What I would like to understand mathematically is the following situation:
- Prepare a quantum state that spans two Hilbert spaces
- Operate on one space with observable operator $\hat{O}$.
- Obtain measurement statistics (probabilities) of the second state after first operation.
I think the procedure would be the following:
Suppose I have a density matrix of a mixed state over two Hilbert spaces:
$$\rho = \sum_{j,k} P_{jk} |\psi_j\rangle \langle \psi_j | \otimes |\phi_k\rangle \langle \phi_k |$$
with $P_{jk}$ being the respective probabilities. Let us suppose now I choose to operate on the first Hilbert space with the observable operator $\hat{O}$. This gives:
$$\rho\hat{O} = \sum_{j,k} P_{jk} |\psi_j\rangle \langle \psi_j |\hat{O} \otimes |\phi_k\rangle \langle \phi_k |$$
I believe that taking the partial trace of this over the first space results in the post-measurement state of space 1, but pre-measurement state of space 2.
$$\rho_2 = \text{Tr}_1(\rho) = \sum_{jk} P_{jk}\langle \psi_j |\hat{O}|\psi_j\rangle |\phi_k\rangle \langle \phi_k |$$
My questions are as follows:
- Due to the operation of $\hat{O}$, this object does not appear to represent a density matrix anymore, as the probabilities are being multiplied by the expectation value of $\hat{O}$ and certainly will not normalize to 1. So what does this represent?
- Is there a way to obtain the measurement probabilities of space 2 after measurement of $\hat{O}$ space 1 in this manner? Or is there a better way to do this?
I hope this makes sense. Thanks!
EDIT (Additional Information):
Based on the comments, I was doing some reading on quantum measurement and found this wiki article, among other resources. I still don't quite understand so let's make my question less general. Let's say my density matrix is written in terms of the Fock state basis:
$$\rho = \sum_{n,m} P_{nm} |n\rangle \langle n | \otimes |m\rangle \langle m |$$
Just to make things simple, lets also suppose the measurement I require is a number operator measurement, such that the eigen vectors will be Fock basis vectors. Let's now say I measure a particle number $p$ from space 1. The projection operator for this measurement would be $\hat{M}=|p\rangle\langle p|$. So from the wiki article, the post measurement state would be:
$$\rho_2 = \frac{\hat{M}\rho \hat{M}^{\dagger}}{\text{tr}(\hat{M}\rho \hat{M}^{\dagger})}$$
This gives (for the numerator):
$$ \sum_m P_{pm}|p\rangle \langle p | \otimes |m\rangle \langle m|$$
I am confused at this point. For the denominator, the equation indicates I trace over both Hilbert spaces? So using the vector $|p,q\rangle$ for the trace calculation, this would give me $P_{pq}$, so the post measurement state would be:
$$ \rho_2 = \frac{1}{P_{pq}}\sum_m P_{pm}|p\rangle \langle p | \otimes |m\rangle \langle m|$$
Am I on the right track here? I am not confident about these calculations…
Best Answer
With respect to the pre-edit question: Note that operation by the operator representing a physical observable does \emph{not} correspond to measurement of that observable, even in the case of pure states. To get the post-measurement reduced density matrix of system 2, you need to act with a \emph{projection operator} that projects onto an eigen-subspace of the operator $\hat{0}$. (More generally, a quantum measurement is not necessarily projective, but see below.) Then, you can normalize the resulting density matrix.
With respect to the post-edit question:
The denominator in the expression $$ \rho_2 = \frac{\hat{M}\rho \hat{M}^{\dagger}}{\operatorname{Tr}(\hat{M}\rho \hat{M}^{\dagger})} $$ is just the quantity needed to normalize the new state, since the process of projecting onto the eigensubspace of the measured operator renders the density matrix non-normalized. (This is also true in the more general case where the measurement is not projective.) To compute this quantity, we do \begin{align*} \operatorname{Tr}(\hat{M}\rho \hat{M}^{\dagger}) &= \operatorname{Tr}\left(\sum_m P_{pm}|p\rangle \langle p | \otimes |m\rangle \langle m|\right) \\&= \sum_{jk}(\langle j|\otimes\langle k|) \left(\sum_{m} P_{nm} |n\rangle \langle n | \otimes |m\rangle \langle m |\right) (|j\rangle\otimes|k\rangle) \\&= \sum_{jk} \sum_{m} P_{nm} \langle j|n\rangle \langle n | j\rangle \langle k|m\rangle \langle m | k \rangle \\&= \sum_{m} \sum_{jk} P_{nm} \delta_{jn}\delta_{mk} \\&= \sum_{m} P_{nm}\,. \end{align*} Since this is the trace of the density matrix arrived at after applying the measurement operator, dividing by this quantity makes the resulting density matrix have trace 1, and hence the total probability is 1.