# QM - Simple Questions

1. Aug 1, 2004

### cepheid

Staff Emeritus
Hello,

I've come across some confusing areas in my Quantum Mechanics notes. This is an introductory course, taken over the summer semester, so I don't know my stuff too well yet.

My prof's notes begin with the Bohr model of the atom, stating that the entire model comes from classical physics with the additional condition that angular momentum is quantized:

$$L = n \hbar$$
$$n = 1, 2, ...$$

Directly underneath, he states that this condition can be replaced by

$$\oint \vec{p} \cdot d \vec{r} = nh = 2 \pi \hbar n$$

I'm wondering what this quantity in the path integral represents. If the two conditions are truly equivalent, then it should represent angular momentum. But angular momentum $$\inline \vec{r} \times \vec{p}$$, so how do you arrive at this integral? Also, if the two conditions are supposed to be equivalent, then why does the former have $$\inline n \hbar$$, and the latter $$\inline 2\pi n \hbar$$?

Now we skip a few pages, and we get to the point where he has defined the wave function for an electron as:

$$\psi (x,t) = Ae^{i \frac{px-Et}{\hbar} }$$

And underneath:

Note: $$\frac{\partial^2 \psi}{\partial x^2} = -\hbar^2 p^2 \psi (x,t)$$

I don't know about you, but when I differentiated, I actually got:

$$\frac{\partial^2 \psi}{\partial x^2} = -\frac{p^2}{\hbar^2} \psi (x,t)$$

It's entirely possible that my prof merely needs a new pair of glasses. But if I'm the one in error and am incapable of performing basic tasks such as differentiation, then I'd like to know sooner rather than later

2. Aug 1, 2004

### marlon

I think your prof made a mistake in the derivation, your outcome is right...

The first question is solved in the attached word-doc, ok???

regards marlon

Last edited: Feb 9, 2006
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