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I've been trying to prove a component of a proof the gauge invariance of the schrodinger equation. Specifically the part in the first answer here where this is stated:

$$\big(\frac{\nabla}{i}-q(\vec{A} +\nabla \Lambda)\big)e^{iq\Lambda}\psi = e^{iq\Lambda}\big(\frac{\nabla}{i}-q\vec{A}\big)\psi$$

By the product rule:

$$\big(\frac{\nabla}{i}-q(\vec{A} +\nabla \Lambda)\big)e^{iq\Lambda}\psi = (e^{iq\Lambda} \frac{\nabla}{i} \psi + \psi \frac{\nabla}{i} e^{iq\Lambda} -e^{iq\Lambda}q\vec{A}\psi+ \nabla\Lambda e^{iq\Lambda}\psi) $$

where $$\nabla\Lambda e^{iq\Lambda}\psi = \Lambda e^{iq\Lambda}\nabla\psi + \Lambda\psi\nabla e^{iq\Lambda} + e^{iq\Lambda}\psi\nabla\Lambda $$

and with Lambda as a function of the spatial coordinates: $$\nabla e^{iq\Lambda} = iq e^{iq\Lambda}\nabla \Lambda$$

Then $$\nabla\Lambda e^{iq\Lambda}\psi = \Lambda e^{iq\Lambda}\nabla\psi + iq\Lambda\psi e^{iq\Lambda}\nabla \Lambda+ e^{iq\Lambda}\psi\nabla\Lambda $$

and

$$(e^{iq\Lambda} \frac{\nabla}{i} \psi + \psi \frac{\nabla}{i} e^{iq\Lambda} -e^{iq\Lambda}q\vec{A}\psi+ \nabla\Lambda e^{iq\Lambda}\psi) $$

$$=e^{iq\Lambda} \frac{\nabla}{i} \psi + q\psi e^{iq\Lambda}\nabla \Lambda -e^{iq\Lambda}q\vec{A}\psi+ \Lambda e^{iq\Lambda}\nabla\psi + iq\Lambda\psi e^{iq\Lambda}\nabla \Lambda+ e^{iq\Lambda}\psi\nabla\Lambda $$ $$=\big(e^{iq\Lambda} ( \frac{\nabla}{i} -q\vec{A})\psi\big) + (q\psi e^{iq\Lambda}\nabla \Lambda + \Lambda e^{iq\Lambda}\nabla\psi + iq\Lambda\psi e^{iq\Lambda}\nabla \Lambda+ e^{iq\Lambda}\psi\nabla\Lambda )$$

Somehow the right set of parentheses goes to zero, and I'm wondering what I'm missing.

This is another stack exchange question where the same pattern is used: Gauge Invariance of Schrodinger Equation

pongoS
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1 Answers1

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Your last term $\nabla\Lambda e^{iq\Lambda}\psi)$ in the second line is simply $e^{iq\Lambda}\psi\cdot(\nabla\Lambda)$ and is not the third line where you are acting by nabla on all functions. Maybe you misunderstood the first line where $\nabla\Lambda$ is just a gradient of $\Lambda$.

Vladimir Kalitvianski
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  • I don't understand how to tell that these are separate, I was assuming an operator operators on everything to the right. It makes sense that $\Lambda$ and $e^{iq\Lambda}$ commute, but how do I know that $\nabla \Lambda $ commutes with $e^{i q \Lambda}$ and $\psi$? – pongoS Dec 09 '18 at 07:14
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    In the first line nabla does not act on other functions, just on Lambda. The gradient invariance means exactly that: the wave funstion "phase" transformation is compensated with a gradient "extension" of the vector potential. At early times the "gauge field"s were often called "compensating fields". – Vladimir Kalitvianski Dec 09 '18 at 07:23
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    A deeper thing is that you may solve the Schrodinger equation with any choice of gauge of the vector potential; the resulting wave functions, although different for different gauges, are physically equivalent. – Vladimir Kalitvianski Dec 09 '18 at 07:31