Density operator for one part of a two-party state

[zoemoe]

Homework Statement

Let [itex]|\psi\rangle_{AB}=\sum_{i}\sum_{j}c_{ij}|\varphi_{i}\rangle_{A}\otimes|\psi_{j}\rangle_{B}[/itex] be a normalized two party state, being [itex]\{|{\varphi_{i}}\rangle_{A}\}[/itex] and [itex]\{|{\psi_{j}}\rangle_{B}\}[/itex] basis of H[itex]_{A}[/itex] and H[itex]_{B}[/itex] respectively, with dimensions N and M. Find [itex]\rho_{A}[/itex] and prove that it's positive and its trace is 1.

Homework Equations

Given in the attempt at a solution

The Attempt at a Solution

I first tried to find what was the state corresponding only to A, since I'm given a two party state but the density operator that I have to find relates only to A.
I assumed that since A only "knows" about its part of the state, but not about the B part, then the state from A's point of view should be a mix of the part that A knows about over all of the possibilities for the B part (which A doesn't know), but I wasn't sure about how to express that.
After some searching, I found this expression for the A part of the state:

[itex]|\psi\rangle_{A}=\sum_{j}\sum_{i}|c_{ij}|^{2}\frac{\sum_{i}c_{ij}|\varphi_{i}\rangle_{A}}{\sqrt{\sum_{i}|c_{ij}|^{2}}}[/itex]

And tried to see if it made sense with my "definition" of the state corresponding to A:

*[itex]|c_{ij}|^{2}[/itex] is the probability of the two party state being in [itex]|\varphi_{i}\rangle_{A}\otimes|\psi_{j}\rangle_{B}[/itex], and we sum that over all possible i's (meaing over all of the possibilities for the A part) so that would be the probability of finding [itex]|{\psi_{j}}\rangle_{B}[/itex], and since the very first sum, which affects all the following terms, is over all j's, this matches what I said about a mix over all of the possibilities for B.

*[itex]c_{ij}|{\varphi_{i}}\rangle_{A}[/itex] is the part of the state that A "knows" about; we sum that over all possibilities for A, so up until now I think this adds up to what I thought the state should look like.

*Finally, we have the [itex]\sqrt{\sum_{i}|c_{ij}|^{2}}[/itex] in the denominator that I hadn't accounted for, which I guess is a normalization constant but I'm not sure.

I would like to know if this last term is in fact a normalization constant or something different, and also if my reasoning is correct and, if it isn't, I'd really appreciate if somebody would point out where or why it fails.

Once I know the expression for [itex]|{\psi}\rangle_{A}[/itex] I think I can manage the rest just fine.

Thanks a lot for helping me out.

 

[mark726]

Thank you for your post. Your reasoning seems to be mostly correct, but there are a few things that could use some clarification.

Firstly, the term \sqrt{\sum_{i}|c_{ij}|^{2}} in the denominator is indeed a normalization constant. This is because we want the state |\psi\rangle_{A} to be a valid state, meaning it must have a norm of 1. In other words, the sum of all the probabilities (|c_{ij}|^{2}) must equal 1, so we divide by this normalization constant to ensure that is the case.

Secondly, your reasoning about the state corresponding to A seems to be on the right track. However, I would like to point out that the state |\psi\rangle_{A} is actually a mixed state, as it is a sum of different states (|{\varphi_{i}}\rangle_{A}) with different probabilities (|c_{ij}|^{2}). This means that A does not have knowledge of the exact state, but rather a probabilistic distribution of possible states.

Lastly, to prove that \rho_{A} is positive and has a trace of 1, we can use the definition of the density operator:

\rho_{A} = \sum_{i}\sum_{j}|c_{ij}|^{2}|\varphi_{i}\rangle_{A}\langle\varphi_{i}|_{A}

We can then show that \rho_{A} is positive by taking the inner product with an arbitrary state |\alpha\rangle_{A}:

\langle\alpha|\rho_{A}|\alpha\rangle = \sum_{i}\sum_{j}|c_{ij}|^{2}\langle\alpha|\varphi_{i}\rangle_{A}\langle\varphi_{i}|\alpha\rangle_{A}

Since the inner product is always positive, and the probabilities |c_{ij}|^{2} are also positive, we can conclude that \rho_{A} is positive.

To show that the trace of \rho_{A} is 1, we can simply take the trace of \rho_{A}:

Tr(\rho_{A}) = \sum_{i}\sum_{j}|c_{ij}|^{2}\langle\varphi_{i}|\varphi_{i}\rangle_{A} = \sum_{i}\sum_{j}|c_{