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Hermitian Conjugates of x, i, d/dx, and a+

  1. Nov 8, 2005 #1
    How do you find the hermitian conjugate of x, i, d()/d(x), a+ 'the harmonic oscilator raising operator'?
  2. jcsd
  3. Nov 8, 2005 #2
    1. What has to be if a+ wants to be a hermitian conjugato of the annihilation operator a?
    2. I'm sure you know what happens if you let a+ act on psy(harmonicOscillator) so you also know what happens if the hermitian conjugate acts on this wave function.
    3. You could actually derive this operators from the hermitian polynoms but this is rather a long way. Try to figure out what the annihilation operator must be.
  4. Nov 8, 2005 #3
    Then I guess i isn't an operator, its just a complex number. And if you let i act on a wavefunction you will not get any usefull informations, sinse i*psy=i*psy
  5. Nov 9, 2005 #4


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    Set the problem right, that is, search for dense everywhere domains from a chosen Hilbert space where you can define your initial operators. They can be unbounded. Then simply use the definition of the adjoint of an operator.

  6. Nov 11, 2005 #5
    huh? I'm a bit of a beginner.
  7. Nov 29, 2005 #6

    Tom Mattson

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    Forgive me, but this smells suspiciously like a homework question. So, I'm only going to give you some hints.

    Since [itex]\hat{x}[/itex] is Hermitian, this should be trivial.

    The Hermitian conjugate of any number is just its complex conjugate.

    How is [itex]\frac{d}{dx}[/itex] related to the momentum operator [itex]\hat{p}[/itex]? Use the fact that [itex]\hat{p}[/itex] is Hermitian and the answer to the previous part to get the Hermitian conjugate of this operator.

    How is [itex]\hat{a}^{\dagger}[/itex] related to the position and momentum operators [itex]\hat{x}[/itex] and [itex]\hat{p}[/itex]? Use that relationship, plus the fact that [itex]\hat{x}[/itex] and [itex]\hat{p}[/itex] are themselves Hermitian, to find the Hermitian conjugate of this operator.

    You can easily check your answer for this by using the fact that for any operator [itex]\hat{O}[/itex] the following is true.

    Last edited: Dec 2, 2005
  8. Feb 25, 2007 #7
    Maybe I am missing something obvious but how do you show that x is hermitian
  9. Dec 13, 2009 #8
    Prove X = XT
    1) <x'|X|x> = xDirac'sDelta(x' - x)
    2) Take conjugate
    <x|XT|x'> = x*Dirac'sDelta(x - x') = xDirac'sDelta(x' - x)
    3) the right-hand sides are iqual, so
    so, X = XT
  10. Feb 17, 2010 #9
    Maybe I'm not seeing something. Why did you insert a dirac delta?

    Shouldn't we be showing that <f|xf> = <xf|f> for any L2 integrable function f to show that x is hermitian? I'm not able to convince myself that x=xT is equivalent. It seems like you assumed that x is hermitian and showed that the necessary condition is for x=xT which is only the case for finite dimensional systems. In general we cannot assume that.
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