Assume you have a system consisting of two spinless particles.  Its wave function is Ψ(r₁,r₂).  A  special case of such a wave function is a product function of the form Ψ(r₁,r₂) = Φ(r₁)ζ(r₂) . The state vector corresponding to such a product is denoted by
|Ψ> = |Φ(1)> ⊗ |ζ(2)> = |Φ(1)>|ζ(2)>.

|Ψ> is a vector in the two-particle space E, while |Φ(1)> and |ζ(2)> are vectors in the spaces E₁ and E₂ of particle 1 and particle 2, respectively. 
We say that E is the tensor product of E₁ and E₂ and write E = E₁ ⊗ E₂. 
If {|u_i>} and {|v_j>} are bases for E₁ and E₂ respectively, then the set of all tensor product vectors {|u_i> ⊗ |v_j>} is a basis for E.  Every vector in E is a sum of product vectors.  This, however, does not mean that every vector in E is a product vector, only that it can be written as a linear combination of product vectors.  Not every Ψ(r₁,r₂) is a product function.

The inner product in E satisfies
(<Φ'(1)| ⊗ <ζ'(2)|)(|Φ(1)> ⊗ |ζ(2)>) = <Φ'(1)Φ(1)> <ζ'(2)|ζ(2)>) .
Tensor products occur whenever a system has two or more independent degrees of freedom.

• The space of a single particle with spin S is E = E_space ⊗ E_spin .
• The space of a single particle moving in three dimensions is E = E_x ⊗ E_y ⊗ E_z .
  {|x>}, {|y>}, and {|z>} are bases for E_x , E_y , and E_z respectively.
  |r> = |x> ⊗ |y> ⊗ |z> = |x,y,z> is a basis for E. 
  We have Ψ(r) = Ψ(x,y,z) = <x,y,z|Ψ>,
  |Ψ> = ∫dxdydzΨ(x,y,z)|x,y,z>
  The x, y and z dependence cannot, in general, be factored.

We can regard E = E₁ ⊗ E₂ as a factoring of E, and it is an important fact that a given space can be factored in many different ways.

Certain operators in E are product operators.  If A(1) and B(2) are operators acting in E₁ and E₂, respectively, we define the product operator A(1) ⊗ B(2) through the relation
(A(1) ⊗ B(2))(|Φ(1)> ⊗ |ζ(2)>) = A(1)|Φ(1)> ⊗ B(2)|ζ(2)>.
If there can be no confusion we write A(1) ⊗ B(2) = A(1)B(2).

• The momentum operator for particle 1 is P₁(in E₁) ⊗ I (in E₂).
• The momentum operator for particle 2 is I (in E₁) ⊗ P₂(in E₂)
• The total momentum operator is P₁ + P₂  =  P₁(in E₁) ⊗ I (in E₂) + I (in E₁) ⊗ P₂(in E₂).
• The two particle Hamiltonian is of two interacting spinless particles is
  P₁²/(2m₁) + P₂²/(2m₂) + U(r₁,r₂)
  = P₁²/(2m₁) (in E₁) ⊗ I (in E₂)
  + I (in E₁) ⊗ P₂²/(2m₂) (in E₂)
  + U(r₁,r₂) (in E₁ ⊗ E₂).
  Note: U(r₁,r₂) is not a product operator.
  However, if we write
  P_cm²/(2m) + P₂²/(2μ) + U(r)
  = P_cm²/(2m) (in E_cm) ⊗ I (in E_r)
  + I (in E_cm) ⊗ [P₂²/(2μ) + U(r)] (in E_r),
  with  M = m₁ + m₂ and μ = m₁m₂/(m₁ + m₂), then H is a sum of product operators.
  
  The evolution operator may then be written as
  U(t,0) = exp(-iHt/ħ) = exp(-i(H_cm + H_rel)t/ħ) = exp(-iH_cmt/ħ) ⊗ exp(-iH_relt/ħ).

Problem:

Consider a system of two distinguishable particles with mass m which do not interact and which are both placed in an infinite potential well of width a.  Denote by H(1) and H(2) the Hamiltonians of each of the two particles.  Let {|Φ_n(1)>} and {|Φ_q(2)>} be an eigenbasis of H(1) and H(2) respectively.  A basis for the global system is  {|Φ_n(1)> ⊗ |Φ_q(2)> = |Φ_nΦ_q>}.
(a)  What are the eigenvectors and eigenvalues of the total Hamiltonian?   Give the degree of degeneracy of the two lowest energy levels.

(b)  Assume that the system at time t = 0 is in the state
|Ψ(0)> = (1/√6)|Φ₁Φ₁> + (1/√3)|Φ₁Φ₂> + (1/√6)|Φ₂Φ₁> + (1/√3)|Φ₂Φ₂>.
What is the state of the system at time t?  If H is measured at time t, what results can be found and with what probabilities? 
If H(1) is measured at time t, what results can be found and with what probabilities?

