1
$\begingroup$

I am having some trouble with the notion that the different components of a vector operator can be hermitian in one coordinate system but non-hermitian in another. I have seen e.g. Bra-ket notation in a 3 dimensional system says a euclidean vector operator acting on a state should be thought of as an operator that has euclidean vectors for its eigenvalues. So if we were in an eigenstate of momentum then:

$$ \hat{\underline{p}}|\underline{p}\rangle = \underline{p}|\underline{p}\rangle. $$

My first question is to ask is it ok to think of this as being three copies of the state vector with different scalar eigenvalues in front of each:

$$ \hat{\underline{p}}|\underline{p}\rangle = p_x|\underline{p}\rangle\underline{e}_x + p_y|\underline{p}\rangle\underline{e}_y + p_z|\underline{p}\rangle\underline{e}_z. $$

My main question is about when and how the operator can be decomposed into component terms. It seems that it is acceptable in cartesian coordinates here to say:

$$ \hat{\underline{p}}|\underline{p}\rangle = \hat{p}_x|\underline{p}\rangle\underline{e}_x + \hat{p}_y|\underline{p}\rangle\underline{e}_y + \hat{p}_z|\underline{p}\rangle\underline{e}_z $$

and that I write $$\hat{\underline{p}} = \hat{p}_x\underline{e}_x + \hat{p}_y\underline{e}_y + \hat{p}_z\underline{e}_z.$$

But this unravels for coordinate systems with non-unity Jacobians. Specifically the spherical polar coordinates (Jacobian = $r^2\sin\theta$) I understand that the "radial component" of the euclidean vector operator is not hermitian (nor is the $\theta$ component but the $\phi$ component is hermitian - it all being connected to whether that coordinate appears in the Jacobian or not).

But the vector operator is still hermitian, so it seems reasonable to me to believe that I could have:

$$ \hat{\underline{p}}|\underline{p}\rangle = p_r|\underline{p}\rangle\underline{e}_r + p_\theta|\underline{p}\rangle\underline{e}_\theta + p_\phi|\underline{p}\rangle\underline{e}_\phi $$

where all those eigenvalues are real-valued. But if I try to write $$\hat{\underline{p}} = \hat{p}_r\underline{e}_r + \hat{p}_\theta\underline{e}_\theta + \hat{p}_\phi\underline{e}_\phi$$ I dont think I can equate, e.g. $\hat{p}_r$ to the hermitian operator that returns the radial momentum component (which is not the same operator as the non-hermitian operator that is the radial component of the euclidean momentum operator).

If I call the hermitian operators that correspond to the coordinates of momentum $\hat{p}'_r,\hat{p}'_\theta,\hat{p}'_\phi$ then I'm happy that e.g. $\hat{p}'_r|\underline{p}\rangle=p_r|\underline{p}\rangle$ but I do not believe that $$\hat{\underline{p}} = \hat{p}'_r\underline{e}_r + \hat{p}'_\theta\underline{e}_\theta + \hat{p}'_\phi\underline{e}_\phi.$$ Specifically, when I try to calculate the angular momentum operators from this definition of $\hat{\underline{p}}$ and a definition of $\hat{\underline{r}} = \hat{r}\underline{e}_r$ (i.e. I try to say $\hat{\underline{L}} = \hat{\underline{r}}\times\hat{\underline{p}}$ it doesn't give the answers I expect).

So the question is, what is the right way to write the euclidean momentum operator in spherical polar coordinates and relate it to the operators corresponding to the respective components of momentum?

Improve this question
$\endgroup$
6
  • $\begingroup$ Leaving QM aside, you are completely, unequivocally, securely comfortable with the gradient and integration in classical coordinates, right? You then accept the canonical translation, right? You have an extraneous problem? There is nothing in Sakurai-Napolitano, e.g., that you fail to understand? $\endgroup$
    – Cosmas Zachos
    Mar 5, 2022 at 17:18
  • $\begingroup$ I'm not sure what you mean by this comment? I feel pretty comfortable with all the concepts building up to this point, I am really just trying to get clear what the correct notation is for a "euclidean vector operator" - I started out thinking its an operator that has euclidean vector eigenvalues, but I cannot reconcile $\hat{\underline{p}}|\underline{p}\rangle = (p_r,p_\theta,p_\phi)|\underline{p}\rangle$ with $\hat{\underline{p}} = (\hat{p}_r,\hat{p}_\theta,\hat{p}_\phi)$ because e.g. $\hat{p}_r|\underline{p}\rangle \neq p_r|\underline{p}\rangle$ $\endgroup$
    – user1488777
    Mar 8, 2022 at 21:00
  • $\begingroup$ Well, you won’t have simultaneous eigenstates of positions and coordinates. Just cross multiply the operators, in polar coordinates, and consider thr eigenstates of r and the two angular momentum quantities, Lz and LL, commuting among themselves. You are barking up a tree your text didn’t send you to… $\endgroup$
    – Cosmas Zachos
    Mar 8, 2022 at 21:11
  • $\begingroup$ Tell me if you just want $\mathbf{L} = -i\hbar(\mathbf{r} \times \nabla)= i \hbar \left(\frac{\hat{\boldsymbol{\theta}}}{\sin(\theta)} \frac{\partial}{\partial\phi} - \hat{\boldsymbol{\phi}} \frac{\partial}{\partial\theta}\right) $, which is straightforward. Again, if you are looking at momentum eigenstates, you've gone astray. $\endgroup$
    – Cosmas Zachos
    Mar 8, 2022 at 22:56
  • $\begingroup$ You've picked a representation (the position representation), and my question isn't about what is the angular momentum operator in spherical polar coordinates in the position representation, my question is what is the momentum operator in spherical coordinates WITHOUT picking a representation. My question I think is more a notation one, about how to correctly notate a euclidean vector operator without picking a representation but after having chosen a coordinate for the vector. $\endgroup$
    – user1488777
    Mar 10, 2022 at 10:06

0

Reset to default

Your Answer

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