## Assignment 8, solutions

#### Problem 1:

Using the definition of angular momentum
L = r × p,
(a)  show that [Lx, Ly] = iħ Lz,   [Ly, Lz] = iħ Lx,   [Lz, Lx] = iħ Ly.
(b)  Using this result show that L2 commutes with Lz.

Solution:

• Concepts:
Commutator algebra
• Reasoning:
We use [Ri,Rj] = [Pi,Pj] = 0,  [Ri,Pj] = ihδij.
Two useful identities using commutators are
[A,BC] = B[A,C] + [A,B]C and [AB,C] = A[B,C] + [A,C]B.
• Details of the calculation:
(a) Assume the classical relationship L = r × p.
Lx = ypz - zpy,  Ly = zpx - xpz,  Lz = xpy - ypx.
For the quantum mechanical operators we then have
Lx = YPz - ZPy,  Ly = ZPx - XPz,  Lz = XPy - YPx.
[Lx,Ly] = [YPz - ZPy,ZPx - XPz] = [YPz,ZPx] + [ZPy,XPz] - [YPz,XPz] - [ZPy,ZPx]
0               0
= [YPz,Z]Px + Z[YPz,Px] + [ZPy,X]Pz + X[ZPy,Pz]
0              0
= Y[Pz,Z]Px +  X[Z,Pz]Py = -iħ(YPx - XPy) = iħ Lz.

Similarly for [Ly, Lz] = iħ Lx,   [Lz, Lx] = iħ Ly.

(b)   L2 = Lx2 + Ly2 + Lz2.  If [L2,Lz] = 0 then it is possible to find simultaneous eigenstates.
[L2,Lz] = [Lx2,Lz] +  [Ly2,Lz] + [Lz2,Lz] = Lx[Lx,Lz] + [Lx,Lz]Lx + Ly[Ly,Lz] + [Ly,Lz]Ly
= -iħ(LxLy + LyLx) + iħ(LxLy + LyLx) = 0.

#### Problem 2:

A system has a wave function ψ(x,y,z) = N*(x + y + z)*exp(-r22) with α real.  If Lz and L2 are measured, what are the probabilities of finding 0 and 2ħ2?

Solution:

• Concepts:
The spherical harmonics, the postulates of quantum mechanics
• Reasoning:
If we want to find the probability of measuring the eigenvalue a of an observable A we must express the wave function of the system as a linear combination of eigenfunctions of A.  Then  P(a) = Σi|<ai|ψ>|2.
• Details of the calculation:
Express the wave function in terms of spherical harmonics.

x + y + z = r sinθ cosφ + r sinθ sinφ + r cosθ.
cosφ + sinφ = ½(e+ e-iφ) - i½(e- e-iφ)
= ½(e(1 - i) + e-iφ(1 + i)) = 2(ee-iπ/4 + e-iφeiπ/4).
x + y + z = r sinθ 2(ee-iπ/4 + e-iφeiπ/4) + r cosθ
= r (4π/3)½(-e-iπ/4Y11(θ,φ) + eiπ/4Y1-1(θ,φ) + Y10(θ,φ)).

ψ(x,y,z) = N*rexp(-r22)(4π)½  * (1/3)½(-e-iπ/4Y11(θ,φ) + eiπ/4Y1-1(θ,φ) + Y10(θ,φ))
= f(r) * χ(θ,φ).
0πsinθ dθ∫0dφ |χ(θ,φ))|2 = 1,  ∫0r2 |f(r)|2 dr = 1, for properly chosen N.

χ(θ,φ) = Σl,mdlmYlm(θ,φ), dlm = 0 for l ≠ 1.
Let P(l,m) denote the probability of finding the eigenvalues l(l + 1)ħ2 and mħ .
P(l,m) = |dlm|2.  P(1,1) = 1/3,  P(1,0) = 1/3,  P(1,-1) = 1/3,  P(l,m)l≠1 = 0.
The probabilities of finding Lz = 0 is 1/3 and the probability of finding L2 = 2ħ2 is 1.

