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I'm trying to get a better sense of what causes an increase in the magnitude and phase of the wave function at a given point. Is there a way to rewrite the schroedinger equation such that it explicitly denotes the magnitude and angle of $\Psi(x,t)$? I.e. denote $$\Psi(x,t)=r(x,t)\cdot e^{i\theta(x,t)}.$$ Can we rewrite the schroedinger equation in terms of $r$ and $\theta$ instead of $\Psi$? (Consider only the one-dimensional case for simplicity).

Qmechanic
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user56834
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1 Answers1

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Schrödinger equation written as a product of a real modulus $R$ and a unitary phase $U$

$$\Psi(t,x)= R(t,x) U(t,x)$$ with $ R>0, U^* U = 1$, such that for a scalar wave function $U=e^{i\theta}$

one has

$$ i \hbar \partial_t R(t,x) U(t,x)= - \frac{\hbar ^2}{2m} \Delta (R(t,x) U(t,x)) + V R(t,x) U(t,x) $$

by the product rule

$$i \hbar \left( U \partial_t R + R\partial_t U\right) = -\frac{\hbar^2}{2m} \left(U \Delta R + R \Delta U + 2 \nabla R \cdot \nabla U \right) + V R U $$

By multiplication with the unitary factor $U^{-1} = U^*$ from the left

$$i \hbar \left( \partial_t R + R (U^* \partial_t U) \right) = -\frac{\hbar^2}{2m} \left( \Delta R \ + \ R \ (U^* \Delta U) \ + \ 2 \ \nabla R \cdot (U^* \nabla U) \right) + V R U $$

which can be written with the real phase $\theta(t,r)$

$$U^*\ \nabla U = e^{-i \theta} \ \nabla e^{i \theta} = i \nabla \theta$$

$$U^*\ \nabla \nabla U = e^{-i \theta} \ \nabla^2 e^{i \theta} = e^{-i \theta} \ \nabla \left( i \nabla \theta e^{i \theta} \right) = i \Delta \theta - (\nabla \theta)^2 $$

$$i \hbar \left( \partial_t R + i R \partial_t \theta \right) = -\frac{\hbar^2}{2m} \left( \Delta R \ + \ R \ (i \Delta \theta - (\nabla \theta)^2 ) \ + \ 2 \ i \ \nabla R \cdot ( \nabla \theta) \right) + V R U $$

This nonlinearity in the phase $\theta$ renders the form useless practically.

Roland F
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