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$xp$ is not a hermitian operator and hence doesn't represent an observable. Then, how can we interpret

$$ \langle x p \rangle \text{,} $$

i.e. the expected value of position times momentum?

For some operator $p$ that is hermitian, we can interpret its expected value e.g. as imagining us estimating it from measurements on some number of identically prepared systems.

$\langle x p \rangle$, on the other hand, would denote the expected value of an operator that doesn't represent an observable. An imagined experiment similar to that for $p$ then doesn't seem to help us.

Yet, if we know $\Psi$ we can calculate $\langle x p \rangle$. My question is: how can we interpret the value resulting from such a calculation?

(The question arises from a line in my textbook - "in a stationary state, $\frac{d}{dt} \langle x p \rangle = 0$, ..." without further explanation - why is this obvious?)

Carl
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  • "why is this obvious?" Hint: a stationary state has a simple time dependence: $\Psi(t)\rangle = e^{-iE_{\Psi}t/\hbar}|\Psi(0)\rangle$. – Alfred Centauri Jul 16 '12 at 17:07
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    The expectation value of $xp$ is $i\hbar/2$ plus a real number because $xp-i\hbar/2 = (xp+px)/2$ is Hermitian. It's hard to construct a "device" that measures $xp$ or equivalently the Hermitian $xp-i\hbar/2$ but such "devices" exist in principle. What's the problem here? What do you exactly mean by an interpretation? An observable is an observable that may be in principle measured and its expectation value is the average of many measurements in the same state. And no, the measurement of $xp$ can't be reduced to a measurement of $x$ and $p$ separately - the latter two don't commute. – Luboš Motl Jul 16 '12 at 17:22
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    But the fact that $xp$ can't be measured by measuring $x$ and $p$ isn't anything special about $xp$. $J_z$ may also be written as an "operator function" of $J_x$ and $J_y$. In particular, $J_z=(J_x J_y - J_y J_x)/i\hbar$ but that doesn't mean that one should or could measure $J_z$ by measuring $J_x$ and then $J_y$. That's not possible as those two don't commute and one measurement would distort the state. But that doesn't mean that $J_z$ can't be measured, does it? – Luboš Motl Jul 16 '12 at 17:25
  • Otherwise, the expectation values of all quantities should be constant in a stationary state, by definition of a stationary state. ;-) – Luboš Motl Jul 16 '12 at 17:50
  • The device to directly measure $xp$, as well as $[x,p]$ for single photons turns out to be quite simple and is described here – straups Jul 17 '12 at 06:03

2 Answers2

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The line from your textbook you're puzzling over is in fact quite a bit easier: in a stationary state nothing changes apart from phases that cancel out in any expectation value, and therefore $\frac{d}{dt}\left[\textrm{anything} \right]=0.$

You're right to point out that $xp$ is not a hermitian operator, but that does not mean that the expectation value $\langle xp \rangle$ is meaningless. Specifically, take the uncertainty relation $xp-px=i\hbar$ and substract $xp$ twice: you get $$xp+px=-i\hbar+2xp,$$ which you can rearrange into $$\langle xp\rangle = \left\langle \frac{xp+px}{2}\right\rangle+\frac{i\hbar}{2}.$$ The expectation value is now of the hemitian operator $\frac{1}{2}(xp+px)$, and you can see that your original expectation value has a trivial imaginary part.

If it's a physical interpretation for this quantity you're after, try this question.

Community
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Emilio Pisanty
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Here is how you interpret: the expectation value in a state $|\Psi\rangle$ of products of Hermitian operators like $x$ and $p$ (corresponding to observables) quantifies how correlated (quantum mechanically entangled) the two observables are in that state.

Therefore, the statement $\frac{d}{dt}\langle xp\rangle=0$ in English reads: "the extent to which the two observables 'position' and 'momentum' are entangled correlated does not change with time". Now since stationary states are independent of time (up to a phase), it should be pretty clear that the correlation should not change with time (i.e. it is fixed).

QuantumDot
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    While the expectation value $\langle xp\rangle$ does describe a correlation, it is incorrect to say that the observables are entangled. States can be entangled in quantum mechanics, as well as show non-classical kinds of correlations, but observables can't. – Emilio Pisanty Jul 16 '12 at 23:39
  • @EmilioPisanty, I completely agree, but the partition of system on subsystems is determined by localy accessible observables (http://arxiv.org/abs/quant-ph/0308043). However, the statement in this answer is still incorrect :) – straups Jul 17 '12 at 05:55
  • @EmilioPisanty OK, but if the word "entangled" isn't used, it's still correct to interpret as how correlated the two quantities are? – Carl Jul 17 '12 at 19:03
  • I have corrected the answer to reflect the well-justified objections. – QuantumDot Jul 18 '12 at 04:25
  • @EmilioPisanty I agree that the quantity $\langle xp \rangle$ describes a correlation and not entanglement but I'm not sure I agree that "observables can't [be entangled]". Entanglement is usually described in terms of (non-classical) correlations between observables. With that in mind, it seems to be an arbitrary and unimportant as to whether you say the states are entangled or the observables (in this system) are entangled. It seems to me like you could probably formulate a definition of entanglement entirely based on observables in the system. – aquirdturtle Feb 07 '16 at 00:30
  • Entanglement describes nonclassical correlations between measurement results, and an entangled state can be perfectly defined as a bipartite state that is not separable. How would you even define entangled observables, and what does that definition have to do with the classical definitions of bipartite and multipartite entanglement? It's on you to show that it can be done, really, but as I said I think if you stretch the definition of entanglement so far that it covers $⟨xp⟩$, you'll bend it beyond recognition. – Emilio Pisanty Feb 07 '16 at 00:54