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It is sometimes convenient to write the wavefunction as $$ \Psi(x,t)~=~ e^{\Phi(x,t)} $$ and then work with $\Phi$ instead. This is particularly sensible in the context of the WKB approximation, where $\Phi$ is nice a smooth and slowly changing.

But, does $\Phi$ exist in general? There's a reason it might not. Let's say that $\Psi(x_0,t_0) = 0$. It could be that, near $(x,t) = (x_0,t_0)$, the wavefunction looks like

$$ \Psi(x,t) = e^{i \tan^{-1} \frac{x-x_0}{t-t_0}} \text. $$

(When I write $\tan^{-1}$, I mean something like the C/python "atan2" function. It's just the angle formed by the vector $(x-x_0,t-t_0)$. I hope that's clear.)

So, this wavefunction is smooth and well-behaved at $(x_0,t_0)$, but the imaginary part of its logarithm can't be consistently defined. I guess one could say that $\Phi(x,t)$ has a branch point.

Obviously this can't happen if there wavefunction never vanishes. The only place where I know I can guarantee $\Psi \ne 0$ is the ground state of a bosonic system. Are there other situations in which $\Phi$ can be guaranteed to make sense?

(For that matter... does $\Phi$ have a name?)

Qmechanic
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Scott Lawrence
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1 Answers1

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For what it's worth, concerning the 1D TISE, the WKB wave function $$\psi(x) ~\sim~\frac{1}{\sqrt{p(x)}}\sum_{\pm}C_{\pm} \exp\left(\pm \frac{i}{\hbar}\int_{x_0}^x \! p(x)~\mathrm{d}x\right)$$ typically becomes singular near turning points, cf. e.g. this Phys.SE post.

Qmechanic
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  • Another example: $n=2$ states of hydrogen. There's always a zero of $\Psi$, but depending on $\ell,m$, $\Phi$ can either be well defined or not. – Scott Lawrence Jan 03 '21 at 20:03