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Uncertainty principle in terms of expectations values in Dirac notatio

  1. Feb 25, 2013 #1

    Leb

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    1. The problem statement, all variables and given/known data
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    [itex](\Delta A)^{2} = \langle \psi |A^{2}| \psi \rangle - \langle \psi |A| \psi \rangle ^{2}\\
    \phantom{(\Delta A)^{2} }=\langle \psi | (A - \langle A \rangle )^{2} | \psi \rangle ,[/itex]
    where [itex]\Delta A[/itex] is the uncertainty of an operator A and [itex]\langle A \rangle [/itex] is the expectation value of A.
    2. Relevant equations
    [itex] \langle \psi |A| \psi \rangle = \langle A \rangle [/itex]
    completeness relation: [itex] |\psi \rangle = \sum_{n}b_{n}|\psi_{n}\rangle [/itex]. We can use the completeness relation to expand out any wavefunction and derive an expression for the expectation value in terms of the weights in the expansion.


    3. The attempt at a solution

    I tried expanding the first line and got up to [itex] \sum_{n}\left( b_{n}^{2}a_{n}^{2} - b_{n}^{2}a_{n} \right) [/itex]. To be honest, I am jsut guessing the left result with the square term.
     
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  3. Feb 25, 2013 #2

    dextercioby

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    Re: Uncertainty principle in terms of expectations values in Dirac not

    You don't need the completeness relation. This is a standard calculation, if you assume everything to be well-defined mathematically. Start with the definitions and show where you get 'lost'.
     
  4. Feb 25, 2013 #3

    Leb

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    Re: Uncertainty principle in terms of expectations values in Dirac not

    To be honest I do not know the definition of it. It was just introduced as a notation and that is all I know. I've looked up online and it does not seem to me to have a formal definition, just notation.

    I checked that it has linearity and associativity properties, but do not see how they help.

    I have spent maybe 4 hours on this and am feeling like the idiot I am.

    Anyway, I tried expanding the second line to get

    [itex] \langle \psi|(A-\langle A \rangle )^2|\psi \rangle = \langle \psi|A^2 -2A\langle A \rangle + \langle A \rangle ^2 |\psi \rangle = \langle \psi|A^{2}|\psi \rangle + \langle \psi | \langle A \rangle ^2 |\psi \rangle - \langle \psi | 2A\langle A\rangle | \psi \rangle [/itex]
    Realizing that <A> should be a number (found it somewhere in these forums). How do I now get rid of -2A<A> and get - instead of +?
     
    Last edited: Feb 25, 2013
  5. Feb 25, 2013 #4

    dextercioby

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    Re: Uncertainty principle in terms of expectations values in Dirac not

    For A being a (compact self-adjoint) linear operator acting everywhere on a complex (separable) Hilbert space, the definition of [itex] \Delta A [/itex] for a (pure) state [itex] \psi [/itex] is

    [tex] \Delta A =: \sqrt{\left\langle \psi \left| \left(A-\langle A\rangle_{\psi}\hat{1}\right)^2 \right| \psi\right\rangle} [/tex]
     
  6. Feb 25, 2013 #5

    dextercioby

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    Re: Uncertainty principle in terms of expectations values in Dirac not

    The 3rd term is (-2) times the second one. Can you see why ?
     
    Last edited: Feb 25, 2013
  7. Feb 25, 2013 #6

    Leb

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    Re: Uncertainty principle in terms of expectations values in Dirac not

    Because [itex]\langle \psi | 2A\langle A\rangle | \psi \rangle = \langle \psi | 2A\langle\psi | A |\psi \rangle | \psi \rangle = \langle \psi | 2AA\langle\psi |\psi \rangle | \psi \rangle = \langle \psi | 2A^{2} | \psi \rangle = 2A^{2}\langle \psi | \psi \rangle [/itex] ?
    And it does not have to be a_{n}^{2} because we can choose whether we are interested in an eigenfunction or eigenvalue ?


    I think my problem is that I am not comfortable manipulating them.
     
  8. Feb 25, 2013 #7

    dextercioby

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    Re: Uncertainty principle in terms of expectations values in Dirac not

    Not really <A> is a number, so

    <psi|2A<A>|psi> = 2<A> <psi|A|psi> = 2<A> <A> = 2<A>2
     
  9. Feb 25, 2013 #8

    Leb

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    Re: Uncertainty principle in terms of expectations values in Dirac not

    Now I am confused even more... Where was my manipulation wrong ? I thought <A>=<psi|A|psi>... Or was your step simply correct in this particular situation due to insight ?

    Anyway, thank you for your help! At least I know how to solve it. Still not sure how I can manipulate them and why I cannot always just factor out something from inside the bra-ket.
     
    Last edited: Feb 25, 2013
  10. Feb 25, 2013 #9

    dextercioby

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    Re: Uncertainty principle in terms of expectations values in Dirac not

    Yes, but A is an operator, you can't equate A to a number such <psi|A|psi>, nor take it out of the bra-ket.
     
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