4
$\begingroup$

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 $b$.

$\endgroup$
3
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/218983/2451 $\endgroup$
    – Qmechanic
    Commented Aug 29, 2016 at 21:56
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Aug 29, 2016 at 22:01
  • $\begingroup$ Related: physics.stackexchange.com/q/199515 $\endgroup$
    – Blazej
    Commented Aug 30, 2016 at 0:34

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.