Schrodinger Eqn

Introduction to Quantum MechanicsSchrodinger’s Equation

Instructor: Gautam Dutta

DA-IICT

winter 2012

Schrodinger’s Equation


Particles are associated with waves.

Waves of what? – Electric field, Magnetic field, mass density...

Schrodinger: Probability waves

P (x) = e2πih (px−Et) = e

i~(px−Et) (1)

Here ~ = h2π.

Can probability be negative? Experimentally these probabilities must

form an interference pattern.

Electromagnetic waves have amplitude ~E which can be both positive

[1]


and negative but intensity that we measure is |~E|2

We denote the wave amplitude by ψ(x) given by Eq.1 and the

non-negative probability P (x) will be given as

P (x) = |ψ(x)|2 = ψ∗(x)ψ(x)

ψ(x) can be +ve or -ve. We allow it to be complex in general. We

need such a quantity to produce the interference effect.

ψ(x) is called the probability amplitude and P (x) is the probability

density for the particle to be found at x.

[2]


Since it is certain to find the particle somewhere between −∞ and

+∞, it is obvious to expect the following normalization for ψ∫ ∞

−∞ψ∗ψdx = 1

What is the Equation that governs the evolution of ψ Consider

ψ = ei~(px−Et). Then

−~2∂2ψ

∂x2= p2ψ

Also

i~∂ψ

∂t= Eψ

[3]


Classically E = p2/2m for a free particle. So ψ = ei~(px−Et) satisfies

the following equation

i~∂ψ

∂t= − ~2

2m∂2ψ

∂x2(2)

If the particle is under the influence of a pottential V (x) then

E =p2

2m+ V

An appropriate equation for the wavefunction of such a particle would

be

i~∂ψ

∂t= − ~2

2m∂2ψ

∂x2+ V (x)ψ (3)

[4]


Obviously ψ = ei~(px−Et) with a single value of E and p will not

satisfy this equation. A superposition

ψ(x, t) =∫f(p)e

i~(px−Et)dp

might satisfy.

The above Equation was suggested by Erwin Schrodinger for the

evolution of the wavefunction of a particle of mass m in a pottential

V .

It is like the Newton’s Equation in Classical Mechanics. Once we

solve Schrodinger’s Equation and find ψ evrything about the particle

is known that is possible to be known in Quantum Mechanics.

ψ(x, t) gives the probability amplitude of finding the particle at the

[5]


point x at the time t. The probability of finding the particle between

x and x + dx is ψ∗ψdx. So the probability of finding the particle

between x1 and x2 is

P (x1 ≤ x ≤ x2) =∫ x2

x1

ψ∗ψdx

[6]


Operators: How do we get the values of physical quantities like

Energy, Momentum, Position Angular Momentum etc from ψ ?

Here we don’t have a well defined position and momentum of the

particle. So we can’t assign values of physical quantities to the

particle, like energy, angular momentum etc, which are functions of

its position and momentum.

But here we have states of particle defined by the wavefunctions

satisfying the Schrodinger’s equation. The collection of all possible

states of the particle constitute a vector space since the Schrodinger’s

equation is a linear differential equation. We can define linear

operators on this vector space. Certain states may be eigenstates of

[7]


these operators. It can be shown that the eigenstates of an operator

are linearly independent and they constitute a basis of the vector

space. We take the example of a free particle described by the

wavefunction ψ = ei~(px−Et). Then

−i~ ∂∂xψ = pψ

So the free particle states are eigenstates of the operator p = −i~ ∂∂x.

Similarly as we have seen earlier these states are eigenstates of the

energy operator E = i~ ∂∂t.

[8]


How do we construct the position operator?

Consider the fourier transform of the spatial wave function ψ(x)

f(p) =∫ψ(x)e−

i~pxdx; ψ(x) =

∫f(p)e

i~pxdp

Then

pψ(x) = −i~ ∂∂xψ(x) =

∫pf(p)e

i~pxdp

and

i~∂f(p)∂p

=∫xψ(x)e−

i~pxdx

[9]


We have the following pairs of fourier transforms:

f(p) ↔ ψ(x)

pf(p) ↔ −i~ ∂∂xψ(x)

i~∂

∂pf(p) ↔ xψ(x)

In the position representation, −i~ ∂∂x is the momentum operator.

