplification in the tunneling current if the separation isreduced from 2 atoms to 1 atom thickness (take approximate size of an atom to be 3 A)

Solution:

V (x) =

{0 x ≥ 0

V0 x > 0(10.7.1)

We can write the TISE for energy eigenstate with eigenvalue E as

1

ψ(x)

d2ψ(x)

dx2=

{−k2 x < 0

κ2 x ≥ 0(10.7.2)

=⇒ ψ(x) =

{Aeikx +Be−ikx x < 0

Ceκx +De−κx x ≥ 0(10.7.3)

Where

−k2 = −2mE

~2(10.7.4)

κ2 =2m(V0 − E)

~2(10.7.5)

Here V0 = 7 eV, and the energy eigenvalue is E = 5 eVBecause the wavefunction cannot blow up at infinity C = 0If the seperation between the start of the step potential and the microscope is d , then thetunneling current is proportional to |ψ(d)|2

=⇒ Current1 atom

Current2 atom

= e2κd = 78.25 (10.7.6)

Where d = 3 A= thickness of atom

11 Harmonic Oscillator and Degenerate states

11.2. Vibrations of the hydrogen molecule can be modeled as a simple harmonic oscillator with thespring constant k = 1.13× 103 Nm-2 and mass m = 1.67× 10−27 kg.

(a) What is the vibrational frequency of this molecule?

Solution: We know that the hydrogen molecule is diatomic and hence the effectivemass of the Quantum Harmonic Oscillator is µ = mH

2

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k = µω2 (11.2.1)

=⇒ ω =

√k

µ(11.2.2)

= 1.16× 1015 Hz (11.2.3)

(b) What are the energy and the wavelength of the emitted photon when the molecule makestransition between its third and second excited states?

Solution: We know that the energy eigenvalues of the QHO is(n+ 1

2

)~ω. Thus

the energy difference between the third and the second excited states is ~ω

=⇒ Eγ = ~ω = 0.76 eV (11.2.4)

=⇒ λ =2πc

ω= 1.62 µm (11.2.5)

11.4. Determine the expectation value of the potential energy for a quantum harmonic oscillator inthe ground state. Use this to calculate the expectation value of the kinetic energy

Solution: We know the ground state eigenfunction for the harmonic oscillator

ψ(x) =(απ

) 14e−α

x2

2 (11.4.1)

Where

α =mω

~(11.4.2)

We thus calculate the expectation value of Potential Energy as follows

=⇒ 〈V 〉 =

ˆ ∞x=−∞

ψ(x)∗V (x)ψ(x) dx (11.4.3)

=mω2

2

√α

π

ˆ ∞x=−∞

x2e−αx2

dx (11.4.4)

= −mω2

4α

√α

π

ˆ ∞x=−∞

x(−2αxe−αx

2)dx (11.4.5)


= −mω2

4α

√α

π

(xe−αx

2∣∣∣∞

x=−∞−ˆ ∞x=−∞

e−αx2

dx

)(11.4.6)

=mω2

4α=

~ω4

(11.4.7)

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This aligns perfectly with classical results, since the expectation of potential energy isexactly half the total expected energy

=⇒ 〈KE〉 = 〈V 〉 =~ω4

(11.4.8)

11.5. A diatomic molecule behaves like a quantum harmonic oscillator with the force constant k = 12Nm-1 and mass m = 5.6× 10−26 kg

(a) What is the wavelength of the emitted photon when the molecule makes the transitionfrom the third excited state to the second excited state?

Solution: Along the lines of Q11.2b

ω =

√k

µ(11.5.1)

= 1.46× 1013 Hz (11.5.2)

=⇒ λ =2πc

ω= 0.13 mm (11.5.3)

(b) Find the ground state energy of vibrations for this diatomic molecule.

Solution:

E0 =~ω2

= 4.8 meV (11.5.4)

11.6. A two-dimensional isotropic harmonic oscillator has the Hamiltonian

H = − ~2

2m

(∂2

∂x2+

∂2

∂y2

)+

1

2k(x2 + y2)

(a) Show that the energy levels are given by

Enxny = ~ω(nx + ny + 1)

nx, ny ∈ {0, 1, 2 · · · }

ω =

√k

m

Solution: We can attack the problem using variable separation.We assume a form separable in the x, y variables

ψ(x, y) = ψx(x)ψy(y) (11.6.1)

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Putting in TISE

Hψ(x, y) = Eψ(x, y) (11.6.2)

=⇒ − ~2

2mψy(y)

∂2ψx(x)

∂x2− ~2

2mψx(x)

∂2ψy(y)

∂y2+

1

2kx2ψx(x)ψy(x) +

1

2ky2ψx(x)ψy(y)

= Eψx(x)ψy(y) (11.6.3)

=⇒ − ~2

2mψx(x)

∂2ψx(x)

∂x2+

1

2kx2

= −(− ~2

2mψy(y)

∂2ψy(y)

∂y2+

1

2ky2 − E

)(11.6.4)

