First Term Test

PH3101-Quantum Mechanics II

Sep 08, 2017 Time Allowed: 50 min

Answer any two of the questions

Q1. Orbital angular momentum of an electron

An electron is prepared in the orbital angular momentum state

ψ(θ, φ) =

√38

Y1,1(θ, φ) +

√18

Y1,0(θ, φ) −

√12

Y1,−1(θ, φ)

Calculate the expectation values of measurement of the three components of the angular

momentum, Lx, Ly, Lz and the total angular momentum, L2. (10 pt)

Q2. Particle moving in a circle A quantum mechanical particle is moving in a circle

in the xy plane with constant angular momentum, 2~.

1. Calculate the wave function describing the motion of the particle. (4 pt)

2. Write down the time independent Schrodinger equation for the particle. (2 pt)

3. Show that the wave function derived in part 1 is a solution of the time independent

Schrodinger equation. What is the associated eigen energy? (4 pt)

Q3. Spin in a magnetic field

A Cu-based quantum magnet is placed in a strong magnetic field applied along the x-axis.

Cu2+ ions carry a spin S=1/2. The strong field forces all the spins to align along the x-axis

such that a measurement of S x (x-component of the spin angular momentum) gives ~2 with

probability 1. Prove that the expectation value of S z in this state vanishes. What is the

expectation value of S 2z in this state? (10 pt)

- End of Paper -

PH3101

Solution

Q1. Orbital angular momentum

|ψ〉 = ψ(θ, φ) =

√38

Y1,1(θ, φ)+

√18

Y1,0(θ, φ)−

√12

Y1,−1(θ, φ) = α |1, 1〉+β |1, 0〉+γ |1,−1〉

Observe that the state is normalised since 〈ψ|ψ〉 =38

+18

+12

= 1. Tabulating the eigen-

values of the various angular momentum operators, in units of ~ or ~2

Yl,m Y1,1 Y1,0 Y1,−1

Probability 38

18

12

m 1 0 −1

Lz = m~ 1 0 −1

L2 = l(l + 1)~2 2 2 2

L+ = ~√

l(l + 1) − m(m + 1) 0√

2√

2

L− = ~√

l(l + 1) − m(m − 1)√

2√

2 0

〈Lz〉 =

(1 ×

38

+ 0 ×18

+ (−1) ×12

)~ = −

~

8

〈L+〉 = 〈ψ|L+|ψ〉

=[α∗ 〈1, 1| + β∗ 〈1, 0| + γ∗ 〈1,−1|

]L+

[α |1, 1〉 + β |1, 0〉 + γ |1,−1〉

]=

[α∗ 〈1, 1| + β∗ 〈1, 0| + γ∗ 〈1,−1|

][0 × α |1, 2〉 + 2β |1, 1〉 + 2γ |1, 0〉

]~

= 2[α∗ 〈1, 1| + β∗ 〈1, 0| + γ∗ 〈1,−1|

][β |1, 1〉 + γ |1, 0〉

]~

= 2[α∗β + β∗γ

]~ = 2β

[α + γ

]~ = 2

√18

√38

+

√12

~ =

√3 + 24

~Notice that 〈L+〉 ,

(0 ×

38

+ 2 ×18

+ 2 ×12

)~ =

54~. This is because L+ changes the

state by increasing the magnetic number m by one.

〈L−〉 = 〈ψ|L−|ψ〉

=[α∗ 〈1, 1| + β∗ 〈1, 0| + γ∗ 〈1,−1|

]L−

[α |1, 1〉 + β |1, 0〉 + γ |1,−1〉

]=

[α∗ 〈1, 1| + β∗ 〈1, 0| + γ∗ 〈1,−1|

][2α |1, 0〉 + 2β |1,−1〉 + 0 |1,−2〉

]~

= 2[α∗ 〈1, 1| + β∗ 〈1, 0| + γ∗ 〈1,−1|

][α |1, 0〉 + β |1,−1〉

]~

= 2[β∗α + γ∗β

]~ = 2β

[α + γ

]~ = 2

√18

√38

+

√12

~ =

√3 + 24

~2

PH3101

Notice that 〈L−〉 ,(2 ×

38

+ 2 ×18

+ 0 ×12

)~ = ~. This is because L− changes the state

by decreasing the magnetic number m by one. Hence in the state |ψ〉

〈L+〉 = 〈L−〉 =

√3 + 24

~Alternatively, we can use 〈L+〉 = 〈L−〉

∗

Proof Recall that

L± = Lx ± iLy =⇒ 〈L±〉 = 〈Lx〉 ± i 〈Ly〉 =⇒ 〈L+〉 = 〈L−〉∗

as 〈Lx〉 , 〈Ly〉 ∈ R since they are Hermitian (observable) operators.�

As a result,

〈Lx〉 =

⟨L+ + L−

2

⟩=〈L+〉 + 〈L−〉

2=

√3 + 24

~〈Ly〉 =

⟨L+ − L−

2i

⟩=〈L+〉 − 〈L−〉

2i= 0

=⇒ 〈Lx〉 , 〈Ly〉

Since l = 1 for all the spherical harmonics Y1,1,Y1,0,Y1,−1 =⇒ 〈L2〉 = 1(1 + 1)~2 = 2~2

