Is the Proof for Normalization in Quantum Mechanics Valid?

[Kyle.Nemeth]

Homework Statement

In Griffiths Introduction to Quantum Mechanics textbook, he shows that for any wave function that is time-dependent (which implies that the state of any particle evolves with time), the wave function will stay normalized for all future time. There is a step in the proof that I seem to intuitively understand, but seems like it may be invalid as it would violate the product rule of differentiation.

2. Homework Equations
Here is what he does in the book,

\frac {\imath\hbar}{2m}(\Psi^*\frac {\partial^2\Psi}{\partial x^2}-\frac {\partial^2\Psi^*}{\partial x^2}\Psi)=\frac{\partial}{\partial x}[\frac {\imath\hbar}{2m}(\Psi^*\frac {\partial\Psi}{\partial x}-\frac {\partial\Psi^*}{\partial x}\Psi)]

The Attempt at a Solution

Since the partial derivatives are operators, it doesn't make sense to me to have them factored out and sort of "skipping over" the psi* function (well, I mean the partial derivative that's being factored out of that first time with the psi* in it). Is it that the psi* and the derivative being multiplied to it are commutative, so that it might make more sense to factor out a partial derivative after switching their order?

 

[PeroK]

Here is what he does in the book,

\frac {\imath\hbar}{2m}(\Psi^*\frac {\partial^2\Psi}{\partial x^2}-\frac {\partial^2\Psi^*}{\partial x^2}\Psi)=\frac{\partial}{\partial x}[\frac {\imath\hbar}{2m}(\Psi^*\frac {\partial\Psi}{\partial x}-\frac {\partial\Psi^*}{\partial x}\Psi)]

Those two expressions are clearly equal. Simply differentiate the RHS.

 

[vela]

Since the partial derivatives are operators, it doesn't make sense to me to have them factored out and sort of "skipping over" the psi* function (well, I mean the partial derivative that's being factored out of that first time with the psi* in it). Is it that the psi* and the derivative being multiplied to it are commutative, so that it might make more sense to factor out a partial derivative after switching their order?

Your error is in assuming that Griffiths factored the partial derivative out. You're right that you can't do that, but that's not what he's doing. If you differentiate the righthand side, applying the product rule as necessary, you'll see you get the result on the lefthand side.

 

[Kyle.Nemeth]

Ahhhhh, I understand exactly. Thank you guys for the help

 

[Kyle.Nemeth]

Whoops, I have just one more question actually just to clear things up. Am I allowed to do something like this,

\frac{\partial \psi}{\partial x}\frac{\partial^2 \psi^*}{\partial x^2}+\frac{\partial \psi^*}{\partial x} \frac{\partial^2 \psi}{\partial x^2}=\frac{\partial}{\partial x}[\psi]\frac{\partial^2 \psi^*}{\partial x^2}+\frac{\partial}{\partial x}[\psi^*] \frac{\partial^2 \psi}{\partial x^2}=\frac{\partial}{\partial x}[\psi\frac{\partial^2 \psi^*}{\partial x^2}+\psi^* \frac{\partial^2 \psi}{\partial x^2}]

and say that I had factored out a derivative?

 

[PeroK]

Whoops, I have just one more question actually just to clear things up. Am I allowed to do something like this,

\frac{\partial \psi}{\partial x}\frac{\partial^2 \psi^*}{\partial x^2}+\frac{\partial \psi^*}{\partial x} \frac{\partial^2 \psi}{\partial x^2}=\frac{\partial}{\partial x}[\psi]\frac{\partial^2 \psi^*}{\partial x^2}+\frac{\partial}{\partial x}[\psi^*] \frac{\partial^2 \psi}{\partial x^2}=\frac{\partial}{\partial x}[\psi\frac{\partial^2 \psi^*}{\partial x^2}+\psi^* \frac{\partial^2 \psi}{\partial x^2}]

and say that I had factored out a derivative?

You're allowed to do it, in the sense that no one can stop you, but it would be entirely wrong. You seem to have got confused about the derivative being an operator and forgotten how it works.

 

[Kyle.Nemeth]

Well, thank you for portraying to me that I am incorrect in my thinking, as I was clearly under the wrong assumption that what I did was mathematically valid. Would you mind explaining why what I have done is incorrect?

 

[PeroK]

Well, thank you for portraying to me that I am incorrect in my thinking, as I was clearly under the wrong assumption that what I did was mathematically valid. Would you mind explaining why what I have done is incorrect?

If you differentiate the RHS, you will get four terms, two of which have derivatives of the 3rd order.

 

[Kyle.Nemeth]

I understand. Thank you again for your help.