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In the phase-space formulation of QM over continuous variables, how can I determine the probability of obtaining a particular measurement outcome $m$ in the following setting.

Given a quantum state $\rho$ described by the Wigner function $W_{\rho}(q,p)$, one measures the observable $\hat{p}$. What is the probability of obtaining a particular outcome $m\in\mathbb{R}$? I know that in standard QM I need to calculate $$\text{Tr }(\rho \; |m\rangle _p\langle m |)$$ and I have found that such expectation values, in phase space formulation, are usually computed via $$\text{Tr }(\rho A)=\int dq dp W_{\rho}(q,p)A(q,p)$$ where $\hat{A}$ is some observable expressed as a combintation of $\hat{q}$ and $\hat{p}$.

Once this easy quadrature measurement is clarified, I would also like to understand partial measurements, i.e. let $W(q_1,p_1,q_2,p_2)$ be the input state, then what is probability of outcome $m$ when measuring $\hat{p}$ only on the first mode.

Note, I am new to the phase-space formulation. Thus, I would also appreciate a good reference where I could simply read up on this.

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The Wigner map image of your momentum projection operator $|m\rangle \langle m|$ is $$ A(q,p)= 2 \int dy ~ e^{2ipy/\hbar} \langle q+y| m\rangle \langle m|q-y\rangle = \frac {2} { 2\pi} \int dy ~ e^{2ipy/\hbar} e^{-2imy/\hbar} = \delta (p-m), $$ by virtue of $\langle x|p\rangle= \exp (ixp/\hbar)~/\sqrt{2\pi}$, etc.

Consequently, the expectation you presumably arrived at in your intro is $$ \int\! dq ~ W_\rho (q,m). $$

Mutatis mutandis, collapsing the momentum of $\hat p _1$ to m, yields $$\int \! dq_1 dq_2 dp_2 ~ W(q_1,m,q_2,p_2).$$

Here is a concise review with exercises to practice on.

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  • $\begingroup$ Great thanks. I was lacking the Wigner map ingredient. Now, things make a lot more sense. Ultimately, I would like to understand quantum teleportation in this setting. There the question came up, whether the probability of obtaining a particular outcome when working with ideal (inifinitely squeezed states) entangled resources, is actually well defined. It seems like it is uniform over the reals which makes it non-normalizable and thus not a probability distribution. Do you happen to know if this is correct? Thanks for the review, I will check it out and do some calculations myself. $\endgroup$ – Marsl Nov 23 '19 at 13:58
  • $\begingroup$ I know there is a cottage industry in this. Groenewold concludes his seminal thesis paper of 1946 with such, and Bell has written an entire paper on it. But I am not familiar with the rich bibliography on the subject. Yes, sharp eigenstates of momentum or position amount to unnormalized delta function Wigner Functions. $\endgroup$ – Cosmas Zachos Nov 23 '19 at 14:49

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