# Outer product in Hilbert space

• jdstokes
In summary, the conversation discusses the properties of the outer product in a Hilbert space and whether it is a surjection. It is determined that not all linear self-maps of H can be formed by tacking together a suitable ket and bra. The conversation also delves into the definition and construction of the tensor product space and its relationship with the outer product. It is concluded that while all possible tensor products of the basis elements form a basis of the product space, not every element of the product space can be expressed as a tensor product of two vectors from the original space.
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 realize 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.

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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.

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'.

EDIT: oops, I re-read your message, and what I wrote is not the answer to that...
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).

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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).

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.

jdstokes said:
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).

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.

jdstokes said:
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.

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.

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.

vanesch said:
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.

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?

## 1. What is the outer product in Hilbert space?

The outer product in Hilbert space is a mathematical operation that takes two vectors and produces a matrix. It is also known as the tensor product or Kronecker product, and is denoted by the symbol ⊗. The resulting matrix has elements that are the products of each element in the first vector with each element in the second vector.

## 2. How is the outer product different from the inner product in Hilbert space?

The outer product and the inner product are two different mathematical operations in Hilbert space. The outer product takes two vectors and produces a matrix, while the inner product takes two vectors and produces a scalar value. Another key difference is that the outer product is commutative, while the inner product is not.

## 3. What is the significance of the outer product in Hilbert space?

The outer product is significant in Hilbert space because it allows for the creation of new vectors and operators. It is also used in various applications such as quantum mechanics, signal processing, and machine learning. Additionally, the outer product can be used to decompose a matrix into its constituent vectors.

## 4. How is the outer product related to the concept of basis in Hilbert space?

The outer product is closely related to the concept of basis in Hilbert space. In fact, the outer product of two basis vectors will produce a matrix that represents the basis transformation between the two vector spaces. Additionally, the outer product can be used to construct a new basis from a given set of vectors.

## 5. Can the outer product be used to represent higher-order tensors in Hilbert space?

Yes, the outer product can be used to represent higher-order tensors in Hilbert space. By taking the outer product of multiple vectors, a higher-order tensor can be constructed. For example, the outer product of three vectors in a three-dimensional Hilbert space will produce a three-dimensional tensor. This allows for the representation of more complex and multidimensional data in Hilbert space.

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