(c) Show that |Ψ(0)> is a tensor product state. 
Calculate at t = 0  <H₁>, <H₂>, and <H₁H₂>. 
Compare <H₁><H₂> with <H₁H₂>.  Show that this remains valid at later times.

(d)  Let |Ψ(0)> = (1/√5)|Φ₁Φ₁> + √(3/5)|Φ₁Φ₂> + (1/√5)|Φ₂Φ₁>.
Show that this is not a tensor product state.  Calculate at t = 0  <H₁>, <H₂>, and <H₁H₂>.  Compare <H₁><H₂> with <H₁H₂>.

• Solution:
  (a)  {|Φ_n(1)>} and {|Φ_q(2)>} are non-degenerate eigenbases of H(1) and H(2) respectively. 
  Note:  H(1) = H (in E₁) ⊗ I (in E₂), H(2) = I(in E₁) ⊗ H(in E₂).
  The eigenvalues are E_n = n²π²ħ²/(2ma²) and E_q = q²π²ħ²/(2ma²), respectively.
  H|Φ_nΦ_q> = (H(1) + H(2))|Φ_nΦ_q> =  E_nq|Φ_nΦ_q> = [(n² + q²)π²ħ²/(2ma²)]|Φ_nΦ_q>
  The eigenstates of H are {|Φ_nΦ_q>} with eigenvalues (n² + q²)π²ħ²/(2ma²).
  The lowest energy eigenstate has n = q = 1. Therefore
  E₁₁ = 2E₀, with E₀ = π²ħ²/(2ma²).  This energy level is not degenerate.
  The second lowest energy eigenstate has n = 1, q = 2 or n = 2, q = 1. 
  Therefore  E₁₂ = E₂₁ = 5E₀.  This energy level is twofold degenerate.

  (b)  |Ψ(0)> = (1/√6)|Φ₁Φ₁> + (1/√3)|Φ₁Φ₂> + (1/√6)|Φ₂Φ₁> + (1/√3)|Φ₂Φ₂>.
  U(t,0) = exp(-i(H(1) + H(2))t/ħ).
  |Ψ(t)> = (1/√6)exp(-iE₁₁t/ħ)|Φ₁Φ₁> + (1/√3)exp(-iE₁₂t/ħ)|Φ₁Φ₂>
  + (1/√6)exp(-iE₂₁t/ħ)|Φ₂Φ₁> + (1/√3)exp(-iE₂₂t/ħ)|Φ₂Φ₂>.
  = (1/√6)exp(-i2E₀t/ħ)|Φ₁Φ₁> + (1/√3)exp(-i5E₀t/ħ)|Φ₁Φ₂>
  + (1/√6)exp(-i5E₀t/ħ)|Φ₂Φ₁> + (1/√3)exp(-i8E₀t/ħ)|Φ₂Φ₂>.
  If H is measured, then the eigenvalues 2E₀, 5E₀, and 8E₀ can be found.
  P(2E₀) = |<Φ₁Φ₁|Ψ(t)>|² = 1/6.
  P(8E₀) = |<Φ₂Φ₂|Ψ(t)>|² = 1/3.
  P(5E₀) = |<Φ₁Φ₂|Ψ(t)>|² + |<Φ₁Φ₂|Ψ(t)>|² = 1/3 + 1/6 = 1/2.
  If H(1) is measured, then the eigenvalues  E_n=1 = E₀, and E_n=2 = 4E₀ can be found.
  P(E₀) = |<Φ₁Φ₁|Ψ(t)>|² + |<Φ₁Φ₂|Ψ(t)>|²  = 1/6 + 1/3 = 1/2.
  P(4E₀) = |<Φ₂Φ₁|Ψ(t)>|² + |<Φ₂Φ₂|Ψ(t)>|²  = 1/6 + 1/3 = 1/2.