Since LHS is purely a function of x, and RHS of y, and the expression has to be truefor all x, y. Thus is only possible if LHS = RHS = some constant, which we write asEx ≤ E

− ~2

2mψx(x)

∂2ψx(x)

∂x2+

1

2kx2 = Ex (11.6.5)

− ~2

2mψy(y)

∂2ψy(y)

∂y2+

1

2ky2 = E − Ex (11.6.6)

We already know the solutions to (11.6.5) and (11.6.6)

Ex =

(nx +

1

2

)~ω (11.6.7)

E − Ex =

(ny +

1

2

)~ω (11.6.8)

=⇒ E = (nx + ny + 1) ~ω (11.6.9)

We need not worry about the general E eigenstate (which will be a linear combinationof separable E eigenstates), because they by definition will have eigenvalue E, whichwe already found

(b) What is the degeneracy of each level?

Solution: The degeneracy of level E = (n+ 1) ~ω will be the number of solutions tonx + ny = n, which we can easily see will be n+ 1 (nx = 0, 1 · · · , n)Therefore the degeneracy of energy E will be E

~ω

11.7. Consider the Hamiltonian of a two-dimensional anisotropic harmonic oscillator

H =p2

1

2m+

p22

2m+

1

2mω2

1q21 +

1

2mω2

2q22 ω1 6= ω2

(a) Exploit the fact that the Schrodinger eigenvalue equation can be solved by separating thevariables and find a complete set of eigenfunctions of H and the corresponding eigenvalues.

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Solution: Proceeding as exactly as in Q11.6a, we get

− ~2

2mψq1(q1)

∂2ψq1(q1)

∂q21

+1

2mω2

1q21 = Eq1 (11.7.1)

− ~2

2mψq2(q2)

∂2ψq2(q2)

∂q22

+1

2mω2

2q22 = E − Eq1 (11.7.2)

We already know the solution to these differential equations

ψ(n1)q1

(q1) =(mω1

π~

) 14 1√

2n1(n1)!Hn1

(√mω1

~q1)

)e−

mω1~

q212 (11.7.3)

ψ(n2)q2

(q2) =(mω2

π~

) 14 1√

2n2(n2)!Hn2

(√mω2

~q2)

)e−

mω2~

q222 (11.7.4)

Where Hn(x) is the nth Hermite polynomial, and

Eq1 =

(n1 +

1

2

)~ω1 (11.7.5)

Eq2 = E − Eq1 =

(n2 +

1

2

)~ω2 (11.7.6)

=⇒ E =

(n1ω1 + n2ω2 +

ω1 + ω2

2

)~ (11.7.7)

The most general E eigenstate is of course

ψE(q1, q2) =∑

Eq1+Eq2=E

cn1n2ψ(n1)q1

(q1)ψ(n2)q2

(q2) (11.7.8)

(b) Assume that ω1

ω2= 3

4. Find the first two degenerate energy levels. What can one say about

the degeneracy of energy levels when the ratio between ω1 and ω2 is not a rational number

Solution: When ω1

ω2= 3

4, finding out the degeneracy amounts to finding the number

of integer solutions to 3n1 +4n2 = n, which amounts to solving Diophantine equationsFinding the degeneracy for the first two energy levels

First State

E = (6n1 + 8n2 + 7)~ω2

8=

13

8~ω2 (11.7.9)

This has degeneracy 1, as only nx = 1, ny = 0 is possible

Second State

E = (6n1 + 8n2 + 7)~ω2

8=

15

8~ω2 (11.7.10)

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This has degeneracy 1, as only nx = 0, ny = 1 is possible

(11.7.11)

If we are asked to find the first energy that has degeneracy 2, that will be

First Energy(nx = 4, ny = 0 and nx = 0, ny = 3 are possible)

E =31

8~ω2 (11.7.12)

In general, for the ratio not being a rational number, we will not have any degeneracyfor any energy

12 Statistical Mechanics and Density of States

12.1 Statistical Mechanics

12.1.1. A national powerball lottery uses two sets of balls. The first set consists of 59 sequentiallynumbered balls and the second set consists of 35 sequentially numbered balls. Assume equalprobability of choosing any ball and that all the balls are differently numbered. Five ballsare chosen without replacement from the set of 59. Then one ball is chosen from the set of35. Calculate the number of ways these six balls can be chosen (and thus your probabilityof winning the grand powerball prize).

Solution: The ways of choosing 5 out of 59 distinguishable balls without replacement is(595

)

The ways of choosing 1 ball out of 35 distinguishable balls is 35=⇒ the ways of drawing 6 balls is

(595

)× 35 = 175223510

The probability of winning the grand prize is hence 1175223510

= 5.7× 10−9

12.1.2. Suppose we have 20 coins and we flip all of them together

(a) Considering all the coins to be independent of each other, how many possible outcomes(no. of microstates) do you expect with such a flipping?

Solution: Every coin can choose to land as either heads or tails, so there are 220possibilities

(b) How many ways are there for obtaining 12 heads and 8 tails?