By analogy i~ ∂∂p is the position operator in the momentum

representation.

The third fourier transform pair above suggests that in the position

representation the position operator is x. So in the position

representation the position operator is a multiplicative operator i.e

[10]

abhiRAM

Highlight


xψ(x) = xψ(x).

Every measurable physical quantity is associated with an operator.

The form of the operator has to be known in a particular

representation. An Operator can be a differential form or a

multiplicative form depending upon the representation we are working

with. For example in the position representation the momentum

operator is given by the differential form

p = −i~ ∂∂x

while the position operator is given by the multiplicative form

x = x

If the wavefunction of the particle is an eigenfunction of an operator

[11]


then it is easy to find the physical quantity corresponding to that

operator.

Oψ = oψ

ψ describes the state of a particle in which the value of the physical

quantity corresponding to the operator O is o. E.g

pei~(px−Et) = pe

i~(px−Et)

Eei~(px−Et) = Ee

i~(px−Et)

However in general ψ may not be in an eigenstate of momentum and

energy. If ψ(x, t) =∫f(p)e

i~(px−Et)dp then

pψ =∫f(p)pe

i~(px−Et)dp

[12]


Probability density of finding the particle with a momentum p is given

by |f(p)|2 = f(p)∗f(p). The average momentum or the Expectation

value of momentum is

p =∫|f(p)|2pdp

Note that this is equal to∫∞−∞ψ∗pψdx.

Try this exercise.

[13]


Steady state or Stationary statesIf the total energy of a particle is constant in a state then its wave

function Ψ(x, t) can be written as

Ψ(x, t) = e−i~ Etψ(x)

Substituting into Schrodinger’s Equation

− i~∂Ψ∂t

=−12m

~2∂2Ψ∂x2

+ VΨ

∴ Eei~Etψ(x) = e

i~Et

[−~2

2m∂2ψ(x)∂x2

+ V ψ(x)]

[14]


∴

[−~2

2m∂2

∂x2+ V

]ψ(x) = Eψ(x)

Hψ(x) = Eψ(x) (4)

Where the operator

H =−~2

2m∂2

∂x2+ V

is called the Hamiltonian operator and equation4 is the steady state

form of the Schrodinger Equation.

It is an eigenvalue equation and it cannot in general be solved for any

value of E and it depends upon the boundary condition. Only specific

values En are allowed. These values are called the eigenvalues of the

of H. The corresponding solution to the equation are labelled by n.

[15]


Thus

Hψn(x) = Enψn(x)

Quantization comes naturally in Schrodinger’s formalism.

[16]


Application of Schrodinger EquationParticle in a rigid box Consider a particle of mass m confined within

a rigid box of side L. In one dimension the potential in which the

particle is kept can be given as

V = 0 for 0 < x < 1

V = ∞ for x ≤ 0 and x ≥ L

In the region outside the box the particle cannot exist since the

potential is infinite. Within the box the Schrodinger Eqn is

− ~2

2m∂2ψ

∂x2= Eψ

∴∂2ψ

∂x2=

2mE~2

ψ

[17]

abhiRAM

Highlight


ψ(x) = A sin√

2mE~

x+B cos√

2mE~

x

Boundary conditions: ψ = 0 at x = 0=⇒ B = 0

ψ(x) = A sin√

2mE~

x

ψ = 0 at x = L.

ψ(L) = A sin√

2mE~

L = 0

=⇒√

2mE~

L = nπ

∴ E =n2π2~2

2mL2

Normalisation:

[18]

Schrodinger’s Equation∫ ∞

−infty

|ψ(x)|2dx = 1

∴ A2

∫ L

0

sin2 nπx

Ldx = 1

∴ A2L

2= 1 =⇒ A =

√2L

Find the average value of momentum p, p2, position x, energy E

Find (∆p)2 = p2 − p2, (∆x)2 and ∆p∆x

[19]

abhiRAM

Highlight


Physical Interpretation of Energy EigenfunctionIf Ψ(x, t) = e

i~Entψn(x), the particle has a fixed energy En. If we

have several identical systems and we measure the energy of the

particle independently in each of these systems then the energy will

turn out to be En.

