### A single spin ½ particle

#### Problem:

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 = -(gμB/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 Fz = μz dBz/dz.
Angular momentum is quantized.  The possible values that we can measure for the square of the magnitude of the angular momentum are J2 = j(j+1)ħ2.  The possible projections we can measure along any axis, for example the z-axis, are Jz = 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 Jz is quantized, μz is quantized, μz = -gμBm.   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 μz.

#### Problem:

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 Sz of a spin ½ particle is two-dimensional.  We denote the eigenvectors of Sz by |+> and |->.  The matrix of Sz in the {|+>, |->} basis is
.
The matrices of Sx and Sy in the eigenbasis of Sz , {|+>, |->}, are not diagonal.

, .
If n = n(θ,φ),  then the operator Sn defined through
S
n = Sxsinθcosφ + Sysinθsinφ + Szcosθ.
The matrix of Sn is given in terms of the matrices of Sx, Sy, and Sz
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 Sn is
(Sn) = (Sx)sinθcosφ + (Sy)sinθsinφ + (Sz)cosθ
= .

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)|->.
The probability of observing a component of spin angular momentum ħ/2 along n is
P = |n<+|+>|2 = cos2(θ/2).

#### Problem:

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:

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 Sz), and travel along the y-axis with a constant velocity of magnitude v0
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 = 2μB/ħ, where μB = qeħ/(2me) is the Bohr magneton.  For the electron μ = -γS

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 = γS.  In a magnetic field B the Hamiltonian of an atom is H = -μB = -γSB.
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/v0.  During this time interval the Hamiltonian is -γBSy = ω0Sy.
|ψ(L/v0)> = exp(-iω0SyΔt/ħ)|+> = exp(-iω0SyΔt/ħ) (1/√2)(|+>y + |->y).
|±>y = (1/√2)(|+> ± i|->),  |±> = (1/√2)(|+>y ± |->y).
Sy|±>y = ±ħ/2)|±>y.
|ψ(L/v0)> = (1/√2)(exp(-iω0L/(2v0))|+>y + exp(iω0L/(2v0))|->y)
= ½(exp(-iω0L/(2v0))(|+> + i|->) + exp(iω0L/(2v0))(|+> - i|->))
= cos(ω0L/(2v0))|+> + sin(ω0L/(2v0))|->.
Now the atom enters a spin analyzer and are acted on by a force Fz = μzdBz/dz.
Atoms in the |+> state are deflected in the +z direction (μz and dBz/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 cos20L/(2v0) and the number of atoms/s leaving the analyzer in the |-> state is N- = N sin20L/(2v0).
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/v0)> = cos(ω0L/(2v0))|+> + sin(ω0L/(2v0))|->.
For all atoms to exit in the |+>n state we need ω0L/v0 = θ + n2π,  n = integer.
For all atoms to exit in the |->n state we need ω0L/v0 = θ - π + n2π, n = integer.

Physically, the polarization axis of the beam precesses about the magnetic field direction with angular frequency ω0.  After a time interval L/v0 the precession stops, and if we rotate the spin analyzer by the same angle, all the atoms will be transmitted.

#### Problem:

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 <Sx>, <Sy>, and <Sz>.
(c)  Find the root-mean-square deviations ΔSx, ΔSy, and ΔSz.

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 = (<A2> - <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:

<Sx> = <χ|Sx|χ> = (ħ/50)

 -3i 4
 0 1 1 0
 3i 4

= 0.

<Sy> = <χ|Sy|χ> = (ħ/50)

 -3i 4
 0 -i i 0
 3i 4

= -12ħ/25.

<Sz> = <χ|Sz|χ> = (ħ/50)

 -3i 4
 1 0 0 -1
 3i 4

= -7ħ/50.

(c)  Root-mean-square deviations:
ΔSi = (<Si2> - <Si>2)½,
Sx2 = Sy2 = Sz2  = (ħ2/4)I,
<Sx2> = <Sy2> = <Sz2> = ħ2/4,
This yields ΔSx = ħ/2,  ΔSy = 7ħ/50, and  ΔSz = 12ħ/25.

#### Problem:

Two states of a spin ½ particle are represented in the eigenbasis of Sz by
1> = (1/√2)(|+> + i|->),  |ψ2> = (1/√3)(-i|+> + √2|->).
(a)  Find their representation in the eigenbasis of Sy.
(b)  Find the amplitude <ψ12> in the Sz basis and show that this amplitude remains unchanged when calculated in the Sy basis.  (Show your work.)
(c)  The Hamiltonian for the particle is H = ω0Sz.  Find |ψ1(t)>.  At what times t is |ψ1(t)> an eigenvector of Sx?