Solution: We simply need to choose which coins we need as heads, giving us thenumber as

(2012

)

(c) What is the probability of obtaining 12 heads and 8 tails regardless of the order? Theyare called macrostates.

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Solution: The probability is of course Num. of ways 12HAll num of ways

=(20

12)220 = 0.12

12.1.3. Three indistinguishable particles (say electrons) are to be arranged in three different energylevels of energy 0, E and 2E, with respective degeneracies (ignore spin degeneracy) 2, 10 and20. The total energy available is 3E. What are the possible distributions and what are theirprobabilities?

Solution: If the number of particles with energy 0, E, 2E are n0, nE and n2E respectively,then we have EnE+2En2E = 3E =⇒ nE+2n2E = 3, and n0+nE+n2E = 3. Subtractingthe two equations give us n2E = n0. From this we can easily get the possible distributions

(n0, nE, n2E)=(0, 3, 0)

Prob ∝(

10

3

)= 120 (12.1.3.1)

(n0, nE, n2E)=(1, 1, 1)

Prob ∝(

2

1

)(10

1

)(20

1

)= 400 (12.1.3.2)

=⇒ Prob(0, 3, 0) =120

120 + 400= 0.23 (12.1.3.3)

Prob(1, 1, 1) =400

120 + 400= 0.77 (12.1.3.4)

12.1.4. Consider a particle confined to a 3D harmonic oscillator potential, V (x, y, z) = 12mω2(x2 +

y2 + 4z2)

(a) Calculate the ground state energy of the particle.

Solution: The allowed energy levels are E = (nx + ny + 2nz + 2)~ω, for integernx, ny, nz ≥ 0Hence the ground state is 2~ω

(b) What is the degeneracy of the state with energy, E = 7~ω?

Solution: We need to calculate the integer solutions to nx + ny + 2nz = 5. Fornz = 0 there are 5 possibilities, for nz = 1 there are 3, for nz = 2 there is just 1.Hence the total number is 5 + 3 + 1 = 9

12.1.5. A certain thermodynamic system has non-degenerate energy levels, with energies 0, E, 3E,5E and 9E. Suppose that there are four particles, with total energy U = 9E. Identify thepossible distribution of particles and evaluate their microstates when

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(a) The particles are distinguishable

Solution: As before, nE+3n3E+5n5E+9n9E = 9 and n0+nE+n3E+n5E+n9E = 4.=⇒ 2n3E + 4n5E + 8n9E = 5 + n0

We thus have the possibilities

(n0, nE, n3E, n5E, n9E)=(3, 0, 0, 0, 1)

Prob ∝ 1× 4!

3!= 4 (12.1.5.1)

(n0, nE, n3E, n5E, n9E)=(1, 1, 1, 1, 0)

Prob ∝ 1× 4! = 24 (12.1.5.2)

(n0, nE, n3E, n5E, n9E)=(1, 0, 3, 0, 0)

Prob ∝ 1× 4!

3!= 4 (12.1.5.3)

=⇒ Prob(3, 0, 0, 0, 1) =4

4 + 24 + 4= 0.125 (12.1.5.4)

Prob(1, 1, 1, 1, 0) =24

4 + 24 + 4= 0.75 (12.1.5.5)

Prob(1, 0, 3, 0, 0) =4

4 + 24 + 4= 0.125 (12.1.5.6)

(b) The particles are identical bosons

Solution: We have the possibilities

(n0, nE, n3E, n5E, n9E)=(3, 0, 0, 0, 1)

Prob ∝ 1 (12.1.5.7)

(n0, nE, n3E, n5E, n9E)=(1, 1, 1, 1, 0)

Prob ∝ 1 (12.1.5.8)

(n0, nE, n3E, n5E, n9E)=(1, 0, 3, 0, 0)

Prob ∝ 1 (12.1.5.9)

=⇒ Prob(3, 0, 0, 0, 1) =1

3(12.1.5.10)

Prob(1, 1, 1, 1, 0) =1

3(12.1.5.11)

Prob(1, 0, 3, 0, 0) =1

3(12.1.5.12)

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(c) the particles are identical fermions

Solution: We have the possibilities

(n0, nE, n3E, n5E, n9E)=(3, 0, 0, 0, 1)

Prob ∝ 0 (12.1.5.13)

(n0, nE, n3E, n5E, n9E)=(1, 1, 1, 1, 0)

Prob ∝ 1 (12.1.5.14)

(n0, nE, n3E, n5E, n9E)=(1, 0, 3, 0, 0)

Prob ∝ 0 (12.1.5.15)

=⇒ Prob(3, 0, 0, 0, 1) = 0 (12.1.5.16)

Prob(1, 1, 1, 1, 0) = 1 (12.1.5.17)

Prob(1, 0, 3, 0, 0) = 0 (12.1.5.18)

12.1.6. In how many ways three electrons can occupy ten states (include spin degeneracy)? Is thenumber same as the way in which three persons can occupy ten chairs in a room? State thereason. In case the number is different, find the other number also.

