A coordinate representation of density matrix $\rho$ is defined as

$$ \rho (x, x') \equiv \left<x\right| \rho \left|x'\right> .$$

When $x = x'$, this expresses a probability where a particle is in the state $\left|x\right>$.

Question: what does that mean when $x \neq x'$? Is that related to some probability?

According to Feynman (Statistical Mechanics a set of lectures, (p.72)),

$$ I \equiv \int dx _1 \int dx _2 \cdots \int dx _{n - 1} \left<x\right| \rho \left|x_1\right> \left<x_1\right| \rho \left|x_2\right> \cdots \left<x_{n-1}\right| \rho \left|x'\right>$$

can be interpreted as that "the particle travels from $x'$ to $x$ through a series of intermediate steps, $x_1, x_2, \cdots, x_{n-1}$, which define a path". I don't understand this statement.

  • 1
    $\begingroup$ Could you provide a reference with that specific form of the path integral, involving $\rho$? The interpretation about the particle travelling from $x$ to $x'$ is correct, but I have never seen it with $\rho$ before... $\endgroup$ Commented Jul 7, 2019 at 10:13
  • $\begingroup$ @SuperCiocia I added the book name in the post. $\endgroup$
    – ynn
    Commented Jul 7, 2019 at 10:18
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    $\begingroup$ On page 72 I see the usual definition of the path integral with Hamitlonians, not your expression with just the density matrix. The idea of the path integral is to start with an initial state $|x\rangle$, "evolve" it $e^{-iHt}|x\rangle$, and then find the overlap with a different position $\langle x'|e^{-iHt}|x\rangle$ to quantify the probability it has to move from $x$ to $x'$. But you need something to let it "evolve". In your cane $\langle x|\rho|\rangle = \langle x|\Psi\rangle \langle \Psi |x\rangle$ just gives the combined probability of being at $x$ and at $x'$. $\endgroup$ Commented Jul 7, 2019 at 10:35
  • $\begingroup$ @SuperCiocia So you thankfully checked the book? How kind! Your explanation of the path integral seemed clear and straightforward to me. I now understand what the author wanted to say. Thank you. $\endgroup$
    – ynn
    Commented Jul 7, 2019 at 11:00

1 Answer 1


The diagonal entries of the density matrix are called populations and provide information about the probability density of the particles (described by the density matrix), i.e. their probability of "being found" in real space.

This is easily seen from a density matrix $\rho = |\Psi\rangle \langle \Psi|$, and $$\rho(x,x) = \langle x|\rho|x\rangle = \langle x|\Psi\rangle \langle \Psi|x\rangle = | \langle x |\Psi\rangle|^2 = |\psi(x)|^2,$$ which is the usual quantum mechanical probaibility density.

The off diagonal entries of the density matrix are called coherences and provide information about the phase coherence of the system described by $\rho$ between two positions $x$ and $x'$. Is there a fixed phase relationship between $x$ and $x'$, especially as $|x-x'|\rightarrow \infty$? I.e. will constructive & coherent interference occur over a large distance or will it be washed out?

The most famous application of the off diagonal elements of the density matrix is in Off-Diagonal Long-Range Order (ODLRO), which is what manifests in Bose-Einstein Condensates or Superfluids. These are phases where the system breaks a $U(1)$ phase and the wavefunction "picks" a specific phase $\theta$.

The "broken" phase is distinguished from the unbroken phase because it obeys: $$ \lim_{|x-x'|\rightarrow \infty} \rho(x,x') \rightarrow n_0 \neq 0,$$ i.e. phase coherence is preserved over arbitrarily long distances.


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