How do outer products differ from tensor products?

Question

From my naive understanding, an outer product appears to be a particular case of a tensor product, but applied to the "simple" case of tensor products of vectors. However, in quantum mechanics there appears to be a distinction between the two. For example, given two state vectors ("kets"), $\lvert\psi\rangle$ and $\lvert\phi\rangle$, an outer product of the two is given by $$\lvert\psi\rangle\langle\phi\rvert$$ and this corresponds to a representation of some operator acting on vectors in this Hilbert space.

However, if we consider the two kets, $\lvert\psi\rangle$ and $\lvert\phi\rangle$ as representing two single-particle states, then the tensor product of their two single-particle Hilbert spaces gives a two-particle state of the form $$\lvert\psi\rangle\otimes\lvert\phi\rangle\equiv\lvert\psi\rangle\lvert\phi\rangle$$ which gives a new two-particle state in the tensor-product Hilbert space (of the two single-particle Hilbert spaces).

Is the distinction that the outer product maps two vectors in a given Hilbert space to an operator acting on that same Hilbert space, whereas, the tensor product maps two vectors from two different Hilbert spaces to a vector in the (tensor) product Hilbert space?

Answer 1

It's basically the same, it's mainly a matter of implicit identifications.

Namely, in a Hilbert space $H$ there is a natural isomorphism $H\to H^*$ (which is the bra-ket duality), so $H\otimes H\simeq H\otimes H^*$.

Now there is a map $H\otimes H^*\to L(H)$ which corresponds to what you call the outer product.

So for two kets $|\psi\rangle$ and $|\phi\rangle$ in $H$, you can either see $|\psi\rangle\otimes |\phi\rangle$ as an element of $H\otimes H$ and interpret that as a kind of two-particles state, or you can look at its canonical image in $H\otimes H^*$, which the physicists note $|\psi\rangle \langle\phi|$ (which makes sense since $\langle\phi|$ is the element of $H^*$ corresponding to $|\phi\rangle$), and then interpret it as an operator in $L(H)$.