Confusion: deriving momentum expectation value in QM

1. Aug 2, 2008

buffordboy23

On pages 16-17 of Griffith's Intro to QM, he writes the following:

$$\frac{d\left\langle x \right\rangle}{dt}=$$ $$\int x \frac{\partial}{\partial t}|\Psi|^{2} dx = \frac{i\hbar}{2m}\int x \frac{\partial}{\partial x} \left( \Psi^{*}\frac{\partial\Psi}{\partial x}- \frac{\partial\Psi^{*}}{\partial x}\Psi \right) dx$$

This expression can be simplified using integration by-parts:

$$\frac{d\left\langle x \right\rangle}{dt}= - \frac{i\hbar}{2m}\int \left( \Psi^{*}\frac{\partial\Psi}{\partial x}- \frac{\partial\Psi^{*}}{\partial x}\Psi \right) dx$$

(I used the fact that $$\partial x / \partial x = 1$$, and threw away the boundary term, on the ground that $$\Psi$$ goes to zero at (+/-) infinity.)

My two questions

1. I obtained the following intermediate form between these two equations:

$$\frac{d\left\langle x \right\rangle}{dt}= \frac{i\hbar}{2m} \left[ -\int \left( \Psi^{*}\frac{\partial\Psi}{\partial x}- \frac{\partial\Psi^{*}}{\partial x}\Psi \right) dx + x\left( \Psi^{*}\frac{\partial\Psi}{\partial x}- \frac{\partial\Psi^{*}}{\partial x}\Psi \right) \right|^{+\infty}_{-\infty} \right]$$

Is this correct?

EDIT: The second part doesn't quite make sense according to my current arguments. I will have to get back to you all. It was clear before I left my house but apparently not when I got home. Problems with Latex stole my focus. =)

2. Assuming the response is correct, how can the author make his claim that the second term equals 0?

$$x\left( \Psi^{*}\frac{\partial\Psi}{\partial x}- \frac{\partial\Psi^{*}}{\partial x}\Psi \right) = 0$$

I ask this because of the following supposition:

If $$\Psi$$ and $$\Psi^{*}$$ are even functions, then $$\partial \Psi / \partial x$$ and $$\partial \Psi^{*} / \partial x$$ must be odd functions (unless there is some function that defies this rule(?)). Therefore, the products $$x\Psi \partial \Psi / \partial x$$ and $$x\Psi^{*} \partial \Psi / \partial x$$ are even. But if $$f \left( x \right)$$ if an even function, then

$$\int^{+a}_{-a} f \left( x \right) \neq 0$$

always.

Mathematically, his conclusion only makes sense to me if $$\Psi$$ is of the general form

$$\Psi = A \psi \left( x \right) \psi \left( t \right)$$

where

$$\psi \left( x \right) = e^{-ax^{n}}$$

where

$$A$$, $$a$$, and $$n$$

are constants.

From my experience with QM, this general form is a common description for particle wave-functions. Is his claim based on physical grounds, which is analogous to how the potential energy of a gravitational or electromagnetic field equals 0 at infinity and allows him to neglect the other mathematical functions that are in contradiction, such as $$\Psi = Ax$$ ?

2. Aug 2, 2008

Irid

Normalization requires that

$$\int_{-\infty}^{\infty} \Psi^* \Psi \, dx = 1\, .$$

For bound states this means that

$$\lim_{x\rightarrow \pm \infty} \Psi(x) = 0$$

otherwise, the wave function isn't normalizable and doesn't correspond to a physical particle. Thus, your example of $$\Psi = Ax$$ defined over all space isn't a particle. The wave-function could have such form only in some localization of space.

The picture is a little different for scattering states, such as a free particle, but I haven't studied them too well (yet).

3. Aug 3, 2008

buffordboy23

Irid, I liked your description on bound states. It's been a while since I worked with QM, but things are slowly coming back into mind.

Let's talk about the free particle $$V_{0} = 0$$ in regards to Griffith's statement that the second term

$$x\left(\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^{*}}{\partial x}\Psi \right)\right|^{+\infty}_{-\infty} = 0$$

since $$\Psi\rightarrow0$$ at $$\pm\infty$$.

Now, for a free particle, $$\Psi = A \psi \left( x \right) \psi \left( t \right)$$, where $$\psi \left( x \right) = e^{ikx}$$ and $$A$$ and $$k$$ are constants. This wave-function is a complex exponential traveling wave. Mathematically, $$\Psi\Psi^{*}$$ is infinite and flat in extent and the wave-function is not normalizable. Physically, we would say this particle must be localized in some region of space and would use the method of box normalization in determining $$A$$--an example of such a situation is a proton beam from cyclotron. So in these scenarios, Griffith's is still correct in supposing that the second term goes to zero. Does anyone know if any such scenarios exist where we cannot make such an assumption? Basically, this scenario would describe a particle whose wave-function is not normalizable. Does it even make sense to ask such a question?

Last edited: Aug 3, 2008
4. Aug 3, 2008

nrqed

Physically, there is no such thing as a free particle of definite momentum. To be rigorous, one should always work instead with wavepackets which correspond to a superposition of different momenta. When we do scattering problems, we often work with momentum eigenfunctions but rigorously we should always work with wavepackets. However, since wavepackets are linear combinations of momentum eigenstates, it is still useful to do problems with momentum eigenstates because then it is straighforward to then extend the results to wavepackets. This is usually not shown in introductory QM classes.

5. Aug 3, 2008

buffordboy23

Thanks nrqed. So it appears that a wave-function for a particular momentum eigenstate may not necessarily be normalizable, but through the superposition of multiple eigenstates, interference occurs and localization of the particle is revealed.

6. Sep 13, 2009

Proofrific

I have a question about why

$$\left. x \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|^{+\infty}_{-\infty} = 0.$$

I understand that normalization requires that $$\Psi$$ goes to zero at $$\pm \infty$$. But, what about the x in front of the parenthesis? Doesn't it result in $$\infty \cdot 0$$, which is indeterminate?