Answer

〈Lx〉 =

√3 + 24

~ 〈Ly〉 = 0 〈Lz〉 = −~

8〈L2〉 = 2~2

Q2. Particle moving in a circle

(i) Given particle moving in a circle with constant angular momentum 2~ in the xy

plane, the wave function must satisfy

Lzψ = 2~ψ =⇒ −i~∂ψ

∂φ= 2~ψ =⇒

∫∂ψ

ψ= 2i

∫∂φ =⇒ lnψ = 2iφ+C =⇒ ψ(φ) = Ae2iφ

Normalisation, assuming A ∈ R

1 =

2π∫0

ψ∗ψdφ =

2π∫0

|A|2dφ = 2π|A|2 =⇒ A =1√

2π=⇒ ψ(φ) =

e2iφ

√2π

3

PH3101

(ii) Time independent Schrodinger equation (in spherical coordinates):

Hψ = Eψ =⇒

L2z

2mR2

ψ = Eψ =⇒ −~2

2mR2

∂2ψ

∂φ2 = Eψ =⇒∂2ψ

∂φ2 +2mR2E~2 ψ = 0

where m is the mass of the particle and R is the radius of the circular motion. Recall

that L2 = L2z as the particle’s circular motion is in the xy plane with constant polar

angle θ =π

2.

(iii) Substitute the wave function in (i) into the Schrodinger equation in (ii) we have

∂2ψ

∂φ2 =∂2

∂φ2

(e2iφ

√2π

)= (2i)2

(e2iφ

√2π

)= −4ψ

Hence the wave function ψ(φ) =e2iφ

√2π

satisfy the Schrodinger equation with

2mR2E~2 = 4 =⇒ E =

2~2

mR2 = E2 (energy eigenvalue)

as we have shown in Problem Set 3 Question 9 that En =n2~2

2mR2 .

Q3. Spin in a magnetic field

Given P(S x =

~

2

)= 1, this imply that the particle is in the eigenstate of S x. Since S z do

not commute with S x =⇒ 〈S z〉 = 0. Alternatively, the normalised eigenstate of S x with

eigenvalue~

2is

|ψ〉 =1√

2

[∣∣∣∣∣12⟩

+

∣∣∣∣∣−12

⟩]

where∣∣∣∣∣±1

2

⟩are eigenkets of S z with eigenvalues ±

~

2respectively.

Proof

Let |ψ〉 = α

∣∣∣∣∣12⟩

+ β

∣∣∣∣∣−12

⟩whereα, β ∈ C. Since |ψ〉 is an eigenket of S x with eigenvalue

~

2

~

2

[α

∣∣∣∣∣−12

⟩+ β

∣∣∣∣∣12⟩]

=

[S + + S −2

] [α

∣∣∣∣∣12⟩

+ β

∣∣∣∣∣−12

⟩]= S x |ψ〉 =

~

2|ψ〉 =

~

2

[α

∣∣∣∣∣12⟩

+ β

∣∣∣∣∣−12

⟩]4

PH3101

where

S + |l,m〉 = ~√

l(l + 1) − m(m + 1) |l,m + 1〉 = ~√

34 − m(m + 1)

∣∣∣12 ,m + 1

⟩S − |l,m〉 = ~

√l(l + 1) − m(m − 1) |l,m − 1〉 = ~

√34 − m(m − 1)

∣∣∣12 ,m − 1

⟩Since

∣∣∣∣∣±12

⟩are linearly independent kets, comparing their coefficients we have α = β.

Also |ψ〉 is a normalised state =⇒ |α|2 + |β|2 = 1 =⇒ α = β =1√

2assuming α, β ∈ R.

Thus |ψ〉 =1√

2

[∣∣∣∣∣12⟩

+

∣∣∣∣∣−12

⟩]and the expectation of S z in this state is

〈S z〉 = 〈ψ|S z|ψ〉 =12

[⟨12

∣∣∣∣∣ +

⟨−

12

∣∣∣∣∣] S z

[∣∣∣∣∣12⟩

+

∣∣∣∣∣−12

⟩]=~

4

[⟨12

∣∣∣∣∣ +

⟨−

12

∣∣∣∣∣] [∣∣∣∣∣12⟩−

∣∣∣∣∣−12

⟩]= 0

Similarly,

〈S 2z 〉 = 〈ψ|S 2

z |ψ〉 =12

[⟨12

∣∣∣∣∣ +

⟨−

12

∣∣∣∣∣] S 2z

[∣∣∣∣∣12⟩

+

∣∣∣∣∣−12

⟩]=~

4

[⟨12

∣∣∣∣∣ +

⟨−

12

∣∣∣∣∣] S z

[∣∣∣∣∣12⟩−

∣∣∣∣∣−12

⟩]=~2

8

[⟨12

∣∣∣∣∣ +

⟨−

12

∣∣∣∣∣] [∣∣∣∣∣12⟩

+

∣∣∣∣∣−12

⟩]=~2

8× (1 + 1) =

~2

4

Hence the standard deviation of S z is

∆S z =

√〈S 2

z 〉 − 〈S z〉2 =

√~2

4=~

2

5