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

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

the expectation value of position times momentum?

How to interpret the expectation value of a hermitian operator, say $p$, is clear: make several measurements on identically prepared systems and get the expectation value from the measured quantities.

But now we are talking about the expectation value of an operator that doesn't represent an observable. The discussion above hence isn't valid. However, computing $\langle x p \rangle$ knowing $\Psi$ according to the mathematical definition can be done, so $\langle x p \rangle$ can clearly be deduced if we know the wave function - but how can we interpret the quantity we get?

(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?)

  • $\begingroup$ "why is this obvious?" Hint: a stationary state has a simple time dependence: $\Psi(t)\rangle = e^{-iE_{\Psi}t/\hbar}|\Psi(0)\rangle$. $\endgroup$ – Alfred Centauri Jul 16 '12 at 17:07
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    $\begingroup$ 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. $\endgroup$ – Luboš Motl Jul 16 '12 at 17:22
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    $\begingroup$ 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? $\endgroup$ – Luboš Motl Jul 16 '12 at 17:25
  • $\begingroup$ Otherwise, the expectation values of all quantities should be constant in a stationary state, by definition of a stationary state. ;-) $\endgroup$ – Luboš Motl Jul 16 '12 at 17:50
  • $\begingroup$ The device to directly measure $xp$, as well as $[x,p]$ for single photons turns out to be quite simple and is described here $\endgroup$ – straups Jul 17 '12 at 6:03

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).

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    $\begingroup$ 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. $\endgroup$ – Emilio Pisanty Jul 16 '12 at 23:39
  • $\begingroup$ @EmilioPisanty, I completely agree, but the partition of system on subsystems is determined by localy accessible observables (arxiv.org/abs/quant-ph/0308043). However, the statement in this answer is still incorrect :) $\endgroup$ – straups Jul 17 '12 at 5:55
  • $\begingroup$ @EmilioPisanty OK, but if the word "entangled" isn't used, it's still correct to interpret <xp> as how correlated the two quantities are? $\endgroup$ – Carl Jul 17 '12 at 19:03
  • $\begingroup$ I have corrected the answer to reflect the well-justified objections. $\endgroup$ – QuantumDot Jul 18 '12 at 4:25
  • $\begingroup$ @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. $\endgroup$ – aquirdturtle Feb 7 '16 at 0:30

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.

  • $\begingroup$ Nice argument there :-) $\endgroup$ – David Z Jul 16 '12 at 23:49
  • $\begingroup$ Yes, d/dt(anything)=0 for stationary states is (and should have been to me) clear, was just puzzled by the (anything) being something I couldn't interpret. $\endgroup$ – Carl Jul 17 '12 at 19:04
  • $\begingroup$ This anti-commutator is also known as "Berry-Keating hamiltonian" which is connected with the Riemann hypotheses. And btw. mathematicians don't regard it self-adjunct (=hermitian). $\endgroup$ – Raphael J.F. Berger Mar 22 '18 at 17:07

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