Solution: If we count spin degeneracy, then that is equivalent to having 20 degeneratestates. The number of ways then becomes

(203

)= 1140

For the number of ways of 3 people occupying 10 chairs, the people aren’t indistinguish-able (in contrast to electrons) hence we need to permute them also. The number thenbecomes

(103

)3! = 720

12.1.7. The energy of a particle in a 3-D cubical box is given by

Enx,ny ,nz =π2~2

2mL2(n2

x + n2y + n2

z)

If these levels are going to be occupied by electrons, write the energy values correspondingto the five lowest levels, taking into account the spin degeneracy. If three electrons occupythese states, find out the possible distributions which would yield a total energy of 18π2~2

2mL2 .Also find out the probability for each distributions.

Solution: 5 lowest levels energies with respective degeneracies

(nx, ny, nz)=(1, 1, 1). Degeneracy g0 = 1× 2 = 2

E0 =3π2~2

2mL2(12.1.7.1)

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(nx, ny, nz)=(2, 1, 1) or (1, 2, 1) or (1, 1, 2). Degeneracy g1 = 3× 2 = 6

E1 =6π2~2

2mL2(12.1.7.2)


E2 =9π2~2

2mL2(12.1.7.3)


E3 =11π2~2

2mL2(12.1.7.4)

(nx, ny, nz)=(2, 2, 2). Degeneracy g4 = 1× 2 = 2

E4 =12π2~2

2mL2(12.1.7.5)

Again, proceeding as before, 3n0+6n1+9n2+11n3+12n4 = 18 and n0+n1+n2+n3+n4 =3. We can clearly see n3 = 0=⇒ n2 + 2n4 = n0 We can thus find the possibilities

(n0, n1, n2, n3, n4)=(2, 0, 0, 0, 1)

Prob ∝(

2

2

)(2

1

)= 2 (12.1.7.6)

(n0, n1, n2, n3, n4)=(1, 1, 1, 0, 0)

Prob ∝(

2

1

)(6

1

)(6

1

)= 72 (12.1.7.7)

(n0, n1, n2, n3, n4)=(0, 3, 0, 0, 0)

Prob ∝(

6

3

)= 20 (12.1.7.8)

=⇒ Prob(2, 0, 0, 0, 1) =2

2 + 72 + 20= 0.021 (12.1.7.9)

Prob(1, 1, 1, 0, 0) =72

2 + 72 + 20= 0.766 (12.1.7.10)

Prob(0, 3, 0, 0, 0) =20

2 + 72 + 20= 0.213 (12.1.7.11)

12.1.8. Consider a system of five particles trapped in a 1D harmonic oscillator potential.

(a) What are the microstates of the ground state of this system for classical particles,identical Bosons and identical spin half Fermions

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Solution: Let the wavefunction of the system be denoted as ψ(x1, x2, x3, x4, x5).Let the ith normalized QHO state for a single particle be written as ψi(x). We arenow in a capacity to write the microstates (eigenfunctions) having the ground stateenergy of this system

Classical particles (ground state energy eigenvalue 52~ω)

ψ(x1, x2, x3, x4, x5) = ψ0(x1)ψ0(x2)ψ0(x3)ψ0(x4)ψ0(x5) (12.1.8.1)

Bosons (Notice that the function is symmetric to exchange), (ground state energyeigenvalue 5

2~ω)

ψ(x1, x2, x3, x4, x5) (12.1.8.2)

=1√120

∑

(p1,p2,p3,p4,p5)=perm(1,2,3,4,5)

ψ0(xp1)ψ0(xp2)ψ0(xp3)ψ0(xp4)ψ0(xp5) (12.1.8.3)

The case for fermions is a little bit more complex. We need to realize that we nowalso have to take into account the spin configuration. Our system wavefunction thenwill be φ(x1, x2, x3, x4, x5, s1, s2, s3, s4, s5). Let α(s) be the “up” spin wavefunction,and β(s) be the “down” spin wavefunction.To write our ground state wavefunction we need to use slater’s determinants (whichensures antisymmetricity to exchange). There are two possibilities (hence 2 mi-crostates), each with energy eigenvalue 13

2~ω:

2 fermions “up” and “down” in ψ0, 2 “up” and “down” in ψ1, 1 “up” in ψ2

ψ(x1, x2, x3, x4, x5, s1, s2, s3, s4, s5)

=1√120

∣∣∣∣∣∣∣∣∣

ψ0(x1)α(s1) ψ0(x1)β(s1) ψ1(x1)α(s1) ψ1(x1)β(s1) ψ2(x1)α(s1)ψ0(x2)α(s2) ψ0(x2)β(s2) ψ1(x2)α(s2) ψ1(x2)β(s2) ψ2(x2)α(s2)

...ψ0(x5)α(s5) ψ0(x5)β(s5) ψ1(x5)α(s5) ψ1(x5)β(s5) ψ2(x5)α(s5)

