Physical interpretation of the wave function $\psi (x,t)$ is a probability amplitude for location $x$ and the Fourier transform of $\psi (x,t)$ can be interpreted as a probability amplitude for momentum $p$.

Therefore $x$ and $p$ are conjugate variables, in the Fourier sense, and we can actually take the operator $\hat{p}=-i\hbar\nabla$ from the integral of the expectation value of momentum.

My question is that is there a similar expectation value integral calculation for observable $L= x\times p$? The Born rule tells us that the operator is $-i\hbar (x \times \nabla)$.

It can surely be expected that someone (Born, Pauli, Weil?) has actually calculated all the known classical observables via Fourier transformation.

EDIT: As I was informed, the operator for angular momentum is just defined that way.

  • $\begingroup$ Erm... It's not the Born rule that tells us what is the operator, it's the definition basically of what is $L$. Could you elaborate on the $x$ and $p$ example so that it were more clear what do you ask $\endgroup$ – OON Sep 14 '17 at 11:17
  • $\begingroup$ @OON I did some clarifications. $\endgroup$ – Hulkster Sep 14 '17 at 11:34

The angular momentum is conjugated to various angles in the spherical coordinates. Unlike usual coordinate $x$ those angles have limited range. So you may expect that just like in case of the Fourier transform on the limited range you will get not the integral but discrete series. However it doesn't have to look exactly like the usual Fourier series.

Indeed if we introduce the spherical coordinates, \begin{equation} x=r\cos\phi\sin\theta,\quad y=r\sin\phi\sin\theta,\quad z=r\cos\theta \end{equation} then any "good" wavefunction can be represented as a series, \begin{equation} \psi(r,\phi,\theta)=\sum_{l=0}^{+\infty}\sum_{m=-l}^{+l}\psi_{lm}(r)Y_{lm}(\phi,\theta) \end{equation} where $Y_{lm}$ are certain functions known as spherical harmonics that are orthonormal in the following sense, \begin{equation} \int_{0}^{2\pi} d\phi \int_{0}^{\pi} d\theta \sin\theta\, Y_{l'm'}^\ast (\phi,\theta) Y_{lm}(\phi,\theta)=\delta_{ll'}\delta_{mm'} \end{equation} Because of that we find that, \begin{equation} \psi_{lm}(r)=\int_{0}^{2\pi} d\phi \int_{0}^{\pi} d\theta \sin\theta\,Y_{lm}^\ast(\phi,\theta) \psi(r,\phi,\theta) \end{equation}

Now if we take $\hat{L}_z=xp_x-yp_y=-i\partial_\phi$ (I'll take $\hbar=1$) then you'll get that any term in the series is an eigenfunction of this operator, \begin{equation} \hat{L}_z\psi_{lm}(r)Y_{lm}=m\psi_{lm}(r)Y_{lm} \end{equation} If you try to do the same with $\hat{L}_x$ and $\hat{L}_y$ this will not be the case. That's reflection of the fact that those operators don't commute. However if you take $\hat{L}^2=\hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2$ such operator does commute with $\hat{L}_z$ and, \begin{equation} \hat{L}^2\psi_{lm}(r)Y_{lm}=l\psi_{lm}(r)Y_{lm} \end{equation}

Now you may consider $\psi_{lm}(r)$ as a new representation of the wavefunction. It's characterized by continuous parameter $r$ but also two integer parameter $l$ and $m$ with $l\geq 0,|m|\leq l$. Concerning those last two parameters the operators become simply matrices $A_{lm,l'm'}$ with $\hat{L}_z=m\delta_{mm'}$ and $\hat{L}^2=l\delta_{ll'}$. The expectation values can be written as, \begin{equation} \langle O\rangle = \int_{0}^{+\infty} r^2 dr\sum_{lm,l'm'} \psi_{lm}^\ast(r)\Big[\hat{O}_{lm,l'm'}\psi_{l'm'}\Big](r) \end{equation}

  • $\begingroup$ But seems plausible to represent one-dimensional term $xp$ as an expection value integral in terms of $\psi$. $\endgroup$ – Hulkster Sep 14 '17 at 13:35
  • $\begingroup$ @Hulkster well, you didn't clarified that exact point, what do you mean by "representing operator as expectation value". Do you mean representation like that? $K_p(x-y)=\int dp\, p e^{ip(x-y)}$ so that $\int dy K_p(x-y)\psi(y)=\hat{p}\psi(x)$? $\endgroup$ – OON Sep 14 '17 at 13:42
  • $\begingroup$ @Hulkster Please also note that $\hat{x}\hat{p}$ is not Hermitian because $(\hat{x}\hat{p})^\dagger=\hat{p}\hat{x}\neq \hat{x}\hat{p}$ and therefore is not an observable $\endgroup$ – OON Sep 14 '17 at 13:44
  • $\begingroup$ Oh dear! I do remember that in QM exercises we students were ask to form a corresponding operator for $xp$. But indeed, you have a valid point (+1). And what I ment was that if we have the expectation value $<x \times p>$ as somesort of an intergal, then can we extract the corresponding operator, just like in the case of $<p>$. $\endgroup$ – Hulkster Sep 14 '17 at 13:51
  • $\begingroup$ @Hulkster Ok, I'm simply asking you to write how exactly something similar to what you want works in case of $\langle p\rangle$ so we could pinpoint the analog in case of $L$ $\endgroup$ – OON Sep 14 '17 at 13:55

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.