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 {|ui>} and {|vj>} are bases for E1 and E2 respectively, then the set of all tensor product vectors {|ui> ⊗ |vj>} 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.

Examples:
• The space of a single particle with spin S is E = Espace ⊗ Espin .
• The space of a single particle moving in three dimensions is E = Ex ⊗ Ey ⊗ Ez .
{|x>}, {|y>}, and {|z>} are bases for Ex , Ey , and Ez 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.

Example:
Consider the state space of two spinless particles.  We may rewrite Ψ(r1,r2) as a function of
r
cm = (m1r1 + m2r2)/(m1 + m2) and r = r1 - r2.
Ψ(rcm,r) is a vector in the vector space E =  Er1 ⊗ Er2 = Ecm ⊗ Erel spanned by product functions Φ(r1)ζ(r2) or Φ(rcm)ζ(r).

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).

Examples of product operators:
• 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
Pcm2/(2m) + P22/(2μ) + U(r)
= Pcm2/(2m) (in Ecm) ⊗ I (in Er)
+ I (in Ecm) ⊗ [P22/(2μ) + U(r)] (in Er),
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(Hcm + Hrel)t/ħ) = exp(-iHcmt/ħ) ⊗ exp(-iHrelt/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 {|Φn(1)>} and {|Φq(2)>} be an eigenbasis of H(1) and H(2) respectively.  A basis for the global system is  {|Φn(1)> ⊗ |Φq(2)> = |ΦnΦq>}.
(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)  {|Φn(1)>} and {|Φq(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 En = n2π2ħ2/(2ma2) and Eq = q2π2ħ2/(2ma2), respectively.
H|ΦnΦq> = (H(1) + H(2))|ΦnΦq> =  EnqnΦq> = [(n2 + q22ħ2/(2ma2)]|ΦnΦq>
The eigenstates of H are {|ΦnΦq>} with eigenvalues (n2 + q22ħ2/(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  En=1 = E0, and En=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)> = (a11(1)> + a22(1)>) ⊗ (b11(2)> + b22(2)>),
or
|Ψ(0)> = a1b11Φ1> + a1b21Φ2> + (a2b12Φ1> + a2b22Φ2>.
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)> = a11(1)> + a22(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)> = b11(2)> + b22(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 |ai|2 and |bi|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.
Problem:

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
XG = (1/√2)(X1 + X2),  XR = (1/√2)(X1 - X2),
PG = (1/√2)(P1 + P2),  PR = (1/√2)(P1 - P2),
Then [XG,PG] = iħ,  [XR,PR] = iħ,  [XG,PR] = [XR,PG] = 0,
XG2 + XR2 = X12 + X22,  PG2 + PR2 = P12 + P22
H =  ½(PG2 + PR2)/m + ½mω2(XG2 + XR2) + 2gmω2XR2
= ½PG2/m + ½mω2XG2 + ½PR2/m + ½mω2(1 + 4g)XR2
= HG + HR.
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 = EG ⊗ ER .
The eigenfunctions of H are tensor product functions |ΦG>⊗|ΦR>.
HGG> = (nG + ½)ħω|ΦG>,  HRR> = (nR + ½)ħω'|ΦR>.
EnG,nR = (nG + ½)ħω + (nR + ½)ħω' are the eigenvalues of H.

With β = (mω/ħ)1/2 and β' = (mω'/ħ)1/2the corresponding eigenfunctions are
ΨnG,nR,(xG,xR) =  (nG!nR! 2(nG + nR))-1/2 (ββ'/π)1/2HnG(βxG) HnR(β'xR) exp(-½(β2xG2 + β'2xR2)).