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Ladder operator commutator with arbitary function

  1. May 9, 2016 #1
    Hey there!
    1. The problem statement, all variables and given/known data

    I've been given the operators
    [itex]a=\sqrt\frac{mw}{2\hbar}x+i\frac{p}{\sqrt{2m\hbar w}}[/itex] and [itex]a^\dagger=\sqrt\frac{mw}{2\hbar}x-i\frac{p}{\sqrt{2m\hbar w}}[/itex] without the constants and definition of the momentum operator:
    [itex]a=x+\partial_x[/itex] and [itex]a^\dagger=x-\partial_x[/itex] with [itex]\partial_x=\frac{\partial}{\partial x}[/itex]
    I have to calculate [itex][a,f(a^\dagger)][/itex], with f some arbitrary function.

    2. Relevant equations
    Product and chain rule for derivatives.
    Commutator definition.

    3. The attempt at a solution
    [itex][a,f(a^\dagger)]=(x+\partial_x)f(x-\partial_x)-f(x-\partial_x)(x+\partial_x)\\
    =xf(x-\partial_x)+\partial_x f(x-\partial_x) - f(x-\partial_x)x- f(x-\partial_x)\partial_x\\
    =xf(x-\partial_x)+(\partial_x f)(1-{\partial_x}^2)+f(x-\partial_x)\partial_x- f(x-\partial_x)x-f(x-\partial_x)\partial_x\\ =[x,f(a^\dagger)]+(\partial_x f)(1-{\partial_x}^2)[/itex] <-This is where I'm stuck :(

    Can this be simplified further? Is this even correct so far?
    Kind regards
    Alex
     
  2. jcsd
  3. May 9, 2016 #2

    nrqed

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    Are you instructed to use the expressions of ##a## and ##a^\dagger## in terms of x and ##\partial_x## or was this your idea? Because I would be tempted to simply write ##f(a^\dagger)## as a Taylor series and then use the commutator ##[a,(a^\dagger)^n]##.
     
  4. May 9, 2016 #3
    That was my idea. Sorry for putting it at the first point. It would make sense to do it that way. In the problem before on this assignment I calculated [itex][a,(a^\dagger)^n][/itex]. I'll try and do it your way, thanks :). I'll come back if I get stuck again.
     
  5. May 9, 2016 #4
    How does this look? I calculated [itex][a,(a^\dagger)^m]=2^{m-1}a^{m-1}[/itex].
    Then I did the taylor expansion [itex][a,f(a^\dagger)]=[a,\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}[(a^\dagger)^m,a]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}2^{m-1}a^{m-1}[/itex]
     
  6. May 9, 2016 #5

    nrqed

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    Are you sure about this??
     
  7. May 9, 2016 #6
    No, I'm not :) How is this?: [itex][a,(a^\dagger)^m]=2^{m-1}(a^\dagger)^{m-1}[/itex]? I wrote out the commutator for the first few m and found that [itex][a,(a^\dagger)^m]=\sum^{m-1}_{k=0}\binom{m-1}{k}(a^\dagger)^{m-k-1}[a,(a^\dagger)]a^k =(a^\dagger)^{m-1}\sum^{m-1}_{k=0}\binom{m-1}{k}=(a^\dagger)^{m-1}2^{m-1}[/itex]
     
  8. May 10, 2016 #7
    We simplified this commutator in the lecture today and got this: [itex][a,(a^\dagger)^n]=n(a^\dagger)^{n-1}[/itex] Can you see where my mistake in the previous post is?
     
  9. May 10, 2016 #8

    blue_leaf77

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    There is a simpler way though to calculate ##[a,(a^\dagger)^n]## using induction. You can start by calculating this commutator for ##n=0,1,2##. Then infer a pattern for this three values of ##n##'s. Next step is to make this pattern as a hypothesis and proceed through the standard induction procedure.
     