∣∣∣∣∣∣∣∣∣(12.1.8.4)

2 fermions “up” and “down” in ψ0, 2 “up” and “down” in ψ1, 1 “down” in ψ2

ψ(x1, x2, x3, x4, x5, s1, s2, s3, s4, s5)

=1√120

∣∣∣∣∣∣∣∣∣

ψ0(x1)α(s1) ψ0(x1)β(s1) ψ1(x1)α(s1) ψ1(x1)β(s1) ψ2(x1)β(s1)ψ0(x2)α(s2) ψ0(x2)β(s2) ψ1(x2)α(s2) ψ1(x2)β(s2) ψ2(x2)β(s2)

...ψ0(x5)α(s5) ψ0(x5)β(s5) ψ1(x5)α(s5) ψ1(x5)β(s5) ψ2(x5)β(s5)

∣∣∣∣∣∣∣∣∣(12.1.8.5)

(b) Suppose that the system is excited and has one unit of energy (~ω) above the cor-responding ground state energy in each of the three cases. Calculate the number of

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microstates for each of the three cases

Solution:

Classical particles: 5 microstates (each with energy eigenvalue 72~ω)


OR (12.1.8.7)


OR (12.1.8.9)


OR (12.1.8.11)


OR (12.1.8.13)


Bosons: 1 microstate (with energy eigenvalue 72~ω). Notice that even if you have

the second term in each product to be ψ1, that is already covered in the sum sinceit covers all permutations. Hence only 1 microstate possible

ψ(x1, x2, x3, x4, x5) (12.1.8.15)

=1√120

∑

(p1,p2,p3,p4,p5)=perm(1,2,3,4,5)

ψ1(xp1)ψ0(xp2)ψ0(xp3)ψ0(xp4)ψ0(xp5)

(12.1.8.16)

Fermions: 4 microstates (each with energy eigenvalue 152~ω)

2 fermions “up” and “down” in ψ0, 1 “up” in ψ1, 2 “up” and “down” in ψ2

ψ(x1, x2, x3, x4, x5, s1, s2, s3, s4, s5)

=1√120

∣∣∣∣∣∣∣

ψ0(x1)α(s1) ψ0(x1)β(s1) ψ1(x1)α(s1) ψ2(x1)α(s1) ψ2(x1)β(s1)...

ψ0(x5)α(s5) ψ0(x5)β(s5) ψ1(x5)α(s5) ψ2(x5)α(s5) ψ2(x5)β(s5)

∣∣∣∣∣∣∣(12.1.8.17)

2 fermions “up” and “down” in ψ0, 1 “down” in ψ1, 2 “up” and “down” in ψ2

ψ(x1, x2, x3, x4, x5, s1, s2, s3, s4, s5)

=1√120

∣∣∣∣∣∣∣

ψ0(x1)α(s1) ψ0(x1)β(s1) ψ1(x1)β(s1) ψ2(x1)α(s1) ψ2(x1)β(s1)...

ψ0(x5)α(s5) ψ0(x5)β(s5) ψ1(x5)β(s5) ψ2(x5)α(s5) ψ2(x5)β(s5)

∣∣∣∣∣∣∣(12.1.8.18)

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2 fermions “up” and “down” in ψ0, 2 “up” and “down” in ψ1, 1 “up” in ψ3

ψ(x1, x2, x3, x4, x5, s1, s2, s3, s4, s5)

=1√120

∣∣∣∣∣∣∣

ψ0(x1)α(s1) ψ0(x1)β(s1) ψ1(x1)α(s1) ψ1(x1)β(s1) ψ3(x1)α(s1)...

ψ0(x5)α(s5) ψ0(x5)β(s5) ψ1(x5)α(s5) ψ1(x5)β(s5) ψ3(x5)α(s5)

∣∣∣∣∣∣∣(12.1.8.19)

2 fermions “up” and “down” in ψ0, 2 “up” and “down” in ψ1, 1 “down” in ψ3

ψ(x1, x2, x3, x4, x5, s1, s2, s3, s4, s5)

=1√120

∣∣∣∣∣∣∣

ψ0(x1)α(s1) ψ0(x1)β(s1) ψ1(x1)α(s1) ψ1(x1)β(s1) ψ3(x1)β(s1)...

ψ0(x5)α(s5) ψ0(x5)β(s5) ψ1(x5)α(s5) ψ1(x5)β(s5) ψ3(x5)β(s5)

∣∣∣∣∣∣∣(12.1.8.20)

(c) Suppose that the temperature of this system is low, so that the total energy is low (butabove the ground state), describe in a couple of sentences, the difference in the behaviorof the system of identical bosons from that of the system of classical particles

Solution: For a given total energy, because of the distinguishability of classicalparticles, there are far many more possible microstates than in the case of bosons.There are far more ways of sending a particle up in energy for classical case than forbosonic case.Hence the system of bosons will have more particles in lower energies than classicalsystem

12.2 Density of States and Fermi Energy

12.2.1. A system has one state with energy 0, four states with energy 2E and eight states with energy3E. Six electrons are to be distributed among these states such that their total energy is12E. Consider a configuration (j,m, n) in which j electrons are in 0 energy state, m electronsare in 2E energy state and n electrons are in 3E state

(a) Calculate the total number of microstates for the configuration (1, 3, 2)

Solution: We are dealing with fermions here. Number of electrons having eachenergy is (n1, n2, n3) = (1, 3, 2). Degeneracy of each energy is (including spin degen-

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eracy) (g1, g2, g3) = (1× 2, 4× 2, 8× 2)

=⇒ Nmicro(1, 3, 2) =3∏

i=1

(gini

)(12.2.1.1)

=2!

