B How are outer products quantum states?

[Rodia]

In my textbook, quantum states are infinite dimensional vectors. But I was watching a lecture on QM and the professor referred to $|v> <u|$ as itself being a quantum state. Also I saw online people saying the same thing.

Are tensor products just things that tell you whether or not the two particles that created it are entangled? So $<p|q>$ is not technically a state, but rather it is just a matrix which tells you that particle 1 is in state p and particle 2 is in state q? But then, if you have a situation where the two particles are entangled, then how do you ever measure them?

By the way if this is the wrong forum to post in, sorry, it's hard to tell. 'Irreducible tensor products' don't appear in my text until a lot later and I don't know some of the notation. I noticed that $|v><u| w> = |v>a$ for a scalar $'a'$. Whereas $<w| v><u|= b<u|$. So that's how you can tell if there's entanglement, but then what?

If I just don't have enough knowledge for this question let me know.

 

[stevendaryl]

An expression such as $|u\rangle \langle v|$ does not mean entanglement. It's an operator (or, you can think of it as an infinite square matrix).

There is a notion of "state" that uses such objects, and that is density matrices. A density matrix is an operator of the form: $\sum_{\alpha \beta} \rho_{\alpha \beta} |\psi_\alpha\rangle \langle \psi_\beta|$, where $|\psi_\alpha$ is a complete set of orthonormal states. Sometimes people use "state" to mean the density matrix. You can think of it as a notion of "state" that incorporates both quantum mechanics and classical uncertainty.

But it doesn't have anything to do with entanglement. At least not directly.

 

[Truecrimson]

$\langle p|q \rangle$ is a number. Also $|v\rangle \langle u|$ cannot be a state (density matrix) unless $u = v$. (A density matrix must be positive and therefore Hermitian.)

So let's say that you have $|u\rangle \langle u|$ where $|u\rangle$ is normalized. What's the point of using this description? One point is that a vector $e^{i\varphi}|u\rangle$ for an arbitrary phase $\varphi$ physically describes the same state; the overall phase cannot be observed. Writing the outer product eliminates this unphysical phase: $e^{i\varphi}|u\rangle \langle u|e^{-i\varphi} = |u\rangle \langle u|$ while retaining all other information that $|u\rangle$ has.

The other point is what @stevendaryl brought up: now you can "mix" quantum states, adding them in a way that cannot be done in the vector representation. $|u\rangle \langle u| + |v\rangle \langle v|$ is very different from $|u+v \rangle \langle u+v|$! This new way of adding states represent two phenomena which turns out to be intimately related: classical uncertainty and entanglement. (Note that the formula for a density matrix in @stevendaryl's comment has off-diagonal terms, but it can always be diagonalized in some basis by the spectral theorem.)

 

[Truecrimson]

Late correction: I meant to write $(|u\rangle + |v\rangle) (\langle u| + \langle v|)$, not $|u+v\rangle \langle u+v|$ in the last reply.