  10. May 10, 2016 #9
    That's how we did in the lecture. I don't understand though why I get a different result. I did like this, for [itex]n=3[/itex] using a commutation identity: [itex][a,(a^\dagger)^3]=[a,(a^\dagger)^2]a^\dagger+a^\dagger[a,(a^\dagger)^2]\\=([a,a^\dagger]a^\dagger+a^\dagger[a,a^\dagger])a^\dagger+a^\dagger([a,a^\dagger]a^\dagger+a^\dagger[a,a^\dagger])\\=(a^\dagger+a^\dagger)a^\dagger+a^\dagger(a^\dagger+a^\dagger)=2(a^\dagger)^2+2(a^\dagger)^2=4(a^\dagger)^2[/itex]
    This is how our prof did: [itex][a,(a^\dagger)^3]=a(a^\dagger)^3-(a^\dagger)^3a=(1+aa^\dagger)(a^\dagger)^2-(a^\dagger)^3a\\=(a^\dagger)^2+a^\dagger a a^\dagger a^\dagger-(a^\dagger)^3a\\=(a^\dagger)^2+a^\dagger(1+a^\dagger a)a^\dagger - (a^\dagger)^3a\\=(a^\dagger)^2+(a^\dagger)^2+(a^\dagger)^2aa^\dagger-(a^\dagger)^3a\\=2(a^\dagger)^2+(a^\dagger)^2(1+a^\dagger a)-(a^\dagger)^3a\\=3(a^\dagger)^2+(a^\dagger)^3a-(a^\dagger)^3a\\=3(a^\dagger)^2[/itex] which generalizes to [itex][a,(a^\dagger)^n]=n (a^\dagger)^{n-1}[/itex]
    Why do I get some other result, can't I use the identity [itex][A,BC]=B[A,C]+[A,B]C[/itex]? Or is it something else that I'm doing wrong.
     
  11. May 10, 2016 #10

    blue_leaf77

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    You made mistake in the first line already. There you have ##[a,(a^\dagger)^3] = [a,a^\dagger(a^\dagger)^2]##. Denoting ##A = a##, ##B = a^\dagger##, and ##C = (a^\dagger)^2##, how do they transform according to the identity ##
    [A,BC]=B[A,C]+[A,B]C##?
     
  12. May 10, 2016 #11
    Oh that's silly of me it has been bugging me the whole day, thank you :) [itex][a,a^\dagger (a^\dagger)^2]=a^\dagger[a,(a^\dagger)^2]+[a,a^\dagger](a^\dagger)^2[/itex]
    Well then using the correct result I get this for the original question:
    [itex][a,f(a^\dagger)]=[a,\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}[(a^\dagger)^m,a]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}m (a^\dagger)^{m-1}\\=\sum^\infty_{m=1}\frac{f^{(m)}(0)}{(m)!} (a^\dagger)^{m-1}\\=(a^\dagger)^{-1}\sum^\infty_{m=0}\frac{f^{(m)}(0)}{(m)!} (a^\dagger)^{m}-f(0)(a^\dagger)^{-1}\\=(a^\dagger)^{-1} f(a^\dagger)-f(0)(a^\dagger)^{-1}[/itex] Doesn't seem right though. Can I even simplify that series?
     
  13. May 10, 2016 #12

    blue_leaf77

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    Are the operators in the commutator above in the right order?

    Stop right there, how will you simplify ##\frac{m}{m!}##?
     
  14. May 10, 2016 #13
    [itex][a,f(a^\dagger)]=[a,\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}[a,(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}m (a^\dagger)^{m-1}=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{(m-1)!} (a^\dagger)^{m-1}[/itex] Can I return this to the original taylor approximation so that I can write [itex]f(a^\dagger)[/itex] instead of the series?
     
  15. May 10, 2016 #14

    blue_leaf77

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    Does the summation index really start from zero in the last expression?
    If you compare the last expression (after fixing the starting summation index) with the Taylor expansion of ##f(a^\dagger)##, which is
    $$
    \sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}(a^\dagger)^m,
    $$
    what do you think about the relation of them?
     
  16. May 10, 2016 #15

    nrqed

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    Yes, you should reexpress it in terms of ##f(a^\dagger)## but it is not the function itself you get, right? It is something else...And the end result is quite interesting.
     
  17. May 10, 2016 #16
    Well it can't start at 0 since [itex](-1)![/itex] isn't defined. The next reasonable index is 1. Ah, it's the derivative of the function at [itex]a^\dagger[/itex]. What I get is the shifted series. Am I right?
     
  18. May 10, 2016 #17
    [itex][a,f(a^\dagger)]=[a,\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}[a,(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}m (a^\dagger)^{m-1}=\sum^\infty_{m=1}\frac{f^{(m)}(0)}{(m-1)!} (a^\dagger)^{m-1}=\sum^\infty_{m=1}\frac{f^{(m)}(0)}{(m-1)!} (a^\dagger)^{m-1}=\sum^\infty_{m=0}\frac{f^{(m+1)}(0)}{(m)!} (a^\dagger)^{m}=\sum^\infty_{m=0}\frac{f'^{(m)}(0)}{(m)!} (a^\dagger)^{m}=f'(a^\dagger)[/itex]
     
  19. May 10, 2016 #18

    blue_leaf77

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    Yes.
     
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