# Homework Help: Properties of Wave Functions and their Derivatives

1. Jul 21, 2016

### Kyle.Nemeth

1. The problem statement, all variables and given/known data
I am unsure if the first statement below is true.

2. Relevant equations
$$\frac{\partial \psi^*}{\partial x} \frac{\partial^2 \psi}{\partial x^2}=\frac{\partial^2 \psi}{\partial x^2}\frac{\partial \psi^*}{\partial x}$$ Assuming this was true, I showed that $$\int \frac{\partial \psi^*}{\partial x} \frac{\partial^2 \psi}{\partial x^2}dx=\int \frac{\partial \psi}{\partial x} \frac{\partial^2 \psi^*}{\partial x^2}dx$$

3. The attempt at a solution
I am unsure whether it is true due to the fact that $$\psi$$ and $$\psi^*$$ may be complex.

2. Jul 21, 2016

### IanBerkman

The first statement is true. Is your question not
$$\frac{\partial\psi^*}{\partial x}\frac{\partial^2\psi}{\partial x^2} = \frac{\partial\psi}{\partial x}\frac{\partial^2\psi^*}{\partial x^2}?$$

3. Jul 21, 2016

### Kyle.Nemeth

That might be my question (although I didn't realize it). Here is what I did,

$$\int \psi^* \frac{\partial^3 \psi}{\partial x^3}dx=-\int \frac{\partial^2 \psi}{\partial x^2} \frac{\partial \psi^*}{\partial x}dx$$ 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 $$-\int \frac{\partial^2 \psi}{\partial x^2} \frac{\partial \psi^*}{\partial x}dx=-\int \frac{\partial \psi^*}{\partial x} \frac{\partial^2 \psi}{\partial x^2}dx$$ From here, I used integration by parts by letting $$u=\frac{\partial \psi^*}{\partial x}$$ and $$dv=\frac{\partial^2 \psi}{\partial x^2}dx$$ to arrive at the second statement I made in the OP (In the relevant equations section).

4. Jul 22, 2016

### IanBerkman

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 $$\int \frac{\partial^2 \psi}{\partial x^2}\frac{\partial \psi^*}{\partial x}dx=-\int \frac{\partial \psi}{\partial x}\frac{\partial^2 \psi^*}{\partial x^2}dx$$

5. Aug 25, 2016

### Kyle.Nemeth

I apologize for not replying to this post. Thanks IanBerkman, you've been very helpful.