1!1!× 8!

3!5!× 16!

2!14!(12.2.1.2)

= 13440 (12.2.1.3)

(b) Find the ratio of probability of occurrence of a configuration (2, 0, 4) to that of a con-figuration (1, 3, 2)

Solution: Proceeding as above

Nmicro(2, 0, 4) =3∏

i=1

(gini

)(12.2.1.4)

=2!

2!0!× 8!

0!8!× 16!

4!12!(12.2.1.5)

= 1820 (12.2.1.6)

=⇒ Prob(2, 0, 4)

Prob(1, 3, 2)=Nmicro(2, 0, 4)

Nmicro(1, 3, 2)=

1820

13440= 0.135 (12.2.1.7)

12.2.2. The spin independent energy levels of a carbon nanotube are described by those of a 1Dinfinite potential well. In a carbon nanotube of length 1 µm, electrons occupy all the energylevels up to 0.1 eV

(a) Calculate the number of electrons in the carbon nanotube

Solution: The spin independent part of the wavefunction of each particle is descibedby 1D particle in a box wavefunctions.But each wavefunction for a particular spin independent part will have 2 possibilitiesfor the spin part, “up” and “down”Hence the degeneracy of each energy level is 2We know that the electrons occupy all energy levels upto 0.1 eV. This is only possibleat T = 0, and if the fermi energy is Ef = 0.1 eVThe number of particles will then simply be the number of states below 0.1 eV, times2.If EF is the nth state then EF = n2 ~2π2

2mL2 =⇒ N = 2n = 2×√

2mEF~π L = 1034

(b) Calculate the density of states g(E) at the Fermi energy EF in units of [(eV)-1(µm)-1]

Solution: We simply need the number of states having energies between EF and

EF +dE. For each n between√

2mEF~π L and

√2m(EF+dE)

~π L =√

2mEF~π L+ 1

2

√2mEF~π L dE

EF,

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we have 2 states.Thus we have

√2m

~π√EFLdE states between EF and EF + dE

Thus we have√

2m~π√EF

dE states between EF and EF +dE per unit length of container

Thus the density of states is g(EF ) dE =√

2m~π√EF

dE

Thus g(EF ) = 5173 (eV)-1(µm)-1

12.2.3. Consider a non-interacting Fermi gas of N particles in 2D, confined in a square area A = L2

(a) Derive a formula for the density of states g(E)

Solution: Looking at k-space (the state space indexed by (kx, ky)), we know thatthe valid states are arranged in a grid/lattice, with the minimal k-space volumebetween 4 closest grid points being π2

L2 .

Therefore the number of gridpoints enclosed inside a k-space volume Vk is VkL2

π2

We want the number of states between E and E+dE. This corresponds to a quarterring (because all gridpoints in second, third, fourth quadrants are duplicates of first

quadrant states) with radius k =√

2mE~2 and thickness dk =

√m

2~2EdE.

This corresponds to a k-space volume Vk = πk dk2

= πm2~2 dE

Hence the number of states (including spin) having energies between E and E + dE

is 2VkL2

π2 = mL2

~2πdE

Hence the density of states is g(E) dE = m~2π

dE

(b) Find the Fermi energy EF (in terms of N and A) and show that the average energy perparticle E

Nat T = 0 is EF

2

Solution: As we know, the Fermi energy is simply defined as the highest occupiedenergy at T = 0. We also know that fFD,T (E) = 1

eE−EFkBT +1

, which at T = 0 becomes

fFD,0(E) =

{1 E < EF

0 E ≥ EF(12.2.3.1)

=⇒ N

A=

ˆ ∞E=0

fFD,0(E)g(E) dE (12.2.3.2)

=

ˆ EF

E=0

g(E) dE (12.2.3.3)

=

ˆ EF

E=0

m

~2πdE (12.2.3.4)

=mEF~2π

(12.2.3.5)

=⇒ EF =Nπ~2

mA(12.2.3.6)

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Let us find the average energy per particle

N

A〈E〉 =

ˆ ∞E=0

EfFD,0(E)g(E) dE (12.2.3.7)

=

ˆ EF

E=0

Em

π~2dE (12.2.3.8)

=mE2

F

2π~2(12.2.3.9)

=NEF2A

Using (12.2.3.6) (12.2.3.10)

=⇒ 〈E〉 =EF2

(12.2.3.11)

12.2.4. Consider a particle confined to a potential, V (x, y, z) = 12mω2(x2 + y2 + 4z2)

(a) Find the degeneracy of the state with energy E = 7~ω

Solution: Same as 12.1.4b

(b) For E = n~ω (n� 1), calculate the density of states g(n).

