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I Harmonic oscillator ladder operators

  1. Dec 2, 2016 #1
    The ladder operators of a simple harmonic oscillator which obey

    $$[H,a^{\dagger}]=\hbar\omega\ a^{\dagger}$$.

    ---

    I would like to see a proof of the relation

    $$\exp(-iHt)\exp(a^{\dagger})\exp(iHt)|0\rangle=\exp(a^{\dagger}e^{-i\omega t})|0\rangle\exp(i\omega t/2).$$

    Thoughts?
     
  2. jcsd
  3. Dec 2, 2016 #2

    DrClaude

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    Staff: Mentor

    As always, try Taylor expansion.
     
  4. Dec 2, 2016 #3
    I did this.

    ##\exp(-iHt)\exp\left( ca^{\dagger}\right)\exp(iHt)|0,0\rangle##

    ##= \exp\left( c[a^{\dagger},H]\right)|0,0\rangle##

    ##= \exp\left( ca^{\dagger}\exp(-i\omega t)\right)\exp(i\omega t)|0,0\rangle##

    Where have I gone wrong?
     
  5. Dec 2, 2016 #4

    DrClaude

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    I don't understand how you go from the first line to the second.
     
  6. Dec 2, 2016 #5
    I used Heisenberg's equation of motion for the ladder operator in the Heisenberg picture.
     
  7. Dec 2, 2016 #6

    vanhees71

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    The Heisenberg equations of motion are a good idea, but it's perhaps easier to use the Trotter product formula,

    http://theory.gsi.de/~vanhees/faq/quant/node99.html

    The link is in German, but the formula density should be high enough to get what you need. Note that in your case you can resum the resulting series explicitly.
     
  8. Dec 2, 2016 #7
  9. Dec 2, 2016 #8

    stevendaryl

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    Is this a homework problem?
     
  10. Dec 2, 2016 #9
    No, it's not! This is from a paper that I am reading.
     
  11. Dec 2, 2016 #10
    Sorry, I made a mistake. This is what I am trying to prove.

    ##\exp(-iHt)\exp(a^{\dagger})\exp(iHt)|0\rangle=\exp(a^{\dagger}e^{-i\omega t})|0\rangle##
     
  12. Dec 2, 2016 #11

    stevendaryl

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    Okay, so it might be slightly easier to work with the "number" operator [itex]N[/itex], which is related to [itex]H[/itex] via:

    [itex]H = (N+1/2) \omega[/itex] (in units where [itex]\hbar = 1[/itex])

    In terms of the number operator, [itex][N, a^\dagger] = a^\dagger[/itex].

    So we want to show that:

    [itex]e^{-i (N+1/2) \omega t} e^{a^\dagger} e^{+i (N+1/2) \omega t} |0\rangle = exp(a^\dagger e^{-i\omega t}) |0\rangle[/itex]

    Since [itex]e^{-i/2 \omega t}[/itex] and [itex]e^{+i/2 \omega t}[/itex] are just numbers, not operators, they commute, and they combine to form 1. So we need to show:

    [itex]e^{-i N \omega t} e^{a^\dagger} e^{+i N \omega t} |0\rangle = exp(a^\dagger e^{-i\omega t}) |0\rangle[/itex]

    Here, we can write [itex]e^{i N \omega t} = 1 + i N\omega t + 1/2 (i N \omega t)^2 + ...[/itex]. When this acts to the right on [itex]|0\rangle[/itex], we just get [itex]|0\rangle[/itex], because [itex]N |0\rangle = 0[/itex]. So we want to show:

    [itex]e^{-i N \omega t} e^{a^\dagger} |0\rangle = exp(a^\dagger e^{-i\omega t}) |0\rangle[/itex]

    Now, we write [itex]e^{a^\dagger}[/itex] as [itex]\sum_n (a^\dagger)^n/n![/itex]

    We can prove (by induction) that if [itex]|n\rangle[/itex] is the normalized state in which [itex]N |n\rangle = n |n \rangle[/itex], then [itex](a^\dagger)^n |0\rangle = \sqrt{n!} |n\rangle[/itex]

    So we have, on the left-hand side:

    [itex]e^{-i N \omega t} \sum_n \frac{\sqrt{n!}}{n!} |n\rangle [/itex]

    Now, you can bring the first exponent inside the sum to get:

    [itex]\sum_n \frac{\sqrt{n!}}{n!} e^{-i N \omega t} |n\rangle [/itex]

    Since [itex]N |n\rangle = n |n\rangle[/itex], this just becomes:

    [itex]\sum_n \frac{\sqrt{n!}}{n!} e^{-i n \omega t} |n\rangle [/itex]
    [itex]= \sum_n \frac{\sqrt{n!}}{n!} (e^{-i \omega t})^n |n\rangle [/itex]

    Now, we can undo the application of [itex](a^\dagger)^n[/itex]:

    [itex] \sum_n \frac{\sqrt{n!}}{n!} (e^{-i \omega t})^n |n\rangle = \sum_n (e^{-i \omega t})^n (a^\dagger)^n/n!|0\rangle[/itex]
    [itex] = \sum_n (a^\dagger e^{-i \omega t})^n/n!|0\rangle[/itex]
    [itex] = exp(a^\dagger e^{-i \omega t})|0\rangle[/itex]
     
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