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Find annihilation operator

  1. Nov 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Given the Hamiltonian [itex]H(t) = \frac{P^2}{2m} + \frac{1}{2}mw^2X^2 + b(XP+PX)[/itex] from some [itex]b>0[/itex]. Find an annihilation operator [itex]a_b[/itex] s.t. [itex][a_b,a_b^{\dagger}]=1[/itex] and [itex]H = \hbar k (a_b^{\dagger}a_b+\frac{1}{2})[/itex] for some constant [itex]k[/itex]. Hint: [itex][P + aX,X]=[P,X], \forall a[/itex].


    2. Relevant equations
    none


    3. The attempt at a solution
    I am not sure how to go about this problem. I played around with the commutators but can't seem to get it. Any help is appreciated thanks.
     
  2. jcsd
  3. Nov 10, 2013 #2
    I recently started self-studying Quantum Mechanics so I am not really sure for my answer, but I can find an operator that works for some cases.
    I assumed that [itex]\displaystyle{a_b=AX+BP}[/itex] for some constants [itex]\displaystyle{A,B \in \mathbb{C}^*}[/itex]. Then [itex]\displaystyle{a_{b}^{\dagger}=A^{*}X+B^{*}P}[/itex], because [itex]\displaystyle{X}[/itex] and [itex]\displaystyle{P}[/itex] are Hermitian operators.

    Now we can use the fact that [itex]\displaystyle{a_b}[/itex] is annihilation operator and get some equations for [itex]\displaystyle{A,B}[/itex]. I did that the obvious way and I found an operator which works, but without knowing if it is unique. Also, this operator works only if [itex]\displaystyle{k^2+4b^2=w^2}[/itex] (this comes for the equations). But this limits the values of [itex]\displaystyle{b}[/itex], because [itex]\displaystyle{k \in \mathbb{R}\Rightarrow b\leq \frac{w}{2}}[/itex].

    Maybe another person can help us more.
     
  4. Nov 10, 2013 #3
    What do you mean by the obvious way, I am not really sure what the obvious way to start this is.
     
  5. Nov 10, 2013 #4
    Please someone, help. I am so stuck.
     
  6. Nov 11, 2013 #5
    If you use [itex]\displaystyle{[a_b,a_{b}^{\dagger}]=1}[/itex] you get an equation for [itex]\displaystyle{A,B}[/itex] (and their conjugates). Then you use [itex]\displaystyle{a_{b}^{\dagger}a_b=\frac{H}{\hbar k}-\frac{1}{2}}[/itex].

    But now it's not so trivial to find the equations for [itex]\displaystyle{A,B}[/itex], because there are also [itex]\displaystyle{X}[/itex] and [itex]\displaystyle{P}[/itex] in the equation. By the obvious way I meant that you simply equate coefficients of the same variables. For example, if you had:
    [tex]\displaystyle{CX^2+DP=EX^2+FP\Rightarrow \begin{Bmatrix}
    C=E\\
    D=F
    \end{Bmatrix}}[/tex]
    (I am not sure if I am losing some solutions with the above method)

    After doing this you have some equations for [itex]\displaystyle{A,B}[/itex]. If you find a solution that verifies all of them then the coresponding operator certainly works. But from these equations you get also the restriction I mentioned in the previous post. So I am not sure if my method is a general way to solve the problem. That's why I asked for help from someone else.
     
    Last edited: Nov 11, 2013
  7. Nov 11, 2013 #6

    hilbert2

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    Gold Member

    I tried to solve this in the same way as Stealth95, using trial ##a=AP+BX##, with ##A## and ##B## complex numbers. I got a nonlinear system of equations for the real and imaginary parts of ##A## and ##B## and I had to solve it with Mathematica. Maybe there's some other way that is easier to do by hand.
     
  8. Nov 11, 2013 #7
    Yes, the system is nonlinear. But based on the fact that the problem asks to find an annihilation operator I wasn't very strict with Maths. So I assumed that:
    [tex]\displaystyle{B=\frac{i}{\sqrt{2mk\hbar }}}[/tex]
    and then I used one of the equations to find [itex]\displaystyle{A}[/itex]. Note that the value I chose for [itex]\displaystyle{B}[/itex] is "stolen" from the annihilation operator for the harmonic oscillator potential.
    Although guessing solutions is not a good way to solve systems, luckily in our case the solution I get verifies all the equations of the system so it gives an operator (only when [itex]\displaystyle{k=\sqrt{w^2-4b^2}}[/itex] ofcourse, as I mentioned above).
     
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