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Consider the partial differential equation $$i \frac{\partial \psi}{\partial t} - \left(i\nabla + \mathbf{A} \right)^2 \psi = 0 \tag{1}$$ for the scalar function $\psi(x,y,z)$ and the vector function $\mathbf{A} (x,y,z)$ (note $\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y} , \frac{\partial}{\partial z}\right)$).

Suppose we know $\psi$ solves this PDE. Now, suppose we now perform a transformation of $\mathbf{A}$ by adding to $\mathbf{A}$ the gradient of some known scalar function $\lambda(x,y,z)$, i.e.: $$\mathbf{A} \rightarrow \mathbf{A} + \nabla \lambda \tag{2} $$ Then it is straightforward to show that the function $$\psi' = e^{i \lambda} \psi \tag{3}$$ satisfies the transformed PDE, i.e.: $$i \frac{\partial \psi'}{\partial t} - \left(i\nabla + \mathbf{A} + \nabla \lambda \right)^2 \psi' = 0 \tag{4}$$

My question has two parts: first does (4) have any solutions other than (3), i.e. is (3) a unique solution of (4)? Second, if there are other solutions to (4), what are they?

I tried to find other solutions by supposing $\psi' = f \psi$, where $f$ is an unknown function to be determined, and then plugging this into (4) and solving it. This gives a new, complicated, differential equation for $f$. However, so far my attempts at solving this differential equation have failed, so I am looking for alternative approaches.

Any ideas would be appreciated. Thanks

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  • $\begingroup$ I'm pretty sure the answer is no if your potential is nice (i.e. you have uniqueness for equation (1))$. Since your transformed solution is merely a rotation of your previous one in the complex plane $\endgroup$ – Jeb Jun 20 '16 at 20:28

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