I'm doing Griffiths' Introduction to Quantum Mechanics. In a question, it introduces the modified Schrodinger equation in which the Hamiltonian, $$ H~=~-\frac{\hbar^2}{2m}\nabla^2 + V $$is replaced by the classical Hamiltonian, $$ H~=~\left[\frac{1}{2m}(\frac{\hbar}{i}\nabla-q\vec {\mathbf A})^2 +q\phi\right]$$ which I expanded out to $$ H= \left[ \frac{1}{2m}\left\{-\hbar^2\nabla^2+q^2A^2+i\hbar q(\vec\nabla\cdot\vec {\mathbf A}+\vec {\mathbf A}\cdot\vec \nabla)\right \}+q\phi \right]$$

Here $\vec{\mathbf A}$ is the vector potential for the magnetic field $(\vec{\mathbf B}=\vec\nabla \times\vec{\mathbf A}) $ and $\phi$ is the scalar potential for the electric field $(\vec{\mathbf E}=-\nabla\phi-\frac{\partial\vec{\mathbf A}}{\partial t})$.

Now the first part of the question requires one to calculate $[H,\vec {\mathbf r}]$. But that requires me to take the Laplacian of $ \vec {\mathbf r}\psi$ which is a vector function (due to the first term in the expression). Is this legal?

I know that one way around this is to work component wise. But how to make this work?


closed as off-topic by Aaron Stevens, Jon Custer, Cosmas Zachos, Phonon, Yashas May 31 at 15:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Jon Custer, Cosmas Zachos, Phonon, Yashas
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Without expanding the Hamiltonian, use $[A^2,B]= A[A,B]+ [A,B]A$ and CCRs. Notice also that $\vec{r}$ commutes with both $\vec{A}$ and $\phi$... $\endgroup$ – Valter Moretti May 29 at 12:19
  • $\begingroup$ Canonical commutation relations $[p_j, r_k]= -i\hbar \delta_{jk}I$, where $p_j = -i\hbar \nabla_j$. $\endgroup$ – Valter Moretti May 29 at 12:22
  • $\begingroup$ @ValterMoretti Yeah, realized that. $\endgroup$ – Anonymous_original May 29 at 12:22
  • $\begingroup$ @ValterMoretti Yes, but how to set this right? $\endgroup$ – Anonymous_original May 29 at 12:23
  • 1
    $\begingroup$ @Anonymous_original As already suggested you need not really represent the momentum operator as Laplacian: the Hamiltonian is of the form $H=(p-qA)^2 + V$, therefore the commutator can be expanded as function of $[x,p]$ and $[x, A], [x, V]$. $\endgroup$ – gented May 29 at 12:29

Your problem (as I understand it) boils down to calculating the vector Laplacian for the vector field function $\mathbf{r} \psi$. This calculation can be performed in any coordinate system you choose; if you know how to apply divergences, gradients, and curls to an arbitrary vector field in that coordinate system, then the identity $$ \nabla^2 \mathbf{A} \equiv \nabla \left( \nabla \cdot \mathbf{A} \right) - \nabla \times \left( \nabla \times \mathbf{A} \right) $$ allows for the calculation of the vector Laplacian.

In Cartesian components, this reduces to the particularly nice $$ \nabla^2 \mathbf{A} = \left( \nabla^2_s A_x \right) \hat{x} + \left( \nabla^2_s A_y \right) \hat{y} + \left( \nabla^2_s A_z \right) \hat{z}, $$ where $\nabla^2_s$ is the familiar "scalar Laplacian". It should therefore be possible to calculate $\nabla^2 (\mathbf{r} \psi)$ by setting $A_x = x \psi$ in the above, and similarly for $A_y$ and $A_z$. The result will then be expressible in terms of vectors and gradients, which you can then put back into a coordinate-invariant form and re-express in terms of spherical coordinates if you like.

Alternately, if you want to stay in spherical coordinates the whole time, you can use the above identity to express $\nabla^2 \mathbf{A}$ in terms of the spherical coordinates of $\mathbf{A}$. In your case, you will have $\mathbf{A} = \mathbf{r} \psi$, and so $A_r = r \psi$ while $A_\theta = A_\phi = 0$. However, the vector Laplacian is not nearly as nice in spherical coordinates as it is in Cartesian coordinates; if you want to see the full expression, it can be found in this table on Wikipedia. When the dust settles, though, you should find an expression in terms of $\psi$, the spherical coordinates $\{r, \theta, \phi\}$, and the derivatives of the former with respect to the latter.

Both of the above methods are equivalent, and should yield the same result; try them both out if you need to while away a rainy afternoon. And, of course, both are equivalent to the method (suggested by gented in the comments) of expressing $H = \frac{1}{2m}(\mathbf{p} - q \mathbf{A})^2 + q \phi$ and using the canonical commutation relations $[x_i, x_j] = [p_i, p_j] = 0$, $[x_i, p_j] = \delta_{ij}$. I suspect that after going through the Laplacian calculations above, you will appreciate the elegance of the operator method.

  • $\begingroup$ That's what I was looking for. Thanks. I do appreciate the methods suggested by others and will try to use them (whenever need be). It was just that I wanted to know how this particular approach would work (if needed in the future) for the sake of completeness. $\endgroup$ – Anonymous_original May 29 at 13:20

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