Homework Help: Easy QM question

StNowhere · Sep 17, 2005

I guess this isn't really homework; I had homework problems from this book assigned in my QM classes in college, but this one was never assigned, probably because it's too trivial and I'm just trying too hard. Recently, though, I've decided to go through all of my old physics textbooks and do every last problem in them. I'm not too far along yet.

BTW, this problem comes from Griffiths' Intro to QM, 2nd ed. (1.16)

Problem states: show that d/dt of the integral from neg. to pos. inf. of Psi(1)*Psi(2) dx = 0 , where Psi(1) and Psi(2) are any (normalizable) solutions to Schrodinger. [Of course, by Psi(1)*, I mean complex conj. of Psi(1), which basically makes this problem "show that the time derivative of the inner product of two normalizable SE solutions is 0."]

I think I know of a way to do it using a method not mentioned in the first chapter of Griffiths, but I'm assuming he wouldn't have put it at the end of the first chapter unless you could solve it using those methods. That's what I'm looking for.

So far, this seems similar to the example of proving the time derivative of the inner product of a wave function with itself is zero, and so I've been trying to use the same method, but I keep ending up with nasty potentials that I'm not sure I can just blink away.

Maybe I'm trying to be a bit too pedantic, or maybe I'm just missing the point altogether. As I'm working through these on my own, I don't expect answers, but a nudge would be helpful.

inha · Sep 17, 2005

The first thought that came to my mind is that the solutions to the SE are orthogonal and the integral is zero for any [tex]\psi_n[/tex] and [tex]\psi_m[/tex] but orthogonality isn't presented at that point in Griffiths. If you follow how he proves that

[tex]d/dt \int dx\psi \psi*=0[/tex]

you can solve the problem with the same method and the potentials will cancel out.

MalleusScientiarum · Sep 17, 2005

The time evolution of a state is given by
[tex] \Psi(\mathbf{x},t) = \exp \left \{ \frac{-\imath}{\hbar} H t \right \} \Psi(\mathbf{x},0) [/tex]
Plug that into the above equation and see what happens.

StNowhere · Sep 17, 2005

I was pretty sure I had already tried that, but let me try to clarify my position just to make sure. (Beware my use of LaTeX, I'm not very good with it.)

Limits of integration suppressed: they are [tex]-\infty[/tex] to [tex]\infty[/tex]

Problem starts out: Show that [tex]d/dt \int \psi_1^*\psi_2 \ dx = 0[/tex] given that [tex]\psi_1[/tex] and [tex]\psi_2[/tex] are (normalizable solutions to the Schrodinger equation.

Here's what I've done so far.

Bring [tex]d/dt[/tex] inside the integral (as partial deriv.) to make the integral
[tex]\int \partial / \partial t \ (\psi_1^*\psi_2) \ dx[/tex]

The integrand expands under the product rule to become
[tex]\psi_1^* * (\partial\psi_2/\partial t) \ + \ (\partial\psi_1^*/\partial t) * \psi_2[/tex].

Griffiths has the following formulae:
[tex]\frac{\partial\psi_2}{\partial t} = \frac{\imath \hbar}{2m} \frac{\partial^2 \psi_2}{\partial x^2} - \frac{\imath}{\hbar} V_2 \psi_2[/tex] and

[tex]\frac{\partial\psi_1^*}{\partial t} = -\frac{\imath \hbar}{2m} \frac{\partial^2 \psi_1^*}{\partial x^2} + \frac{\imath}{\hbar} V_1 \psi_1^*[/tex]

Substituting these into the integrand expression yields:
[tex]\psi_1^* \ * \ (\frac{\imath \hbar}{2m} \frac{\partial^2 \psi_2}{\partial x^2} - \frac{\imath}{\hbar} V_2 \psi_2) \ + \ (-\frac{\imath \hbar}{2m} \frac{\partial^2 \psi_1^*}{\partial x^2} + \frac{\imath}{\hbar} V_1 \psi_1^*) \ * \ \psi_2[/tex]

I expanded the integrand expression and got
[tex]\frac{\imath \hbar}{2m} \psi_1^*\frac{\partial^2 \psi_2}{\partial x^2} - \frac{\imath}{\hbar}V_2\psi_1^*\psi_2 - \frac{\imath \hbar}{2m} \frac{\partial^2 \psi_1^*}{\partial x^2}\psi_2 + \frac{\imath}{\hbar}V_1\psi_1^*\psi_2[/tex]

Reordering and regrouping, and replacing the integral, I got to
[tex]d/dt \int \psi_1^*\psi_2 \ dx = \int [\frac{\imath \hbar}{2m} \ (\psi_1^* \ * \ \frac{\partial^2 \psi_2}{\partial x^2} - \frac{\partial^2 \psi_1^*}{\partial x^2} \ * \ \psi_2) - \frac{\imath}{\hbar}(V_2 - V_1) \psi_1^*\psi_2] \ dx[/tex]

I reason that the first term should be zero (double integration by parts), but I need the second term to be zero, and I can't justify that one (I can't just assume the same potential for two separate SE solutions, can I?)

Where should I go from here? Should I take a different tack altogether (remember, I'm trying to solve this the way I'm assuming Griffiths intended; so far, I don't have time evolution or orthogonality, so I'm using this to go about it) or am I wrong in my assumptions or reasoning? This is driving me crazy, so please help soon.

inha · Sep 18, 2005

I don't think Griffiths meant that they're solutions to two different potentials. If that was the case the problem probably would state so.

George Jones · Sep 18, 2005

I agree with inha - [itex]\psi_1[/itex] and [itex]\psi_2[/itex] are arbitrary quantum states of the *same* physical system.

It doesn't make sense to say, for example, that [itex]\psi_1[/itex] is a harmionic oscillator state and [itex]\psi_2[/itex] is a particle in a box state.

If [itex]\psi_1[/itex] and [itex]\psi_2[/itex] are solutions to different Schrodinger equations, then I think cooking up examples where the result is false shouldn't be difficult. In fact, choosing appropriate stationary states in my example seems to give a counterexample to the completely general result.

Regards,
George

StNowhere · Sep 18, 2005

You know, now that I think about it, that does make a lot of sense. I had a feeling one of my assumptions was wrong. Ahh. Now I can move on with a clear conscience. Thank you all very much.