Solution: The energy can be E = (nx + ny + nz + 2)~ωWe need to find the number of states with energies between nω and (n+ dn)ω.To do that, let us plot the valid states in n space.

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Here each of the points corresponds to a state having energy (nx + ny + 2nz + 2)~ω

Let us plot the surface made by all points such that (nx + ny + 2nz + 2)~ω isconstant = E = n~ω

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E~ω − 2

E2~ω − 1

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As we can see, this forms a triangular pyramid of volume 13Base Area × Height =

16

(E~ω − 2

)2 ( E2~ω − 1

)= (n−2)3

12

Thus the volume between the two surfaces corresponding to n and n+dn is (n−2)2

4dn

This volume will contain (n−2)2

4dn points corresponding to states

Thus the number of states having energies between n~ω and (n+ dn)~ω is (n−2)2

4dn

g(n) dn = (n−2)2

4dn ≈ n2

4dn

12.2.5. Consider a gas of N identical bosons confined by an isotropic 3D harmonic potential. Theenergy levels in this potential are ε = nhf , where n is a non-negative integer and f is classicaloscillation frequency. The degeneracy of the level is (n+1)(n+2)

2. Find the density of states for

atoms confined by this potential. You may assume n� 1.

Solution: We are already given the degeneracy (a nice exercise is to prove that the givendegeneracy expression is correct)We need to find the density of states, or number of states with “near” degeneracy, thatis, the number of states with energies between nhf and (n+ dn)hf .

We know that all the energies between nhf and (n+dn)hf will have degeneracy (n+1)(n+2)2

,since dn� nThe number of integers between n and n + dn are dn. Thus there are dn valid energylevels between nhf and (n+ dn)hf . Thus total number of states is (n+1)(n+2)

2dn

Thus the density of states is g(n) dn ≈ n2

2dn.

The density of states in energy form is g(E) dE = E2

2(hf)3 dE

12.2.6. Mass Ms of the Sun is 2× 1030 kg. Ignore electron spin in this problem.

(a) Estimate the number of electrons (Ns) in the Sun

Solution: Since the sun is entirely composed of ionized Hydrogen (basically proton)and a corresponding electron which are almost free, the number of protons (and thusNs) in the sun is Ms

MH= 1.2× 1057

(b) Consider a white dwarf star having Ns electrons, contained in a sphere of radius 2× 107

m. Assuming these electrons to be nonrelativistic, estimate the Fermi energy (in eV) ofthis star.

Solution: We will model the white dwarf as a region of constant potential. Hencethe energy structure will be same as that of 3D particle in a boxWe have our possible energy levels as E = ~2π2

2mL2 (n2x + n2

y + n2z). Thus each point

(nx, ny, nz) in n space corresponds to a state.Now, the Fermi energy EF is defined by the energy such that the number of statesbelow that energy is N

V

The volume contained within an eighth sphere in n space with radius n =√

2mEFL~π

is πn3

6= (2mEF )

32

6~3π2 V .

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Thus the number of states and hence electrons (at T = 0) below EF is N = (2mEF )32

6~3π2 V

Thus EF = (6π2NV

)23

~2

2m= 62.5 KeV

(c) Consider another star of volume V . All its electrons, N in number, are extremelyrelativistic such that the electron rest mass energy mec

2 � pc where p is the momentumof the electron. Obtain an expression for the Fermi energy of this star.

Solution: Here we have the possible energy levels as E = ~kc = π~cL

√n2x + n2

y + n2z.

The volume contained within an eighth sphere in n space with radius n = EF~πcL is

πn2

6=

E3F

6~3π2c3V .

Thus the number of states and hence electrons (at T = 0) is N =E3F

6~3π2c3V

Thus EF = (6π2NV

)13~c

12.2.7. Use Bose-Einstein Statistics and the density of state expression, with suitable modifications,if any, to derive Planck’s formula of black body radiation.

Solution: As done in Q12.2.3a, we can derive the density of states as (use the relationk = 2π

cν)

g(ν) dν =8π

c3ν2 dν (12.2.7.1)

We can use the formula fBE(E) = 1

eE

kBT −1

to find the average energy contribution of each

energy level

(12.2.7.2)

U(ν) dν = hνfBE(hν)g(ν) dν (12.2.7.3)

=8πν2

c3

hν

ehνkBT − 1

dν (12.2.7.4)

12.2.8. Show that the kinetic energy of a 3D gas of N free electrons at 0 K is 35NEF

Solution: Since there is no potential, the kinetic energy is the total energyWe only need to find the total energy at T = 0

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First we need the density of states, which we can calculate along similar lines to 12.2.3a

g(E) dE =

√2m3

π2~3

√E dE (12.2.8.1)

