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Unclear about the Heisenberg uncertainty principle

  1. Jun 5, 2007 #1
    Hi everybody,

    I have a slightly noobish question and I've searched for the answer unsuccesfully so I'm posting it here.

    I'm was reading up on the uncertainty priciple when I ran into two phrases that I don't quite understand. They are <A²> and <A>². Am I too apply this as [tex]\int \psi^* A^2 \psi d\tau[/tex] and [tex]\int(\psi^*A\psi)^2 d\tau[/tex]?

    Without understanding this the heisenberg uncertainty principle makes not much sense to me...
    Last edited: Jun 5, 2007
  2. jcsd
  3. Jun 5, 2007 #2

    Meir Achuz

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    Your <A^2> is correct, but [tex]<A>^2=[\int\psi^* A\psi d\tau]^2.[/tex]
  4. Jun 6, 2007 #3
    So [tex]\Delta A = \sqrt{\int{\psi^* A^2 \psi d\tau} - (\int{\psi^* A \psi d\tau})^2}[/tex]?
  5. Jun 6, 2007 #4


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    Yes. If you're wondering where that formula comes from, one starts by defining [itex]\Delta A[/itex] as the square root of the expectation value of the square of the deviation of A from <A>, i.e. as the standard deviation of A:

    [tex]\Delta A = \sqrt { < A - <A> >^2}[/tex]

    By multiplying out the square and collecting terms one arrives at

    [tex]\Delta A = \sqrt { <A^2> - <A>^2 }[/tex]

    which is usually easier to calculate.
  6. Jun 8, 2007 #5
    I start by defining delta (x) as the size of the single QM object (eigenvalue=1 of the s-aj number operator, which commute with the Hamiltonian). And I am wondering how it happens that it is closely related (but not identical) to the standard deviation of X which is the macroscopic quantity. Then I ask what the standard deviation of X describes in the classical statistical mechanics: the size of the statistical sub ensemble or the size of the volume (ball) occupied by that sub ensemble. Please explain all that.

    Regards, Dany.
    Last edited: Jun 8, 2007
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