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I have just found this good site for Q&A on physics. What I want to question is about quantum mechanics written by Griffiths, Section 10.2.3 specifically, which is Aharonov-Bohm Effect.

According to the book, The Hamiltonian $H$ can be expressed in term of $\varphi$(Electric potential), $\vec{A}$(Magnetic potential).

$$H = \frac{1}{2m}\left(\frac{\hbar}{i}\nabla-q\vec{A}\right)^2+q \varphi$$

and $\vec{A} = \frac{\Phi }{2\pi r}\hat{\phi}$ (you'd better see the book, I think.)

and also $\varphi$ is zero!.

so $$H = \frac{1}{2m}\left[-~\hbar^{2}\nabla ^{2}+q^2A^2+2i\hbar q\vec{A}\cdot\nabla\right]$$

I can't understand why factor 2 is multiplied to the third term. So could someone show me how?

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  • $\begingroup$ It's just the square of a sum: $(a+b)^2 = a^2+b^2+2ab$. $\endgroup$
    – Javier
    Commented Nov 15, 2016 at 14:12
  • $\begingroup$ Thanks for your answer. but I will show How I derived it. $(\frac{\hbar}{i}\triangledown-q\vec{A})^2$ = $(\frac{\hbar}{i}\triangledown-q\vec{A})\cdot (\frac{\hbar}{i}\triangledown-q\vec{A})$. =$[-\hbar^2\triangledown^2-\frac{\hbar}{i}\triangledown\cdot(q\vec{A})-\frac{q\hbar}{i}{}\vec{A}\cdot\triangledown+q^2A^2]$ here look at the second term. it is (ignore the coeffcients) $-q\triangledown \cdot\vec{A}-\vec{A}\cdot\triangledown q$. using gauge condition , the divergence of A is zero, and the gradient of q is zero because it is just a charge(Constant scalar). $\endgroup$
    – COMPLEX
    Commented Nov 15, 2016 at 14:27
  • $\begingroup$ Right, of course, I was a bit too hasty to answer :) $\endgroup$
    – Javier
    Commented Nov 15, 2016 at 15:11

1 Answer 1

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To see how to obtain the expression stated in the book, it helps to introduce a 'dummy' wavefunction $\psi$ to keep track of our math:

$$H\psi = \frac{1}{2m}(\frac{\hbar}{i}\triangledown-q\vec{A})^{2}\psi$$

Expanding this equation yields:

$$ H\psi = \frac{1}{2m}(\frac{\hbar}{i}\triangledown\cdot\frac{\hbar}{i}\triangledown-\frac{\hbar}{i}\triangledown\cdot(q\vec{A})-q\frac{\hbar}{i}\vec{A}\cdot\triangledown + q^{2}A^{2})\psi$$

$$ H\psi = \frac{1}{2m}(-\hbar^2\triangledown^2 -q\frac{\hbar}{i}\vec{A}\cdot\triangledown + q^{2}A^{2})\psi-\frac{1}{2m}\frac{\hbar}{i}q\triangledown\cdot(\vec{A}\psi)$$

The last term on the right can be expanded as:

$$ \triangledown\cdot(\vec{A}\psi)=\psi\triangledown\cdot\vec{A}+\vec{A}\cdot\triangledown\psi$$

Since the form of the vector potential $\vec{A}$ is given, we can compute the divergence of this potential. In spherical coordinates, it is : $$ \triangledown\cdot\vec{A} = \frac{1}{r \sin \theta }\frac{\partial}{\partial \phi}\frac{\Phi }{2\pi r}=0$$

Substituting the expression back into the original equation yields:

$$ H\psi = \frac{1}{2m}(-\hbar^2\triangledown^2 -q\frac{\hbar}{i}\vec{A}\cdot\triangledown + q^{2}A^{2})\psi-\frac{1}{2m}\frac{\hbar}{i}q\vec{A}\cdot\triangledown\psi$$

and after factorizing and removing the 'dummy' wavefunction, we finally arrive at the expression:

$$ H = \frac{1}{2m}(-\hbar^2\triangledown^2 -2q\frac{\hbar}{i}\vec{A}\cdot\triangledown + q^{2}A^{2})$$

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  • $\begingroup$ If you want to apply the operator to functions (as you correctly do) you should do it from the very first line. The symbol $\nabla$ stand-alone makes no sense (plus the momentum acts as derivative only if represented on wave functions, otherwise it could be anything else). $\endgroup$
    – gented
    Commented Nov 15, 2016 at 16:37
  • $\begingroup$ Thanks for your answer. That approach is exactly accurate than my approach. It helps me reconsider operators accurately. $\endgroup$
    – COMPLEX
    Commented Nov 21, 2016 at 15:29

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