Drive frequency for second order quantum transitions

Question

Summary

$ \newcommand{\ket}[1]{\left \lvert #1 \right \rangle} \newcommand{\bra}[1]{\left \langle #1 \right \rvert} \newcommand{\braket}[2]{\left \langle #1 | #2 \right \rangle} \newcommand{\bbraket}[3]{\left \langle #1 | #2 | #3 \right \rangle} % $How do we show, from first principles, that the drive frequency $\omega_d$ needed to excite a two-photon transition from state $\ket{i}$ to state $\ket{f}$ is $\omega_d = (\omega_f - \omega_i) / 2$?

Setup

Consider a Hamiltonian $$H(t) = H_0 + V(t)$$ where the time dependent $H_0$ is considered "unperturbed" and $V(t)$ is considered a perturbation. Let us further suppose that $$V(t) = V_0 \, v(t) \, \mathcal{O} \, $$ where $V_0$ captures the magnitude of the perturbation, $v(t)$ captures the time dependence, and $\mathcal{O}$ is a time-indepdendent operator.

It's often useful to work in the interaction picture. In that picture, the Schrodinger equation becomes $$ i \hbar \partial_t \ket{\Psi'(t)} = V'(t) \ket{\Psi'(t)}$$ where $V'(t) \equiv U^{-1}(t) V(t) U(t)$, $U(t)$ is the propagator associated with $H_0$, and primes on the states indicates that the interaction picture states are related to the Schrodinger picture ones via $\ket{n'} = U^{-1}(t)\ket{n(t)}$.

Dyson series

We can solve the Schrodinger equation formally as $$\ket{\Psi'(t)} = \ket{\Psi'(0)} -\frac{i}{\hbar}\int_0^t dt' V'(t') \ket{\Psi'(t')} \, .$$ Plugging this equation into itself gives $$ \ket{\Psi'(t)} = \ket{\Psi'(0)} -\frac{i}{\hbar} \int_0^t dt' \, V'(t') \left[ \ket{\Psi'(0)} - \frac{i}{\hbar} \int_0^{t'} dt'' \, V'(t'') \ket{\Psi'(t'')} \right] \, . $$ So far, this equation is exact. Now suppose we iterate again, but this time make an approximation by keeping only the term where $\ket{\Psi'(t'')} = \ket{\Psi'(0)}$. Let's also hit the whole thing with $\bra{f}$ to compute the transition amplitude to $\ket{f}$. The result is $$ \braket{f}{\Psi'(t)} \approx \underbrace{\langle f \ket{\Psi'(0)}}_{0^\text{th}\text{ order}} - \underbrace{ \frac{i}{\hbar} \int_0^t dt' \, \langle f | V'(t') \ket{\Psi'(0)} }_{1^\text{st} \text{ order}} - \underbrace{\frac{1}{\hbar^2} \int_0^t \int_0^{t'} dt' \, dt'' \, \langle f | V'(t') V'(t'') \ket{\Psi'(0)} }_{2^\text{nd} \text{ order}} \, . $$

Using the following facts

• $V'(t) = U^{-1}(t)V(t)U(t)$

• For an eigenstate $\ket{n}$ of $H_0$ we have $U(t) \ket{n} = \exp(-i \omega_n t) \ket{n}$

• $\sum_n \ket{n} \bra{n} = \text{identity}$

we can reduce the second order term to \begin{align} \braket{f}{\Psi'(t)} = & - \left( \frac{V_0}{\hbar} \right)^2 \sum_n \int_0^t \int_0^{t'} dt' \, dt'' \\ & v(t') v(t'') \bbraket{f}{\mathcal{O}}{n} \bbraket{n}{\mathcal{O}}{i} \\ & \exp \left[ i \left( \omega_f t' - \omega_n t' + \omega_n t'' - \omega_i t'' \right) \right] \end{align} where we've used a shortcut notation $\ket{i} \equiv \ket{\Psi'(0)}$.

Question

I've heard that if $v(t)$ is a sinusoidal drive, e.g. $v(t) = \cos(\omega_d t)$, then the second order term we've computed has large magnitude when $\omega_d = (\omega_f - \omega_i)/2$. How can we show this?

Is there some way we can do either the sum or the integrals in some meaningful limits, e.g. when $t \rightarrow \infty$?