  (c) If |Ψ(0)> is a tensor product state, then it can be written as
  |Ψ(0)> = (a₁|Φ₁(1)> + a₂|Φ₂(1)>) ⊗ (b₁|Φ₁(2)> + b₂|Φ₂(2)>),
  or
  |Ψ(0)> = a₁b₁|Φ₁Φ₁> + a₁b₂|Φ₁Φ₂> + (a₂b₁|Φ₂Φ₁> + a₂b₂|Φ₂Φ₂>.
  For |Ψ(0)> to be a tensor product state we need
  a₁b₁ = 1/√6,   a₁b₂ = 1/√3,   a₂b₁ = 1/√6,   a₂b₂ = 1/√3.
  Therefore a₁ = a₂,  b₁ = (1/√2)b₂.
  For a normalized solution exists if we require 
  |a₁|² + |a₂|² = |b₁|² + |b₂|² = 1.
  a₁ = a₂ = 1/√2,  b₁ = 1/√3,  b₁ = √(2/3).
  A solution exists, |Ψ(0)> is a tensor product state.
  |Ψ(0)> = |ζ(1)> ⊗ |γ(2)>. 
  |ζ(1)> = a₁|Φ₁(1)> + a₂|Φ₂(1)> = (1/√2)|Φ₁(1)> + (1/√2)|Φ₂(1)>
  <H(1)> = <ζ(1)|H(1)|ζ(1)> = |a₁|²E₀ + |a₂|²4E₀ =  (5/2)E₀.
  |γ(2)> = b₁|Φ₁(2)> + b₂|Φ₂(2)> = (1/√3)|Φ₁(2)> + √(2/3)|Φ₂(2)>
  <H(2)> = <γ(2)|H(2)|γ(2))> = |b₁|²E₀ + |b₂|²4E₀ = 3E₀.
  H(1)H(2)|Ψ(0)> = H(1)|ζ(1)> ⊗ |H(2)γ(2)>. 

  <Ψ(0)|H(1)H(2)|Ψ(0)> = <ζ(1)|H(1)|ζ(1)><γ(2)|H(2)γ(2)> = <H(1)><H(2)>.
  The system is in a product state.  Measuring H(1) does not disturb γ(2) and vice versa.
  This remains valid at later times.
  The results depend on |a_i|² and |b_i|².  The time dependence is in the phases and cancels out.

  (d) For |Ψ(0)> to be a tensor product state we need
  a₁b₁ = 1/√5,   a₁b₂ = √(3/5),   a₂b₁ = 1/√5,   a₂b₂ = 0.
  a₂b₂ = 0 --> a₂ = 0 or b₂ = 0.
  If a₂ = 0 then a₂b₁ = 0 and if b₂ = 0 then a₁b₂ = 0. This is inconsistent with what we need, no solution exists, |Ψ(0)> is not a product state.
  <H(1)> = (1/5)E₀ + (3/5)E₀ + (1/5)4E₀ = (8/5)E₀.
  <H(2)> = (1/5)E₀ + (3/5)4E₀ + (1/5)E₀ = (14/5)E₀.
  <H(1)H(2)> =  (1/5)E₀² + (3/5)4E₀² +( 1/5)4E₀² = (17/5)E₀² ≠ <H(1)><H(2)>.
  |Ψ(0)> is a linear combination of product states.  H(1) acting on |ζ(1)> does not disturb |γ(2)> in each of the terms, but it changes the relative weight of the terms.

In one dimension, consider two particles of mass m, coordinates x₁ and x₂, momenta p₁ and p₂, and potential energy
U(x₁,x₂) = ½mω²x₁² + ½mω²x₂² + gmω²(x₁ - x₂)².
Find the eigenvalues and eigenfunctions of the Hamiltonian H of the system.

• Solution:
  We have H =  ½(P₁² + P₂²)/m + ½mω²(X₁² + X₂²) + gmω²(X₁ - X₂)².
  Let
  X_G = (1/√2)(X₁ + X₂),  X_R = (1/√2)(X₁ - X₂),
  P_G = (1/√2)(P₁ + P₂),  P_R = (1/√2)(P₁ - P₂),
  Then [X_G,P_G] = iħ,  [X_R,P_R] = iħ,  [X_G,P_R] = [X_R,P_G] = 0,
  X_G² + X_R² = X₁² + X₂²,  P_G² + P_R² = P₁² + P₂², 
  H =  ½(P_G² + P_R²)/m + ½mω²(X_G² + X_R²) + 2gmω²X_R²
  = ½P_G²/m + ½mω²X_G² + ½P_R²/m + ½mω²(1 + 4g)X_R²
  = H_G + H_R.
  H is the Hamiltonian of two non interacting fictitious particle of mass m in harmonic oscillator potentials with frequency ω and ω' = (1 + 4 g)^1/2ω, respectively. 
  The state space E is the tensor product space E = E_G ⊗ E_R .
  The eigenfunctions of H are tensor product functions |Φ_G>⊗|Φ_R>.
  H_G|Φ_G> = (n_G + ½)ħω|Φ_G>,  H_R|Φ_R> = (n_R + ½)ħω'|Φ_R>.
  E_nG,nR = (n_G + ½)ħω + (n_R + ½)ħω' are the eigenvalues of H.

  With β = (mω/ħ)^1/2 and β' = (mω'/ħ)^1/2the corresponding eigenfunctions are
  Ψ_nG,nR,(x_G,x_R) =  (n_G!n_R! 2^{(nG + nR)})^-1/2 (ββ'/π)^1/2H_nG(βx_G) H_nR(β'x_R) exp(-½(β²x_G² + β'²x_R²)).