10
$\begingroup$

$xp$ 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?)

$\endgroup$
5
  • $\begingroup$ "why is this obvious?" Hint: a stationary state has a simple time dependence: $\Psi(t)\rangle = e^{-iE_{\Psi}t/\hbar}|\Psi(0)\rangle$. $\endgroup$ Commented Jul 16, 2012 at 17:07
  • 7
    $\begingroup$ 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. $\endgroup$ Commented Jul 16, 2012 at 17:22
  • 3
    $\begingroup$ 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? $\endgroup$ Commented Jul 16, 2012 at 17:25
  • $\begingroup$ Otherwise, the expectation values of all quantities should be constant in a stationary state, by definition of a stationary state. ;-) $\endgroup$ Commented Jul 16, 2012 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
    Commented Jul 17, 2012 at 6:03

2 Answers 2

11
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Nice argument there :-) $\endgroup$
    – David Z
    Commented Jul 16, 2012 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
    Commented Jul 17, 2012 at 19:04
  • 1
    $\begingroup$ 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). $\endgroup$ Commented Mar 22, 2018 at 17:07
1
$\begingroup$

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).

$\endgroup$
6
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Jul 16, 2012 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
    Commented Jul 17, 2012 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
    Commented Jul 17, 2012 at 19:03
  • $\begingroup$ I have corrected the answer to reflect the well-justified objections. $\endgroup$
    – QuantumDot
    Commented Jul 18, 2012 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. $\endgroup$ Commented Feb 7, 2016 at 0:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.