Explain why, in a Stern-Gerlach (SG) apparatus, a beam consisting of neutral particles in different spin states is split into different beams.

Solution:

- Concepts:

The Stern-Gerlach experiment - Reasoning:

We are asked to explain the physics that determines the outcome af a Stern-Gerlach type experiment. - Details of the calculation:

If the neutral atoms have nonzero angular momentum**J**, they have a nonzero magnetic moment**m**= -(gμ_{B}/ħ)**J**. In an external magnetic field a magnetic moment can experience a force and a torque. The torque tries to align the magnetic moment with the magnetic field. The force tries to pull an aligned dipole into regions where the magnitude of the magnetic field is larger and push an anti-aligned dipole into regions where magnitude the magnetic field is smaller. If the magnetic field is pointing into the z-direction, the force on a magnetic dipole in that field is given by F_{z}= μ_{z}dB_{z}/dz.

Angular momentum is quantized. The possible values that we can measure for the square of the magnitude of the angular momentum are J^{2}= j(j+1)ħ^{2}. The possible projections we can measure along any axis, for example the z-axis, are J_{z}= mħ. Here j is a non-negative integer or half integer, and for a given j, m can take on values from -j to j in integer steps. Since J_{z}is quantized, μ_{z}is quantized, μ_{z}= -gμ_{B}m. If we passed a beam of randomly oriented atoms with angular momentum**J**through an inhomogeneous magnetic field, we would expect different atoms to be deflected by different discrete amounts depending on the quantized value of μ_{z}.

An electron is in a state with z-component of spin angular momentum
ħ/2. An observation designed to measure the
component of spin angular momentum along an arbitrary direction
**n** is
made. What is the probability of observing a component of spin angular momentum
ħ/2 along **n**?

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. We denote the eigenvectors of S_{z}by |+> and |->. The matrix of S_{z}in the {|+>, |->} basis is

.

The matrices of S_{x}and S_{y}in the eigenbasis of S_{z}, {|+>, |->}, are not diagonal., .

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

S**n**= S_{x}sinθcosφ + S_{y}sinθsinφ + S_{z}cosθ.

The matrix of S_{n}is given in terms of the matrices of S_{x}, S_{y}, and S_{z}.

We can diagonalize it and find its eigenvectors |+>_{n}and |->_{n}.

The probability of observing a component of spin angular momentum ħ/2 along**n**is P = |_{n}<+|+>|^{2}. -
Details of the calculation:

The matrix of S_{n}is

(S_{n}) = (S_{x})sinθcosφ + (S_{y})sinθsinφ + (S_{z})cosθ

= .

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)|->.

The probability of observing a component of spin angular momentum ħ/2 along**n**is

P = |_{n}<+|+>|^{2}= cos^{2}(θ/2).

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).

Consider a beam of N silver atoms per second in their ground state. The
atoms are polarized in the |+> state, (i.e. the spin-up state of S_{z}),
and travel along the y-axis with a constant velocity of magnitude v_{0}.

The atoms traverse a region of space of length L which contains a uniform static
magnetic field of strength B, directed along the y-axis, i.e. the direction of
travel.

Note: The translational motion of the center of mass of the atoms is treated
classically.

(a) Upon leaving the region of length L, the atoms enter a
spin analyzer (a Stern-Gerlach device) with its magnetic field directed along
the positive z-axis and the field gradient pointing in the -z direction. What
is the number of atoms per second in either of the beams emerging from the
analyzer?

(b) How should the analyzer be oriented, (i.e. what should
be the direction of the magnetic field of the analyzer), so that only one beam
emerges from the analyzer? Interpret this result physically.

Recall that the spin gyromagnetic factor for the electron is given by γ = 2μ_{B}/ħ,
where μ_{B} = q_{e}ħ/(2m_{e}) is the Bohr magneton.
For the electron **μ** = -γ**S**.

Solution:

- Concepts:

The two dimensional state space of a spin ½ particle, the evolution operator, the postulates of Quantum Mechanics - Reasoning:

Each atom has a magnetic moment**μ**= γ**S.**In a magnetic field**B**the Hamiltonian of an atom is H = -**μ**∙**B**= -γ**S**∙**B**.

