cepheid
Aug1-04, 07:03 AM
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. :redface:
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 :biggrin:
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. :redface:
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 :biggrin: