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More QM trouble

  1. Jul 29, 2005 #1
    More QM trouble:(

    Hi there,
    I'm having a bit of trouble with something dealing with functions of operators and commutators. It's two different examples actually. For the first one I have that A and B are hermitian operators and their expected values with respect to a normalized state vector /S> are <A>=<S/A\S>
    where [tex]
    \Delta
    [/tex]A=sqrt(A-<A>), and similarly for B.
    Now here's the thing I'm having trouble with. They're trying to derive the uncertainty relation in this book, and for part of it they say that [[tex]
    \Delta
    [/tex]A,[tex]
    \Delta
    [/tex]B]=[A,B]
    (these are the commutators)
    I know it's probably something really obvious but for some reason I don't see it. I tried actually writing out the full expression for the commutator but I don't see why they would be equal.

    As for the second problem I'm having, it has to do with functions of operators. Again I have 2 operators (this time not necessarily hermitian) that do not commute, [A,B] not equal to 0, which implies that [B,F(A)] (some function of A) is also not equal to 0. So here comes the parts I'm not sure of, they say that e^A*e^Bnot equal to e^(A+B), which I don't see why. How could I show this using a taylor expansion. I tried but didn't really get what I should have.
    They then proceed to say that e^A*e^b = e^(A+B)*e^[A,B]/2

    and also,
    e^A*B*e^-A= B+ [A,B] + 1/2![A,[A,B]]+ 1/3![A,[A,[A,B]]]+...

    neither of which I fully understand...I mean, I can see that the last one is a taylor series but I don't fully understand either of the last two things.
     
    Last edited: Jul 29, 2005
  2. jcsd
  3. Jul 29, 2005 #2

    OlderDan

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  4. Jul 29, 2005 #3
    Forgive me for not noticing that thread. Suprisingly that is the problem I was looking at. The thing is it didn't really answer my question. I'm sorry I know that there's probably something really stupid that I'm missing but I don't see this
     
    Last edited: Jul 29, 2005
  5. Jul 29, 2005 #4

    Gokul43201

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    Notice that [itex]\langle X \rangle[/itex] is just a scalar and [itex]IX = XI = X [/itex], so all the other terms cancel off nicely.
     
    Last edited: Jul 29, 2005
  6. Jul 29, 2005 #5
    Damnit, I knew it was something really simple. Those are things that always get me, the obvious or somewhat obvious things :cry:

    Is it something just as obvious that I missed concerning my second question with the exponentials?
     
  7. Jul 29, 2005 #6

    Gokul43201

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    First, Taylor expand the exponentials :

    [tex]a^A \cdot e^B = (I + A + \frac {A^2}{2!} + \frac {A^3}{3!} + ...)(I + B + \frac {B^2}{2!} + \frac {B^3}{3!} + ...)[/tex]
    [tex]=I + A + B + AB + \frac {A^2 + B^2}{2!} + ...
    = I + (A+B) + \frac {A^2 + B^2 + 2AB}{2!} + ... [/tex]

    But notice that, for instance
    [tex](A+B)^2 = A^2 + B^2+AB+BA = A^2 + B^2+2AB+BA-AB=(A+B)^2+[B,A] \neq (A+B)^2 [/tex]

    Use this above and hopefully, the result will follow. :smile:
     
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