What to recall about motion in a central potential

1

I. What to recall about motion in a central potentialII. Example and solution of the radial equation for a particle trapped within radius “a”III. The spherical square well

(Re-)Read Chapter 12 Section 12.3 and 12.4

2

I. What to recall about motion in a central potentialV =V (|rr1 −

rr2 |) between m asses m 1 and m 2

1 24 4 34 4

Recall the tim e-independent Schrodinger Equation:H

p12

2m1+

p22

2m2+V (|rr1 −

rr2 |)⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥Ψ =EΨ

↓6 74 4 4 4 4 44 84 4 4 4 4 4 4{ Ψ =EΨ

To convert this into a separable PDE, define1M

=1m 1

+1m 2

M =m 1 + m 2

R =m 1

rr1 + m 2rr2

Mrr=rr1 −

rr2rP=

rp1 +rp2

rpm=

rp1m 1

−rp2

m 2

Then you get

H =p2

2M+

p2

2m+V (|rr |)

Ignore center of mass motion

Focus on this “Hμ”

rr1 −

rr2

rr1

rr2

m1 m2

O

3

Hm =p2

2m+V (r)

=−h2∇2

2m+V (r)

In spherical coordinates:

∇2 =1r2

r2∂∂r

⎛⎝⎜

⎞⎠⎟ +

1

r2 sinq

∂∂q

sinq∂∂q

⎛⎝⎜

⎞⎠⎟ +

1

r2 sin2q

∂2

∂f 2

1

r2−L2

h2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1 24 4 4 4 4 44 34 4 4 4 4 4 4

Plug this ∇2 into Hm

Write out Hm |Ψ⟩ =E |Ψ⟩

Project into ⟨r,q,f | space:

−h2

2m1r2

∂∂r

r2∂∂r

⎛⎝⎜

⎞⎠⎟−

L2

h2r2⎡⎣⎢

⎤⎦⎥+V (r)

⎧⎨⎩⎪

⎫⎬⎭⎪⟨r,q,f |Ψ⟩ =E⟨r,q,f |Ψ⟩

Guess that ⟨r,q,f |Ψ⟩ =R(r)f(q,f )Separate the equation, the constant of separation turns out to be l(l+1).

f(q,f ) turns out to be Ψlm (q,f )

Then the R equation is:

1R

ddr

r2dRdr

⎛⎝⎜

⎞⎠⎟−

2mr2

h2 V (r)−E[ ] =l(l+1) "Form 1" of the radial equation

the angular momentum operator

4

You can get an alternative completely equivalent form of this equation if you derivem ≡rRThen you get

−h2

2md 2udr2

+ V +h2l(l+1)2mr2

⎡⎣⎢

⎤⎦⎥u=Eu "Form 2" of the radial equation

*Choose either Form 1 or Form 2 depending upon what V is--pick whichever gives and easier equation to solve*Rem em ber the boundary conditions on R:rR(r→ ∞)→ 0 BC1rR(r→ 0)→ 0 BC2Procedure for finding the total Ψ(

rr,t) for a system in a central potential:

(i) Get V (r)(ii) Plug it into the radial equation (either Form 1 or Form 2), solve for R and the energies Ei

(iii) Multiply that R by Ψlm (q,f ) and e

−iEith to get Ψ(r,t)=RΨe

−iEith

5

II. Example-Solution of the radial equation for a particle trapped within radius "a"

V= 0 for r < a∞ for r > a

⎧⎨⎩

This is also called a "spherical box"Recall the radial equation in Form 1 (without the substitution u=rR):

1r2

ddr

r2ddr

⎛⎝⎜

⎞⎠⎟ +

2mh2 E −V (r)−

h2l(l+1)2mr2

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥R =0

Since V =∞ for r > a, the wave function cannot have any portion beyond r > a. So just solve the equation for r < a.Plug in V =0 (r < a )

1r2

ddr

r2ddr

⎛⎝⎜

⎞⎠⎟ +

2mEh2 −

l(l+1)r2

⎡⎣⎢

⎤⎦⎥R =0

expand this: call this "k2"

1r2

r2d 2

dr2+ 2r

ddr

⎛⎝⎜

⎞⎠⎟

d 2

dr2+2rddr

+ k2 −l(l+1)

r2⎡⎣⎢

⎤⎦⎥R =0

6

Define r ≡kr, so 1r=kr

Then ddr

=drdr

ddr

=kddr

d 2

dr2=

ddr

kddr

⎛⎝⎜

⎞⎠⎟ =

drdr

ddr

kddr

⎛⎝⎜

⎞⎠⎟ =k2 d 2

dr2

Plug this in:

k2 d 2

dr2 +2k2

rddr

+ k2 −l(l+1)k2

r2

⎡⎣⎢

⎤⎦⎥R =0

d 2

dr2 +2r

ddr

+ 1−l(l+1)r2

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥R =0

The solution of this equation is R(r) = Cjl(r) + Dnl(r)

jl(r)≡p2r

⎛⎝⎜

⎞⎠⎟1/2

Jl+

12

(r)

"spherical Bessel function" "ordinary Bessel function of half-odd integer order"

exam ples: j0(r)=sinrr

j1(r)=sinrr2 −

cosrr

Normalization not yet specified

Spherical Neuman function, Irregular @ r=0, so get D=0

7

I. Particle in 3-D spherical well (continued)II. Energies of a particle in a finite spherical well

Read Chapter 13

8

Now apply the BC: the wave function R must = 0 @ r = a

jl (ka) =0 whenever a Bessel function = 0 its argum ent (here: (ka)) is called a "zero" of the spherical Bessel function and these colum ns are labelled by n n=1 n=2 n=3 etc.

S jl=0 for ka = 3.14 6.28 9.42 ...P jl=1 for ka = 4.49 7.73 ... ...

D jl=2 for ka = 5.76 9.10 ... ... F jl=3 ...etc.

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

Sum m arize:

Ψ3−Dcentral potential

tim e-independent = R ⋅Ψlm

spherical harm onic where R=C⋅jl

Because each jl has zeros at several n's, we have to specify n too, so R ≡R nl

In spectroscopy the rows are labelled by:

9

Because k2 ≡ 2mEh2 , the boundary condition that jl(ka) = 0 gives the allowed energies.

Recipe to find an allowed energy:(i) Pick the n, l levels that you want. (Exam ple: pick n = 2, l = 0)

(ii) Find the zero of that Bessel function kan =2jl =0

⎛⎝⎜

⎞⎠⎟=6.28

⎛⎝⎜

⎞⎠⎟

(iii) Plug into jl(ka) = 0

kan =2jl =0

⎛⎝⎜

⎞⎠⎟=6.28

use k2 ≡ 2mEh2

k2a2 = (6.28)2

2mEl=0

n=2

h2 a2 = (6.28)2

El=0n=2 =

(6.28)2h2

2ma2

10

I. Energies of a particle in a finite spherical square well

11

III. Energies of a particle in a finite spherical square well

Consider

V(r) = −V 0 for r < a 0 for r > a

⎧⎨⎪⎩⎪

This actuall looks very m uch like the potential between 2 nucleons in the nucleusSolve this analogously to 1-D square well procedure(i) Define regions I and II(ii) Plug in V into radial equation in each region

Pick Form 2, so solve for u=rR(iii) Apply BC @ r = 0, a, ∞ This leads to quantized E (we'll stop here)(iv) Norm alize if you want the exact form of Ψ Carry out this procedure, first for l = 0 only, then for general l:

V(r)

a

0

-V0

rI II

0

12

Case I: Find allowed energies of a particle in an l = 0 state of a finite spherical square well

(i) define regions I and II

Notice that the particle in the well will have E < 0 just because of the way we defined the potential*This is just a convention (i.e. a choice of the origin for the V scale), but since it is common we will use it.Notice this is a different convention that the one we used for the 1-D square well

That was:

but the forms of the solution inside and outside of the well (sinkx or e−kx) are unaffected by the choice of origin.

