# Is the conservation of probability in the Schroedinger's equation unique?

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?

• Hint: you know the greens function ("the solution") for $V=0$. Try squaring that and look if its norm is constant – Bort Feb 2 '16 at 14:52
• @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! – Zach466920 Feb 2 '16 at 15:00
• 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 $2a\Psi^2$? – ZeroTheHero Jul 27 '17 at 23:37
• Related 144832. – Cosmas Zachos Jul 28 '17 at 1:05

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.