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I have a question about a problem I was set recently.

The exercise reads:

The Hamiltonian of an harmonic oscillator driven by a classical force is $H=H_0+H_1$ with $$H_0=\hbar \omega \left( a^\dagger a+\frac{1}{2} \right) \text{ and } H_1=-i\left( \alpha a^\dagger - \alpha^\ast a\right).$$ Use the interaction picture and express the evolution operator $U\left( t\right) = \exp \left( -\frac{i}{\hbar}H_1t\right)$ using the displacement operator $D(\alpha )=\exp(\alpha a^\dagger -\alpha^\ast a)$

The solution reads:

Moving to an interaction picture using $\psi^S=U_0\psi^I$ with $U_0=\exp(-\frac{i}{\hbar}H_0t)$, one has $$H^I_1=U_0^\dagger H_1U_0=-i\alpha e^{i\omega t}a^\dagger+i\alpha^\ast e^{-i\omega t}a$$ and \begin{align} U^I & = \exp(i\omega t\;a^\dagger a)\exp[-\frac{1}{\hbar}(\alpha a^\dagger -\alpha^\ast a)t]\exp(-i\omega t\;a^\dagger a) \\ & =\exp[-\frac{1}{\hbar}(\alpha e^{i\omega t}a^\dagger -\alpha^\ast e^{-i\omega t} a)t] \\ & =D(\frac{1}{\hbar}\alpha e^{i\omega t}t). \end{align}

My question(s):

Why does $U(t)$ have this form? I understand $U^I$ satisfies $$i\hbar \partial_t U^I=H^I_1U^I$$ but $$[H^I_1(t_1), H^I_1(t_2)]=|\alpha |^2(e^{i\omega \Delta t}-e^{-i\omega \Delta t})=2|\alpha |^2\cos(\omega \Delta t)\neq 0.$$ Is the condition (to commute for different times) not necessary to have an exponential of $H^I$ as a solution to this equation?

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    $\begingroup$ I don't understand the downvote here, this is a good conceptual question which obviously satisfies the homework policy. $\endgroup$ – Mark Mitchison May 15 '15 at 10:29
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The essential mathematical point here is that $[H_1^I(t_1),H_1^I(t_2)]$ is a $c$-number, i.e. just a normal number, not an operator.

Now, it is not true that $U^I = e^{iH_0t/\hbar} e^{-iH_1t/\hbar}e^{-iH_0t/\hbar}$, this is not a solution to $i\hbar\partial_tU^I = H_1^I U^I$ unless $[H_1,H_0]=0$. The solution instead is usually written as $$ U^I(t) = T\exp \left( -\frac{i}{\hbar} \int_0^t\mathrm{d}s\, H_1^I(s)\right),$$ where $T\exp$ denotes the time-ordered exponential. This simply means that, in the Taylor expansion of the exponential, the product of operators $H_1(s)H_1(s')H_1(s'')\ldots$ must be always ordered so that $s>s'>s''$ etc.

In this case it is better to use an alternative expression for the solution, using the Magnus expansion, which looks like $$ U^I(t) = \exp \left(-\frac{i}{\hbar} \int_0^t\mathrm{d}s\,H_1^I(s) - \frac{1}{\hbar^2} \int_0^t\mathrm{d}s\,\int_0^s\mathrm{d}s'\,[H_1^I(s),H_1^I(s')]+\cdots \right ).$$ Notice that this is a normal exponential, not a time-ordered one. The higher-order terms which I have left out look something like integrals of $\sim [H_1^I(s),[H_1^I(s'),H_1^I(s'')]].$ In general, the $n^\mathrm{th}$ order term has $n-1$ nested commutators. Now the key is to realise that for your system, $[H^I(s),[H_1^I(s'),H_1^I(s'')]] = 0$, and all higher terms with even more nested commutators also vanish.

The first-order term corresponds to a displacement in phase space, but with a time-dependent displacement that rotates in the $x-p$ plane. The second term gives a phase factor arising from the fact that displacements in different directions do not commute. In many cases this phase factor leads merely to an unobservable global phase, and so it is possible to neglect it.

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    $\begingroup$ On the other hand, if you allow for state-dependent forces (i.e. an ion in its ground state is pulled to the left, and excited states go to the right) then this phase is crucial, and is e.g. the fundamental building block of the Mølmer-Sørensen gate. $\endgroup$ – Emilio Pisanty May 19 '15 at 14:04

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