A beam of electrons in an eigenstate of S_{z} 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 S_{z}of a spin ½ particle is two-dimensional. It can be spanned by the eigenstates of S_{z}by {|+>, |->} or the eigenstates of S_{n}by {|+>_{n}, |->_{n}}.

If**n**=**n**(θ,φ), then the operator S_{n}defined through∙

S**n**= S_{x}sinθcosφ + S_{y}sinθsinφ + S_{z}cosθ. - Details of the calculation:

The eigenvectors of S_{n}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 S_{n}. We want to find the probabilities of finding the system in the states |+>_{n}and |->_{n}.

P(|+>_{n}) = |_{n}<+|+>|^{2}= cos^{2}(θ/2).

P(|->_{n}) = |_{n}<-|+>|^{2}= sin^{2}(θ/2).

The ratio of the intensities of the emerging beams is

P(|+>_{n})/P(|->_{n}) = cot^{2}(θ/2)

**Problem 2:**

A quantum system can exist in two states |ψ_{1}>
and |ψ_{2}>, which are eigenstates of the
Hamiltonian with eigenvalues E_{1} and E_{2}.

An observable A
has eigenvalues ±1 and eigenstates |ψ_{±}>= (1/√2)(|ψ_{1}>
± |ψ_{2}>).

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 c_{1}|ψ_{1}>+ c_{2}|ψ_{2}>.
If p_{n} denotes the
probability that the measurement at t = nT gives the result A = 1, show that p_{n+1}
= ½(1 - cosα) + p_{n} cosα,
where

α/T = (E_{1} - E_{2})/ħ

and deduce that p_{n} = ½ (1 - cos^{n}α)
+ ½|c_{1 }+ c_{2}|^{2} cos^{n}α.

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 a_{n}is given by

P(a_{n}) = ∑_{i=0}^{gn}|<u_{n}^{i}|Ψ>|^{2}, where {|u_{n}^{i}>} (i=1,2,...,g_{n}) is an orthonormal basis in the eigensubspace Ε_{n}associated with the eigenvalue a_{n}. - Details of the calculation:

(a) The probability of measuring A = 1 at t = 0 is

P_{0}= ½|(<ψ_{1}| + <ψ_{2}|)(c_{1}|ψ_{1}> + c_{2}|ψ_{2}>)|^{2}= ½|c_{1}+ c_{2}|^{2}.

(b) Before the first measurement at t = 0, |ψ(0)> = c_{1}|ψ_{1}> + c_{2}|ψ_{2}>. After each measurement the state is either |ψ_{+}> or |ψ_{-}>. The probability that after the nth measurement the state is |ψ_{+}> is P_{n}, the probability that the state is |ψ_{-}> is 1 – P_{n}.

(i) Assume that after the nth measurement the system is in state |ψ_{+}>.

After a time interval T we have |ψ> = 2^{-½}(exp(-E_{1}T/ħ)|ψ_{1}> + exp(-E_{2}T/ħ)|ψ_{2}>).

The probability of measuring A = 1 is |<ψ_{+}|ψ>|^{2}= (1/4)|exp(-E_{1}T/ħ) + exp(-E_{2}T/ħ)|^{2}.

(ii) Assume that after the nth measurement the system is in state |ψ_{-}>.

After a time interval T we have |ψ> = 2^{-½}(exp(-E_{1}T/ħ)|ψ_{1}> - exp(-E_{2}T/ħ)|ψ_{2}>).

The probability of measuring A = 1 is |<ψ_{+}|ψ>|^{2}= (1/4)|exp(-E_{1}T/ħ) - exp(-E_{2}T/ħ)|^{2}.

The total probability of measuring A = 1 is

P_{n+1 }= P_{n}(1/4)|exp(-E_{1}T/ħ) + exp(-E_{2}T/ħ)|^{2}+ (1 – P_{n}) (1/4)|exp(-E_{1}T/ħ) - exp(-E_{2}T/ħ)|^{2}

= P_{n}½(1 + cos((E_{1}– E_{2})T/ħ)) + (1 - P_{n}) ½(1 - cos((E_{1}– E_{2})T/ħ))

= P_{n}½(1 + cos(a)) + (1 - P_{n}) ½(1 - cos(a)) = ½(1 - cosa) + P_{n}cosa,

(c) P_{1}= ½(1 - cosa) + P_{0}cosa.

