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Difficult quantum mechanics question

  1. Oct 22, 2007 #1
    1. The problem statement, all variables and given/known data

    I need to prove:

    [tex]a'(z)=S(z)aS^\dagger(z)[/tex]

    Where [tex]S(z)[/tex] is the squeeze operator and [tex]a'(z)[/tex] is the pseudo-lowering operator.

    2. Relevant equations

    [tex]S(z)=e^{\frac{1}{2}({z^\ast}a^2-za^{\dagger 2})}[/tex]

    [tex]e^x=\sum_n{\frac{x^n}{n!}}[/tex] ; I don't think 2 simultaneous Taylor series expansions will get you very far here...

    -------------------------------------------------------------------

    [tex]a'={\mu}a+{\nu}a^\dagger[/tex]

    [tex]\mu=cosh(r)[/tex]

    [tex]\nu=sinh(r)e^{i\theta}[/tex]

    [tex]cosh(r)=F(\frac{1}{2}},\frac{x^2}{4}})[/tex]

    [tex]sinh(r)=xF(\frac{3}{2}},\frac{x^2}{4}})[/tex]

    where [tex]F(a,x)[/tex] is a power series expansion

    3. The attempt at a solution

    From the second set of equations (which operate on the left side of the original problem), I can make a'(z) into something that looks like a Taylor series expansion. I think you should use the operator expansion theorem now... except I don't know what it is, concretely, or how to use it. Please explain briefly or even better link to a site that provides an explanation (I looked hard: apparently QM isn't that popular).

    This is due on Wed and this site is a last resort. Please respond soon.
     
  2. jcsd
  3. Oct 23, 2007 #2
    How are tau and theta defined?
     
  4. Oct 23, 2007 #3
    theta is the angle that the squeeze state makes with one of the quadrature axes... I think it can be explained away by saying its absolute value is zero, but it's probably not that important. There is no tau; if you mean the 'r' in cosh(r), it is the displacement of the squeeze state or something, but it doesn't really matter because it goes away in the next step.

    I have almost given up on figuring this out in time... please help.
     
  5. Oct 23, 2007 #4
    Well, here's some things that might help:
    [tex]S(z)S^{\dagger} (z)= I^{\hat}[/tex]
    so S(z) is unitary
    [tex]e^{A+B}=e^A e^Be^{-1/2[A,B]}=e^B e^A e^{1/2 [A,B]}[/tex]
    provided that
    [tex][A,[A,B]]=[B,[A,B]]=0[/tex]
    and given the eigen value equation
    [tex]\hat{A}|a>=a|a>[/tex]
    Then
    [tex]e^{\hat{A}}|a> = e^a |a> [/tex]
    I tried it for a while and then I realized I had my own home work to do. Best of luck!

    Edit - This should help you out
    [tex]\langle\beta|\alpha\rangle=e^{-{1\over2}(|\beta|^2+|\alpha|^2-2\beta^*\alpha)}[/tex]
    for where [itex]|\alpha\rangle[/itex] and [itex]\beta\rangle[/itex]are eigenvectors of the [itex]\hat{a}[/itex] operator

    So try inner producting the original expression with arbitrary state vectors beta and alpha and see what you get.
     
    Last edited: Oct 23, 2007
  6. Oct 23, 2007 #5
    Did you try this formula?

    [tex] e^A B e^{-A} = B + [A,B] +\frac{1}{2!}[A,[A,B]] + \frac{1}{3!}[A,[A,[A,B]]] + \ldots [/tex]

    Eugene.
     
  7. Oct 23, 2007 #6
    Ohhhhhhhh. Don't listen to me. Listen to that guy /\/\/\
     
  8. Oct 23, 2007 #7
    YES!!! That's the operator expansion theorem that I couldn't find! Thank you so much meopemuk. And I truly appreciate your efforts [tex]\mu 3[/tex]. Meopemuk, do you know this equation off the top of your head or can you recommend a book? My quantum optics prof. assumes we know QM but we're all engineers and very much do not.

    Thanks again and I'm very open to reading recs.
     
  9. Oct 26, 2007 #8
    Hi guruoleg,

    this is probably the single most useful formula in quantum mechanics, because expressions like [itex] \exp(A)B\exp(-A) [/itex] are all over the place. It can be proven by power expansion of [itex] \exp(A) [/itex], which is kind of tedious. A more intelligent proof goes like this:

    Take the derivative of the left hand side

    [tex] \frac{d}{dx}B(x) = \frac{d}{dx} e^{Ax}Be^{-Ax} = Ae^{Ax}Be^{-Ax} - e^{Ax}Be^{-Ax}A = [A,e^{Ax}Be^{-Ax} ][/tex]

    So, [itex]B(x) [/itex] is a solution of the differential equation

    [tex] \frac{d}{dx}B(x) = [A,B(x) ][/tex]

    Next notice that the right hand side

    [tex] B + x[A,B] +\frac{x^2}{2!}[A,[A,B]] + \frac{x^3}{3!}[A,[A,[A,B]]] + \ldots [/tex]

    is also a solution of this differential equation. We still need to show that both sides are equal at x=0, which is trivial.

    Eugene.
     
    Last edited: Oct 26, 2007
  10. Oct 27, 2007 #9
    Won't simple Taylor expansion do the trick?

    i.e.,first order, [tex]e^{A}=1+A+...[/tex]

    [tex]e^{-A}=1-A+...[/tex]

    and

    [tex]e^A B e^{-A}=B+AB-BA-ABA+...=B+[A,B]+...[/tex]

    Edit: Lol, I just saw it's already been written!
     
  11. Oct 27, 2007 #10

    f95toli

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    Gold Member

    This expanision is known as the Baker-Hausdorff formula and as Meopemuk has already pointed out it is extremely useful. You should be able to find it in any QM book.
     
  12. Nov 4, 2007 #11
    Thank you for all your help :) I figured the problem out.

    ^-^
     
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