# Substitute Schrödinger Equation by complex analysis, was the substitution correct?

Known that

$$\frac{\partial \Psi}{\partial t}=\frac{i\hbar }{2m}\frac{\partial^2\Psi}{\partial x^2}-\frac{i}{\hbar}V\Psi \tag 1$$

By taking the complex conjugate of the equation, obtain

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

where the derivative of the complex conjugate equal to complex conjugate of the derivative.

However, if we substitute $\Psi$ by $\Psi^*$ directly, we obtain $$\frac{\partial \Psi^*}{\partial t}=\frac{i\hbar }{2m}\frac{\partial^2\Psi^*}{\partial x^2}+\frac{i}{\hbar}V\Psi^* \tag 3$$

Thus if we compare the last two equation, then we obtain the following function that $$\partial_x^2\Psi=0 \tag 4$$ and $$\partial_t\Psi^*=\frac{i}{\hbar} V\Psi^*. \tag 5$$

My question was that:

1. Was equation 3 correct? I think it does because if the function was continuous differentialble in $C$, then by complex analysis we could rewrite the function as $\Psi(x+iy)=u(x,y)+iv(x,y)$.

2. What happened in equation 4 and equation 5? was they correct? especially, why $\partial_x^2\Psi=0$ the particle was not accelerate?

(This was not a homework question, because usually people only consider eq. 2 which was in the textbook.)

• I recommend that you edit your post to reword all the 'is my equation correct?' stuff for the reason that 'check my work' questions are considered off-topic and are generally down-voted and closed. – Alfred Centauri Apr 2 '18 at 0:43
• i do not understand what you are doing. if you just replace psi by psi* then the sign of the last term in eq. 3 is incorrect. but in any case it is not because psi satisfies an equation that psi* satisfies the same equation. – Oбжорoв Apr 2 '18 at 18:11
• I wasn’t sure either (that’s why I was asking the question ), $\Psi^*$ was the complex conjugate of the equation. In eq3 i simply replaced $\psi$ by $pis^*$ which could be done through complex analysis if assume equation was continuous and différentiable. – J C Apr 2 '18 at 18:15

The fact that some function $\Psi$ satisfies a differential equation does not imply that $\Psi^*$ satisfies the equation too, so everything after equation 2 is nonsensical.
It is, of course, possible that $\Psi^*$ also satisfies the differential equation, but that would be a special case. I'm not sure why you would think otherwise. For example, the equation $$y' + iy = 0$$ is solved by functions of the form $Ce^{-ix}$ but not by their complex conjugates.
• In a strange way I thought the same. However, I couldn't give me a proper proof. Because the complex conjugate of derivative=derivative of complex conjugate.(I have checked that in Mathstachexchange.) Thus $(\partial_t \Psi(x,t))^*=\partial_t (\Psi(x,t))^*$. If you couldn't accept that just yet, including me, then a usual trick was that $\Psi(x,t)$ could always be written as the combination of its real and imaginary part. i.e. $\Psi(t)=f(x,t)+ig(x,t)$ where $f,g,$ were real functions. Thus you could put it back into the equation – J C Apr 5 '18 at 12:32
• , even If $V$ dependent of $x,t$ it still worked. (If you couldn't write it analytically, you cold write it numerically, both worked,) – J C Apr 5 '18 at 12:32
• Your case was exactly what I said and even better, we have analytic solution to $f,g$ by using Euler identity, and if you check it's the same. – J C Apr 5 '18 at 13:02
• @JC I'm perfectly aware that the derivative of the conjugate is equal to the conjugate of the derivative, but that doesn't mean that the conjugate satisfies the same differential equation. $e^{-ix}$ satisfies the equation I wrote, but if you try to plug in $e^{+ix}$ then it doesn't work. – J. Murray Apr 5 '18 at 13:10