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Math Help - Almost Done with Derivation of the Uncertainty Principle, but stuck!

  1. #1
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    Almost Done with Derivation of the Uncertainty Principle, but stuck!

    I would tremendously apreciate if anyone could please shed some light on a step in The uncertainty principle, in which I have gotten stuck.
    In the link provided:
    HOW DID THEY GO FROM Eq. (4.93) to (4.94) [Page #51] ... I can't seem to find how this happened? Thanks.

    http://farside.ph.utexas.edu/teaching/qmech/qmech.pdf
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  2. #2
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    I think I have some idea on how this was obtained. Just iluminate me onto something. If I have B<A> where B and A are operators and <A> refers to the mean value of A. If I do <B<A>> can this equal <B><A> also, Can <<A><B>> = <A><B> ... ? Thank you

    I know I am being very vague with this question, so forgive me if I commited any obvious mathematical flaw.
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  3. #3
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    You are correct.

    \langle B\langle A\rangle\rangle=\int_{-\infty}^{\infty}\psi^{*}B\langle A\rangle\psi\,dx=\langle A\rangle\int_{-\infty}^{\infty}\psi^{*}B\psi\,dx=\langle A\rangle\,\langle B\rangle.

    So what you have is that, since \langle A\rangle is just a number, it comes out of the integral.

    In Equation 4.93, on the RHS, if you expand out the product in-between the wave functions, you get

    AB-\langle A\rangle B-A\langle B\rangle+\langle A\rangle\,\langle B\rangle.

    When you take the expectation of that whole expression, the right three terms end up being like terms. So you just get

    -\langle A\rangle\,\langle B\rangle.

    That is, the positive term cancels out exactly one of the negative terms.

    Make sense?

    Thanks for posting a QM question, by the way. In my opinion, we don't get nearly enough of those around here!
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