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On page 21-4 of vol. III of The Feynman Lectures on Physics, he derives the formula for the probability current density. Starting with the Schrödinger equation, he presents the time derivative of the probability density, and then states,

"The potential terms and a lot of other stuff cancel out. And it turns out that what is left can indeed be written as a perfect divergence.”

However, this is true only if $\psi^*(A\cdot\nabla)\psi = -\psi(A\cdot\nabla)\psi^*$, which does not appear to be the case. What am I missing?

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  • $\begingroup$ Please give us a bit more information about the derivation in that textbook. $\endgroup$ – Alpha001 Aug 12 '18 at 16:12
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Feynman's book can be found at http://www.feynmanlectures.caltech.edu/III_21.html.

You probably leave out some term. Just to give some hints: \begin{equation} \nabla \Bigg( \psi^* \Big(\frac{\hbar}{i} \nabla - q {\bf A} \Big) \psi \Bigg) = (\nabla \psi^*) \frac{\hbar}{i} (\nabla \psi) + \psi^* \frac{\hbar}{i} \nabla^2 \psi - \psi^* q \nabla({\bf A}\psi) - (\nabla\psi^*) q {\bf A} \psi \end{equation} \begin{equation} \nabla \Bigg( \psi \Big(-\frac{\hbar}{i} \nabla - q {\bf A} \Big) \psi^* \Bigg) = -(\nabla \psi) \frac{\hbar}{i} (\nabla \psi^*) - \psi \frac{\hbar}{i} \nabla^2 \psi^* - \psi q \nabla({\bf A}\psi^*) - (\nabla\psi) q {\bf A} \psi^* \end{equation} Now use that $\nabla ({\bf A} \psi) = {\bf A} \nabla \psi$, because $\nabla {\bf A}=0$.

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  • $\begingroup$ Feynman starts with ∂P/∂t = ψ*(∂ψ/∂t) + ψ(∂ψ*/∂t) Using Schrödinger’s equation, ∂P/∂t = -i/ħ [ψ*(1/2m)(-ih∇ - qA) · (-ih∇- qA)ψ + qφψψ - ψ(1/2m)(iħ∇ - qA) · (iħ∇ - qA)ψ - qφψψ*] He then states this is equal to ∂P/∂t = -(1/2m)∇[ψ*(-iħ∇ - qA) ψ + ψ(iħ∇ - qA)ψ*] But I get (leaving out the potential terms that clearly cancel): ∂P/∂t = -i/ħ (1/2m)[ψ*(-ħ2∇2 + q2A2 + iħ∇·qA + iħqA·∇)ψ -ψ(-ħ2∇2 + q2A2 - iħ∇ · qA - qA · iħ∇)ψ*] = -i/ħ (1/2m)[ψ*(-ħ2∇2 + iħ∇·qA + iħqA · ∇)ψ -ψ(-ħ2∇2 - iħ∇ · qA - iħqA · ∇)ψ*] Feynman is correct only if ψ*(A · ∇)ψ = - ψ(A · ∇)ψ* $\endgroup$ – Dennis Witherell Aug 12 '18 at 18:29
  • $\begingroup$ No, that last equation is not needed and is not true in general. As I've written, $\psi^* {\bf A} \nabla \psi = \psi^* \nabla ({\bf A} \psi)$, so if your statement would be true, then no $\bf A$ dependent term could remain... I suggest you to expand Eq. (21.11) and compare to the terms of expanding (21.10). That's why, I have started that in my answer. $\endgroup$ – fermion Aug 12 '18 at 19:50
  • $\begingroup$ If you're going to make a comment on an answer with a lot of math in it you should still format it with mathjax just like in your original post. $\endgroup$ – Triatticus Sep 23 at 15:17
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The question is how we simplify $\partial_t P=\frac{1}{i\hbar}(\psi^\ast\frac{1}{2m}(\frac{\hbar}{i}\nabla-qA)^2\psi+q\phi\psi^\ast\psi-\psi\frac{1}{2m}(-\frac{\hbar}{i}\nabla-qA)^2\psi^\ast-q\phi\psi\psi^\ast)$ to $\partial_t P=-\nabla(\frac{1}{2m}\psi^\ast(\frac{\hbar}{i}\nabla-qA)\psi+\frac{1}{2m}\psi(\frac{\hbar}{i}\nabla-qA)\psi^\ast)$, viz. Eqs. 21-10/11 here. The $\phi$ terms cancel trivially, so the difference between the two right-hand sides is $\frac{z+z^\ast}{2m}$ with $$z:=\frac{\psi^\ast(\frac{\hbar}{i}\nabla-qA)^2\psi}{i\hbar}+\nabla(\psi^\ast(\frac{\hbar}{i}\nabla-qA)\psi)=qA\psi^\ast(\nabla-\frac{iq}{\hbar}A)\psi+(\nabla\psi^\ast)(\frac{\hbar}{i}\nabla-qA)\psi.$$Alternatively, we can add any imaginary number we want to the definition of $z$ without changing $\frac{z+z^\ast}{2m}$, so e.g. we can put$$z=qA\psi^\ast\nabla\psi-qA(\nabla\psi^\ast)\psi,$$which by inspection is imaginary.

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  • $\begingroup$ One other question--doesn't your conclusion assume that ψ∗(−iqA/ℏ)ψ and ψ∗(-iℏ∇)ψ are imaginary numbers, when they are potentially complex? $\endgroup$ – Dennis Witherell Aug 27 '18 at 20:14
  • $\begingroup$ @DennisWitherell No; from $z$ I've deleted the terms $-\frac{iq^2A}{\hbar}\psi^\ast\psi-i\hbar\nabla\psi^\ast\nabla\psi$, both imaginary multiples of an expression of the form $w^\ast w\in\mathbb{R}$. Note $A$ is a real classical field. $\endgroup$ – J.G. Aug 27 '18 at 20:21
  • $\begingroup$ I think I finally see the issue. Since (A ⋅ ∇) is a matrix operator, isn't there a problem going from ψ∗(A⋅∇)ψ to Aψ∗∇ψ? $\endgroup$ – Dennis Witherell Aug 28 '18 at 12:42
  • $\begingroup$ @DennisWitherell No, but one does need care rewriting $\nabla\cdot (A\psi)$ as $A\cdot\nabla\psi+(\nabla\cdot A)\psi$. It comes down to which vector spaces we're operating on. $\endgroup$ – J.G. Aug 28 '18 at 14:27

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