# Mutli-partite states and operators

1. Jul 20, 2006

### MaverickMenzies

Hi, i've seen the following in quantum info textbooks and papers and I was just wondering if anyone knows if it has any phsical interpretation or significance?

The space of operators that act on a HIlbert space is isomorphic to the tensor product of the original Hilbert space with its dual:

H otimes H* ~ L(H)

This means that if we start with a tensor product Hilbert space then it is possible to regard the states in this space as operators acting on another space. I.e. there is a one-to-one relationship between the vectors of this tensor product space and operators acting on a seperate space.

Ok, the axioms of QM require that the tensor product be used when the system has more than one degree of freedom. Does this then mean that perhaps composite quantum systems in certain states can be imagined as acting as observables on other quantum systems?

What do you guys think? Am I reading too much into it?

2. Jul 20, 2006

### Rach3

You're reading too much into it. Operators are uniquely defined by their matrix representation in a basis - which looks like

\begin{align*}\hat{A} &= \sum_{a} \sum_{a'} | a \rangle \langle a | \hat{A} | a' \rangle \langle a' | \\ &= \sum_{a} \sum_{a'} \left( \langle a | \hat{A} | a' \rangle \right) | a \rangle \langle a' | \end{align}

with matrix elements $$\left( \langle a | \hat{A} | a' \rangle \right)$$. So an arbitary operator is a linear combination of operators of the form $$| a \rangle \langle a' |$$ - i.e. tensor products of basis vectors and dual basis vectors - which themselves form a basis for the space of operators.

To check your intuition, note that in finite spaces, the dimension of the space of NxN matrices (operators) is N^2, the square of the dimension N of the vector (Hilbert) space.

Last edited: Jul 20, 2006