The initial state evolves as |ψ(t)> = exp(-iHt/ħ)|ψ(0)>. - Details of the calculation:

(a) Assume an atom enters the field region in the state |+>. It leaves the field region after a time interval Δt = L/v_{0}. During this time interval the Hamiltonian is -γB**S**_{y}= ω_{0}S_{y}.

|ψ(L/v_{0})> = exp(-iω_{0}S_{y}Δt/ħ)|+> = exp(-iω_{0}S_{y}Δt/ħ) (1/√2)(|+>_{y}+ |->_{y}).

|±>_{y}= (1/√2)(|+> ± i|->), |±> = (1/√2)(|+>_{y}± |->_{y}).

S_{y}|±>_{y}= ±ħ/2)|±>_{y}.

|ψ(L/v_{0})> = (1/√2)(exp(-iω_{0}L/(2v_{0}))|+>_{y}+ exp(iω_{0}L/(2v_{0}))|->_{y})

= ½(exp(-iω_{0}L/(2v_{0}))(|+> + i|->) + exp(iω_{0}L/(2v_{0}))(|+> - i|->))

= cos(ω_{0}L/(2v_{0}))|+> + sin(ω_{0}L/(2v_{0}))|->.

Now the atom enters a spin analyzer and are acted on by a force F_{z}= μ_{z}dB_{z}/dz.

Atoms in the |+> state are deflected in the +z direction (μ_{z}and dB_{z}/dz are both negative) and atoms in the |-> state are deflected in the -z direction.

The number of atoms leaving the analyzer in the |+> state is N_{+}= N cos^{2}(ω_{0}L/(2v_{0}) and the number of atoms/s leaving the analyzer in the |-> state is N_{-}= N sin^{2}(ω_{0}L/(2v_{0}).

N_{+}+ N_{-}= N.

(b) We need to rotate the apparatus about the y-axis so it makes an angle θ with the z-axis.

Assume the spin analyzer is oriented so it deflects atoms in the |+>_{n}state in the**n**(θ,0) direction.

|+>_{n }= cos(θ/2)|+> + sin(θ/2)|->,

|->_{n }= -sin(θ/2)|+> + cos(θ/2)|->.

|ψ(L/v_{0})> = cos(ω_{0}L/(2v_{0}))|+> + sin(ω_{0}L/(2v_{0}))|->.

For all atoms to exit in the |+>_{n}state we need ω_{0}L/v_{0}= θ + n2π, n = integer.

For all atoms to exit in the |->_{n}state we need ω_{0}L/v_{0}= θ - π + n2π, n = integer.

Physically, the polarization axis of the beam precesses about the magnetic field direction with angular frequency ω_{0}. After a time interval L/v_{0}the precession stops, and if we rotate the spin analyzer by the same angle, all the atoms will be transmitted.

An electron is in the spin state |χ> = A(3i|+> + 4|->) = A

3i | ||

4 |

.

(a) Determine the normalization constant A.

(b) Find the expectation values <S_{x}>, <S_{y}>, and <S_{z}>.

(c) Find the root-mean-square deviations ΔS_{x}, ΔS_{y}, and ΔS_{z}.

Solution:

- Concepts:

The mean value and the root mean square deviation of an observable - Reasoning:

The expression for the mean value of an observable A in the normalized state |Ψ> is

<A> = <Ψ|A|Ψ>. If |Ψ> is not normalized then < A> = <Ψ|A|Ψ>/<Ψ|Ψ>.

The root mean square deviation ΔA characterizes the dispersion of the measurement around <A>.

ΔA = (<(A - <A>^{2})>)^{½}= (<A^{2}> - <A>^{2}))^{½}.

Here the system is the 2-dimensional state space of a spin-½ particle, and the observables are the Cartesian components of its spin. - Details of the calculation:

(a) Normalization: < χ| χ> = |A|^{2}(9 + 16) = 25 |A|^{2}= 1. A = 1/5.

(b) Expectation values:

<S

_{x}> = <χ|S_{x}|χ> = (ħ/50)-3i 4 0 1 1 0 3i 4 = 0.

<S

_{y}> = <χ|S_{y}|χ> = (ħ/50)-3i 4 0 -i i 0 3i 4 = -12ħ/25.

<S

_{z}> = <χ|S_{z}|χ> = (ħ/50)-3i 4 1 0 0 -1 3i 4 = -7ħ/50.

