Tensor product space

Assume you have a system consisting of two spinless particles.  Its wave function is y(r1,r2). A special case of such a wave function is a product function of the form y(r1,r2)=f(r1)c(r2) . The state vector corresponding to such a product is denoted by

 .

|y > is a vector in the two-particle space E, while |f(1)> and |c(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=E1ÄE2 respectively.  If {|ui>} and {|vj>} are bases for E1 and E2 respectively, then the set of all tensor product vectors {|ui>Ä|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 y(r1,r2) is a product function.

The inner product in E satisfies

.

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=EspaceÄEspin .
The space of a single particle moving in three dimensions is E=ExÄEyÄEz . {|x>}, {|y>}, and {|z>} are bases for Ex , Ey , and Ez respectively. |r>=|x>Ä|y>Ä|z>=|x,y,z> is a basis for E.  We have y(r)=y(x,y,z)=<x,y,z|y>,

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The x, y and z dependence cannot, in general, be factored.

We can regard E=E1ÄE2 as a factoring of E , and it is an important fact that a given space can be factored in many different ways.

Example:

We may rewrite y(r1,r2) as a function of

.

y(rcm,r) is a vector in the vector space E= Er1ÄEr2=EcmÄEr spanned by product functions f(r1)c(r2) or f(rcm)c(r).

Certain operators in are product operators.  If A(1) and B(2) are operators acting in E1 and E2 respectively, we define the product operator A(1)ÄB(2) through the relation

.

If there can be no confusion we write .

Examples:

The momentum operator for particle 1 is P1(in E1) Ä I(in E2).
The momentum operator for particle 2 is I(in E1) Ä P2(in E2)
The total momentum operator is P1+P2 = P1(in E1) Ä I(in E2) + I(in E1) Ä P2(in E2).
The two particle Hamiltonian is

(in E1) Ä I(in E2)

+ I(in E1) Ä (in E2) + V(r1,r2) (in E1 Ä E2).

Note: V(r1,r2) is not a product operator.

However, if we write

(in Ecm) Ä I(in Er)

+ I(in Ecm) Ä [ + V(r)] (in Er ),

with  , then H is a sum of product operators.

The evolution operator may then be written as
.

For further discussion of the tensor product of state see Cohen-Tannoudji, chapter II F and Complement D III.

Problem:

Consider a system of two 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 and be an eigenbasis of H(1) and H(2) respectively.  A basis for the global system is .

(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 .

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 H1 is measured at time t, what results can be found and with what probabilities?

(c) Show that 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 .  Show that this is not a tensor product state.  Calculate at t=0 <H1>, <H2>, and <H1H2>. Compare <H1><H2> with <H1H2>.

Solution:

(a) and are non-degenerate eigenbases of H(1) and H(2) respectively. 

The eigenvalues are .

.

The eigenstates of H are with eigenvalues .

The lowest energy eigenstate has n=q=1. Therefore
. The second lowest energy eigenstate has n=1, q=2 or n=2, q=1. Therefore  . This energy level is twofold degenerate.

(b) .

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(c) is a tensor product state.

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

.

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

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The system is in a product state.  Measuring H(1) does not disturb f(2) and vice versa.

The results depend on .  The time dependence is in the phases and cancels out.

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

.

.  If a2=0 then a2b1=0 and if b2=0 then a1b2=0. This is inconsistent with what we need, no solution exists, |y(0)> is not a product state.

.

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|y(0)> is a linear combination of product states.  H(1) acting on |c(1)> does not disturb |f(2)> in each of the terms, but it changes the relative weight of the terms.