ψn

En

ψn

En

ψn

En

ψn

En

[20]


In general the particle may not be in any eigenstate. In such

a case a measurement in identical systems will give different values

but each measurement will allways give only one of the eigenvalues

E1, E2, E3, .....En, ..... The same applies for any other observable

O that we measure.

ψ

E k

ψ

E E

ψ

E

ψ

4 1 2

[21]


Let

Ψ(x, t) = a1Ψ1(x, t) + a2Ψ2(x, t) + ....+ anΨn(x, t)

Since energy is a physical observable with real values, the energy

eigenfunctions are real.

The average value of the energy in the linearly superposed state

Ψ(x, t) must also be real.

〈E〉 =∫ ∞

−∞Ψ∗(x, t)

i~∂∂t

Ψ(x, t)dx

=n∑

i=1

|ai|2Ei +∑i 6=j

a∗iajEje− i

~(Ei−Ej)t

∫ ∞

−∞ψ∗i (x)ψj(x)dx

[22]


Since 〈E〉 must be a real number in any state, we must have∫ ∞

−∞ψ∗i (x)ψj(x)dx = 0 for Ei 6= Ej

∫∞−∞ψ∗(x)φ(x)dx is defined as the scalar product in the space of the

states of the quantum system.

So we conclude that eigenstates corresponding to different eigenvalues

of a physical observable are orthogonal. We will see that the linear

operators corresponding to physical observable are hermitian.

∫ ∞

−∞Ψ∗Ψ = 1 =

∑i

|ai|2

[23]


Also

〈E〉 =∑

i

|ai|2Ei

These implies the following interpretations:

We measure the energy of the system in this state. We will always

get the values E1, E2...En with the following probabilities

P (E1) = |a1|2

P (E2) = |a2|2

..... = ....

P (En) = |an|2

Mesurement in Quantum Mechanics

[24]


Let ψ(x) be an eigenstate corresponding to a potential V (x) with

energy E. The measuring system will be some potential V ′(x), say

the infinite well potential.

Then the process of measurement spontaneously throws(captures)

the particle in one of the eigenstates of V ′(x) ψ1, ψ2,... ψn

with probability |a1|2, |a2|2, ....|an|2. The process of measurement

inherrently have to disturb the system and we cannot measure the

exact value E in any measurement.

[25]


V V’

ψ

ψ

ψ

ψ

1

2

3E

E 1

E

E 1

2

3

Given system Measuring system

[26]


The Harmonic Oscillator

V(x)

V (x) =12kx2

[27]


H =p2

2m+ V (x)

= − ~2

2m∂2

∂x2+

12kx2 (5)

Time independent Schrodinger’s Eqn:

− ~2

2m∂2ψ(x)∂x2

+12kx2ψ(x) = Eψ(x)

ψ(x) = e−αx2with α =

√mk

2~is a solution with energy

E =~2

√k

m=

12~ω

[28]


In general the solutions which are ”well defined” are given by

ψn(x) = AnHn(√

2αx)e−αx2

where

H0(y) = 1

H1(y) = 2y

H2(y) = 4y2 − 2

H3(y) = 8y3 − 12y

H4(y) = 16y4 − 48y2 + 12

The energy in the state ψn(x) is En = (n+ 1/2)~ω

[29]


V(x)

E

E

E

E

0

1

2

3

The wavefunction exist even in the classically forbidden region.

[30]


V(x)

E 4

ψ4

−A A

The wavefunction exist even in the classically forbidden region.

[31]


Particle in a non-rigid boxFinite potential well.

