Properties of Wave Functions and their Derivatives

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 2K views
Kyle.Nemeth
Messages
41
Reaction score
0

Homework Statement


I am unsure if the first statement below is true.

Homework Equations


[tex]\frac{\partial \psi^*}{\partial x} \frac{\partial^2 \psi}{\partial x^2}=\frac{\partial^2 \psi}{\partial x^2}\frac{\partial \psi^*}{\partial x}[/tex] Assuming this was true, I showed that [tex]\int \frac{\partial \psi^*}{\partial x} \frac{\partial^2 \psi}{\partial x^2}dx=\int \frac{\partial \psi}{\partial x} \frac{\partial^2 \psi^*}{\partial x^2}dx[/tex]

The Attempt at a Solution


I am unsure whether it is true due to the fact that [tex]\psi[/tex] and [tex]\psi^*[/tex] may be complex.
 
Physics news on Phys.org
The first statement is true. Is your question not
$$\frac{\partial\psi^*}{\partial x}\frac{\partial^2\psi}{\partial x^2} = \frac{\partial\psi}{\partial x}\frac{\partial^2\psi^*}{\partial x^2}?$$
 
That might be my question (although I didn't realize it). Here is what I did,

[tex]\int \psi^* \frac{\partial^3 \psi}{\partial x^3}dx=-\int \frac{\partial^2 \psi}{\partial x^2} \frac{\partial \psi^*}{\partial x}dx[/tex] I used the product rule for the integrand on the left hand side of the equation to show this statement. From here (using the assumption I originally asked about in the OP), I claimed that [tex]-\int \frac{\partial^2 \psi}{\partial x^2} \frac{\partial \psi^*}{\partial x}dx=-\int \frac{\partial \psi^*}{\partial x} \frac{\partial^2 \psi}{\partial x^2}dx[/tex] From here, I used integration by parts by letting [tex]u=\frac{\partial \psi^*}{\partial x}[/tex] and [tex]dv=\frac{\partial^2 \psi}{\partial x^2}dx[/tex] to arrive at the second statement I made in the OP (In the relevant equations section).
 
You could also perform this step immediately by $$
\int \frac{\partial}{\partial x}(\frac{\partial \psi}{\partial x}\frac{\partial \psi^*}{\partial x})dx= \int \frac{\partial^2 \psi}{\partial x^2}\frac{\partial \psi^*}{\partial x}dx+\int \frac{\partial \psi}{\partial x}\frac{\partial^2 \psi^*}{\partial x^2}dx$$
Because the L.H.S. is equal to zero at the infinities, we obtain $$
\int \frac{\partial^2 \psi}{\partial x^2}\frac{\partial \psi^*}{\partial x}dx=-\int \frac{\partial \psi}{\partial x}\frac{\partial^2 \psi^*}{\partial x^2}dx$$
 
I apologize for not replying to this post. Thanks IanBerkman, you've been very helpful.