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_xsinθcosφ + S_ysinθsinφ + S_zcosθ.
• 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 = 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 c₁|ψ₁>+  c₂|ψ₂>.  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 = (E₁ - E₂)/ħ
and deduce that  p_n = 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 a_n is given by
  P(a_n) = ∑_i=0^gn|<u_nⁱ|Ψ>|²,  where {|u_nⁱ>} (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
  P₀ = ½|(<ψ₁| + <ψ₂|)(c₁|ψ₁> + c₂|ψ₂>)|² =  ½|c₁ + c₂|².
  (b)  Before the first measurement at t = 0, |ψ(0)> = c₁|ψ₁> + c₂|ψ₂>.   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(-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^-½(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
  P_n+1 = P_n (1/4)|exp(-E₁T/ħ) + exp(-E₂T/ħ)|² + (1 – P_n) (1/4)|exp(-E₁T/ħ) - exp(-E₂T/ħ)|²
  = P_n ½(1 + cos((E₁ – E₂)T/ħ)) + (1 - P_n) ½(1 - cos((E₁ – E₂)T/ħ))
  = P_n ½(1 + cos(a)) + (1 - P_n) ½(1 - cos(a)) = ½(1 - cosa) + P_n 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 P_n = ½(1 - cosⁿa) + P₀cosⁿa.  This holds for P₁ and P₂.
  Then P_n+1 = ½(1 - cosa) + P_n cosa = ½(1 - cosa) + (½(1 - cosⁿa) + P₀cosⁿa)cosa
   = ½(1 – cos⁽ⁿ⁺¹⁾a) + P₀cos⁽ⁿ⁺¹⁾a.
  If  P_n = ½(1 - cosⁿa) + P₀cosⁿa  holds for n, it also holds for n + 1.
  (d)  As as n --> ∞,  cosⁿa  --> 1, P_n --> 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 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) = ω₀(t)S_z.
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^-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 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 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 = ω₀(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)> = ω₀(t)S_z|ψ(t)> = 2^-½ω₀(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∫₀^t'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^-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 = 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.