#### Problem 3:

Consider the Hamiltonian of a spinless particle of charge qe in the presence of a static and uniform magnetic field B = B k.
H = (1/(2m)) (p - qeA(r,t))2.
By using the gauge in which A = 0, demonstrate that the Hamiltonian can be expressed as
H = p2/(2m) - (qe/(2m)) LB + (qe2B2/(8m))(x2 + y2).
Note that the second term corresponds to the linear coupling between the external field and the magnetic moment.

Solution:

• Concepts:
• Reasoning:
p = (ħ/i) is a differential operator in coordinate space.
• Details of the calculation:

(p - qA)∙(p - qeA)ψ = (-ħ2/(2m))2ψ + qe2A2ψ - (ħqe/i)A(r,t)ψ - (ħqe/i)Aψ
= -ħ22ψ + qe2A2ψ  - (2ħqe/i)Aψ.
H =(-ħ2/(2m))2 + (qe2/(2m))A2  + (iħqe/m)A.
For B = B k we can choose A = ½B(-y, x, 0) or A = -½(r×B).
Aψ = -½(r×B)∙∇ψ = ½(r×ψ)∙B = (i/(2ħ)(r×pψ)∙B = (i/(2ħ)LBψ.
Therefore (iħqe/m)A = -(qe/(2m))LB.
(qe2/(2m))A2 = (qe2/(8m))(r2B2 - (rB)2) = (qe2B2/(8m))(x2 + y2).
H = p2/(2m) - (qe/(2m)) LB + (qe2B2/(8m))(x2 + y2).

#### Problem 4:

An electron is a rest in magnetic field B = B0 k + B1(cos(ωt) i + sin(ωt) j).  B0, B1, and ω are constants.  At t = 0 the electron is in the |+> eigenstate of Sz.  Let μ = -γS be the magnetic moment of the electron and  ω0 = γB0, ω1 = γB1 be constants.
(a)  Construct the Hamiltonian matrix for this system and write down the time-dependent Schroedinger equation,

iħ

 ∂α/∂t ∂β/∂t

= H(t)

 α β

.

in matrix form in the {|+>, |->} basis.

(b)  Convert this equation into a "Schroedinger equation" with a time independent "Hamiltonian" by choosing new expansion coefficients
a(t) = exp(iωt/2)α(t), b(t) = exp(-iωt/2)β(t).
(Hint:  Given ∂α/∂t and ∂β/∂t find ∂a/∂t and ∂b/∂t.)
(c)  Find the eigenvalues and eigenvectors if the "Hamiltonian" in part (b).
Hint:  Write

H' = A

 cosθ sinθ sinθ -cosθ

.

The eigenvalues of the matrix are λ = ±1, and the corresponding eigenvectors are
+> =  cos(θ/2)|+> + sin(θ/2)|->,  |ψ-> =  -sin(θ/2)|+> + cos(θ/2)|->.]
(d)  The Schroedinger equation now implies that
U(t,0)|ψ+> = |ψ+>exp(-iAλ+t/ħ), and U(t,0)|ψ-> = |ψ->exp(-iAλ-t/ħ).
Find the probability of finding the electron in the |-> eigenstate of Sz as a function of time.

Solution:

• Concepts:
A two-state system with a time dependent Hamiltonian.
• Reasoning:
The magnetic field component in the xy-plane is rotating with angular speed ω.  Switching expansion coefficient means switching basis vectors.  The  new basis vectors are time dependent.
• Details of the calculation:

a)  H =-μ∙B = -μzBz - μxBx - μyBy = γSzBz + γSxBx + γ1SyBy
= ω0Sz + ω1(Sx +  Sy).

H = (ħ/2)

 ω0 ω1 exp(-iωt) ω1 exp(iωt) -ω0

.

iħ

 ∂α/∂t ∂β/∂t

= (ħ/2)

 ω0 ω1 exp(-iωt) ω1 exp(iωt) -ω0

 α β

.

