(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Given in the attempt at a solution

3. 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.

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# Homework Help: Density operator for one part of a two-party state

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