Given $N$ distinguishable quantum particles in the canonical ensemble, we can estimate the probability of finding one of those, labelled by $j$, in a certain position $x\in\mathbb R$ by computing \begin{equation}\label{1} \frac 1{Z_\beta}\mathrm{tr}\{e^{-\beta\hat H} \hat \delta(\hat x_j-x\hat I)\}\equiv f_j(x) \end{equation} where $Z_\beta=\mathrm{tr}\{e^{-\beta\hat H}\}$. I would assume that \begin{equation} f_j(x) = \int_{\mathbb R^{N-1}}\prod_{\substack{i=1\\i\neq j}}^N\mathrm d x_i\;\lvert \psi(x_1,\cdots x_j\equiv x,\cdots, x_N)\rvert^2 \end{equation} where $\psi(x_1,\cdots, x_N)$ is the stationary solution of the Shroedinger equation (the system is in canonical equilibrium). Is the statement correct? If so, how can I formally prove it?


This is not true (except for at $T=0$). The first indication of the problem is that your first expression depends on temperature, but temperature does not appear in your second expression. More formally, for your second expresstion to make sense your system must have a wavefunction, however in order to write down a wavefunction the system must be in a pure state.

In your first expression the system is in a themal state, with density matrix $\rho = \frac{e^{-\beta H}}{Z_\beta}$, which is not a pure state unless you are at $T=0$

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  • $\begingroup$ I see the point, thanks. I had indeed trouble in finding a correspondence between trajectories averages and phase averages in quantum mechanics. Could we however say that distribution from the trace of the density operator is the ''closest'' analogue to the marginal of $|\psi|^2$ for an ensemble of states? $\endgroup$ – Graz Sep 18 '19 at 12:12
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    $\begingroup$ What do you mean by closest here? I might say that it was the natural generalisation. You could write the expression in an analogous way by writing the density matrix as $\rho(x_1,x_2...x_n,x_1',...x_n')$, fixing your chosen coordinate (in both the row and column indices) and taking the partial trace over the remaining coodinates. $\endgroup$ – By Symmetry Sep 18 '19 at 12:28
  • $\begingroup$ I indeed meant the natural generalization. And I see the connection with the density matrix, thanks. $\endgroup$ – Graz Sep 18 '19 at 12:52

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