Assume you have a system consisting of two spinless particles. Its wave
function is Ψ(**r**_{1},**r**_{2}). A special case of
such a wave function is a product function of the form Ψ(**r**_{1},**r**_{2})
= Φ(**r**_{1})ζ(**r**_{2})
. 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 E_{1} and E_{2} of particle 1 and particle 2,
respectively.

We say that E is the **tensor product** of E_{1}
and E_{2} and write E = E_{1 }⊗ E_{2}.

If {|u_{i}>} and {|v_{j}>} are bases for
E_{1} and E_{2} 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(**r**_{1},**r**_{2}) 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 = E_{1 }⊗ E_{2}
as a factoring of E, and it is an important fact that a given space can be factored in
many different ways.

Consider the state space of two spinless particles. We may rewrite Ψ(

r

Ψ(

Certain operators in E are **product operators**. If A(1)
and B(2) are operators acting in E_{1} and E_{2},
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
**P**_{1}(in E_{1}) ⊗ I (in E_{2}). - The momentum operator for particle 2 is I (in E
_{1}) ⊗**P**_{2}(in E_{2}) - The total momentum operator is
**P**_{1 }+**P**_{2}=**P**_{1}(in E_{1}) ⊗ I (in E_{2}) + I (in E_{1}) ⊗**P**_{2}(in E_{2}). - The two particle Hamiltonian is of two interacting spinless particles is

P_{1}^{2}/(2m_{1}) + P_{2}^{2}/(2m_{2}) + U(**r**_{1},**r**_{2})

= P_{1}^{2}/(2m_{1}) (in E_{1}) ⊗ I (in E_{2})

+ I (in E_{1}) ⊗ P_{2}^{2}/(2m_{2}) (in E_{2})

+ U(**r**_{1},**r**_{2}) (in E_{1}⊗ E_{2}).

Note: U(**r**_{1},**r**_{2}) is not a product operator.

However, if we write

P_{cm}^{2}/(2m) + P_{2}^{2}/(2μ) + U(r)

= P_{cm}^{2}/(2m) (in E_{cm}) ⊗ I (in E_{r})

+ I (in E_{cm}) ⊗ [P_{2}^{2}/(2μ) + U(**r**)] (in E_{r}),

with M = m_{1}+ m_{2}and μ = m_{1}m_{2}/(m_{1}+ m_{2}), 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_{cm}t/ħ) ⊗ exp(-iH_{rel}t/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_{n}Φ_{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)|Φ_{1}Φ_{1}> + (1/√3)|Φ_{1}Φ_{2}>
+ (1/√6)|Φ_{2}Φ_{1}> + (1/√3)|Φ_{2}Φ_{2}>.

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 <H_{1}>, <H_{2}>, and
<H_{1}H_{2}>.

Compare <H_{1}><H_{2}>
with <H_{1}H_{2}>. Show that this remains valid at
later times.

(d) Let |Ψ(0)> = (1/√5)|Φ_{1}Φ_{1}> + √(3/5)|Φ_{1}Φ_{2}>
+ (1/√5)|Φ_{2}Φ_{1}>.

Show that this is not a tensor product state. Calculate at t = 0 <H_{1}>,
<H_{2}>, and <H_{1}H_{2}>. Compare <H_{1}><H_{2}> with <H_{1}H_{2}>.

- 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 E_{1}) ⊗ I (in E_{2}), H(2) = I(in E_{1}) ⊗ H(in E_{2}).

The eigenvalues are E_{n}= n^{2}π^{2}ħ^{2}/(2ma^{2}) and E_{q}= q^{2}π^{2}ħ^{2}/(2ma^{2}), respectively.

H|Φ_{n}Φ_{q}> = (H(1)_{ }+ H(2))|Φ_{n}Φ_{q}> = E_{nq}|Φ_{n}Φ_{q}> = [(n^{2}+ q^{2})π^{2}ħ^{2}/(2ma^{2})]|Φ_{n}Φ_{q}>

The eigenstates of H are {|Φ_{n}Φ_{q}>} with eigenvalues (n^{2}+ q^{2})π^{2}ħ^{2}/(2ma^{2}).