(c) Root-mean-square deviations:

ΔS_{i}= (<S_{i}^{2}> - <S_{i}>^{2})^{½},^{ }S_{x}^{2}= S_{y}^{2}= S_{z}^{2}= (ħ^{2}/4)I,

<S_{x}^{2}> = <S_{y}^{2>}= <S_{z}^{2}> = ħ^{2}/4,

This yields ΔS_{x}= ħ/2, ΔS_{y}= 7ħ/50, and ΔS_{z}= 12ħ/25.

Two states of a spin ½ particle are represented in the eigenbasis of S_{z}
by

|ψ_{1}>_{ }= (1/√2)(|+>
+ i|->), |ψ_{2}>_{ }= (1/√3)(-i|+>
+ √2|->).

(a) Find their representation in the eigenbasis of S_{y}.

(b) Find the amplitude <ψ_{1}|ψ_{2}> in the S_{z}
basis and show that this amplitude remains unchanged when calculated in the S_{y }
basis. (Show your work.)

(c) The Hamiltonian for the particle is H = ω_{0}S_{z}.
Find |ψ_{1}(t)>. At what times t is |ψ_{1}(t)> an
eigenvector of S_{x}?

Solution:

- Concepts:

Change of representation, the evolution operator - Reasoning:

We are switching from the eigenbasis of S_{z}to the eigenbasis of S_{y}and verify that inner products do not depend on the choice of representation.

We use the evolution operator to find the state of the system at time t and then check if it is an eigenvector of S_{x}. -
Details of the calculation:

(a) The eigenvectors of S_{y}are |+>_{y }= (1/√2)(|+> + i|->) and |->_{y }= (1/√2)(|+> - i|->).

Therefore: |+> = (1/√2)(|+>_{y }+ |->_{y}) and |-> = (-i/√2)(|+>_{y }- |->_{y}).

|ψ_{1}> = |+>_{y}.

|ψ_{2}> = (-i/√6)(|+>_{y}+ |->_{y}) - (i/√3)(|+>_{y}- |->_{y}) = (-i)(0.986|+>_{y }+ 0.169|->_{y }(b) <ψ_{1}|ψ_{2}> = (1/√6)(-i<+|+> - i√2<-|->) = -i/√6 - i√3^{ }= -i0.986 in the S_{z}basis.

<ψ_{1}|ψ_{2}> = (-i/√6)(_{y}<+|+>_{y}) - (i/√3)(_{y}<+|+>_{y}) = -i/√6 - i√3^{ }= -i0.986 in the S_{y}basis.

(c) |ψ_{1}(t)> = (1/√2)(exp(-iω_{0}t/2)|+> + iexp(iω_{0}t/2)|->).

= (ħ/4)i(-exp(-iω_{0}t) + exp(iω_{0}t)) = -(ħ/2)sin(ω_{0}t).

If <S_{x}> = ±(ħ/2), then |ψ_{1}(t)> is an eigenvector of S_{x}. This happens when ω_{0}t = nπ/2, n = odd.

At t = 0 the x-component of the spin of a spin ½ particle is measured and
found to be ħ/2. At t = 0 the particle is therefore in the |+>_{x}
eigenstate of the S_{x} operator. The particle is confined to a region
with a uniform magnetic field **B** = B_{0}**k**, its Hamiltonian
is H = ω_{0}S_{z}. The eigenstates of H are |+> and |->,

H|+> = (ħω_{0}/2)|+> and H|-> = -(ħω_{0}/2)|->.

|+>_{x} can be written as a linear combination of eigenstates of H.

(a) Find the probability of measuring S_{x} = ħ/2 at t = T.

(b) What is the mean value of S_{x}, <S_{x}>, at t = T?

(c) Find the probability of measuring S_{z} = ħ/2 at t = T.

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}S_{z}. The evolution operator is exp(-iHt/ħ). We use the evolution operator to find the state of the system at time T and then answer the questions using the postulates of Quantum Mechanics. - Details of the calculation:

(a) The matrix of S_{x}in the eigenbasis of S_{z}is

.