V(r)

a

0

-V0

rI II

0

0

+V0

13

Because E is intrinsically negative, we can write: E = -|E| when we wishRecall Form 2 of the radial equation:

−h2

2md 2udr2

+ V +h2l(l+1)2mr2

⎡⎣⎢

⎤⎦⎥u=Eu

when l = 0 this is:

−h2

2md 2udr2

+V u=Eu

This is exactly the sam e form as for the 1-D square well, so the wavefunctions "in" will have the sam e form as the "Ψ's" for the 1-D square well(ii) Plug in V :Make a table as we did in Chapter 4:

Region IV = -V 0

Region IIV = 0

General tim e-independent Radial equation:−h2

2md 2udr2

+V 0

⎡⎣⎢

⎤⎦⎥u=Eu

−h2

2md 2udr2

⎡⎣⎢

⎤⎦⎥u=Eu

Solve Radial equation to get:

where:

"allowed region solution:"uI = Asink1r+ Bcosk1r

k1 ≡2m(V 0 + E)

h

= 2m(V 0−|E |)

h

"forbidden region solution:"

uII = Ce+k2r + De−k2r

k2 ≡ 2m(−E)

h

= 2m |E |h

14

(iii) Apply BC's:

BC1: rR(r → ∞)→ 0 u(r→ 0)→ 0 So B=0BC2: rR(r→ ∞)→ 0 u(r→ ∞)→ 0 So C=0BC3: uI(r=a) = uII(r=a)

Asink1a = De−k2a "equation 1"

BC4: duI

dr(r=a) =

duII

dr(r=a)

k1Acosk1a = -k2De−k2a "equation 2"

Divide equation 1equation 2 :

k1 cotk1a = -k2 This has the sam e form (variation in som e factors or 2) as the Class 2 (or odd) solutions to the 1-D square well.Multiply both side by a: k1acotk1a = -k2a

-cotk1a =k2ak1a

15

Define l ≡2mV 0a

2

h2

Define y ≡ k1a = a 2m(V 0−|E |)

h Notice that

k2a = a 2m |E |

h=

(2mV 0a2 )−[2m(V 0−|E |)a2 ]

h2

= l −y2

-coty= l −y2

y Plot both sides versus y.

Intersection points are solution to the equation:

(use the identity that -cotx = +tan(p2+x))

y0 π/2 π 3π/2 2π 5π/2 3π 7π/2 4π

LHS(-cot)

16

y0 π/2 π 3π/2 2π 5π/2 3π 7π/2 4π

How the value of λ affects the plot:l −y2

y for large λ

l −y2

y for small λ

*Notice if l is very sm all there m ay be NO interactions

proportional to V 0a2

well depth (well width)2

To find E:Locate graphically, or com putationally, the y-coordinates of the points where:

-coty=l −y2

y

Call these yi Exam ple:

Then plug in the definition of y:

yi = ah

2m(V 0−|Ei |)

Plug in these values of a, u, V 0 , h, to solve for Ei

y1 y2 y3

17

Alternatively if you measure the Ei (for example the bound state energies of a deuteron),

you can work backwards to find V0a2 .

Recall Case 1 was for l = 0 only. Now,Case 2: Find the allowed energies for finite spherical square well for a particle in an arbitrary l state.

(i) Regions I II(ii) In this case Form 1 of the Radial equation is easier to solve:

1r2

ddr

r2 ddr

⎛⎝⎜

⎞⎠⎟ +

2mh2 E −V (r)−

h2l(l+1)2mr2

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥R =0

expand the derivative and write E = -|E|

d 2

dr2+2rddr

+2mh2 −|E |−V (r)−

h2l(l+1)2mr2

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥R =0

Make Table:

V(r)

a

0

-V0

rI II

0

18

Region IV = -V0

Region IIV = 0

General time-independent Radial equation:d 2

dr2 +2rddr

+2mh2 V 0−|E |−

l(l+1)r2

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥R =0

d 2

dr2+2rddr

−2m |E |h2 −

l(l+1)r2

⎡⎣⎢

⎤⎦⎥R =0

kI ≡2m(V 0−|E |)

hrI ≡kIr (just like p.6 in notes)

d 2

drI2 +

2rI

ddrI

+ 1−l(l+1)rI

2

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥R =0

RlI(rI)=Aljl(rI)+ Bljl(rI)

kII ≡2m |E |h

rII ≡ikIIr

To solve this define:and

Then the radial equation becomes:

The solution is:

19

I. Energies of a particle in a finite spherical square well, continuedII. The Hydrogen Atom

20

If rII ≡ikIIr

Then 1r=ikIIrII

ddr

=drII

drd

drII

=ikIId

drII

d 2

dr2=

ddr

ikIId

drII

⎛⎝⎜

⎞⎠⎟ =

drII

drd

drII

ikIId

drII

⎛⎝⎜

⎞⎠⎟ =ikIIikII

d 2

drII2 =−kII

2 d 2

drII2

Plug these in: then the Radial Equation becom es

−kII2 d 2

drII2 +

2ikIIrII

ikIId

drII

−kII2 +

l(l+1)kII2

rII2

⎡⎣⎢

⎤⎦⎥R =0

Multiply through by (-1):

d 2

drII2 +

2rII

ddrII

+1−l(l+1)rII

2

⎡⎣⎢

⎤⎦⎥R =0

Again the solution is R lII(rII)=Cljl(rII)+ Dlnl(rII)

Since Region II does not include r=0, we do no thave to discard the Neum ann for nl.

Since we have not yet fixed C and D, we can rewrite R lII as a different form of linear com bination of jl and nl.

21

In particular we could define

h l(1) ≡jl + inl and

hl(2) ≡jl −inl "The spherical Hankel functions"

Then

R lII(rII)= ′Clhl

(1)(rII)+ ′Dlhl(2)(rII)

(iii) Now apply the boundary conditions:

BC1: rR(r→ 0)→ 0 This is satisfied by R lI =Ajl(rI)= Ajl(kIr)

BC2: rR(r→ ∞)→ 0 Exam ine the hl(r→ ∞):

jl(r → ∞)→1rcos r −

(l+1)p2

⎡⎣⎢

⎤⎦⎥

nl ≡(−1)l+1p2r

⎛⎝⎜

⎞⎠⎟12

J−l−

12

(r)

r→ ∞⏐ →⏐ ⏐1rsin r −

(l+1)p2

⎡⎣⎢

⎤⎦⎥

Plug these in:

22

I. Energies of a particle in a finite spherical square well (continued)II. The Hydrogen Atom

23

hl(1)(rII) r→ ∞⏐ →⏐ ⏐

1ikIIr

cos ikIIr−(l+1)p

2⎡⎣⎢

⎤⎦⎥+ i

1ikIIr

sin ikIIr−(l+1)p

2⎡⎣⎢

⎤⎦⎥

=1

ikIIrcos ikIIr−

(l+1)p2

⎡⎣⎢

⎤⎦⎥+ isin ikIIr−

(l+1)p2

⎡⎣⎢

⎤⎦⎥

⎧⎨⎩⎫⎬⎭

Ignore the phases for now

Recall eiq =cosq + isinq Here "q" is ikIIr

:1

ikIIrei(ikIIr){ }

:1

ikIIre−kIIr{ } This → 0 as r→ ∞ so it satisfies the BC

hl(2)(rII) r→ ∞⏐ →⏐ ⏐

1ikIIr

cos ikIIr−(l+1)p

2⎡⎣⎢

⎤⎦⎥−i

1ikIIr

sin ikIIr−(l+1)p

2⎡⎣⎢

⎤⎦⎥

:1

ikIIre+kIIr{ } This blows up as r→ ∞ so we m ust set its coefficient ′D =0

BC3: R I(r=a)=R II(r=a)

Ajl(kIa)= ′C hl(1)(kIIa)

BC4: ∂R I

∂r(r=a)=

∂R II

∂r(r=a)

Addr

jl(kIa)= ′Cddr

hl(1)(kIIa)

24

Divide BC4BC3

to eliminate the normal Bat. coefficients:

ddr

jl (kIa)

jl (kIa)=

ddr

h(1)l(kIIa)

h(1)l(kIIa)

For a specific V 0 , a, u, and l, one can solve this for the bound states Ei

I. The Hydrogen AtomWhat this m eans is "the eigen functions and eigen energiespossible for the electron in a one-electron atom "This is just the central potential problem again, now for

V =−Ze2

r

So we know that Ψelectron will have the form Ψ : R(r)Ψlm

because of this (l, m ), also subscript the ′Ψ :Ψlm

So the m ost general Ψ m ust be a linear com bination of all possible Ψlm's, so

Ψ = Ψlm = R(r)Ψl

m (q,f )l,m∑

l,m∑

# protons in nucleus

Charge of electron

energy levels

25

The Ylm's are standard functions so we can ignore them for now. We will find R and then just multiply by Yl

m later.To find R, recall the Radial Equation (Form 2):

−h2

2md 2udr2

+ V +h2l(l+1)2mr2

⎡⎣⎢

⎤⎦⎥u=Eu (where u≡rR)

Consider bound states, so se E=-|E|

Plug in V =−Ze2

r−h2

2md 2udr2

+h2l(l+1)2mr2

−Ze2

r+ |E |

⎡⎣⎢

⎤⎦⎥u=Eu

To sim plify the form of the equation, m ultiply through by −2mh2 and define

r ≡8m |E |h2

⎛⎝⎜

⎞⎠⎟

12

r

So

1r=

8m |E |h2

⎛⎝⎜

⎞⎠⎟

12 1r

and

ddr

=drdr

ddr

=8m |E |h2

⎛⎝⎜

⎞⎠⎟

12 ddr

Defined as E<0

26

and

d 2

dr2 =ddr

8m |E |h2

⎛⎝⎜

⎞⎠⎟

12 ddr

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=drdr

ddr

8m |E |h2

⎛⎝⎜

⎞⎠⎟

12 ddr

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=8m |E |h2

⎛⎝⎜

⎞⎠⎟

12 d 2

dr2

Plug these in:

8m |E |h2

⎛⎝⎜

⎞⎠⎟d 2

dr2 −l(l+1)r2

8m |E |h2

⎛⎝⎜

⎞⎠⎟ +

2mh2

Ze2

r8m |E |h2

⎛⎝⎜

⎞⎠⎟

12

−2m |E |h2

⎡

⎣⎢⎢

⎤

⎦⎥⎥u=0

divide through by 8m |E |h2

⎛⎝⎜

⎞⎠⎟ :

d 2

dr2 −l(l+1)r2 +

2mZe2

h2

8m |E |h2

⎛⎝⎜

⎞⎠⎟

12

1r−14

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥u=0

Ze2

hm

2 |E | call this "l"

d 2

dr2 −l(l+1)r2 +

lr−14

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥u=0

27

Procedure to solve: this is similar to the one for the analytic (non-a+a) solution of the harm onic oscillator.Recall the 2 radial BC's: BC1: u(r→ ∞)→ 0; BC2: u(r→ 0)→ 0.

(i) consider the case where r → ∞

This elim inates the 1r,

1r2 term s:

Then the equation is approxim ately:

d 2udr2 −

u4=0 (for u(r → ∞) only)

u : Ae−r2 + Be

+r2

Apply BC1: Recall a physically acceptable Ψ (or u) m ust → 0 for r→ ∞, r → ∞ so B m ust = 0.

(ii) Consider the case where r → 0

Then the 1r2 term dom inates the

1r and the

14 term s, so the equation is approxim ately:

d 2

dr2 −l(l+1)r2

⎡⎣⎢

⎤⎦⎥u=0

u=Cr−l + Drl+1

solution

solution

28

I. The Hydrogen Atom (continued)II. Facts about the principal quantum number nIII. The wavefunction of an e- in hydrogen

29

Apply BC2: u(r → 0)→ 0, soC=0

(iii) Now insist that the exact solution of the equation have a form which will join up with these asym ptotic cases sm oothly.

Assum e u(all r) = e−r2

⎛⎝⎜

⎞⎠⎟rl+1H(r)

H(r)= airi

i∑

H ≠a ham iltonian herePlug this into the full equation:

d 2

dr2 −l(l+1)r2 +

lr−14

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥e

−r2

⎛⎝⎜

⎞⎠⎟rl+1H(r)=0

ddr

e−r2

⎛⎝⎜

⎞⎠⎟ rl+1 dH

dr+ (l+1)rlH

⎡⎣⎢

⎤⎦⎥−12e

−r2

⎛⎝⎜

⎞⎠⎟rl+1H(r)

⎧⎨⎪⎩⎪

⎫⎬⎪⎭⎪

ddr

e−r2

⎛⎝⎜

⎞⎠⎟rl+1 dH

dr+ e

−r2

⎛⎝⎜

⎞⎠⎟(l+1)rlH −e

−r2

⎛⎝⎜

⎞⎠⎟rl+1 H

2

⎧⎨⎪⎩⎪

⎫⎬⎪⎭⎪

=e−r2

⎛⎝⎜

⎞⎠⎟ rl+1 d

2Hdr2 + (l+1)rl dH

dr⎡⎣⎢

⎤⎦⎥−12e

−r2

⎛⎝⎜

⎞⎠⎟rl+1 dH

dr

+ (l+1)e−r2

⎛⎝⎜

⎞⎠⎟ lrl−1H + rl dH

dr⎡⎣⎢

⎤⎦⎥−12e

−r2

⎛⎝⎜

⎞⎠⎟(l+1)rlH

−12e

−r2

⎛⎝⎜

⎞⎠⎟ (l+1)rlH + rl+1 dH

dr⎡⎣⎢

⎤⎦⎥+14e

−r2

⎛⎝⎜

⎞⎠⎟rl+1H

30

and divide out e−r2

⎛⎝⎜

⎞⎠⎟:

[rl+1 d2Hdr2 + (l+1)rl dH

dr−12rl+1 dH

dr

+ (l+1)lrl−1H + (l+1)rl dHdr

−12(l+1)rlH

− 12(l+1)rlH −

12rl+1 dH

dr+14rl+1H −l(l+1)rl−1H + lrlH −

14rl+1H] = 0

Collect term s:

rl+1 d2Hdr2 + (l+1)rl −

12rl+1 + (l+1)rl −

12rl+1⎡

⎣⎢⎤⎦⎥dHdr

+ (l+1)lrl−1 −12(l+1)rl −

12(l+1)rl +

14rl+1 −l(l+1)rl−1 + lrl −

14rl+1⎡

⎣⎢⎤⎦⎥H =0

Collect term s:

rl+1 d2Hdr2 + 2(l+1)rl −rl+1⎡⎣ ⎤⎦

dHdr

+ −(l+1)rl + lrl⎡⎣ ⎤⎦H =0

Divide through by rl :

rd 2Hdr2 + 2l+ 2 −r[ ]

dHdr

− l+1−l[ ]H =0

*Notice that for a given value of l, this is an eigenvalue equation for H with eigenvalue = l. Recall H here is part of a wavefunction and not a ham iltonian

31

Plug in H= airi∑

dHdr

= iairi−1∑

d 2Hdr2 = i(i−1)air

i−2∑

r i(i−1)airi−2∑ + (2l+ 2 −r) iair

i−1∑ −(l+1−l) airi∑ =0

i(i−1)airi−1∑ + (2l+ 2) iair

i−1∑ − (l+1−l + i)airi∑ =0

i(i−1+ 2l+ 2)airi−1∑ − (l+1−l + i)air

i∑ =0

i(i + 2l+1)airi−1∑ − (l+1−l + i)air

i∑ =0

Set coefficients of each power of r separately = 0:

32

Set coefficients of each power of r separately = 0:

r0 1(2l+2)a1 −(l+1−l + 0)a0 =0

r1 2(2l+2)a2 −(l+1−l +1)a1 =0

r2 3(2l+4)a3 −(l+1−l + 2)a2 =0

r3 4(2l+5)a4 −(l+1−l + 3)a3 =0

r4 5(2l+6)a5 −(l+1−l + 4)a4 =0

rn (n+1)(2l+ n+2)an+1 −(l+1−l + n)an =0

an+1 =l+1+ n( )−l⎡⎣ ⎤⎦

(n +1)(2l+ n + 2)an Recursive realtion for the ai

This relationship between the ai is like the one for an exponential function. To see this com pare:

The "H(r)" series "The er series", er =ri

i, so∑ ai =

1i!

an+1

an

=l+1+ n−l

(n +1)(2l+ n + 2)

an+1

an

=

1(n +1)!

1n!

=n!

(n +1)!=

1n +1

when n→ large when n→ large

≈n

n⋅n:1n :

1n

33

So if H(r) is not truncated, what we have so far is

u(r) = e−r2 rl+1H(r)

= e−r2 rl+1e+r

= e+r2 rl+1 r→ ∞⏐ →⏐ ⏐ ∞

To force u to be a physically acceptable wavefunction, truncate the series HPick som e i whose ai is the highest non-zero ai

call it im ax

Recall ai+1 =l+1+ i−l[ ]

(i +1)(2l+ i + 2)ai

when i=im ax

ai =ai m ax

ai+1 =0

0 = l+1+im ax −l

l=l+1+im ax

Since i and l and 1 are all integers, l m ust be an integer. Renam e it "n"

34

I. Facts about the principal quantum number n (continued)II. The wavefunction of an e- in hydrogenIII. Probability current for an e- in hydrogen

35

Facts about l=n1. "n" is called the Principal Quantum Num ber2. Recall the definition

n=l=Ze2

hm

2 |En |, so

En =−|En | = −Z2e4

h2

m2n2 These are the allowed bound state energies of the Coulom b potential.

3. n = l+im ax +1, so n ≥l+1 (im ax =0,1,...) l≤n-1, or lm ax =n−14. Notice no m atter how large n is, its En will always be slightly < 0, so at lease weakly bound.

So this potential V =−Ze2

r has an infinite num ber of bound energy states (unlike the square well).

5. Energy degeneracies: (i) due to l→ The energy depends only on n, but for each n, there are n possible values of l

l = 0, 1,..., n-1 (ii) due to m → For each l, m can be -l, -l+1, ..., 0, ..., l-1, l (iii) total due to m and l is then

(2l+1)l=0

n−1

∑ =n2 (this ignores spin for now)

num ber m values

36

6. A (2l +1)-fold degeneracy of m-values is characteristic of a spherically symmertric potential.

II. The wavefunction of an e− in hydrogen

Recall we found that

Ψe inhydrogen

tim e-independent =R ⋅Ψlm

to solve the R equation we used Form 2,

so we defined r=8m |E |h2 r=2

2m |E |h2 r=2kr

and u=rR

R=1ru=

1r

e−r2 rl+1H(r)⎡

⎣⎤⎦

airi

i

im ax

∑

=2kr

e−r2 rl+1H(r)⎡

⎣⎤⎦

=e−r2 rl2k air

i

i

im ax

∑ These turn out to be fam ous functions

37

the associated Lguerre Polynomials, LqP (r)

H(r)=-Ln+l2l+1(r)

LqP(x)=

dP

dxP Lq(x)

Lq(x)≡exd q

dxq (xqe−x)

Exam ple of Lq's and LqP's:

L0(x)=1 ′L1(x)=−1L1(x)=1−x ′L2(x)=−4 + 2x

L2(x)=2 −4x + x2 ′L3(x)=2

Sum m ary:

R(r)=Ce−r2 rl -Ln+l

2l+1(r)( )

Norm alize:

1= |Ψ |2∫ dV 0L= |Ψlm |2 dW |R |2 r2 dr∫∫

autom atically=1 because the Ψ's are norm alized

Normalization not determined yet

38

| R(r)|2 r2 dr∫ recall r=2kr, so r2dr=

1(2k)3

r2dr

C2

(2k)3Ln+l2l+1(r)⎡⎣ ⎤⎦∫

2e−rr2lr2dr

2n (n + l)![ ]

3

(n −l−1)!

So C = (2k)3(n −l−1)!2n (n + l)![ ]

3

⎧⎨⎪⎩⎪

⎫⎬⎪⎭⎪

12

Plug in:

R=-(2k)3(n−l−1)!2n (n + l)![ ]

3

⎧⎨⎪⎩⎪

⎫⎬⎪⎭⎪

12

rle−r2 Ln+l

2l+1(r) where r=2kr=8m |E |h2 r

*It is com m on to derive a0 ≡−h2

me2, the Bohr radius.

Then r=8m |E |h2 r can be sim plified:

Recall E=−Z2e4mh22n2 , so |E|=

Z2e4mh22n2

r=8mh2

Z2e4mh22n2 r

39

I. Facts about Ψe in hydrogen

II. Probability current for an e- in hydrogen

Read 2 handoutsRead Chapter 14

40

r =2μ Ze2r

h2n=

2Zra0n

Specific Rnl (r)'s are listed in Goswami Eq. 13.23

In general, Rnl (r) =2Zna0

⎛⎝⎜

⎞⎠⎟

3(n − l −1)!

2n (n + l )![ ]3

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

12

2Zra0n

⎛⎝⎜

⎞⎠⎟

l

e−Zra0 n Ln+l

2l +1 2Zra0n

⎛⎝⎜

⎞⎠⎟

Facts about the Ψ e in hydrogen :

(i) They are totally orthogonal:

Ψ ′n ′l ′m Ψ nl m dVolume = δ ′n n δ ′l l δ ′m m∫

due to the eimϕ in the Ylm

due to the orthogonality of Legendre Polynomials Pl

recall Ylm =(Normalization constant) ⋅eimϕ ⋅Pl

m

The Laguerre Polynomials are orthogonal in n(ii) Recall what a complete set of ′l functions is: it can act as a basis for its space

i.e., any possible wavefunction in that space can be written as a linear combination of the elements of the basis.

41

The comlete set of eigen functions of the hydrogen atom look like:

So the bound states do not form a complete set by themselves

(iii) Notice all the Rnl : r l

So for l >0, Rnl (r =0)=0

No probability of finding the e− at the origin in these states for l=0, R nl(r=0)=constant

So the ground state e− has a spherical probability distribution which includes the

origin. So the ground state e− has finite probability to be found inside the nucleus.

(iv) To calculate the probability of finding the e− at a specific r, calculate

Probability(r)=r2 |R nl |2

This is sim ilar to P(x)=|Ψ(x)|2 . The r2 adjusts to spherical coordinates.

E>0 (scattering) states(we will study these in Chapter 23)

E<0 (bound) states

V =−Zer2

42

When you calculate r2 | R |2 you find

(v) to calculate the probability of finding the e− at a particular q,

calculate Probability(q)=|Ψlm (q,f )|2 sinqdqdf

since the f appears in eimf it will disappear from the probability

etc.

2p

2s3p

3s

1s(n =1,l=0)Th

e n=

1 sh

ell

The

n=2

shel

l

The

n=3

shel

l

43

You find:

Prob(l =0,m=0) = s-orbital

Prob(l =1) = p-orbital

Prob(l =2) = d-orbital

etc.

(vi) Recall Parity Rs the operation that inverts the Ψ through the origin

So RΨ(r,q,f )=R(r)(-1)lΨlm (p −q,p +f )

unchanged by R

So R(Ψe in hydrogen )=(−1)lΨ

44

Review SyllabusRead Goswami section 13.3 and chapter 14 plus the preceding chapter as necessary for reference

45

Read Chapter 14

I. Probability current for an e- in hydrogenII. The effect of an EM field on the eigen functions and eigen values of a charged particleIII. The Hamiltonian for the combined system of a charged particle in a general EM field

46

I. Probability current for an e− in Hydrogen

Recall the definition of probability current:vJ=

h2mi

Ψ*v∇Ψ−Ψ

v∇Ψ*( )

Recall this describes the spatial flow of probability→ NOT necessarily the m otion of the point of m axim um probability→ definitely not the m otion of a particle whose probability of location is related to Ψ

Calculate J for the e− in Hydrogen:

Plug in Ψnlm =Rnl(r)⋅Ψlm (q,f )⋅e

−iEnth

call this Q(q)⋅eimf , where Q(q) is pure real

and v∇spherical

coordinates=r

∂∂r

+q1r∂∂q

+ f1

rsinq∂∂f

you get:

J=h2mi

R*q*e−imf eiEt

h( )∇ Rqeimfe−iEt

h( )− Rqeimf e−iEt

h( )∇ R*q*e−imf eiEt

h( )⎡⎣⎢

⎤⎦⎥

Notice since R * =R

and q* = q,

that R*q*∇Rq −Rq∇R*q* =0so consider only the f part of the equation

47

So vJ=f

h2mi

R*q*e−imf eiEt

h( )1

rsinq∂∂f

Rqeimf e−iEt

h( )− Rqeimfe−iEt

h( )1

rsinq∂∂f

R*q*e−imf eiEt

h( )⎡⎣⎢

⎤⎦⎥

=fh2mi

R 2q 2 e−imf1

rsinq∂∂f

eimf −eimf1

rsinq∂∂f

e−imf⎛⎝⎜

⎞⎠⎟

=fhR 2q 2

2mie−imf

1rsinq

imeimf −eimf1

rsinq(−im )e−imf⎛

⎝⎜⎞⎠⎟

=fhR 2q 2mmrsinq

=f|Ψ |2 mhmrsinq

Note this current is(i) tim e-independent(ii) circulating around the z-axis (not f ) but rem aining sym m etric about it

(iii) NOT the sam e as an orbiting e−

This circulating vJ is related to the m agnetic dipole m om ent of the e−

To show this recall from EM:

48

If you have a physical charged current density vJe =

vIA

Consider a differentially sm all piece of it which is located at a distance vr from the origin.This piece form s an elem ent of a current loop which is at least m om entarily circulating relative to the origin.This is physically identical to a m agnetic dipole whose m agnetic dipole m om ent vm is given by:

vm ≡12

vr×volum ewhere Jecirculates

∫vJe(

vr)dV olume

we can convert our probability current vJ into a physical charged current

vJe by m ultiplying by the charge:

vJe =

−ec

⎛⎝⎜

⎞⎠⎟vJ

Then the m agnetic dipole m om ent of the e− is given by

vm ≡12

vr×volum e∫

−ec

⎛⎝⎜

⎞⎠⎟vJdV olume

Notice since vJ=|J|f , vm m ust be |vm |z

So we only need vr×vJ( )z =rJsinq

A I →

vr

O

vm

vrR × J

J

q

49

So we want

m=m z =−e2c

rJsinq dV olume∫

=−e2c

r|Ψ |2 mhmrsinq

sinq dV olume∫

=−emh2mc

|Ψ |2 dV olume∫ by norm alization

so m z =−eh2mc

⋅m

Since m h is the eigen value of Lz, m m ust be the eigen value of som e operator "−e2mc

Lz"

call this the z com ponent m z m agnetic m om ent operator M

Then M=−e2mc

L

angular momentum operator

magnetic momentum operator

quantum number “m”

This factor is called the Bohr magneton

Magnitude of the z-component of magnetic dipole moment of the e-

50

I. The effect of an EM field on the eigen functions and eigen values of a charged particleII. The Hamiltonian for the combined system of a charged particle in a general EM fieldIII. The Hamiltonian for a charged particle in a uniform, static B field

Read Chapter 15

51

The effect of and EM field on the eigen functions and eigen values of a charged particle

General plan:(i) We know the Hamiltonian of a free particle of momentum, vp:

H=p2

2m(ii) Find the H for that sam e particle with charge q in an EM field (

vE,

vB)

we will find that H :p2

2m+ [stuff] ⋅B+[stuf ′f]B2

we will study each kind separately

II. The Ham iltonian for the com bined system of a charged particle in a general EM fieldProcede to find H:

(i) Recall that classically, H ≡ pi&xi −i=1

3

∑ L

where the xi = canonical coordinatespi = canonical m om enta

⎧⎨⎪⎩⎪

(ii) Find L. Recall that equations of m otion m ust be obtainable from it via

∂L∂xi

−ddt

∂L∂&xi

⎛⎝⎜

⎞⎠⎟ =0

52

I. The Hamiltonian for a charged particle in an EM fieldII. The Hamiltonian for a charged particle in a uniform, static B fieldIII. The normal Zeeman effect

53

(iii) Plug in L to get H, then convert everything possible to operatorsCarry this out:To find L, recall that usuallyL=T-V

Here T=12mv2 (use m for m ass everywhere)

What is v? Recall the EM field has 2 kinds of potential:

scalar potential f and vector potential vA

How to com bine them ? (we can't just say "V =f +vA")

It turns out the L=12mv2 −qf +

qcvA⋅

vV

To dem onstrate this, we will show that

∂L∂xi

−ddt

∂L∂&xi

⎛⎝⎜

⎞⎠⎟ =0 (Lagrange's equation)

successfully produces the known Lorentz Force Law vF=q

vE+

qvvc

×vB

To plug into this, we need:∂L∂xi

=−q∂f∂xi

+qcvvg∂A∂xi

To find ∂L∂xi

, notice we can expand:

54

L =12m &xi

2∑ −qf +qc

Ai&xi∑

So ∂L∂&xi

=m&xi +qcAi [Note this is the definition of the canonical m om entum pi ]

Then ddt

∂L∂&xi

⎛⎝⎜

⎞⎠⎟ =m&&xi +

qcddt

Ai

∂A∂t

+∂xj

∂t∂A∂xjj

∑

∂A∂t

+ vv⋅∇vAi

Plug all this into Lagrange's Equation:

-q∂f∂xi

+ qvvc∂vA

∂xi=m&&xi +

qc

∂Ai

∂t+ vv⋅∇

vAi

⎛⎝⎜

⎞⎠⎟

Reorder:

m&&xi =-q∂f∂xi

−qc∂Ai

∂t+qvvc

∂vA

∂xi−qcvv⋅∇

vAi

This equation concerns com ponent i (i=1, 2, or 3) of a vector equation. Generalize to the full vector equation.

mva(=vF)=q −∇f −

1c∂A∂t

⎛⎝⎜

⎞⎠⎟ +

qc

v∇(vv⋅

vA)−(vv⋅∇)

vA⎡⎣ ⎤⎦

To understand this recall: vE

55

So qc

∇(vv⋅A)- (v⋅∇)A⎡⎣ ⎤⎦ u× (y× z)=y(u⋅z)−z(u⋅y)Plug in v ∇

vA

=∇(vv⋅A)- (v⋅∇)A

qc

v× (∇×A)⎡⎣ ⎤⎦

So we have vF=q

vE+

qcvv×

vB

So we know we have the right LagrangianNow find H= pi&xi −L∑= m&xi +

qcAi

⎛⎝⎜

⎞⎠⎟∑ &xi −

12m &x2

i −qf +qc

Ai&xi∑∑⎛⎝⎜

⎞⎠⎟

=m2∑ &x2

i + qf

=m&xi( )

2

2m∑i

+ qf

Now use again the definition of canonical m om enta

pi =m&xi +qcAi

m&xi( )2= pi −

qcAi

⎛⎝⎜

⎞⎠⎟∑2

∑

vp-qc

vA

⎛⎝⎜

⎞⎠⎟2

canonical momentum

56

H=

vp −qvAc

⎛⎝⎜

⎞⎠⎟2

2m+ qf

By convention, if the particle is: then we use:

an e− (negative) q=-e

e+ (positive) q=+e

II. The Ham iltonian for a charged particle in a uniform static vB field

Recall general H=12m

vp−qvAc

⎛⎝⎜

⎞⎠⎟2

+ qf

=12m

vp2 −qc

vA⋅vp+ vp⋅

vA( )+

q2

c2vA2⎛

⎝⎜⎞⎠⎟+ qf

we want to sim plify thisRecall:

[f(x),p]=ih∂f∂x

(see next page)

Generalize to 3D: [vf(vr),p]=ih

v∇⋅

vf

Plug in A, then:[A,p]= ih

v∇⋅

vA

now recall that for a static vB field,

v∇⋅

vA=0

Show this:

Recall in general B can be produced by (i) a uniform current and (ii) a ∂E∂t

.

Consider the static case, so A=f(I) only

57

Recall if a current vI flows along a segment d ′l of path C, then the vector potential

vA at distance R from C is

vA(x)=

m0

4pId ′lR

this is perm iability, not m ass

vAdue to

total C

(x)=m0

4pId ′vlRC

—∫

v∇⋅

vA=

v∇

m0

4pId ′vlRC

—∫⎡⎣⎢

⎤⎦⎥=

m0I4p

v∇d ′vlRC

—∫ Recall ∇⋅u

vV( )=

vV ⋅∇u+u

v∇⋅

vV( )

1R d ′vl

d ′vl ⋅

v∇

1R

⎛⎝⎜

⎞⎠⎟ +

1R

v∇d ′

vl

0 since ∇ operating on unprim ed coordinates only, and d ′vl is constructed of prim ed coordinates

v∇

1R

⎛⎝⎜

⎞⎠⎟ =

v∇

1

(xi′∑ −xi)

⎛

⎝⎜⎜

⎞

⎠⎟⎟ =−

v∇

1R

⎛⎝⎜

⎞⎠⎟

Rewrite:v∇⋅

vA=

−m0I4p

v′∇

1R

⎛⎝⎜

⎞⎠⎟d ′

vl

C—∫

Recall Stoke's Theorem : for any vector vv,

vvdvl

C—∫ =

v∇× vv( )dArea

surface enclosedby C

∫

vx

O ′vx

R =v′x −vx d ′l vA

C ∇=

∂∂x

etc. not ∂∂ ′x

58

I. The Hamiltonian for an e− in a uniform , static B field (continued)II. The norm al Zeem an effect

III. Response to the e− to the B2 term

IV . Sum m ary of e− response to static uniform BzV . The discovery of spin

59

v∇⋅

vA=

−m0 I4p

′∇ × ′∇1R

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢⎢

⎤⎦⎥⎥⋅dArea

S∫

but the curl of a divergence always = 0So,

v∇⋅

vA=0 for static B

Return to A, p⎡⎣ ⎤⎦=ihv∇⋅

vA

0

So vA and vp com m ute.

So (vA⋅vp+vp⋅

vA)=2

vA⋅vp

Then He in static B =12m

p2 −qc2vA⋅vp+

q2

c2A2⎡

⎣⎢

⎤⎦⎥+ qf

Now further restrict this static (no tim e dependence) B field to be also uniform:call it

vB=|B|z

constant, 1 directional, no position dependenceIn general for any B,vB=

v∇×

vA

Bxx + Byy+ Bzz =∂Az

∂y−∂Ay

∂z

⎛⎝⎜

⎞⎠⎟x + +

∂Ax

∂z−∂Az

∂x⎛⎝⎜

⎞⎠⎟y+

∂Ay

∂x−∂Ax

∂y

⎛⎝⎜

⎞⎠⎟z

When vB=|B|z, this reduces to:

0 + 0 + Bz =0+0+∂Ay

∂x−∂Ax

∂y

⎛⎝⎜

⎞⎠⎟z

60

There is more than 1 solution to this. One is:

Ax =−12B⋅y

Ay =12B⋅x

Ax =0

⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

This can be written precisely as:vA=

12

vB× vr

Plug this into vA⋅vp:

12

vB× vr( )⋅

vp

Recall the vector identity: vu⋅(vy⋅vz)=(vu⋅vy)⋅vz12

vB⋅ vr× vp( )

vL= angular m om entum

Now we have:

He in static B =12m

p2 −qc212

vB⋅

vL+

q2

c2A2⎡

⎣⎢

⎤⎦⎥+ qf

Plug in A2 =12

vB× vr( )

⎡⎣⎢

⎤⎦⎥2

61

Recall vB × vr=Brsinq and

vB⋅vr=Brcosq

SovB× vr( )

2=B2r2sin2q =B2r2(1−cos2q)=B2r2 −

vB⋅vr( )

2

So

A2 =14

B2r2 −vB⋅vr( )

2⎡⎣⎢

⎤⎦⎥

Plug this into H

He in uniform static B =p2

2m−

q2mc

vB⋅

vL+

q2

8mc2B2r2 −

vB⋅vr( )

2

( )+ qf

note qvL

2mc is the m agnetic m om ent

vM of the particle

Since we choose the coordinate system so that vB=Bz,

vB⋅

vL=BLz

and r2B2 − vr⋅vB( )

2=(x2 + y2 + z2 )B2 −(x2Bx

2 + y2By2 + z2Bz

2)=(x2 + y2 )B2

Then

He in uniform static B =p2

2m−

q2mc

BLz +q2

8mc2B2(x2 + y2 )( )+ qf

For an electron, q=-e, so

He in uniform static B =p2

2m−eBLz

2mc+e2B2

8mc2(x2 + y2 )−ef

"m" is the reduced m ass of the system *If you want the answer in m ks units, set c=1

62

I. The normal Zeeman effectII. Response to the e− to the B2 term

III. Sum m ary of e− response to static uniform BzIV . The discovery of spin

63

We will study the effect of each term separately upon the e's wavefunction and energy.

II. The Normal Zeeman Effect

Recall He in uniform static B in z =p2

2m+eLzB2mc

+e2B2(x2 + y2)

8mc2−ef

= "H0"+"H1" + "H2" -ef

com pare relative sizes of H1 and H2

H2

H1

=

e2B2(x2 + y2)8mc2

eLzB2mc

=eB(x2 + y2)

4cLz

Plug in e = 1.6x10−19 C

B = 1 tesla = 104 gauss

(x2 + y2)~(5x10−11)2 m 2 (Bohr radius)2

c = 1 (unitless) to convert Gaussian→ MKS units

Lz ~mh~h = 1x10−34 Joule-seconds

Then,H2

H1

~(1.6x10−19)(1 tesla)(5x10−11)2m 2

4(1)(1x10−34J−sec)=10−6

So H2 = H1 for B<105 −106T

m agnetic field at earth's surface is ~0.5x10−4 T superconducting m agnets ~ 10 TSo, in a norm al situation:

64

He in uniform static B in z ≈p2

2m+eLzB2mc

−ef

Suppose f=zer, so

-ef=−ze2

rThen the e is in the Coulom b Field of its own nucleus

This gives it En =−mz2e4

2h2n2

plus the extra vB field

Recall each energy level "n" is degenerate, all of its l and m levels have the sam e energyHow does this H1 affect the e's energy levels?

H1Ψnlm =eLzB2mc

Ψnlm =eB2mc

⎛⎝⎜

⎞⎠⎟LzΨnlm

m hΨnlm quantum num ber "m" NOT m ass

call this collection of constants "wL",

the Larm or frequency

65

So the presence of H1 means that the different "m" levels are no longer degenerate;

each has its own energy given by:

Em =−mz2e4

2h2n2+w Lmh

So without B with B

−mz2e4

2h2n2+w Lh m=+1

−mz2e4

2h2n2 m=0

E of all levels −mz2e4

2h2n2−wLh m=-1

with sam e n m=-2 etc.

The fact that a m agnetic field can cause the levels designated by "m" change energy causes "m" to becalled "The m agnetic quantum num ber"

II. Response of the e- to the ~B2 term

Recall H = p2

2m+eBLz

2mc+e2B2(x2 + y2)

8mc2−ef

=pz2

2m+eBLz

2mc+px2 + py

2

2m+e2B2(x2 + y2 )

8mc2−ef

ignore for now (let→ 0)this is identical to the Harm onic Oscillator:

H2DHo

=px2

2m+

py2

2m+12k1x

2 +12k2y

2 where k1 =k2 =e2B2

4mc2

E

66

From Chapter 9 (2-D systems)H2 D

HOΨ =(nx + ny+1)hwΨ

w=km

=eB

2c m

1

m=

eB2cm

=wLarm or

III. Sum m ary of e response to static uniform BzSo far we found that

H = pz2

2m+eBLz

2mc+px2 + py

2

2m+e2B2(x2 + y2)

8mc2−ef

Recall: a particular Ψ is sim ultaneously an e function of 2 operators vΨ and

vZ only if [

vΨ,Z]=0.

Notice [pz,Lz] =0 since Lz =xpy−ypx

[pz,H2DHO

] =0

[Lz,H2DHO

] =0

So all the term s of H have the sam e e function

call it "Ψnmk"

Plug it in:

Set = 0 for nowH2D, HO

67

I. The Discovery of SpinII. Filtering particles with a Stern-Gerlach apparatusIII. Experiments with filtered atoms

68

HΨnmk =pz2

2mΨnmk +

eBLz

2mcΨnmk + H2D

HOΨnmk

Recall: pzΨ =hkΨ

LzΨ =mhΨ

H2DHOΨ =(nx + ny+1)hw LΨ

HΨnmk =h2k2

2m+ mhwL + (nx + ny+1)hw L

⎡⎣⎢

⎤⎦⎥Ψnmk

But in general HΨ =EΨ, so this m ust be "E"

Enmk =h2k2

2m+ (m + nx + ny+1)hw L

III. The discovery of spin

Suppose you wanted to m easure the total angular m om entum of a particle

call it vJ as in Ch. 11 (Note: this J is not a current)

W e showed in Chapter 13 that angular m om entum ∝ m agnetic m om ent

−e

vL

2mc =

vM

Now call this "vJ" to be general, i.e. to allow for m ore than just orbital angular m om entum

69

So vJ=

−2mcvM

e

So we want to design an apparatus to m easure vM

Recall from E&M that if a m agnetic dipole vM is in a m agnetic field

vB it feels a force on it

which depends on the relative orientation of vM and

vB:

vF=

v∇(

vM⋅

vB) (stored energy e=

vM⋅

vB then

vF=

v∇e)

Expand vM=Mxx + Myy+ Mzz

Then vF=

v∇ MxBx +M yBy+MzBz

⎡⎣ ⎤⎦

Design and apparatus in which B=Bz is in the z direction only

Then vF=

v∇ MzBz⎡⎣ ⎤⎦

Since M is a fundam ental property of a particle, it has no dependence on z

To get v∇ MzBz

⎡⎣ ⎤⎦≠0, m ust have Bz = f(z)

Then F=Mz

∂B∂z

Now if a particle with m om ent Mz is in the apparatus, it will feel a force ∝ Mz.

If the particle is m oving through the apparatus, the force will deflect its path from a straight lineand its deflection will m easure Mz

deflection ∝ Mz z

F ∝ Mz

particle

apparatus

70

An apparatus like this is called a Stern-Gerlach ExperimentNotice that if you pass 1 particle through the SG, you find out its specific M z .

If you accumulate a large number of identical particles, you can find out whatare all the possible M z values that they can have.

Recall m can take only quantized values. But are there any restrictions on M z ?

∝ the apparent orientation of the object Mz =m cosq

particle's m agnetic dipole

Stern and Gerlach m ad a beam of silver atom s.For each atom g the nucleus has

vM ≈0

g all e− but the outerm ost one are paired, so their cum ulative vM ≈0

g the outerm ost e was in the s-suborbital of its shell

l=0, m=0

g the atom s were cooled so that it was unlikely the outerm ost e− could acquire

enough therm al energy to m ove a higher -l

higher m

⎧⎨⎪⎩⎪

⎫⎬⎪⎭⎪ sub-orbital

When the atom s passed through the SG, they all deflected, but each ended up in one of only 2 possible spots:

z q M z

M

71

I. The Discovery of SpinII. Filtering particles with a Stern-Gerlach apparatusIII. Experiments with filtered atoms

Read handout from Feynman lectures and Chapter 15

72

spot 1

spot 2

smear that didn't appear

⎛⎝⎜

⎞⎠⎟

What this meant:

1. the vM cannot have arbitrary orientation: otherwise instead of 2 spots there would have been a continuous

smear reflecting that all possible M z states were present

2. each outer e− had a non-zero Mz which was not related to its orbital angular m om entum

that had been arranged to have l=m=0 call this "non-orbital angular m om entum " = spin. Its quantum num ber is s and its "orbital angular m om entum " m = m s

3. Recall for orbital angular m om entum , the possible values m can take are -l, l+1, ..., 0, ..., l−1, l so the num ber of possible values of m is (2l+1)4. here, experim entally it was found that the num ber of possible m s values is 2

so (2s+1) = 2

s = 12

*Conclusion: Every particle has, in addition to orbital angular m om entum , another property which is m athem atically like angular m om entum but which is not due to any kind of rotation. This new kind of angular m om entum is called spin.

73

Recall that regular angular momentum is quantized in the direction its allowed to have, so that

Lz = (integer m )*h

the possible num ber of orientations is 2l+1

m sz is also quantized, but since s for an e− is always 1

2,

the num ber of possible m s orientations is always only 2:

+ 12 and - 1

2

I. Filtering particles with a Stern-Gerlach apparatus

Recall if you have a collection of atom s with different m z values,

if you pass them through a Stern-Gerlach device, they will separate into different beam s, one beam for each value.

Notice if you obstruct all but 1 output path, you can produce a beam that ispurely com posed of particles with 1 definite m z value.

z 2h

1h

0h

−1h etc.

quantum # m

if their l =0 this isdue to their spin z

mz =+1

mz =0

mz =−1

etc.

particles with allpossible m z

74

pure mz =+1

Example:

pure mz =+1

When a beam is put into a definite state like this, it (the output of the SG) is called a "prepared beam"or a "filtered beam" or a "polarized beam"

Now make a slightly modified SG that can return the polarized beam to the original axis of travel

modified SG device

this barrier is moveable so wecould select any m z value

75

Make up a symbol for the modified SG device:

+1 0 −1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ this shows what is blocked

S this gives a particular device a nam e in case m ore than 1 is in series

Make up sym bols for the prepared states:

what com e out of S|+ >|0 >|- >Now im agine placing several SG's in series.Exam ple:

beam → +1 0 −1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 −1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

S ′S

If this S is the initial state we are forcing the particles to have (labelled |+ >, etc.) then this ′Sis the final state we are checking to see IF they have label final states with bras.This one is < -1| Other possibilities for this system are < 0 | or < +1|

*Note S can represent both the device thatprepares a particle's state, or the state itself

76

I. Filtering particles with a SG (continued)II. Experiments with filtered atomsIII. SG in seriesIV. Basis states and interference

77

Examples of some diferent possible results of putting 2 SG's in series:

Configuration: Result exiting ′S A sym bolic way to represent this:

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ all the pure m z =+1 exit final

stateinitialstate

=fraction S that pass S

S ′S +1 +1 =1

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ nothing exits −1 +1 =1

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ all the pure m z =−1 exit −1−1 =1

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ nothing exits 0 −1 =1

*Notice we draw the S, ′S in the order in which the beam reaches them , but we order final initial from right to left.

W e could sum m arize all possibilities in a m atrix as we have done before:

initial state:

final state +1 0 −1

+1 1 0 0

0 0 1 0

−1 0 0 1

beam →

beam →

beam →

beam →

78

All these examples have

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ S ′S prepares 1 "analyzes" state state

Suppose S could prepare several states with definite fractions, so

initial =a + + b 0 + c −

Then the am plitude for having a particular final state exit would befinalinitial =a final+ + b final0 + c final−

= a if final = +1

= b if final = 0

= c if final = −1

Then the probability of observing a particular final state is

finalinitial2= a

2, b

2, or c

2

We always assum e that finalinitial2

final∑ =1

a2+ b

2+ c

2=1

(This is norm alization.)

particles with unknown states

79

II. Experiments with filtered atoms*The purpose of these examples is to show you how different basis sets could actually be realized in nature.*Suppose we put 2 SG filters in series, but one is tilted with respect to the other:

+S ≠ +T , so +T +S ≠1

However, since +S is not orthogonal to +T , it is also not true that +T +S =0

It turns out that +T +S = som e am plitude "a" where 0 ≤a≤1 and a=f(a)

There are also specific am plitudes for all of the following possibilities:

+T +S +T 0S +T −S

0T +S 0T 0S 0T −S

−T +S −T 0S −T −S

Note: for norm alization, the square of the 1 in each colum n m ust sum to 1.

*Keep in m ind that the m atrix of possibilities does not have to be 3x3. It is in general nxn, where n is the num ber of states the beam particles can have.*

beam →

S call this "T"

a

x

y

z

80

III. SG filters in series

The message of this section is,once a particle goes through a filter it loses all information about the orientation of previous filters it passes through.That is not the same as saying, "each filter analyzes, or measures the state of the particle, and the measurementprocess places the particle in and eigenstate of that measurement"

aligned along one of its basis states of that SG filter

To see this, consider 3 consecutive SG filters:

Suppose the not only have relative angles, but also have their blocking pads in different places:

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

Notice S and ′S represent the sam e basis (which has 3 states), and T is a different one (which also has 3 states).

beam →

S T

a

′S

81

I. SG filters in series (continued)II. Basis states and interferenceIII. Describing a measurement matrix

82

You might guess that a particle got to here, S T S

it would get to here

with 100% probability, because it would "remember" that is had been +S earlier. It does NOT.

The T filter places it into a 0T state, which does not have 100% overlap with a +S .

Dem onstrate that the fraction of particles that pass through T and ′S depends only on T and ′S (not S)

Com pare:

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ S T ′S S T ′S

Am plitude to exit ′S is:

+ ′S 0T 0T 0S 0 ′S 0T 0T 0S

Ratio of am plitudes is:

LHSRHS

=+ ′S 0T 0T 0S

0 ′S 0T 0T 0S

independent of state of S.

83

I. Basis states and interferenceII. Describing a measurement with a matrixIII. Sequential measurements

Read Goswami 11.2

84

So the presence of S affects the absolute number of particles that get to T (and then have the chance to reach ′S ),but once they are at T, having passed through S does not affect their chance to pass through ′S .

IV . Basis states and interference

Consider several experim ents:

Experim ent 1: +1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

N particles survive⏐ →⏐ ⏐ ⏐

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ aN⏐ →⏐

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ b∝N⏐ →⏐ ⏐

S T ′S

Experim ent 2: +1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ N ⏐ →⏐ ⏐ ⏐

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ aN⏐ →⏐

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ g∝N⏐ →⏐ ⏐

S T ′S

Experim ent 3:

repeat Exp. 1 but

rem ove blocks from T

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ N ⏐ →⏐ ⏐ ⏐

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ N⏐ →⏐

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ N⏐ →⏐

S T ′S

Experim ent 4:

repeat Exp. 2 but

rem ove blocks from T

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ N ⏐ →⏐ ⏐ ⏐

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ N⏐ →⏐

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ 0⏐ →⏐

S T ′S

b = + ′S 0T

2

g = 0 ′S 0T

2

a = 0T +S

2

85

Conclusions:g Experiment 2 produces more final state particles than Experiment 4:Inserting blocks in T must eliminate destructive interference (or produce constructive interference) in this case.g Experiment 1 produces less than Experiment 3:Inserting blocks in Tmust produce destructive interference in this case.

