How Do Quantum Harmonic Oscillators Change Frequency?

[center o bass]

I'm working on a problem with a Harmonic oscillator which suddenly goes from the frequency ω_a to ω_b and I'm trying to find the expansion coefficients in

$$|0 \rangle _a = \sum_{n = 0}^\infty \alpha_n |n\rangle _b$$

where |0>_a is the ground state right before the change in frequency. The raising and lowering operators of the oscillators with different frequency are related by

$$a = cb + sb^\dagger, \quad a^\dagger = sb + cb^\dagger$$
and
$$b=ca - sa^\dagger, \quad a^\dagger = - sa + ca^\dagger$$

and they can also be expressed as related by the unitary transformation

$$U = e^{-\frac{\xi}{2}(a^2 - (a^\dagger)^2)} = e^{\frac{\xi}{2}(b^2 - (a^\dagger)^b)}.$$

where $s = \sinh \xi, \quad c = \cosh \xi$. I tried to find the coefficients by observing that

$$\alpha_n = \langle n|_b 0\rangle_a = \langle 0|_b \frac{b^n}{\sqrt{n!}}|0\rangle_a$$

but I failed to proceed from here. The correct answer is supposed to be that

$$\alpha_{2n} = (-\frac{s}{2c})^n \frac{\sqrt{2n!}}{n!} \alpha_0$$

and

$$\alpha_0^{-2}= \sum_{n=0}^\infty (\frac{s}{2c})^{2n}\frac{(2n)!}{(n!)^2}.$$

Could someone help me in showing these results? :)

 

[eljose79]

Hello there,

First of all, it's great to see that you are working on a problem with a Harmonic oscillator. It is a fundamental system in quantum mechanics and has many interesting applications.

Now, let's get to the problem at hand. You are trying to find the expansion coefficients in the equation |0>_a = ∑_(n=0)^∞ α_n|n>_b, where |0>_a is the ground state right before the change in frequency. The raising and lowering operators of the oscillators with different frequency are related by the equations given in the forum post.

To proceed, let's first express the unitary transformation U in terms of the raising and lowering operators a and b. We can do this by using the Baker-Campbell-Hausdorff formula, which states that e^(A+B) = e^Ae^B e^(-[A,B]/2), where [A,B] is the commutator of A and B. Applying this formula, we get:

U = e^(-ξ/2 (a^2-(a^†)^2)) = e^(-ξ/2 (b^2-(b^†)^2)) e^(-ξ[a,a^†]/2) e^(-ξ[b,b^†]/2)

= e^(-ξ/2 (b^2-(b^†)^2)) e^(-ξ/2) e^(-ξ/2 (b^2-(b^†)^2))

= e^(-ξ (b^2-(b^†)^2)) e^(-ξ/2) = e^(-ξ (b^2-(b^†)^2))

= e^(ξ(b^†)^2) e^(-ξ b^2)

Now, we can use the fact that b=ca-sa^† and b^†=sa+ca^† to express b^2 and (b^†)^2 in terms of a and a^†. We get:

b^2 = (ca-sa^†)^2 = c^2a^2 - 2csa^† - s^2a^†a

(b^†)^2 = (sa+ca^†)^2 = s^2a^†a + 2csa^† + c^2a^2