P_{2}= ½(1-cosa) + P_{1}cosa. = ½(1 - cosa) + ½(1 - cosa)cosa + P_{0}cos^{2}a

= ½(1 - cosa^{2}) + P_{0}cos^{2}a.

Assume P_{n}= ½(1 - cos^{n}a) + P_{0}cos^{n}a. This holds for P_{1}and P_{2}.

Then P_{n+1}= ½(1 - cosa) + P_{n}cosa = ½(1 - cosa) + (½(1 - cos^{n}a) + P_{0}cos^{n}a)cosa

= ½(1 – cos^{(n+1)}a) + P_{0}cos^{(n+1)}a.

If P_{n}= ½(1 - cos^{n}a) + P_{0}cos^{n}a holds for n, it also holds for n + 1.

(d) As as n --> ∞, cos^{n}a --> 1, P_{n}--> P_{0}. (Quantum Zeno Effect)

Continuously measuring suppresses the evolution of the system.

**Problem 3:**

Consider a spin ½ particle with magnetic moment **m **= γ**S**. Let
|+> and |-> denote the eigenvectors of S_{z} and let the state of the
system at t = 0 be |ψ(0)> = |+>.

(a) At t = 0 we measure S_{y }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) = ω_{0}(t)S_{z}.

Assume ω_{0}(t) = 0 for t < 0 and for t > T, and increases linearly from
0 to ω_{0} when 0 < t < T. Show that at time t the state vector can be
written as

|ψ(t)> = 2^{-½}[exp(iθ(t))|+> + iexp(-iθ(t))|->]

and calculate the real function θ(t).

(c) At time t = τ > T, we measure S_{y}. What results can we find and
with what probability?

Determine the relation that must exist between ω_{0} 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 S_{z}of a spin ½ particle is two-dimensional. We denote the eigenvectors of S_{z}by |+> and |->. The Hamiltonian of the system is H = ω_{0}(t)S_{z}. 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^{-½}[exp(iθ(t))|+> + iexp(-iθ(t))|->].

iħ∂|ψ(t)>/∂t = iħ2^{-½}[i (∂θ/∂t) exp(iθ(t))|+> + (∂θ/∂t) exp(-iθ(t))|->]

= -ħ2^{-½}[(∂θ/∂t) exp(iθ(t))|+> - i(∂θ/∂t) exp(-iθ(t))|->].

H|ψ(t)> = ω_{0}(t)S_{z}|ψ(t)> = 2^{-½}ω_{0}(t)(ħ/2)[exp(iθ(t))|+> - iexp(-iθ(t))|->].

For ψ(t)> to be a solution to the Schroedinger equation we need ω_{0}(t)/2 = -∂θ/∂t,

-dθ = ½ω_{0}(t)dt, θ(t') - θ(0) = -½∫_{0}^{t'}ω_{0}(t)dt.

θ(0) = θ(t < 0) = 0. Let α = ω_{0}/T, then ω_{0}(t) = αT.

θ(t') = -αt'^{2}/4 for (0 < t' < T), θ(t') = -αT^{2}/4 for ( t' > T).

(c) |ψ(τ)> = 2^{-½}[exp(iθ(τ))|+> + iexp(-iθ(τ))|->].

Possible result of measuring S_{y}are ±ħ/2.

P(+ħ/2) = |_{y}<+|ψ(τ)>|^{2}= ¼|exp(iθ(τ)) + exp(-iθ(τ)|^{2}= cos^{2}(θ(τ)).

P(-ħ/2) = |_{y}<-|ψ(τ)>|^{2}= sin^{2}(θ(τ)).

θ(τ) = -αT^{2}/4. We need αT^{2}/4 = nπ/2 or ω_{0}= n2π/T to be sure of the result.

The physical interpretation is a precession, similar to Larmor precession.