This interference of amplitudes is similar to what happens in a double slit experiment with light:

light⏐ →⏐ ⏐ light⏐ →⏐ ⏐

2 slits 1 slit(like "no blocks" in T) (loss of one slit like adding a block in T)gives m inim um gives m axim um

output at 20o output at 20o

20o 20o

86

Write the amplitudes: Condense the rotation:Experiment 4: 0 ′S +T +T +S 0 ′S +T +T +S

all T∑ =0

+ 0 ′S 0T 0T +S

+ 0 ′S −T −T +S

0

Experim ent 3: + ′S +T +T +S + ′S +T +T +Sall T∑ =1

+ + ′S 0T 0T +S

+ + ′S −T −T +S

1

Facts about all of this:1. Experim ent 3 would have the sam e result ifT is present but all open or T is not present at all

0 ′S T T +ST∑ = 0 ′S +S

⇒ So T TT∑ =1 If T includes all possible im term ediate states

it is a basis, a com plete set

87

2. Experiment 3 would have the same result if T were replaced by some other filter "R" tipped atan angle other than a, as long as R were also unblocked.Now

⇒ So a choice of basis is not unique.

3. All of this only works if the states within a basis are orthogonal, for exam ple Ti Tj =d ij

To see this, go back to

0 ′S T T +ST∑ = 0S +S

Note that ′S and S are really the sam e basis, so delete the prim e

0S T T +ST∑ = 0S +S

Renam e +S = j generic

0S = c generic

T T = i i any basis

beam →

S T

a

′S

beam →

S R

d

′S

88

c i i ji∑ = c j

Now since j is generic, it could be a m em ber of the basis set, j

c i i ji∑ = c j

This can only be true if i j =d ij

4. Revising the order of a process (i.e. exchanging the initial and final states) is the sam e as takingthe com plex conjugate of its am plitude.Show this:2 colum ns:Colum n 1: If a particle starts in som e state, it m ust end up in one of the possible final states (i.e. it cannot get lost). So, for exam ple:

+T +S2+ 0T +S

2+ −T +S

2=1

Expand:

+T +S*+T +S + 0T +S

*0T +S + −T +S

*−T +S =1 "Equation A"

Colum n 2:

Now also recall that if a state (say +S ) is norm alized, for exam ple:

+S +S =1

we can insert T TT∑ =1

89

I. Describing a measurement with a matrixII. Sequential measurementsIII. Relating matrix notation and Dirac notationIV. Spinors

90

Example:

Suppose:+S = an electron

A = and interaction or m easurem ent

+R = and up quark

Finding +R A+S tells us the probability that this interaction

converts e → w, which is fundam ental inform ation about the interaction.

91

+S T T +ST∑ =1

+S +T +T +S + +S 0T 0T +S + +S −T −T +S =1 "Equation B"

Com pare Equation A and Equation BBoth have RHS=1, so their LHS's m ust be equal.This can only be true if

Si Tj = Tj Si

*

V . Describing a m easurem ent by a m atrix

Com m on question in physics:

g A system begins in som e initial state, say +S (The full set of possible states is "S")

g Som ething happens to it (a m easurem ent--call it "A" or and interaction force)

g What is the probability that it will end up in any particular final state, say +R (the full set of possible final states is "R")

So we want +R A+S . (see previous page)

How to calculate this?

Here +S and +R are bras and kets in Hilbert space, it is hard to calculate with them without choosing a basis.

But what basis is best for them ? What if calculating with each are easier if they are in different basis?

(This could happen, for exam ple, if +S is a state with rectangular sym m etry and +R is a state with spherical sym m etry.)

92

How to handle this:

We have:

+R A+S = +1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

A

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ S R

Inserting an unblocked T

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ anywhere has no effect:

So

+R A+S = +1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

A

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

+1 0 -1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ S T T R

Make a table:The am plitude for going from +S→ T T→ A→ T T→ +Ris given by

Ti +S Tj ATii∑ +R Tj

j∑

93

I. Describing a measurement matrix (continued)II. Sequential measurementsIII. Relating matrix notation to Dirac notation

94

Reordering:

+R A+S = +R Tj Tj ATi Ti +Sij∑

Renam e +S = j generic state

+R = c generic state

Ti , Tj → i , j m em bers of any basis

Then

c Aj = c j jA i i ji, j∑

What this m eans:

Suppose that j and c can be written in term s of bases with 3 basis states.

Then i=(1,2,3) and j=(1,2,3)

So there are only 9 possible am plitudes jA i

For exam ple: i→ + 0 -j↓

+ + A+ + A0 + A−

0 0 A+ 0 A0 0 A−

- − A+ − A0 − A−

(Notice order the colum ns and rows in descending order of the eigenvalues of the quantum num ber involved.)

95

And there are only 3 amplitudes i j

And there are only 3 am plitudes c j

So a total of 9+3+3=15 pieces of inform ation are required.

Once they are plugged into the sum of the RHS, you get the LHS, which is a very general peice of

inform ation: "How does the m easurem ent A relate the states j and c ?"

operator

V I. Sequential m easurem entsSuppose "m easurem ent A" really involves "first m easure B, then C"

Exam ple: to find out the m ass of a fundam ental particles you could m easure first its vp, then its vv, then

calculate m=pv

Get p from tracking the curvature of its path in a vB field:

curvature k ∝Bp

Then get vv by putting it through a "speed trap": m easure its tim es t1 and t2 crossing 2 points

separated by length l, then com pute v=l

t2 −t1

So procedurally the m easurem ent would be:

{ } → A{ } → { } = { } → B{ } → C{ } → { }

j c j c

96

{ } A{ } { } = { } B{ } { } C{ } { }

j c j T c

Sym bolically:

c Aj = c C Tj Tj B jTj

∑ notice generator order is right-to-left

since Tj unblocked

each of these is a m atrix in which the ket=initial state labels colum ns bra= final state, labels rows

The sum over the Tj Tj represents the norm al procedure for m atrix m ultiplication.

Show this:In norm al m atrix m ultiplication the (c,j)th elem ent of m atrix A is the sum of the elem ent-by-elem entproduct of the m atrix elem ents in the c-th row of B and the j-th row of C:

Exam ple: for c=2, j=3

A =B⋅C

97

(1,1) (1,2) (1,3) ... (1,5)(2,1) (2,2) (2,3) ... (2,5)

MM

(5,1) (5,2) (5,3) ... (5,5)

⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

=(2,1) (2,2) (2,3) (2,4) (2,5)

⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

g

(1,3)

(2,3)

(3,3)

(3,4)

(3,5)

⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

m atrix A m atrix B m atrix CA23 =B21C13 + B22C23 + B23C33 + B24C43 + B25C53 = B2Tj

CTj3Tj

∑

I. Relating m atrix notation and Dirac notationRecall we have

initial state = i

finale state = f

operator = AThen

f A i m eans:

g som ething ("A") is close to state i

g that event changes the state to som ething else (call is state Ai )

g we want to know how Ai com pares to the state f

g the overlap of them is given by f Ai

c

f A i

Another way to say this is "what is the probablility that Ai is identical

to f ?"

98

Now recall that since the elements a of a basis are com plete,

a aa∑ =1

This is true for any basis so also true for the basis b

b bb∑ =1

This allows is to write

f A i = f b b Aa a ib∑

a∑

Exam ple: Suppose that i and f can each take on 3 values. (Ex: +1, 0, -1)

Then

g f A i is a 3x3 m atrix

g if the a 's span the space in which i exists there need be only 3 values of a

g if the b 's span the space in which f exists there need be only 3 values of b

Recall that these are matrices

Recall that this symbol means "Hilbert space state i

projected into the a basis"

Recall f b = b f*

so this is the com plex conjugate

of Hilbert space state f

projected into the b basis.

99

I. SpinorsII. The matrices and eigenspinors of Sx and Sy

100

So the equation looks likef A i = f b

a ,b∑ b Aa a i

Norm al rules of m atrix m ultiplication state that the vector to the right of the m atrixm ust be a column vectorand the vector to the left of the m atrix m ust be a row vector

So translating from Dirac notation to m atrix notation gives:

⎛

⎝⎜⎜

⎞

⎠⎟⎟ = ( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

3x3

3-component vector

3x3 3-component vector

f A i

f b

b Aa

a i

a i

f b