Can an annihilation operator be found for this Hamiltonian?

[wileecoyote]

Homework Statement

Given the Hamiltonian H(t) = \frac{P^2}{2m} + \frac{1}{2}mw^2X^2 + b(XP+PX) from some b>0. Find an annihilation operator a_b s.t. [a_b,a_b^{\dagger}]=1 and H = \hbar k (a_b^{\dagger}a_b+\frac{1}{2}) for some constant k. Hint: [P + aX,X]=[P,X], \forall a.

Homework Equations

none

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.

 

[Stealth95]

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 \displaystyle{a_b=AX+BP} for some constants \displaystyle{A,B \in \mathbb{C}^*}. Then \displaystyle{a_{b}^{\dagger}=A^{*}X+B^{*}P}, because \displaystyle{X} and \displaystyle{P} are Hermitian operators.

Now we can use the fact that \displaystyle{a_b} is annihilation operator and get some equations for \displaystyle{A,B}. 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 \displaystyle{k^2+4b^2=w^2} (this comes for the equations). But this limits the values of \displaystyle{b}, because \displaystyle{k \in \mathbb{R}\Rightarrow b\leq \frac{w}{2}}.

Maybe another person can help us more.

 

[wileecoyote]

Now we can use the fact that \displaystyle{a_b} is annihilation operator and get some equations for \displaystyle{A,B}. I did that the obvious way and I found an operator which works

What do you mean by the obvious way, I am not really sure what the obvious way to start this is.

 

[wileecoyote]

Please someone, help. I am so stuck.

 

[Stealth95]

What do you mean by the obvious way, I am not really sure what the obvious way to start this is.

If you use \displaystyle{[a_b,a_{b}^{\dagger}]=1} you get an equation for \displaystyle{A,B} (and their conjugates). Then you use \displaystyle{a_{b}^{\dagger}a_b=\frac{H}{\hbar k}-\frac{1}{2}}.

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

After doing this you have some equations for \displaystyle{A,B}. 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.

 

[hilbert2]

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.

 

[Stealth95]

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.

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:
\displaystyle{B=\frac{i}{\sqrt{2mk\hbar }}}
and then I used one of the equations to find \displaystyle{A}. Note that the value I chose for \displaystyle{B} 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 \displaystyle{k=\sqrt{w^2-4b^2}} ofcourse, as I mentioned above).