Problem 1:

A beam of electrons in an eigenstate of Sz with eigenvalue ½ħ is fed into a Stern-Gerlach apparatus, which measures the component of spin along an axis at an angle θ to the z-axis and separates the particles into distinct beams according to the value of this component.  Find the ratio of the intensities of the emerging beams.

Solution:

• Concepts:
The two dimensional state space of a spin ½ particle, the postulates of quantum mechanics
• Reasoning:
The electron is a spin ½ particle. The state space corresponding to the observable Sz of a spin ½ particle is two-dimensional.  It can be spanned by the eigenstates of Sz by {|+>, |->} or the eigenstates of Sn by {|+>n, |->n}.
If n = n(θ,φ),  then the operator Sn defined through
S
n = Sxsinθcosφ + Sysinθsinφ + Szcosθ.
• Details of the calculation:
The eigenvectors of Sn are
|+>n = cos(θ/2)exp(-iφ/2)|+> + sin(θ/2)exp(iφ/2)|->,
|->n = -sin(θ/2)exp(-iφ/2)|+> + cos(θ/2)exp(iφ/2)|->,
or
|+> = cos(θ/2)|+>n - sin(θ/2)|->n
|-> = sin(θ/2)|+>n + cos(θ/2)|->n.
At t = 0 the system is in the state |+>.  The Hamiltonian is proportional to Sn.  We want to find the probabilities of finding the system in the states |+>n and |->n.
P(|+>n) = |n<+|+>|2 = cos2(θ/2).
P(|->n) = |n<-|+>|2 = sin2(θ/2).
The ratio of the intensities of the emerging beams is
P(|+>n)/P(|->n)  = cot2(θ/2)

Problem 2:

A quantum system can exist in two states |ψ1> and |ψ2>, which are eigenstates of the Hamiltonian with eigenvalues E1 and E2
An observable A has eigenvalues ±1 and eigenstates |ψ±>= (1/√2)(|ψ1> ± |ψ2>).
This observable is measured at times t = 0, T, 2T, ... .  The normalized state of the system at t = 0, just before the first measurement, is c11>+  c22>.  If pn denotes the probability that the measurement at t = nT gives the result A = 1, show that pn+1 = ½(1 - cosα) + pn cosα, where
α/T = (E1 - E2)/ħ
and deduce that  pn = ½ (1 - cosnα) + ½|c1 + c2|2 cosnα.
What happens in the limit as n --> ∞ with nT = t fixed?

Solution:

• Concepts:
Fundamental assumptions of QM
• Reasoning:
When a physical quantity described by the operator A is measured on a system in a normalized state |Ψ>, the probability of measuring the eigenvalue an is given by
P(an) = ∑i=0gn|<uni|Ψ>|2,  where {|uni>} (i=1,2,...,gn) is an orthonormal basis in the eigensubspace Εn associated with the eigenvalue an.
• Details of the calculation:
(a)  The probability of measuring A = 1 at t = 0 is
P0 = ½|(<ψ1| + <ψ2|)(c11> + c22>)|2 =  ½|c1 + c2|2.
(b)  Before the first measurement at t = 0, |ψ(0)> = c11> + c22>.   After each measurement the state is either |ψ+> or |ψ->.  The probability that after the nth measurement the state is |ψ+> is Pn, the probability that the state is |ψ-> is 1 – Pn.
(i)  Assume that after the nth measurement the system is in state |ψ+>.
After a time interval T we have  |ψ> = 2(exp(-E1T/ħ)|ψ1> + exp(-E2T/ħ)|ψ2>).
The probability of measuring A = 1 is |<ψ+|ψ>|2 = (1/4)|exp(-E1T/ħ) + exp(-E2T/ħ)|2.
(ii)  Assume that after the nth measurement the system is in state |ψ->.
After a time interval T we have  |ψ> = 2(exp(-E1T/ħ)|ψ1> - exp(-E2T/ħ)|ψ2>).
The probability of measuring A = 1 is |<ψ+|ψ>|2 = (1/4)|exp(-E1T/ħ) - exp(-E2T/ħ)|2.
The total probability of measuring A = 1 is
Pn+1 = Pn (1/4)|exp(-E1T/ħ) + exp(-E2T/ħ)|2 + (1 – Pn) (1/4)|exp(-E1T/ħ) - exp(-E2T/ħ)|2
= Pn ½(1 + cos((E1 – E2)T/ħ)) + (1 - Pn) ½(1 - cos((E1 – E2)T/ħ))
= Pn ½(1 + cos(a)) + (1 - Pn) ½(1 - cos(a)) = ½(1 - cosa) + Pn cosa,
(c)  P1 = ½(1 - cosa) + P0 cosa.
P2 = ½(1-cosa) + P1 cosa. = ½(1 - cosa) + ½(1 - cosa)cosa + P0 cos2a
= ½(1 - cosa2) + P0 cos2a.
Assume Pn = ½(1 - cosna) + P0cosna.  This holds for P1 and P2.
Then Pn+1 = ½(1 - cosa) + Pn cosa = ½(1 - cosa) + (½(1 - cosna) + P0cosna)cosa
= ½(1 – cos(n+1)a) + P0cos(n+1)a.
If  Pn = ½(1 - cosna) + P0cosna  holds for n, it also holds for n + 1.
(d)  As as n --> ∞,  cosna  --> 1, Pn --> P0.  (Quantum Zeno Effect)
Continuously measuring suppresses the evolution of the system.