The lowest energy eigenstate has n = q = 1. Therefore

E_{11}= 2E_{0}, with E_{0}= π^{2}ħ^{2}/(2ma^{2}). This energy level is not degenerate.

The second lowest energy eigenstate has n = 1, q = 2 or n = 2, q = 1.

Therefore E_{12}= E_{21}= 5E_{0}. This energy level is twofold degenerate.(b) |Ψ(0)> = (1/√6)|Φ

_{1}Φ_{1}> + (1/√3)|Φ_{1}Φ_{2}> + (1/√6)|Φ_{2}Φ_{1}> + (1/√3)|Φ_{2}Φ_{2}>.

U(t,0) = exp(-i(H(1)_{ }+ H(2))t/ħ).

|Ψ(t)> = (1/√6)exp(-iE_{11}t/ħ)|Φ_{1}Φ_{1}> + (1/√3)exp(-iE_{12}t/ħ)|Φ_{1}Φ_{2}>

+ (1/√6)exp(-iE_{21}t/ħ)|Φ_{2}Φ_{1}> + (1/√3)exp(-iE_{22}t/ħ)|Φ_{2}Φ_{2}>.

= (1/√6)exp(-i2E_{0}t/ħ)|Φ_{1}Φ_{1}> + (1/√3)exp(-i5E_{0}t/ħ)|Φ_{1}Φ_{2}>

+ (1/√6)exp(-i5E_{0}t/ħ)|Φ_{2}Φ_{1}> + (1/√3)exp(-i8E_{0}t/ħ)|Φ_{2}Φ_{2}>.

If H is measured, then the eigenvalues 2E_{0}, 5E_{0}, and 8E_{0}can be found.

P(2E_{0}) = |<Φ_{1}Φ_{1}|Ψ(t)>|^{2 }= 1/6.

P(8E_{0}) = |<Φ_{2}Φ_{2}|Ψ(t)>|^{2 }= 1/3.

P(5E_{0}) = |<Φ_{1}Φ_{2}|Ψ(t)>|^{2}+ |<Φ_{1}Φ_{2}|Ψ(t)>|^{2 }= 1/3 + 1/6 = 1/2.

If H(1) is measured, then the eigenvalues E_{n=1}= E_{0}, and E_{n=2}= 4E_{0}can be found.

P(E_{0}) = |<Φ_{1}Φ_{1}|Ψ(t)>|^{2 }+ |<Φ_{1}Φ_{2}|Ψ(t)>|^{2 }= 1/6 + 1/3 = 1/2.

P(4E_{0}) = |<Φ_{2}Φ_{1}|Ψ(t)>|^{2 }+ |<Φ_{2}Φ_{2}|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)> = (a_{1}|Φ_{1}(1)> + a_{2}|Φ_{2}(1)>) ⊗ (b_{1}|Φ_{1}(2)> + b_{2}|Φ_{2}(2)>),

or

|Ψ(0)> = a_{1}b_{1}|Φ_{1}Φ_{1}> + a_{1}b_{2}|Φ_{1}Φ_{2}> + (a_{2}b_{1}|Φ_{2}Φ_{1}> + a_{2}b_{2}|Φ_{2}Φ_{2}>.

For |Ψ(0)> to be a tensor product state we need

a_{1}b_{1}= 1/√6, a_{1}b_{2}= 1/√3, a_{2}b_{1}= 1/√6, a_{2}b_{2}= 1/√3.

Therefore a_{1}= a_{2}, b_{1}= (1/√2)b_{2}.

For a normalized solution exists if we require

|a_{1}|^{2}+ |a_{2}|^{2}= |b_{1}|^{2}+ |b_{2}|^{2}= 1.

a_{1}= a_{2}= 1/√2, b_{1}= 1/√3, b_{1}= √(2/3).

