# Trying to understand post-measurement density matrices in a state that spans 2 Hilbert spaces

What I would like to understand mathematically is the following situation:

1. Prepare a quantum state that spans two Hilbert spaces
2. Operate on one space with observable operator $$\hat{O}$$.
3. 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:

1. 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?
2. 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!

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...

• What do you mean with Prepare a quantum state that spans two Hilbert spaces exactly? Do you mean that you have a bipartite Hilbert space, e.g. something like $\mathscr H = \mathscr H_1 \otimes \mathscr H_2$? What do you mean with your second question? Could you elaborate? Dec 8, 2021 at 21:49
• Yes bipartite, apologies for my confusing choice of words. For my 2nd question, I essentially want to know how to obtain the probabilities of the 2nd measurement after the first. In other words, I want to investigate how the measurement of the particle in space 1 affects the subsequent measurements of the particle in space 2. I want to calculate a new $P_{k}$ if that makes sense. Dec 8, 2021 at 22:00
• Note that operation by the operator representing a physical observable does 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 projection operator that projects onto an eigen-subspace of the operator $\hat{O}$. Then, you can normalize the resulting density matrix and trace over system 1 to get a density matrix for system 2. Dec 9, 2021 at 4:05
• So I'm not sure what you're doing. In your post it's not quite clear. Are you measuring $\hat{O}$, or are you just making a new operator by multiplying $\rho$ and $\hat{O}$? If it's the first, then see my previous comment. But it it's the second, it's not clear what that means physically, because the operation of an observable on a state rarely has a direct physical meaning (and it certainly doesn't correspond to measurement of that observable). Dec 9, 2021 at 4:07
• @march Thanks so much for your comments. It has cleared some things up for me, but I am still confused. I have added more information in my original question to see if I am on the right track. Dec 9, 2021 at 20:43

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.
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.