Unclear about the Heisenberg uncertainty principle

  • Thread starter Nerro
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  • #1
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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...
 
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Answers and Replies

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

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.
 
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