Problem 3:

Consider a spin ½ particle with magnetic moment m = γS.  Let |+> and |-> denote the eigenvectors of Sz and let the state of the system at t = 0 be |ψ(0)> = |+>.
(a)  At t = 0 we measure Sy and find +½ħ.  What is the state vector |ψ(0)> immediately after the measurement?
(b)  Immediately after this measurement we apply a uniform, time-dependent field parallel to the z-axis.
The Hamiltonian operator becomes H(t) = ω0(t)Sz.
Assume ω0(t) = 0 for t < 0 and for t > T, and increases linearly from 0 to ω0 when 0 < t < T.  Show that at time t the state vector can be written as
|ψ(t)> = 2[exp(iθ(t))|+> + iexp(-iθ(t))|->]
and calculate the real function θ(t).
(c)  At time t = τ > T, we measure Sy.  What results can we find and with what probability?
Determine the relation that must exist between ω0 and T in order for us to be sure of the result.  Give a physical interpretation.

Solution:

• Concepts:
The two dimensional state space of a spin ½ particle, the evolution operator, the postulates of Quantum Mechanics
• Reasoning:
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 Hamiltonian of the system is H = ω0(t)Sz.  We verify that the given |ψ(t)> is a solution to the Schroedinger equation and then answer the questions using the postulates of Quantum Mechanics.
• Details of the calculation:
(a)  After the measurement |ψ(0)> = |+>y = (1/√2)(|+> + i|->).
(b)  Assume |ψ(t)> = 2[exp(iθ(t))|+> + iexp(-iθ(t))|->].
iħ∂|ψ(t)>/∂t = iħ2[i (∂θ/∂t) exp(iθ(t))|+> + (∂θ/∂t) exp(-iθ(t))|->]
= -ħ2[(∂θ/∂t) exp(iθ(t))|+> - i(∂θ/∂t) exp(-iθ(t))|->].
H|ψ(t)> = ω0(t)Sz|ψ(t)> = 2ω0(t)(ħ/2)[exp(iθ(t))|+> - iexp(-iθ(t))|->].
For ψ(t)> to be a solution to the Schroedinger equation we need ω0(t)/2 = -∂θ/∂t,
-
dθ = ½ω0(t)dt,  θ(t') - θ(0) = -½0t'ω0(t)dt.
θ(0) = θ(t < 0) = 0.   Let α = ω0/T, then ω0(t) = αT.
θ(t') = -αt'2/4  for (0 < t' < T),  θ(t') = -αT2/4  for ( t' > T).
(c)  |ψ(τ)> = 2[exp(iθ(τ))|+> + iexp(-iθ(τ))|->].
Possible result of measuring Sy are ±ħ/2.
P(+ħ/2) = |y<+|ψ(τ)>|2 = ¼|exp(iθ(τ)) + exp(-iθ(τ)|2 = cos2(θ(τ)).
P(-ħ/2) = |y<-|ψ(τ)>|2 = sin2(θ(τ)).
θ(τ) = -αT2/4.  We need  αT2/4 = nπ/2  or ω0 = n2π/T  to be sure of the result.
The physical interpretation is a precession, similar to Larmor precession.