# Time integral of a time-dependent Displacement operator

The diplacement operator on a bosonic mode with creation and annihilation operators, $$\hat{a}^\dagger,\hat{a}$$, is usually defined as $$\hat{D}(\alpha)=\exp(\alpha \hat{a}^\dagger - \alpha^*\hat{a})$$.

Suppose that $$\alpha$$ is time-dependent, then my question is how can one find the time integral of the displacement operator?

$$\int_0^t d\tau \hat{D}(\alpha(\tau)) = \int_0^t d\tau\exp(\alpha(\tau) \hat{a}^\dagger - \alpha^*(\tau)\hat{a})$$

My first thought is to write out the explicit defintion of the exponenital and stick to normal ordering, then use the binomial expansion to brute force compute it. So in the end integrating terms like $$(\alpha(\tau) \hat{a}^\dagger - \alpha^*(\tau)\hat{a})^k$$ using the binomial expansion.

I'm looking for tips if this is a correct way to approach this or possibly existing results before I spend time on this and it turns out to be an exercise in futility.

Edit: This question was closed because it needed more details or clarity - which I'm not sure how would affect the technical part of the question. The background for this question is that I am interested in calculating a first order Dyson series term of a time evolution operator (in the interaction picture) which has the form $$\hat{U}_I^{(1)}(t,0)=-\frac{i}{\hbar}\int_0^t d\tau \hat{H}^{(I)}(\tau)$$ where $$\hat{H}^{(I)}(t)$$ is the effective interaction picture Hamiltonian. In this first order term, there is a term involving exactly this integral $$\int_0^t d\tau \hat{D}(\alpha(\tau))$$. This is the background.

As it seems the general case cannot be explicitly tackled, I'm interested in the case where $$\alpha(t)$$ is linear in time, $$\alpha(t)=\alpha_0 t$$, and the case where it's as sinusoidal function in time, $$\alpha(t)=\alpha_0\cos(\omega t +\phi)$$. Any help for these two cases will be plenty.

• "Suppose that α is time-dependent... how can one find the time integral of the displacement operator?" It depends on the specific form of the time dependence. Also, the notation in your second equation does not make sense, you are integrating over $\tau$ on the LHS, but you still have $\tau$ dependence on the RHS.
– hft
Commented Jan 9, 2023 at 17:40
• @hft The specifics will depend on the specifics, as always, but it's a legitimate question to ask for methods to deal with problems with this structure. Commented Jan 9, 2023 at 17:53
• The final equation is badly formatted but I think the intent is quite clear. I'll edit. @LostInEuclid please check that I don't change your intent. Commented Jan 9, 2023 at 17:54
• @EmilioPisanty the final edit is the intent thanks for pointing this out, forgot the integral as a typo. Commented Jan 9, 2023 at 18:04
• @hft I'm interested in general methods to tackle this problem for arbitrary time dependence. But if you have insights only on particular specifications of it, I would be interested first in a linear dependence on time, i.e. $\alpha(t)=\alpha_0 t$, and second in a periodic time dependence, i.e. $\alpha(t)=\alpha_0\cos(\omega t + \phi)$. Commented Jan 9, 2023 at 18:06

Concretely, you can write thanks to closure relations : $$\hat{D}(\alpha(t)) = \int_\mathbb{C}\frac{\mathrm{d}^2\mu}{2\pi i}\int_\mathbb{C}\frac{\mathrm{d}^2\nu}{2\pi i}\,D_{\mu\nu}(\alpha(t))|\mu\rangle\langle\nu|$$ where $$\mathrm{d}^2\beta := \mathrm{d}\beta^*\wedge\mathrm{d}\beta = 2i\,\mathrm{d}\Re(\beta)\wedge\mathrm{d}\Im(\beta)$$ and $$\begin{array}{rcl} D_{\mu\nu}(\alpha) &=& \langle\mu|\hat{D}(\alpha)|\nu\rangle \\ &=&\displaystyle e^{-\frac{1}{2}|\nu|^2}\sum_{n\ge0}\frac{\nu^n}{\sqrt{n!}}\langle\mu|\hat{D}(\alpha)|n\rangle \\ &=&\displaystyle e^{-\frac{1}{2}|\nu|^2}\sum_{n\ge0}\frac{\nu^n}{n!}\langle\mu|\hat{D}(\alpha)(\hat{a}^\dagger)^n|0\rangle \\ &=&\displaystyle e^{-\frac{1}{2}|\nu|^2}\sum_{n\ge0}\frac{\nu^n}{n!}\langle\mu|\hat{D}(\alpha)(\hat{a}^\dagger)^n\hat{D}(\alpha)^\dagger\hat{D}(\alpha)|0\rangle \\ &=&\displaystyle e^{-\frac{1}{2}|\nu|^2}\sum_{n\ge0}\frac{\nu^n}{n!}\langle\mu|(\hat{a}^\dagger+\alpha^*)^n|\alpha\rangle \\ &=&\displaystyle e^{-\frac{1}{2}|\nu|^2}\sum_{n\ge0}\frac{\nu^n}{n!}(\mu^*+\alpha^*)^n\langle\mu|\alpha\rangle \\ &=&\displaystyle \exp\left(-\frac{1}{2}|\mu|^2-\frac{1}{2}|\nu|^2-\frac{1}{2}|\alpha|^2+\mu^*\alpha+\nu(\mu+\alpha)^*\right) \end{array}$$ where we used the facts that $$\hat{D}(\alpha)^\dagger\hat{a}\hat{D}(\alpha) = \hat{a}+\alpha$$ and $$\langle\mu|\alpha\rangle = e^{-\frac{1}{2}|\mu|^2-\frac{1}{2}|\alpha|^2+\mu^*\alpha}$$. Then you can integrate the components $$D_{\mu\nu}(\alpha)$$ with respect to time; in the case where $$\alpha(t) = at+b$$, you will be facing a gaussian integral.