# Schrödinger Equation Solution's Conjugate

I've read a few other discussions of this topic (for example Confusion of Schrödinger equation and complex conjugates and Schrödinger's Equation and its complex conjugate), but I'm still confused about the relationship between a solution to the Schrödinger equation and that solution's conjugate.

Specifically, why is it the case that $$i\hbar\frac{\partial}{\partial t}\psi^*=\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi^*-V\psi^*$$ instead of $$i\hbar\frac{\partial}{\partial t}\psi^*=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi^*+V\psi^*~?$$

On the one hand I understand that if $z_1=z_2$, then $z_1^*=z_2^*$. The Schrödinger equation states the equivalence of two complex numbers. The first is

$$i\hbar\frac{\partial}{\partial t}\psi=i\hbar\frac{\partial}{\partial t}(\psi_x+i\psi_y)=-\hbar\frac{\partial}{\partial t}\psi_y+i\hbar\frac{\partial}{\partial t}\psi_x$$

and the second is

$$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi+v\psi=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_x+v\psi_x+i(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_y+v\psi_y)$$

If these two complex numbers are equal, then so are their conjugates

$$-\hbar\frac{\partial}{\partial t}\psi_y-i\hbar\frac{\partial}{\partial t}\psi_x=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_x+v\psi_x-i(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_y+v\psi_y)$$

So

$$-i\hbar\frac{\partial}{\partial t}(\psi_x-i\psi_y)=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}(\psi_x-i\psi_y)+V(\psi_x-i\psi_y)$$

and

$$i\hbar\frac{\partial}{\partial t}\psi^*=\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi^*-V\psi^*$$

On the other hand $i\hbar\frac{\partial}{\partial t}\psi^*$ can be calculated directly:

$i\hbar\frac{\partial}{\partial t}\psi^*=i\hbar\frac{\partial}{\partial t}(\psi_x - i\psi_y)$

$=i\hbar\frac{\partial}{\partial t}\psi_x-i(i\hbar\frac{\partial}{\partial t}\psi_y)$

$=(-\frac{\hbar^2 }{2m}\frac{\partial^2}{\partial x^2}\psi_x + V\psi_x)-i(-\frac{\hbar^2 }{2m}\frac{\partial^2}{\partial x^2}\psi_y + V\psi_y)$

$=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}(\psi_x-i\psi_y)+V(\psi_x-i\psi_y)$

$=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi^*+V\psi^*$

This contradicts the earlier result. I'm well aware this second argument is wrong, but I can't seem to find my mistake. Any help would be appreciated!

Your mistake is the following: you let $\psi_x$ and $\psi_y$ be the real and imaginary parts of the Schrodinger equation and assert that each individually obeys the Schrodinger equation, $$i \hbar \frac{\partial}{\partial t} \psi_x = H \psi_x, \quad i \hbar \frac{\partial}{\partial t} \psi_y = H \psi_y.$$ You can tell this isn't right by noting that the left-hand side is pure imaginary but the right-hand side is real. Instead start with the full Schrodinger equation, $$i \hbar \frac{\partial}{\partial t} (\psi_x + i \psi_y) = H (\psi_x + i \psi_y).$$ Taking the real and imaginary parts of both sides gives instead $$- \hbar \frac{\partial}{\partial t} \psi_y = H \psi_x, \quad \hbar \frac{\partial}{\partial t} \psi_x = H \psi_y.$$ This gives the right answer, thanks to the extra sign.