The eigenvectors of S_{x}are |+>_{x}and |->_{x}.

|±>_{x}= (1/√2)(|+> ± |->).

|ψ(t)> = U(t,0)|ψ(0)> = U(t,0)|+>_{x}= (1/√2)exp(-iHt/ħ)(|+> + |->).

|ψ(t)> = (1/√2)(exp(-iω_{0}t/2)|+> + exp(iω_{0}t/2)|->).

P(S_{x}=ħ/2,T) = |_{x}<+||ψ(t)>|^{2}= ¼|(<+| + < -|)(exp(-iω_{0}T/2)|+> + exp(iω_{0}T/2)|->)|^{2 }= ¼|exp(-iω_{0}T/2)|+> + exp(iω_{0}T/2)|^{2}= cos^{2}(ω_{0}T/2).(b) P(S

_{x}=-ħ/2,T) = 1 - cos^{2}(ω_{0}T/2) = sin^{2}(ω_{0}T/2).

<S_{x}(T)> = (ħ/2)[cos^{2}(ω_{0}T/2) - sin^{2}(ω_{0}T/2)] = (ħ/2)cos(ω_{0}T).(c) P(S

_{z}=ħ/2,T) = |<+|ψ(t)>|^{2}= ½|<+|(exp(-iω_{0}t/2)|+>|^{2}= ½.

The probability of measuring S_{z}= ħ/2 is independent of t.

An elementary spin-½ particle with magnetic moment μ_{B}
is in it's lower level state in a magnetic field **B** parallel to z-axis. At time
t = 0 the magnetic field **B** is flipped to point parallel to x-axis.

(a) Find the time-dependent spin wave function of the
particle for t > 0.

(b) Find the rotation frequency for the magnetic moment of
the particle.

Solution:

- Concepts:

The two dimensional state space of a spin ½ particle, the evolution operator, the postulates of Quantum Mechanics, the sudden approximation - Reasoning:

The state space corresponding to the observable S_{x}of a spin ½ particle is two-dimensional. We denote the eigenvectors of S_{x}by |+>_{x}and |->_{x}. After the magnetic field has suddenly been flipped the particle is not in an eigenstate of S_{x}, but in a linear combination of eigenstates.

|+> = 2^{-½}(|+>_{x}+ |->_{x}).

The Hamiltonian of the system is H = ω_{0}S_{x}. The evolution operator is exp(-iHt/ħ). We use the evolution operator to find the state of the system at times t > 0, and then answer the questions using the postulates of Quantum Mechanics. - Details of the calculation:

(a) Choose {|+> and |->}, the eigenstates of S_{z}, as the basis states. Initially, these are the eigenstates of H.

B = B**k**. H = -**m**∙**B**= -m_{z}B = -γS_{z}B = ω_{0}S_{z}. Let ω_{0}= -γB.

At t = 0 the Hamiltonian becomes H = ω_{0}S_{x}and the eigenstates of the Hamiltonian become |+>_{x}and |->_{x}.

|+> = 2^{-½}(|+>_{x}+ |->_{x}).

|ψ(0)> = |+>.

|ψ(t)> = U(t,0) |+> = 2^{-½}U(t,0) (|+>_{x}+ |->_{x})

= 2^{-½}(exp(-iω_{0}t/2)|+>_{x}+ exp(+iω_{0}t/2)|->_{x})

= 2^{-1}[exp(-iω_{0}t/2)(|+> + |->) + exp(+iω_{0}t/2)(|+> - |->)]

= cos(ω_{0}t/2)|+> + isin(ω_{0}t/2)|->.

(b) P(S_{z}=ħ/2,t) = cos^{2}(ω_{0}t/2), P(S_{z}=-ħ/2,t) = sin^{2}(ω_{0}t/2).

<S_{z}>(t) = (ħ/2)( cos^{2}(ω_{0}t/2) - sin^{2}(ω_{0}t/2)) = (ħ/2)cos(ω_{0}t).

The eigenvectors of S_{y}are |+>_{y }= (1/√2)(|+> + i|->) and |->_{y }= (1/√2)(|+> - i|->).

<S_{y}>(t) = (ħ/2)|_{y}<+|ψ(t)>|^{2}- (ħ/2)|_{y}<-|ψ(t)>|^{2}= (ħ/2)sin(ω_{0}t).

<S_{x}>(t) = 0.

We have Larmor precession with frequency |ω_{0}| about the x-axis.

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.