Properties of Wave Functions and their Derivatives

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Homework Help Overview

The discussion revolves around the properties of wave functions and their derivatives, particularly focusing on the validity of certain mathematical statements involving complex wave functions.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions the truth of a specific mathematical statement involving derivatives of wave functions, expressing uncertainty due to the complex nature of the functions. Other participants engage by reformulating the question and providing alternative mathematical approaches, including the use of integration by parts and the product rule.

Discussion Status

Participants are exploring various interpretations of the original question and discussing different mathematical techniques. Some have provided insights and alternative methods, but there is no explicit consensus on the validity of the original statement.

Contextual Notes

There is a mention of assumptions regarding the properties of complex wave functions, which may influence the discussion. The context of homework constraints is implied, as participants are navigating through mathematical reasoning without providing direct solutions.

Kyle.Nemeth
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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.
 
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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.
 

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