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In the classic free particle equation (one spatial dimension),

$$\imath\frac{\partial \psi}{\partial t} = -\frac{\hbar}{2m}\frac{\partial^2 \psi}{\partial x^2}$$

if we expand the complex function as $\psi=a+ib$, then

$$\frac{\partial a}{\partial t} = -\frac{\hbar}{2m}\frac{\partial^2 b}{\partial x^2} $$

$$\frac{\partial b}{\partial t} = \frac{\hbar}{2m}\frac{\partial^2 a}{\partial x^2}. $$

If above is correct, time variations in $a$ are related to shape variations in $b$ and vise verse. It seems somewhat similar to the connection between electrical and magnetic fields (changes in one create changes in another).

Are there any discussions about possibility of a field that interacts with $a$ only but not $b$ or other way around? Basically considering $a$ and $b$ as some physical fields? Even $a$ interacting with $b$.

Qmechanic
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paradox
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    Related: http://physics.stackexchange.com/q/218983/2451 – Qmechanic Aug 29 '16 at 21:56
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    All physical observable are unchanged under a gauge transformation $\psi \rightarrow e^{\imath \alpha}\psi$. This transformation will mix your $a$ and $b$, so 1) the initial choice of $a$ and $b$ is arbitrary and 2) any field which couples exclusively to $a$ or $b$ will break this invariance. – By Symmetry Aug 29 '16 at 22:01
  • Related: http://physics.stackexchange.com/q/199515/ – Blazej Aug 30 '16 at 00:34

1 Answers1

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There is no way in standard quantum mechanics to have something "couple" to only the imaginary or only the real part of the wavefunction, since multiplication by a constant complex number does not change the physical state, so you can always make the "gauge choice" to have one of these components vanish at a point. Furthermore, you can typically choose a complete basis of purely real eigenfunctions, see this answer by EmilioPisanty, so it's very difficult to see any observable acting only on imaginary parts.

There is a much more meaningful way to split the time-dependent wavefunction, namely into magnitude and phase as $\psi(x,t) = \sqrt{\rho(x,t)}\mathrm{e}^{\mathrm{i}S(x,t)}$, where $\rho$ is the probability density and $S$ fulfills a modified Hamilton-Jacobi equation obtained from the Schrödinger equation by using the continuity equation for $\rho$ and the probability current $j = \frac{\rho}{m}\frac{\partial S}{\partial x}$:

$$ -\partial_t S = \frac{\partial_x S}{2m} + V(x) - \frac{\hbar^2}{4m\sqrt{\rho}}\partial_x\left(\frac{\partial_x \rho}{\sqrt{\rho}}\right)$$

This is one possible starting point for semi-classical approximations like the WKB method, since neglecting the term of order $\hbar^2$ gives back exactly the Hamilton-Jacobi equation of classical mechanics for a standard Hamiltonian.

ACuriousMind
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