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?

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    $\begingroup$ 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 '19 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 '19 at 12:22
  • $\begingroup$ @ValterMoretti Yeah, realized that. $\endgroup$ – Anonymous_original May 29 '19 at 12:22
  • $\begingroup$ @ValterMoretti Yes, but how to set this right? $\endgroup$ – Anonymous_original May 29 '19 at 12:23
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    $\begingroup$ @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 '19 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.

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  • $\begingroup$ 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 '19 at 13:20

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