(b)  i∂α/∂t = (ω0/2)α + (ω1/2)exp(-iωt)β,
i∂β/∂t = -(ω0/2)β + (ω1/2)exp(iωt))α.
Assume a(t) = exp(iωt/2)α(t), b(t) = exp(-iωt/2)β(t),
Then, with Δω = ω0 - ω, we have
i∂a/∂t = -(Δω/2)a + (ω1/2)b,
i∂b/∂t = (ω1/2)a + (Δω/2)b.

iħ

 ∂a/∂t ∂b/∂t

= H'

 a b

,

with

H' = (ħ/2)

 -Δω ω1 ω1 Δω

.

(c)  The eigenvalues of H' are E'± = ±(ħ/2)(ω12 + Δω2)½ obtained from

(ħ/2)

 -Δω-λ ω1 ω1 Δω-λ

= 0.

Let

H' = A

 cosθ sinθ sinθ -cosθ

.

Then A = (ħ/2)(ω12 + Δω2)½.
The eigenvectors of H' are
+> =  cos(θ/2)|+> + sin(θ/2)|->,  |ψ-> =  -sin(θ/2)|+> + cos(θ/2)|->,
where tanθ = - ω1/Δω.
(d) P-+(t) = |<-|ψ(t)|2>
|+> = cos(θ/2)|ψ+> - sin(θ/2)|ψ->,  |-> = sin(θ/2)|ψ+> + cos(θ/2)|ψ->.
|ψ(0)> = |+>.  |ψ(t)> = cos(θ/2)exp(-iE+t/ħ)|ψ+> - sin(θ/2)exp(-iE-t/ħ)|ψ->.
<-|ψ(t)> = sin(θ/2)cos(θ/2)(exp(-iE+t/ħ)| - exp(iE+t/ħ)),
|<-|ψ(t)|2> = ¼sin2(θ)4sin2(E+t/ħ) = (ω12/(ω12 + Δω2))sin2((ω12 + Δω2)½t/2).
[1/sin2θ = 1/tan2θ + 1].

#### Problem 5:

For any two quantum-mechanical operators A and B, the uncertainty principle says that
<(ΔA)2><(ΔB)2>  ≥ ¼|<[A,B]>|2.  Consider a spin ½ particle.  Show that for the spin operators Sx and Sy the eigenstate |+> of the Sz operator is a minimum uncertainty state.

Solution:

• Concepts:
The two dimensional state space of a spin ½ particle
Mean value <A> = <ψ|A|ψ> and root mean square deviation
ΔA = (<(A - <A>2)>)½ = (<A2> - <A>2))½.
• Reasoning:
We are asked to evaluate <(ΔSx)2><(ΔSy)2> and show that it is equal to ¼|<[Sx,Sy]>|2.
• Details of the calculation:

The state space corresponding to the observable Sz of a spin ½ particle is two-dimensional.  We denote the eigenvectors of Sz by |+> and |->.  The matrices of Sz,  Sx, and Sy in the eigenbasis of Sz , {|+>, |->} are given below.

Sz = (ħ/2)

 1 0 0 -1

.

Sx = (ħ/2)

 0 1 1 0

.

Sy = (ħ/2)

 0 -i i 0

.

The matrices of  Sz2, Sx2, and Sy2 are

Sz2 = (ħ2/4)

 1 0 0 1

.

Sx2 = (ħ2/4)

 1 0 0 1

.

Sy2 = (ħ2/4)

 1 0 0 1

.

[Sx,Sy] = iħSz.  These are the commutation relations for angular momentum.

The column matrices representing |+> and |-> are

 1 0

and

 0 1

respectively.

For the eigenstate |+> we have:

<S>z = (ħ/2)

 1 0

 1 0 0 -1

 1 0

= ħ/2.

<Sx> = 0,  <Sy> = 0,  <Sz2> = ħ2/4,  <Sx2> = ħ2/4,  <Sy2> = ħ2/4.
Therefore
(ΔSx)2 = <Sx2> - <Sx>2 = ħ2/4,  (ΔSy)2 = <Sy2> - <Sy>2 = ħ2/4.
<(ΔSx)2><(ΔSy)2> = ħ4/16.
<[Sx,Sy]> = <iħSz> = iħ2/2.
<(ΔSx)2><(ΔSy)2> = ¼|<[Sx,Sy]>|2.

The equality holds, |+> is a minimum uncertainty state.