Asin(kx) +B Cox(kx)

V

L

e k(L−x)kxe

Properties of wavefunctions1) ψ should be continuous

2) lim|x|→∞ψ(x) = 0

[32]


3)∫∞−∞ |ψ|

2dx = 14) The first spatial derivative of ψ must be continuous wherever V

is finite. Let us divide the whole space into three distinct regions.

Region 1: −∞ < x < 0; V (x) = V0

Region 2: 0 < x < L; V (x) = 0Region 3: L < x <∞; V (x) = V0

We solve the Schrodinger’s Equation in the three regions and

match Ψ(x) and its derivative at the boundaries of the regions.

In region 2 we have

− ~2

2m∂2Ψdx2

= EΨ

[33]


Here the soln is

Ψ2(x) = A2ek2x +B2e

−k2x k2 =

√−2mE

~2

where A2 and B2 are arbitrary constants to be determined from the

boundary conditions.

In region 1 the time independent Eqn is

− ~2

2m∂2Ψdx2

+ V0Ψ = EΨ

Here the soln is


−k1x k1 =

√−2m(E − V0)

~2

[34]


Similarly in region 3 we have


−k3x k3 =

√−2m(E − V0)

~2

Note that k2 is imaginary, while k1 and k3 are real.

As x → −∞, B1e−k1x → ∞. So we put B1 = 0 As x →

∞, A3ek3x →∞. So we put A3 = 0 So we have

Ψ1(x) = A1ek1x


−k2x

Ψ3(x) = B3e−k3x

[35]


At x = 0 we have Ψ1(0) = Ψ2(0). This gives

A1 = A2 +B2 (6)

At x = L we have Ψ2(L) = Ψ3(L). This gives

A2ek2L +B2e

−k2L = B3e−k3L (7)

We match the first derivative of Ψ at the two boundaries of the well,

Ψ′1(0) = Ψ′

2(0) and Ψ′2(L) = Ψ′

3(L) This gives

A1k1 = (A2 −B2)k2 (8)

−B3k3e−k3L = (A2e

k2L −B2e−k2L)k2 (9)

[36]


From Eq.6 and Eq.8 we get

A2 =12

(1 +

k1

k2

)A1 B2 =

12

(1− k1

k2

)A1

From 7 and 9 we have

B3 =k2 + k1

k2 − k3A1e

(k2+k3)L

So we have

Ψ1(x) = A1ek1x

Ψ2(x) =A1

2

(1 +

k1

k2

)ek2x +

A1

2

(1− k1

k2

)e−k2x

Ψ3(x) = A1k2 + k1

k2 − k3ek2Lek3(L−x)

[37]


A1 can be obtained by normalizing Ψ(x).

k1

k2=√V0 − E

i√E

= −i√V0

E− 1

∴ 1 +k1

k2=

√V0

Ee−iα where tanα =

√V0

E− 1

1− k1

k2=

√V0

Eeiα

k2 + k1

k2 − k3= e−2iα

[38]


Now Ψ2 can be written as

Ψ2(x) = A1

√V0

Ecos

(√2mE~

x− α

)

At x = L, Ψ2(L) = Ψ3(L). This gives√V0

Ecos

(√2mE~

L− α

)= ei(−2α+

√2mE~ L) (10)

The imaginary part of the R.H.S must equal to 0. So we have

√2mE~

L− 2α = nπ

[39]


So Eqn.(10) becomes

cos

(√2mE2~

L+nπ

2

)= (−1)n

√E

V0

The energy values satisfying the above transcendental equation gives

the discrete energies that the particle can have within the box.

Though n can be any integer, distinct values of energy are obtained

only for n = 0, 1, 2and3. Beyond this we get the cosine curves are

the same and hence we get the same energies. We can plot the

two graphs (−1)n√

EV0

and cos(√

2mE2~ L+ nπ

2

)as a function of

√E

and look for their intersection. All the solutions will lie for E < V0.

