In quantum mechanics, any density matrix (or density operator) is Hermitian. Observables are also represented by Hermitian operators. So it follows that a density matrix can also be interpreted as an observable.

What is the physical meaning of this observable?


In the simple case of a pure state, the density matrix is the projection $\rho_\psi=\lvert\psi\rangle\langle\psi\rvert$, for some $\psi\in \mathscr{H}$ (the Hilbert space). Measuring it on another general state $\rho$ (i.e. calculating $\mathrm{Tr}[\rho_\psi\rho]$) gives you the probability of $\psi$ being the "configuration" (or Hilbert space vector) corresponding to $\rho$.


Not every Hermitian operator is an observable. The identity on a space is certainly Hermitian, but it is no observable in any physical sense - its eigenvalue is $1$, and that's it. The eigenvalue of scalar operators gives you precisely zero information about the state considered, and usually, we would like an observable to at least give us some information about the state we measure.

This is a subtlety that is usually glossed over in most courses, but if you think about canonical quantization, it is quite clear that not every Hermitian operator on the space of states will be induced by an observable on the classical phase space.

The density matrix is similar - yuggib's interpretation is correct, but this is quite unlike our usual observables, where the eigenvalues of the observable would correspond to some (classical) property of the state being measured. I'm reluctant to unequivocally pronounce it an "observable" or not on these grounds.

  • $\begingroup$ It really depends on what would you call observable. I agree with you that it is debatable; however would you define an observable the spectral projection (of an observable)? In my opinion it does not differ much from a density matrix (both in meaning and mathematical form). $\endgroup$ – yuggib Jul 20 '14 at 14:22
  • $\begingroup$ @yuggib: Right. As so often, I'm partial to the nLab approach of generating the $C^*$-algebra of observables from smooth functions on the classical phase space. In this approach, it would be unequivocally clear that operators that do not come from functions on the phase space (and such an operator is the density matrix, I think), are not to be called observables. However, as you have shown, it is possible to take another, broader, stand on observables. $\endgroup$ – ACuriousMind Jul 20 '14 at 14:29
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    $\begingroup$ Why is the identity not an observable? I can even give you a "measurement device" for it: Just write an "1" on the display. And while I'm not familiar with the nLab approach, I'd be very surprised if the identity would not correspond to the constant function with value 1, which clearly is a smooth function. $\endgroup$ – celtschk Jul 20 '14 at 14:42
  • $\begingroup$ @celtschk: Alright, if you see it that way, the identity is observable - but is it really meaningful to call something an observable if its observed value does not depend on the state measured at all? And you are right, the constant function with value $1$ seems to me to produce indeed the identity operator. I've spoken too fast. It all boils down to what you consider to be "physically observable", see also this old question, where they could not reach a real consensus. $\endgroup$ – ACuriousMind Jul 20 '14 at 15:09

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