# Algebraic trouble in gauge invariance of Schrodinger equation

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

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 $$\Lambda$$.
• 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 $\psi$? Dec 9, 2018 at 7:14