A solution exists, |Ψ(0)> is a tensor product state.

|Ψ(0)> = |ζ(1)> ⊗ |γ(2)>.

|ζ(1)> = a_{1}|Φ_{1}(1)> + a_{2}|Φ_{2}(1)> = (1/√2)|Φ_{1}(1)> + (1/√2)|Φ_{2}(1)>

<H(1)> = <ζ(1)|H(1)|ζ(1)> = |a_{1}|^{2}E_{0}+ |a_{2}|^{2}4E_{0}= (5/2)E_{0}.

|γ(2)> = b_{1}|Φ_{1}(2)> + b_{2}|Φ_{2}(2)> = (1/√3)|Φ_{1}(2)> + √(2/3)|Φ_{2}(2)>

<H(2)> = <γ(2)|H(2)|γ(2))> = |b_{1}|^{2}E_{0}+ |b_{2}|^{2}4E_{0}= 3E_{0}.

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)>.

(d) For |Ψ(0)> to be a tensor product state we need

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.

a_{1}b_{1}= 1/√5, a_{1}b_{2}= √(3/5), a_{2}b_{1}= 1/√5, a_{2}b_{2}= 0.

a_{2}b_{2}= 0 --> a_{2}= 0 or b_{2}= 0.

If a_{2}= 0 then a_{2}b_{1}= 0 and if b_{2}= 0 then a_{1}b_{2}= 0. This is inconsistent with what we need, no solution exists, |Ψ(0)> is not a product state.

<H(1)> = (1/5)E_{0}+ (3/5)E_{0}+ (1/5)4E_{0}= (8/5)E_{0}.

<H(2)> = (1/5)E_{0}+ (3/5)4E_{0}+ (1/5)E_{0}= (14/5)E_{0}.

<H(1)H(2)> = (1/5)E_{0}^{2}+ (3/5)4E_{0}^{2}+( 1/5)4E_{0}^{2}= (17/5)E_{0}^{2}≠ <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 x_{1} and x_{2}, momenta p_{1}
and p_{2}, and potential energy

U(x_{1},x_{2}) = ½mω^{2}x_{1}^{2} + ½mω^{2}x_{2}^{2}
+ gmω^{2}(x_{1} - x_{2})^{2}.

Find the eigenvalues and eigenfunctions of the Hamiltonian H of
the system.

- Solution:

We have H = ½(P_{1}^{2}+ P_{2}^{2})/m + ½mω^{2}(X_{1}^{2}+ X_{2}^{2}) + gmω^{2}(X_{1}- X_{2})^{2}.

Let

X_{G}= (1/√2)(X_{1}+ X_{2}), X_{R}= (1/√2)(X_{1}- X_{2}),

P_{G}= (1/√2)(P_{1}+ P_{2}), P_{R}= (1/√2)(P_{1}- P_{2}),

Then [X_{G},P_{G}] = iħ, [X_{R},P_{R}] = iħ, [X_{G},P_{R}] = [X_{R},P_{G}] = 0,

X_{G}^{2}+ X_{R}^{2}= X_{1}^{2}+ X_{2}^{2}, P_{G}^{2}+ P_{R}^{2}= P_{1}^{2}+ P_{2}^{2},

H = ½(P_{G}^{2}+ P_{R}^{2})/m + ½mω^{2}(X_{G}^{2}+ X_{R}^{2}) + 2gmω^{2}X_{R}^{2 }= ½P_{G}^{2}/m + ½mω^{2}X_{G}^{2}+ ½P_{R}^{2}/m + ½mω^{2}(1 + 4g)X_{R}^{2 }= 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)^{1/2}ω, 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)^{1/2}the corresponding eigenfunctions are

Ψ_{nG,nR,}(x_{G},x_{R}) = (n_{G}!n_{R}! 2^{(nG + nR)})^{-1/2 }(ββ'/π)^{1/2}H_{nG}(βx_{G})^{ }H_{nR}(β'x_{R})^{ }exp(-½(β^{2}x_{G}^{2}+ β'^{2}x_{R}^{2})).