# Outer product in Hilbert space

1. Feb 13, 2008

### jdstokes

A question arose to me while reading the first chapter of Sakurai's Modern Quantum Mechanics. Given a Hilbert space, is the outer product $\mathcal{H}\times \mathcal{H}^\ast \to End(\mathcal{H}); (| \alpha\rangle,\langle \beta|)\mapsto | \alpha\rangle\langle \beta|$ a surjection? Ie, can any linear self-map of H be formed by tacking together a suitable ket and bra?

After thinking about this a bit longer I realise the answer is no. If we think about a n-dimensional Hilbert space (n < oo), then the outer product operation corresponds to matrix multiplication of a column vector with a row vector. Clearly not all n x n matrices can be formed in this way. I'm not sure quite how many matrices you can cover in this manner, however.

Last edited: Feb 13, 2008
2. Feb 13, 2008

### jdstokes

I'd like to rephrase my question, what is the outer product of $\mathcal{H}$ with $\mathcal{H}^\ast$ equal to, where $\mathcal{H}$ is any Hilbert space.

3. Feb 13, 2008

### vanesch

Staff Emeritus
The simplest thing to imagine is that you take a basis of H, call it B, and you take a basis of H', call it B'. You now consider the set product B x B' (that is, all possible couples with first element in B and second element in B'). Well, B x B' is a basis of H (x) H'.

Of course, that's not the nicest mathematical definition, because you should now show that you always span the same space, independent of the choices of B and B' and so on. But it comes closest to the PHYSICAL reason of why H (x) H' is the hilbert space of a combined system.

Indeed, a basis of the hilbert space of a system gives you normally an exhaustive list of mutually exclusive states of that system, and the whole statespace is then obtained by a superposition of all of these states.

Well, the exhaustive list of mutually exclusive states of a combined system is of course the set product of the mutually exclusive states of system 1 and the mutually exclusive states of system 2 (for each state of system 1, we can have all the states of system 2).

This then constructs the basis B x B'. And applying superposition gives you the spanned vector space, H (x) H'.

I think that the tensor product (outer product) of H with H* IS equal to End(H). Indeed, thinking of each End(H) as a matrix, you can obtain it by a superposition of matrices which are 0 everywhere except for one single element, where they are 1, and that's nothing else but the representation of (0, 0, 0,....0,1,0,....0) x (0,0,...0,1,0,....0)' (or |alpha> < beta| if you want to, where alpha and beta are basis vectors).

Last edited: Feb 13, 2008
4. Feb 13, 2008

### jdstokes

Are you sure about this? It seems rather strange to me that we can construct the entire $M_n(\mathbb{C})$ an ($n^2$-dimensional space) by tensor multiplying $\mathbb{C}^n$ and $(\mathbb{C}^n)^\ast$ (both n-dimensional).

I guess what you're doing is kind of forbidden by my construction $\mathcal{H}\times \mathcal{H}^\ast \to End(\mathcal{H}); (| \alpha\rangle,\langle \beta|)\mapsto | \alpha\rangle\langle \beta|$. I wasn't considering the possibility of adding up various outer products of vectors (only one multiplication was allowed, no addition).

5. Feb 13, 2008

### jdstokes

Another way of asking my question is:

What is the set $\{ v \cdot m \in M_n(\mathbb{C}) | v\in \mathbb{C}^{n\times 1}, w \in \mathbb{C}^{1\times n} \}$ equal to? It's certainly not $M_n(\mathbb{C}) = End(\mathbb{C}^n)$.

That notation indicates column and row vectors, respectively. The dot indicates matrix multiplication.

6. Feb 13, 2008

### vanesch

Staff Emeritus
But the tensor product of an n-dimensional space with an m-dimensional space is an nxm dimensional space ! That is exactly because not all elements of the tensor product space are of the form a (x) b. In fact, only a small subset is, and that subset is not a linear space.

7. Feb 13, 2008

### vanesch

Staff Emeritus
This set is the set of product states. It is not a linear space, as it is not closed under addition. But it spans the tensor product space. That is, the closure of this set under addition is the tensor product space.

8. Feb 13, 2008

### Ben Niehoff

It is true that all possible tensor products of the basis elements form a basis of the product space. It is not true, however, that every element of the product space can be expressed as a tensor product of two vectors from the original space. In general, elements of the product space are linear combinations of tensor products of vectors in the original space.

For example, take a 3-dimensional vector space V spanned by {i, j, k}. Clearly, VxV is spanned by the nine tensor products {ii, ij, ik, ji, jj, jk, ki, kj, kk}. But any two particular members of V have only six degrees of freedom between them: (a,b,c) and (d,e,f). Therefore, not every member of VxV can be decomposed into a tensor product between two members of V.

9. Sep 23, 2011

### karthik5

Given, the tensor product space, can one find this subset which can be written in the form a(x)b?

I think this question makes sense when the initial two spaces ( n-dimensional space and m-dimensional space) are subspace of two larger spaces. And hence, the tensor product space is also a subspace in an larger space.

for eg. The two initial spaces are 2 dimensional subspaces in 3 dimensional space. So the tensor product will span a 4d subspace in a 9 dimensional space. Given this tensor product space, is it possible to find the subset which can be written in the form a(x)b?