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Something in the derivation

  1. Oct 22, 2012 #1
    Hello guys,

    Check the attachment please.
    How is it what's written after "HENCE"?
    Where did 2*A^*<A> term go?
    the same thing for 2*B^*<B>??
     

    Attached Files:

  2. jcsd
  3. Oct 22, 2012 #2

    dextercioby

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    It was absorbed generating a term similar to the one already there, namely <A>^2 or <B>^2.

    <A>^2 - 2<A><A> + <A^2> = <A^2> + <A>^2 - 2 <A>^2 = <A^2> - <A>^2 and the same for B.
     
  4. Oct 22, 2012 #3
    thanks for your reply,

    but you wrote: <A>^2 - 2<A><A> + <A^2>
    it is supposed to be: <A>^2 - 2<A>A + A^2

    see figure please
     
  5. Oct 22, 2012 #4

    Cthugha

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    If you have a look at the figure you have posted, you will see that you are supposed to take the expectation value of that second line you just gave (<A>^2 - 2<A>A + A^2). Taking the expectation value of that line gives you the first line you just gave (<A>^2 - 2<A><A> + <A^2>).
     
  6. Oct 22, 2012 #5
    how come?
    i didn't get it, mind elaborating?
     
  7. Oct 22, 2012 #6
    The key is to remember that [itex]\langle A \rangle \equiv \langle \psi | A | \psi \rangle[/itex].

    So we have:
    [tex]\langle \psi | (\Delta A)^2 | \psi\rangle = \langle \psi | \hat{A}^2 - 2\hat{A}\langle \hat{A} \rangle + \langle \hat{A} \rangle ^ 2 | \psi \rangle \\
    = \langle \psi | \hat{A}^2 | \psi \rangle - 2\langle \hat{A} \rangle \langle \psi | \hat{A} | \psi \rangle + \langle \hat{A} \rangle ^2 \langle \psi | \psi \rangle \\
    = \langle \hat{A}^2\rangle - 2\langle \hat{A}\rangle \langle \hat{A} \rangle + \langle \hat{A} \rangle^2 \\
    = \langle \hat{A}^2\rangle - \langle \hat{A}\rangle ^2[/tex]

    Does that make sense?
     
  8. Oct 23, 2012 #7
    Yes Very!
    Thank you very much Chopin.
     
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