Beyond this the absolute value of R.H.S is greater than one and

hence it can’t intersect the sine or cosine curve on the L.H.S. The

[40]


number of discrete energy solutions we have depends upon the height

of the well V0 and the width L of the well. For every period of cosine

cycle we have 2 possible energies. So for n = 0, 1, 2, 3) we will have

2 × 4 = 8 distinct energy eigenvalues. The cosine cycle completes

when√E changes by 4π~√

2mL. So the number of discrete eigenvalues

less than V0 is given by

8×√V0

4π~√2mL

=2√

2mV0L

π~

When E > V0, the system is not bounded and we can have a valid

wavefunction for the particle for any continuous value of energy.

[41]


Figure 1: Eigen values of energy are given by the Eqn. cos�√

2mE2~ L + nπ

2

�= (−1)n

qEV0

. We

plot the two functions on the r.h.s and the l.h.s as a function of√

E. The solid curves are for even

n while the dashed curves are for odd n. The intersections give the values of E. There will be no

bound states beyond E = V0 .

[42]


Examples

1. Schrodinger Equation is a linear differential Equation. Show that

if Ψ1 and Ψ2 are two solutions of the Schrodinger Equation then

Ψ = a1Ψ1 + a2Ψ2

is also a solution of Schrodinger Equation.

2. The angular momentum along z direction is given by Lz = xpy −ypx. Treating px and py as operators viz..

px = −i~ ∂∂x

andpy = −i~ ∂∂y

[43]


show that

Lz = −i~ ∂

∂Φ

3. Solve the Schrodinger Equation for a free particle which is in a

stationary state with energy E. Show that

ψ(x) = Aei~√

2mEx +Be−i~√

2mEx

is a solution to the steady state Schrodinger Equation. Find the

average momentum in this state.

[44]


Two and three dimensionsSteady state Schrodinger Eqn:

p2

2mψ(x, y, z) + V (x, y, z)ψ(x, y, z) = Eψ(x, y, z)

p2 = px2 + py

2 + pz2

∴ Eψ(x, y, z) = − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)ψ(x, y, z)

+V (x, y, z)ψ(x, y, z)

= − ~2

2m∇2ψ(x, y, z) + V (x, y, z)ψ(x, y, z)

[45]


The momentum operator is a vector operator and is given by

~p = −i~~∇ = −i~(i∂

∂x+ j

∂

∂y+ k

∂

∂z

)

[46]


Two Dimensional box

V= 8

V= 8

V= 8

V= 8

V=0

a

b

x

y

V (x, y) = 0 0 ≤ x ≤ a and 0 ≤ y ≤ b

= ∞ elsewhere

[47]


− ~2

2m

(∂2

∂x2+

∂2

∂y2

)ψ(x, y) = Eψ(x, y)

Assume variable separable form

ψ(x, y) = ψ(x)ψ(y)

ψ(x, y) = A sin(nxπx

a) sin(

nyπy

b)

E =

(n2

x

a2+n2

y

b2

)~2π2

2m

[48]


For 3-dim box of size (a,b,c)

ψ(x, y, z) = A sin(nxπx

a) sin(

nyπy

b) sin(

nzπz

c)

E =

(n2

x

a2+n2

y

b2+n2

z

c2

)~2π2

2m

[49]


V=0

V= 8

a

V=0

a

b

V=0

a

b

c

[50]


Spherically symmetric potential

V (r, θ, φ) = V (r)

E.g Hydrogen atom

V (r) = − Ze2

4πε0r2Harmonic Oscillator

V (r) = kr2

Particle in spherically symmetric box

V (r) = 0, when 0 ≤ r ≤ a

= ∞. when r > a

[51]


a−a

V=0

V= 8Quantum Dot

[52]


Schrodinger’s Equation in Sphericle polar co-ordinate:

Eψ(r, θ, φ) = − ~2

2m∇2ψ(r, θ, φ) + V (r)ψ(r, θ, φ)

= − ~2

2m

[1r2∂

∂r

(r2∂

∂r

)+

1r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1r2 sin2 θ

∂2

∂φ2

]ψ(r, θ, φ)

+V (r)ψ(r, θ, φ)

Comparing with the classical Equation

E =p2

r

2m+

L2

2mr2+ V (r)

[53]


L2 = −~2

[1

sin θ∂

∂θ

(sin θ

∂

∂θ

)+

1sin2 θ

∂2

∂φ2

]Exercise: Starting with ~L = ~r× ~p show that L

2is given by the above

operator.

[54]


Separate r and the angular variables

ψ(r, θ, φ) = R(r)Y (θ, φ)

After rearrangement the Sch Eqn gets decoupled into r and θ, φ

− ~2

2m1R

∂

∂r

(r2∂R

∂r

)+ r2V (r)− r2E

− ~2

2m1Y

[1

sin θ∂

∂θ

(sin θ

∂

∂θ

)+

1sin2 θ

∂2

∂φ2

]Y = 0

[55]


Multiplying this by 2m2 gives

− ~2 1R

∂

∂r

(r2∂R

∂r

)+ r22m(V (r)− E)

−~2 1Y

[1

sin θ∂

∂θ

(sin θ

∂

∂θ

)+

1sin2 θ

∂2

∂φ2

]Y = 0

The angular equation can have acceptable solutions only if it is equal

to a constant given by ~2l(l + 1) where l is an integer.[1

sin θ∂

∂θ

(sin θ

∂

∂θ

)+

1sin2 θ

∂2

∂φ2

]Y = −l(l + 1)Y

Y (θ, φ) are called spherical harmonics. Again by separation of

[56]


variables θ and φ it can be shown that

Y (θ, φ) = Θ(θ)Φ(φ) = Pml (θ)e±imφ

e±imφ is the solution to the φ part of the above equation after

separation into θ and φ

∂2Φ∂φ2

= −m2Φ

The periodic boundary condition along φ requires m to be an integer.

The functions Pml (θ) are known as Associated Lagendre polynomials

given as

Pml (θ) = (−1)m(1− cos2 θ)m/2 dm

d cosm θPl(cos θ)

[57]


Here Pl(cos θ) is the Legendre Polynomial of degree l.

So 0 ≤ m ≤ l.

Operating Lz on ψ(r, θ, φ) gives (check this)

Lzψ = −i~∂ψ∂φ

= ~mψ

Also

L2ψ = ~2l(l + 1)ψ

So the energy eigen states of any spherically symmetrical pottential

are also eigenstates of Lz and L2

[58]


−h v/ l(l+1) −h

z

m

L and L2z

[59]


/l(l+1)v h−

h−

h−

h−

h−

h−

z

m=2

1

0

−1

−2

l=2

Eg. Show that the wave function given by R(r)P 11 (cos θ)eiφ,

R(r)P1(cos θ) and R(r)P 11 (cos θ)e−iφ are eigenstates of L

2and Lz

operators and find the corresponding eigenvalues. (P1(cos θ) = cos θ)

[60]


Summary

• Given any system specified by a potential V (r) we can find

energy eigen states of the system which are also called stationary

states. These states are physicaly acceptable solutions to the time

independent Schrodinger Equation.

HΨ(~r) = EΨ(~r)

The probability of finding the particle at the position ~r will be given

as |Ψ(~r)|2 ∫allspace

|Ψ(~r)|2dV = 1

[61]


• Physical Quantities like momentum, angular momentum etc. are

replaced by their corresponding operators. A given stationary state

Ψ may or may not be also an eigenstate of these operators. If they

are then the physical quantity corresponding operator has a well

defined value in the state Ψ given by the eigenvalue. Otherwise we

can only talk about the average value of that Physical quantity.

• Any general state Ψ of a particle can be expressed

as a linear combination(superposition) of the statinary

states Ψ1,Ψ2,Ψ3,.....,Ψn with energy eigenvalues E1, E2, .....En

respectively.

Ψ = a1Ψ1 + a2Ψ2 + .....+ anΨn

If we measure its energy the probability that the particle will be

[62]


found in state Ψi with energy Ei is given by |ai|2.The average energy of the particle will be given by∫

V

Ψ∗HΨdV =∑

i

|ai|2Ei

[63]

Schrodinger Eqn

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