4
$\begingroup$

Consider the famous problem of measuring both the position and momentum of an electron. We start with two photographs at different times, then worry about the momentum of the photon, the wavelength of the light, the aperture of the camera etc. We end up with the classical uncertainty relations; well actually a bit more: we have an idea of the signs of the errors (e.g. we know on which side the electron was hit by the photon).

Now suppose that we have an idea that the measurements are made by application of SOME operators (not necessarily the standard ones) on a Hilbert space. Can we deduce anything like the canonical commutation relations between position and momentum operators? In other words, is the implication between commutation and order of measurement only one way (the commutation relations are consistent with the measurements), or can we give a limited implication the other way?

The interest is with novel quantum systems - if there is an idea how measurement theory works, then is this any guide to the algebra of observables? (As well as being of general historical or philosophical interest in whether quantum theory might have taken another turn using different operator representations.)

$\endgroup$
  • $\begingroup$ Well, you can derive $\Delta A \cdot \Delta B\geq \frac{1}{2} \left<\left[A,B\right]\right>$ from the assumption that a measurement corresponds to expectation values of operators on a Hilbert space. Hence, if your "measurement theory" tells you that the product of the uncertanties is bounded by something, you will get some information on the commutator. Note that this has to hold for all possible measurements on all possible wave functions (The LHS of the inequality depends on the wave function!). $\endgroup$ – Toffomat Apr 19 '17 at 10:54
  • $\begingroup$ Thanks - that looks interesting. Where does the inequality come from? I guess that may be the limit of what can be done in general. $\endgroup$ – Edwin Beggs Apr 19 '17 at 13:51
  • $\begingroup$ For the opposite way: commutation relations $\Rightarrow$ uncertainty, see physics.stackexchange.com/q/10362/2451 , physics.stackexchange.com/q/24116/2451 and links therein. $\endgroup$ – Qmechanic Apr 19 '17 at 14:08
  • $\begingroup$ These notes explain your issue in all detail (they prove $\Delta A \Delta B \leq 1/2 \langle[A,B]\rangle$ damtp.cam.ac.uk/user/eal40/teach/QM2012/7comm.pdf Proof starts at eqn 331 Good luck If you really want me to I can summarize it but you should be able to get the point from the notes (I think they are Tong's) $\endgroup$ – gertian Apr 19 '17 at 14:41
2
+50
$\begingroup$

The uncertainty principle between two observables is related to their commutator in a general and profund way. The generalized uncertainty principle can be proved quite generally using simple matrix algebra and the Cauchy-Schwartz Inequality:

I) Supose we have two hermitian operators (aka observables) $\hat{A}$ and $\hat{B}$. The possible results of a measurment are their eivenvalues and the dispersion in the measurment is:

\begin{equation} ({\Delta\hat{A}})^2 = \langle\hat{A}^2\rangle-\langle\hat{A}\rangle^2 \end{equation}

We can calways chose a new reference system to set $<\hat{A}>=0$ so we get:

\begin{equation} ({\Delta\hat{A}})^2 = \langle\hat{A}^2\rangle = \int\psi^*\hat{A}^2\psi dx =\langle\psi|\hat{A}^2|\psi\rangle \end{equation}

And obviusly the same holds for $\hat{B}$.

II) Using the Cauchy-Schwartz Inequality:

\begin{equation} \langle\psi|\hat{A}\hat{A}|\psi\rangle\langle\psi|\hat{B}\hat{B}|\psi\rangle \geqslant |\langle\psi|\hat{A}\hat{B}|\psi\rangle|^2 \end{equation}

One can inmediately obtain:

\begin{equation} ({\Delta\hat{A}})^2({\Delta\hat{B}})^2 \geqslant |\langle\psi|\hat{A}\hat{B}|\psi\rangle|^2 \end{equation}

III) Now,we can reduce the term in the right

\begin{equation} |\langle\psi|\hat{A}\hat{B}|\psi\rangle| \geqslant |Im\Big[\langle\psi|\hat{A}\hat{B}|\psi\rangle \Big] = |\frac{1}{2i}\Big[\langle\psi|\hat{A}\hat{B}|\psi\rangle - \langle\psi|\hat{A}\hat{B}|\psi\rangle^* \Big]| \end{equation}

Where I have used that the modulus of a complex number is bigger than its Imaginary part and then I used the fact that if $f= Re(f)+i Im(f)$ then $Im(f)=\frac{1}{2i}(f-f^*)$.

IV) Because $\hat{A}$ and $\hat{B}$ are observables then $\langle\psi|\hat{A}\hat{B}|\psi\rangle^*=\langle\psi|(\hat{A}\hat{B})^{\dagger}|\psi\rangle=\langle\psi|(\hat{B}\hat{A})|\psi\rangle$

V) Finally, using this result we can rewrite the inequality as:

\begin{equation} ({\Delta\hat{A}})^2({\Delta\hat{B}})^2 \geqslant |\frac{1}{2i}\Big[\langle\psi|\hat{A}\hat{B}|\psi\rangle - \langle\psi|\hat{B}\hat{A}|\psi\rangle \Big]| =|\frac{1}{2i}\Big[\langle\hat{A}\hat{B}\rangle - \langle\hat{B}\hat{A}\rangle \Big] |=|\frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle| \end{equation}

So the dispersion in any two hermitian operators is related to their commutator

\begin{equation} ({\Delta\hat{A}})^2({\Delta\hat{B}})^2 \geqslant|\frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle| \end{equation}

I suppose that you could measure the dispersion of two observables with increasing accuracy to find some upper limmit on their commutator.

Note that this work for any two observables you like to use, not just canonical ones like X and P.

$\endgroup$
  • $\begingroup$ From context, I can tell that by $\Delta\hat{A}^2$ you mean $\left(\Delta\hat{A}\right)^2$ and not $\Delta\left(\hat{A}^2\right)$, but it would be nice if you would make the difference explicit, especially for questions like this one, which attract many novice physicists. $\endgroup$ – Bob Knighton Apr 22 '17 at 22:58
  • 1
    $\begingroup$ the problem is the bound is in general state dependent so you'd get a best some information about average values of the commutators, not much more. Lunless you can guarantee you are in a specific eigenstate of this commutator, this isn't much... and then forget about off-diagonal elements... $\endgroup$ – ZeroTheHero Apr 22 '17 at 23:01
  • $\begingroup$ @BobKnighton you are absolutely right, let me edit it $\endgroup$ – P. C. Spaniel Apr 22 '17 at 23:15
  • $\begingroup$ In the V) step the square of right hand side is missing $\endgroup$ – pointguard0 Feb 6 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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