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QM - Simple Questions

  1. Aug 1, 2004 #1


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    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:

    [tex]L = n \hbar [/tex]
    [tex]n = 1, 2, ...[/tex]

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

    [tex]\oint \vec{p} \cdot d \vec{r} = nh = 2 \pi \hbar n[/tex]

    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 [tex]\inline \vec{r} \times \vec{p}[/tex], so how do you arrive at this integral? Also, if the two conditions are supposed to be equivalent, then why does the former have [tex]\inline n \hbar[/tex], and the latter [tex]\inline 2\pi n \hbar[/tex]?

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

    [tex]\psi (x,t) = Ae^{i \frac{px-Et}{\hbar} }[/tex]

    And underneath:

    Note: [tex]\frac{\partial^2 \psi}{\partial x^2} = -\hbar^2 p^2 \psi (x,t)[/tex]

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

    [tex]\frac{\partial^2 \psi}{\partial x^2} = -\frac{p^2}{\hbar^2} \psi (x,t)[/tex]

    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:
  2. jcsd
  3. Aug 1, 2004 #2
    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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