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(a) Any textbook gives the interpretation of the density matrix in a single continuous basis $|\alpha\rangle$:

  • The diagonal elements $\rho(\alpha, \alpha) = \langle \alpha |\hat{\rho}| \alpha \rangle$ give the populations.

  • The off-diagonal elements $\rho(\alpha, \alpha') = \langle \alpha |\hat{\rho}| \alpha' \rangle$ give the coherences.

(b) But what is the physical interpretation (if any) of the density matrix $\rho(\alpha, \beta) = \langle \alpha |\hat{\rho}| \beta \rangle$ for a double continuous basis $|\alpha\rangle$, $|\beta\rangle$?

I know that when the double basis are position and momentum then $\rho(p, x)$ is interpreted as a pseudo-probability. I may confess that I have never completely understood the concept of pseudo-probability [*], but I would like to know if this physical interpretation as pseudo-probability can be extended to arbitrary continuous basis $|\alpha\rangle$, $|\beta\rangle$ for non-commuting operators $\hat{\alpha}$, $\hat{\beta}$ and as probability for commuting ones.

[*] Specially because $\rho(p, x)$ is bounded and cannot be 'spike'.

EDIT: To avoid further misunderstandings I am adding some background. Quantum averages can be obtained in a continuous basis $| \alpha \rangle$ as

$$\langle A \rangle = \int \mathrm{d} \alpha \; \langle \alpha | \hat{\rho} \hat{A} | \alpha \rangle$$

(a) Introducing closure in the same basis $| \alpha \rangle$

$$\langle A \rangle = \int \mathrm{d} \alpha \int \mathrm{d} \alpha' \; \langle \alpha | \hat{\rho} | \alpha' \rangle \langle \alpha' | \hat{A} | \alpha \rangle = \int \mathrm{d} \alpha \int \mathrm{d} \alpha' \; \rho(\alpha,\alpha') A(\alpha',\alpha)$$

with the usual physical interpretation for the density matrix $\rho(\alpha,\alpha')$ as discussed above.

(b) Introducing closure in a second basis $| \beta \rangle$, we obtain the alternative representation

$$\langle A \rangle = \int \mathrm{d} \alpha \int \mathrm{d} \beta \; \langle \alpha | \hat{\rho} | \beta \rangle \langle \beta | \hat{A} | \alpha \rangle = \int \mathrm{d} \alpha \int \mathrm{d} \beta \; \rho(\alpha,\beta) A(\beta,\alpha)$$

When the two basis are momentum $| p \rangle$ and position $| x \rangle$ the density $\rho(p,x)$ is the well-known Wigner function whose physical interpretation is that of a pseudo-probability. My question is about the physical interpretation of $\rho(\alpha,\beta)$ in two arbitrary basis $| \alpha \rangle$, $ | \beta \rangle$.

glS
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juanrga
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  • One interpretation is that they are complex probabilities, see http://iopscience.iop.org/1367-2630/14/4/043031/pdf/1367-2630_14_4_043031.pdf. – Piotr Migdal Nov 07 '13 at 15:55
  • @PiotrMigdal Excellent article! The idea of a joint probability makes sense. I will study it more. – juanrga Dec 14 '13 at 13:57

3 Answers3

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Probabilities have a physical meaning only in a context where measureemnt is possible. Between states of a pointer basis in a measurement context, the matrix elements of a density matrix have the standard probabilistic meaning.

In any other basis, they are just mathematical expressions intermediate to other calculations of interest. (I wouldn't give a penny for attempts to interpret these in terms of nonphysical pseudo-probabilities.)

Arnold Neumaier
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  • Thank you! The part about measurements seems very related to my remark about the spikeness of the Wigner function, but my aim was not to discuss here the concept of pseudo-probability but to know if it can be extended beyond momentum-position basis. Moreover your answer seems to avoid the cases when the operators commute. My belief is that those would be true probabilities instead of pseudo. Am I mistaken? – juanrga Nov 19 '12 at 19:06
  • @juanrga: Yes, but you called them $x$ and $p$, which don't commute. - You had asked about a pseudo-probability interpretation. This is somewhat meaningful in the Wigner case because one can take the calssical limit where phase space probabilities result. But for general bases, one has no such limiting interpretation. – Arnold Neumaier Nov 19 '12 at 19:10
  • Well I referred to "arbitrary continuous basis $\alpha$ and $\beta$" and said I was interested in both cases when the operators commute and when do not. I asked about a pseudo-probability interpretation for non-commuting operators and about the interpretation "as probability for commuting ones". I used $x$ and $p$ only as illustrative example. I do not follow your argument about the classical limit (Wigner function does not reduce to classical phase space probabilities), but thank you. – juanrga Nov 19 '12 at 20:34
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Your statement on a pseudo-probability interpretation is slightly off, in that you most certainly did not write the Wigner quasi-probability distribution in your edit.

Taking your two bases to be $x$ and $p$ for specificity, what you wrote, $ρ(x,p)$, up to a phase: $e^{(ixp/ħ)}$, is the "standard ordering prescription" for quasi-distribution functions, often misnamed the “Mehta prescription”, introduced by Terletski in 1937, and Blokhintsev in 1940, cf. Exercise 0.19 of Ref 1 available online here.

In fact, your expectation value in b) is precisely Blokhintsev's eqn (11') via his (5), or Terletsky's eqn (18)!

Your $ρ(x,p)$ is, nevertheless, connected to the Wigner quasi-probability distribution through the kernel provided in that exercise, --and in Moyal's original paper where it was discovered in the 40s.

You are quite right that the Wigner distribution, and hence the standard one, are bounded, so they preclude full localization and violate the Kolmogorov axiom about distinct disjoint alternatives for different points in phase space, its arguments, and so their interpretation is somewhat metaphorical, as anyone familiar with phase-space quantization keeps in mind, day and night and during the crepuscular hours of complexity.

Your quest for a physical interpretation is a bit open-ended, though: People in quantum optics routinely write formally similar expressions with creation ad annihilation operators, or Bargmann-Segal space integrals and they emphasize the obvious analogy with phase space you wrote. Why don't you just follow the formulas and get correct results instead of stretching fanciful and dangerous analogies? Blokhintsev paid dearly for trying to do that, indeed...

Ref 1: Thomas L. Curtright, David B. Fairlie, & Cosmas K. Zachos, A Concise Treatise on Quantum Mechanics in Phase Space, World Scientific, 2014,

Community
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Cosmas Zachos
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How is $\langle \alpha | \hat \rho | \beta \rangle$ different from $\langle \alpha | \hat \rho | \alpha^\prime \rangle$? Both representations are basis-independent, that is, you can choose any basis of your choice (position, momentum, you-name-it).

If your question is referring to the fact that it is sometimes useful to use two indices rather than one to enumerate the states (such as spin and momentum), realising that you can easily combine them into one index (which is then possibly multi-dimensional) and that you can similarly split any single basis index in multiple indices should resolve your problem.

Claudius
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  • $|\alpha \rangle$ and $|\alpha' \rangle$ are two elements of the same basis. $|\alpha \rangle$ and $|\beta \rangle$ are two elements of two different basis. – juanrga Nov 17 '12 at 12:04
  • Isn’t the Dirac representation meant to be basis-independent? Not to mention that having two different bases in the (matrix multiplication) expression doesn’t make much sense either to me. – Claudius Nov 17 '12 at 13:53
  • (i) Basis-independence does not mean that the physical interpretation remains unchanged in any representation and I am precisely asking about the physical interpretation of the density matrix $\rho(\alpha, \beta)$. (ii) It must not make much sense for you, but as stated in the original question $\rho(p, x)$ is interpreted as a pseudo-probability in the textbooks. – juanrga Nov 17 '12 at 17:21