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Proof in heisenbergs uncertainty relation involving bra-ket

  1. Mar 31, 2010 #1
    hey, can someone show me the step between these two lines of equations please:

    [tex](\Delta A)^2=<\psi|A^2|\psi>-<\psi|A|\psi>^2[/tex]
    [tex]=<\psi|(A-<A>)^2|\psi>[/tex]

    where A is an operator and [tex]\psi[/tex] is the wavefunction and [tex]<A>[/tex] is the expectation value of A
     
  2. jcsd
  3. Mar 31, 2010 #2

    Doc Al

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    Staff: Mentor

    Just expand this expression, realizing that <A> is just a number. ([tex]<A> = <\psi|A|\psi>[/tex])
     
    Last edited: Mar 31, 2010
  4. Mar 31, 2010 #3
    thanks for quick reply! however im still not getting it. could you write it out expliitly please?
     
  5. Mar 31, 2010 #4

    Doc Al

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    [tex]<\psi|(A-<A>)^2|\psi> = <\psi|A^2 -2A<A> + <A>^2 |\psi>[/tex]

    I'll let you do the rest.
     
  6. Mar 31, 2010 #5

    Fredrik

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    Just a little LaTeX tip: Use \langle and \rangle instead of < and >. (Doc AI's answer is good, so I have nothing to add, except the complete solution, but you should try it yourself first. Note: "just a number" really means "just a number times the identity operator". OK, I guess I did have something to add :smile:).
     
  7. Apr 1, 2010 #6

    Doc Al

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    I thought those bras and kets looked a bit off. :uhh: (Thanks, Fredrik!)
     
  8. Apr 1, 2010 #7
    Must... not... say... I... prefer... bras... off... must... not say... to staff...ARGGGH.. Too late. :smile: Be gentle!
     
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