The Schroedinger's equation can be viewed as a diffusion equation with imaginary constants $a$ and $b$ satisfying,

$$\quad \Psi_t=a \cdot \Delta \Psi-b \cdot V(x,t) \cdot \Psi \tag{1} $$ However if $a$ and $b$ are positive real coefficients, we get the standard diffusion equation.

Now it's standard fair to prove, $$\cfrac{d}{dt} \int |\Psi|^2 \ dr^3=0 \tag{2} $$ if $a$ and $b$ are imaginary. Is this true for the standard diffusion equation?

My (educated) guess is no. For the one dimensional case, the derivative can be brought inside and we get, $$\int 2 \cdot \Psi_t \cdot \Psi \ dr^3 $$ Using the known expression for $\Psi_t$ we get, $$\int \left(2 \cdot a \cdot \Psi_{xx} \cdot \Psi-2 \cdot b \cdot V \cdot \Psi^2\right) \ dr^3 \tag{3}$$

Using integration by parts and noting that $\Psi$ needs to go to zero at infinity (this is self evident right?) we get, $$\int \left(2 \cdot a \cdot \Psi^2-2 \cdot b \cdot V \cdot \Psi^2\right) \, dr^3 \tag{4}\, .$$

The first term is positive definite. The second term could easily be positive as well, so in general, the integral is time dependent.

Can a general proof for or against this be shown? In addition, assuming my argument is correct, are there cases where the integral in $(2)$ isn't time dependent?

  • 1
    $\begingroup$ Hint: you know the greens function ("the solution") for $V=0$. Try squaring that and look if its norm is constant $\endgroup$ – Bort Feb 2 '16 at 14:52
  • $\begingroup$ @Bort Thanks. The absolute value squared of the green's function for the Schroedinger equation is $1$. However, this isn't the case for the diffusion equation green's function. So I'm left to note that the integral would thus be time dependent. That does help, thanks! $\endgroup$ – Zach466920 Feb 2 '16 at 15:00
  • 1
    $\begingroup$ Could there be a typo in the first term of (4)? Going from (3) to (4) via integration by parts would produce $2a\Psi_x^2$ rather than $2a\Psi^2$? $\endgroup$ – ZeroTheHero Jul 27 '17 at 23:37
  • $\begingroup$ Related 144832. $\endgroup$ – Cosmas Zachos Jul 28 '17 at 1:05

You are misreading the celebrated analytic continuation to diffusion of the Schroedinger equation. This is probably possible by notational overkill induced by superfluous dimensions potentials, and absorbable constants that should not deny you any firm ground to stand on, but evidently do.

The crucial piece to appreciate is that, for diffusion, (in 1d, rescaled, free flow), $$ (\partial_t -\partial^2_x)~ \phi (x,t)=0, $$ which follows from the continuity equation (conservation of matter/substance whose density is $\phi$), you have, ipso facto, $$ \partial_t \int dx ~\phi =0, $$ the time derivative being a total divergence.

When you complexify/analytically continue, $\phi (x,it)\equiv \psi(x,t)$, to the free Schroedinger equation, $$ (-i\partial_t -\partial_x^2)~\psi(x,t)=0, $$ the complex conjugate such combines with it to yield $$ \partial_t (\psi^* \psi) =(\psi^* \partial_x^2 \psi -\psi \partial_x ^2\psi^*)=\partial_x(\psi^* \partial_x \psi -\psi \partial_x \psi^*), $$ also a total divergence, to which you may add inessential cancelling potential terms for free. The corresponding continuity equation then automatically produces $$ \partial_t \int dx ~ \psi^* \psi =0 . $$ by inspection.

Why on earth would you expect to replicate this structure on the real φ, so without the analyticity structure option of doubling up ? The answer for the analog of its square, (2), is of course not .

If you are keen on structure, reassure yourself by trying this.


For time independent potentials there is a field of mathematics studying this. The answer is no for physically reasonable potentials. Look for Schrodinger semigroup.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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