Solution:

• Concepts:
Change of representation, the evolution operator
• Reasoning:
We are switching from the eigenbasis of Sz to the eigenbasis of Sy 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 Sx.
• Details of the calculation:
(a)  The eigenvectors of Sy are |+>y = (1/√2)(|+> + i|->) and |->y = (1/√2)(|+> - i|->).
Therefore: |+> = (1/√2)(|+>y + |->y) and |-> = (-i/√2)(|+>y - |->y).
1> = |+>y.
2> = (-i/√6)(|+>y + |->y) - (i/√3)(|+>y - |->y) = (-i)(0.986|+>y + 0.169|->y
(b)  <ψ12> = (1/√6)(-i<+|+> - i√2<-|->) = -i/√6 - i√3 = -i0.986 in the Sz basis.
12> = (-i/√6)(y<+|+>y) - (i/√3)(y<+|+>y) = -i/√6 - i√3 = -i0.986 in the Sy basis.
(c)  |ψ1(t)> = (1/√2)(exp(-iω0t/2)|+> + iexp(iω0t/2)|->).

= (ħ/4)i(-exp(-iω0t) + exp(iω0t)) = -(ħ/2)sin(ω0t).
If <Sx> = ±(ħ/2), then |ψ1(t)> is an eigenvector of Sx.  This happens when ω0t = nπ/2, n = odd.

#### Problem:

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 Sx operator.  The particle is confined to a region with a uniform magnetic field B = B0k, its Hamiltonian is H = ω0Sz.  The eigenstates of H are |+> and |->,
H|+> = (ħω0/2)|+> and H|-> = -(ħω0/2)|->.
|+>x can be written as a linear combination of eigenstates of H.
(a)  Find the probability of measuring Sx = ħ/2 at t = T.
(b)  What is the mean value of Sx, <Sx>, at t = T?
(c)  Find the probability of measuring Sz = 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 Sz of a spin ½ particle is two-dimensional.  We denote the eigenvectors of Sz by |+> and |->.  The Hamiltonian of the system is H = ω0Sz.  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 Sx in the eigenbasis of Sz is
.
The eigenvectors of Sx 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(-iω0t/2)|+> + exp(iω0t/2)|->).
P(Sx=ħ/2,T) = |x<+||ψ(t)>|2 = ¼|(<+| + < -|)(exp(-iω0T/2)|+> + exp(iω0T/2)|->)|2
= ¼|exp(-iω0T/2)|+> + exp(iω0T/2)|2 = cos20T/2).

(b) P(Sx=-ħ/2,T) = 1 - cos20T/2) = sin20T/2).
<Sx(T)> = (ħ/2)[cos20T/2) - sin20T/2)] = (ħ/2)cos(ω0T).

(c) P(Sz=ħ/2,T) = |<+|ψ(t)>|2 = ½|<+|(exp(-iω0t/2)|+>|2 = ½.
The probability of measuring Sz = ħ/2 is independent of t.

#### Problem:

An elementary spin-½ particle with magnetic moment μB 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 Sx of a spin ½ particle is two-dimensional.  We denote the eigenvectors of Sx by |+>x and |->x.  After the magnetic field has suddenly been flipped the particle is not in an eigenstate of Sx, but in a linear combination of eigenstates.
|+> = 2 (|+>x + |->x).
The Hamiltonian of the system is H = ω0Sx.  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 Sz, as the basis states.  Initially, these are the eigenstates of H.
B = Bk.  H = -mB = -mzB = -γSzB = ω0Sz.  Let ω0 = -γB.
At t = 0 the Hamiltonian becomes H = ω0Sx and the eigenstates of the Hamiltonian become |+>x and |->x.
|+> = 2 (|+>x + |->x).
|ψ(0)> = |+>.
|ψ(t)> = U(t,0) |+> = 2 U(t,0) (|+>x + |->x)
= 2 (exp(-iω0t/2)|+>x + exp(+iω0t/2)|->x)
= 2-1[exp(-iω0t/2)(|+> + |->) + exp(+iω0t/2)(|+> - |->)]
= cos(ω0t/2)|+> + isin(ω0t/2)|->.
(b)  P(Sz=ħ/2,t) = cos20t/2),  P(Sz=-ħ/2,t) = sin20t/2).
<Sz>(t) = (ħ/2)( cos20t/2) - sin20t/2)) = (ħ/2)cos(ω0t).
The eigenvectors of Sy are |+>y = (1/√2)(|+> + i|->) and |->y = (1/√2)(|+> - i|->).
<Sy>(t) = (ħ/2)|y<+|ψ(t)>|2 - (ħ/2)|y<-|ψ(t)>|2 =  (ħ/2)sin(ω0t).
<Sx>(t) = 0.
We have Larmor precession with frequency |ω0| about the x-axis.

#### Problem:

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) = -½∫0t'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.