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Properties of Wave Functions and their Derivatives

  1. Jul 21, 2016 #1
    1. The problem statement, all variables and given/known data
    I am unsure if the first statement below is true.

    2. Relevant 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]

    3. 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.
     
  2. jcsd
  3. Jul 21, 2016 #2
    The first statement is true. Is your question not
    $$\frac{\partial\psi^*}{\partial x}\frac{\partial^2\psi}{\partial x^2} = \frac{\partial\psi}{\partial x}\frac{\partial^2\psi^*}{\partial x^2}?$$
     
  4. Jul 21, 2016 #3
    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).
     
  5. Jul 22, 2016 #4
    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 $$
    \int \frac{\partial^2 \psi}{\partial x^2}\frac{\partial \psi^*}{\partial x}dx=-\int \frac{\partial \psi}{\partial x}\frac{\partial^2 \psi^*}{\partial x^2}dx$$
     
  6. Aug 25, 2016 #5
    I apologize for not replying to this post. Thanks IanBerkman, you've been very helpful.
     
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