=⇒ Etot = V

ˆ EF

E=0

Eg(E) dE (12.2.8.2)

=

√2m3V

π2~3

2

5E

52F (12.2.8.3)

But

N = V

ˆ EF

E=0

g(E) dE (12.2.8.4)

=⇒ N =

√2m3V

π2~3

2

3E

32F (12.2.8.5)

=⇒ Etot = KE =3

5NEF (12.2.8.6)

12.2.9. (a) Using the Fermi Dirac (FD) Statistics, find the probability that a state is occupied ifits energy is higher than εF by 0.1kBT , 1.0kBT , 2.0kBT and 10.0kBT , where εF is theFermi Energy. How good is the approximation in neglecting 1 in the denominator ofFDS for an energy equal to 10kBT .

Solution:

fFD(0.1kBT ) =1

e0.1 + 1= 0.475 (12.2.9.1)

fFD(1.0kBT ) =1

e1.0 + 1= 0.269 (12.2.9.2)

fFD(2.0kBT ) =1

e2.0 + 1= 0.119 (12.2.9.3)

fFD(10.0kBT ) =1

e10.0 + 1= 4.5398× 10−5 (12.2.9.4)

If we neglect 1 in the denominator for 10kBT , we get

fFD(10.0kBT ) = e−10 = 4.5399× 10−5 (12.2.9.5)

Hence the approximation is correct to within 0.002% margin of error

(b) In the Fermi Dirac distribution, substitute ε = εF + δ. Compute δ for the probabilityof occupancy equal to 0.25 and 0.75.

Solution:

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fFD(ε) = 0.25

1

eδ

kBT + 1= 0.25 (12.2.9.6)

=⇒ eδ

kBT = 3 (12.2.9.7)

=⇒ δ = kBT ln 3 (12.2.9.8)

fFD(ε) = 0.75

1

eδ

kBT + 1= 0.75 (12.2.9.9)

=⇒ eδ

kBT =1

3(12.2.9.10)

=⇒ δ = −kBT ln 3 (12.2.9.11)

(c) Show that for a distribution system governed by FD distribution, the probability ofoccupation of a state with energy higher than εF by an amount ∆E is equal to theprobability that a state with energy lower than εF by ∆E is unoccupied.

Solution:

Probunocc(εF −∆E) = 1− 1

e−∆EkBT + 1

(12.2.9.12)

=e−∆EkBT

e−∆EkBT + 1

(12.2.9.13)

=1

e∆EkBT + 1

(12.2.9.14)

= Probocc(εF + ∆E) (12.2.9.15)

12.2.10. The Fermi energy of Cu is 7.04 eV. Calculate the velocity and de Broglie wavelength ofelectrons at the Fermi level of Cu. Can these electrons be diffracted by a crystal?

Solution:

v =

√2EFm

= 1.6× 106 ms-1 (12.2.10.1)

λ =h

mv= 4.63 A (12.2.10.2)

Since the wavelength is comparable to atomic spacing, these electrons can be diffractedby a crystal

12.2.11. Show that the fraction of electrons within kBT of the Fermi energy is 1.5kBTεF

, under the

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assumption that the temperature is so low that the probability of occupancy of levels is notaltered from the one at T = 0 K. Calculate numerically the value of this fraction for copper(εF = 7.04 eV) at 300 K and 1360 K (approximate melting point of Cu). This fraction is ofinterest because it is a rough measure of the percentage of electrons excited to higher energystates at a temperature T. Find roughly the electronic contribution to specific heat of Cuusing this expression.

Solution: First we need the density of states, which we can calculate along similar linesto 12.2.3a

g(E) dE =

√2m3

π2~3

√E dE (12.2.11.1)

Using this we can find εF in terms of number of electrons N by inspecting fFD,0(E)

N = V

ˆ ∞E=0

fFD,0(E)g(E) dE (12.2.11.2)

= V

√2m3

π2~3

ˆ εF

E=0

√E dE (12.2.11.3)

=

√8m3ε3

F

3π2~3V (12.2.11.4)

We now find the fraction of electrons within kBT of ε

f = VfFD,0(ε−F )g(εF )kBT

N(12.2.11.5)

=3kBT

2εF(12.2.11.6)

For Cu this can be calculated to be

f300 K = 0.55% (12.2.11.7)

f1360 K = 2.5% (12.2.11.8)

Thus as we increase the temperature a little bit from T = 0, this fraction of electronsjump from levels below εF to levels above εF . The electron at εF −kBT jumps to ε+kBT ,gaining 2kBT . The electron at εF does not gain any energy. Hence on an average eachelectron gains 3

2kBT The total energy gained is this

Eelectronic =Nf

V

3kBT

2(12.2.11.9)

=9Nk2

BT2

4V εF(12.2.11.10)

=⇒ Cvelectronic =9

2RkBT

εF(12.2.11.11)

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PH 107 : Quantum Physics Solution Booklet - CSE-IITB

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