Answer 1

$ \newcommand{\ket}[1]{\left \lvert #1 \right \rangle} \newcommand{\bra}[1]{\left \langle #1 \right \rvert} \newcommand{\braket}[2]{\left \langle #1 | #2 \right \rangle} \newcommand{\bbraket}[3]{\left \langle #1 | #2 | #3 \right \rangle} % $Your final expression factorizes quite nicely as \begin{align} \braket{f}{\Psi'(t)} = & - \left( \frac{V_0}{\hbar} \right)^2 \sum_n \bbraket{f}{\mathcal{O}}{n} \bbraket{n}{\mathcal{O}}{i} \\ & \qquad \times \int_0^t \mathrm dt' v(t') \exp\left(i(\omega_f-\omega_n)t'\right) \\ & \qquad \quad \times \int_0^{t'} \mathrm dt'' v(t'') \exp\left(i(\omega_n-\omega_i)t''\right) , \end{align} and here (i) the first integral is essentially trivial, yielding some exponentials times a sinc function, while (ii) the second one is slightly more involved but still yields immediately to symbolic integration in Mathematica.

I don't see much point in repeating the full result here, but the crucial aspect is that if you fix the system frequencies ($\omega_i$, $\omega_f$ and $\omega_n$) and you scan over the pump frequency $\omega_p$, then you will get some form of bump around $(\omega_i+\omega_f)/2$ and this bump will become sharper and sharper in the $t\gg 2\pi/\omega$ limit:

(In this sample, $\omega_i=0$ and $\omega_f=1$, without loss of generality, and then $\omega_p=0.55+0.001i$, with the imaginary part added just to regularize the calculation. The sidebands are due to the precise location of $\mathrm{Re}(\omega_p)$, but the bright center at $\omega_p=0.5$ is the two-photon resonance process.)

If you want a more enlightening analytical result, it's better to drop the counter-rotating terms and just use an exponential form for the potential, $v(t) = e^{-i\omega_p t}$, under which the integrals will simplify significantly to $$ \frac{1}{\omega_i-\omega_n+\omega_p} \left[ \frac{e^{i(\omega_f-\omega_i-2\omega_p)t}-1}{\omega_f-\omega_i-2\omega_p} - \frac{e^{i(\omega_f-\omega_n-\omega_p)t}-1}{\omega_f-\omega_n-\omega_p} \right] , $$ which contains an explicit term in $\operatorname{sinc}((\omega_f-\omega_i-2\omega_p)t/2)$; this will converge at $t\to\infty$ to a Dirac delta on precisely the energy-conservation condition $2\omega_p=\omega_f-\omega_i$.

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Finally, if you have multiple intermediate levels $|n⟩$ contributing to the process, then the total amplitude will be the interference of all the individual amplitudes, but the energy-conserving term in $\operatorname{sinc}((\omega_f-\omega_i-2\omega_p)t/2)$ will factor out, and the process you're interested in will have an amplitude that reads \begin{align} \braket{f}{\Psi'(t)} \approx & - \left( \frac{V_0}{\hbar} \right)^2 \frac{e^{i(\omega_f-\omega_i-2\omega_p)t}-1}{\omega_f-\omega_i-2\omega_p} \sum_n \frac{\bbraket{f}{\mathcal{O}}{n} \bbraket{n}{\mathcal{O}}{i}}{\omega_i-\omega_n+\omega_p} , \end{align} in a form that should be pretty recognizable from multiple other places in perturbation theory. Here you get a sum of two-photon matrix elements over various intermediate states, divided by their detuning, with a global sinc-like energy-conservation factor multiplying the whole sum.

Answer 2

Emilio has already put a nice answer (it makes me feel that in my comment I made the problem sound much more complicated then it needs to be!). When I read the question, I understood it as asking, when you are driving a two-photon transition, if it is better to use $\omega_d \approx \frac{\omega_f - \omega_i}{2}$ than use other combinations such as, e.g. $\omega_{d_1} \approx \frac{\omega_f - \omega_i}{8}$ and $\omega_{d_2} \approx \frac{3(\omega_f-\omega_i)}{8}$, and I don't think Emilio addressed this point explicitly so I would like to add my explanation. I don't think much of what I write below (borrowing heavily from the book "Atom-Photon Interactions") will be different from what Emilio has already said, but maybe the presentation looks more formal.

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(Starting from page 28, equation (21) from API), we start with the expression for the second-order contribution to the transition amplitude $\mathcal{P}_{fi}^{(2)}$:

$\mathcal{P}_{fi}^{(2)} = \left(\frac{1}{i\hbar}\right)^2 \underset{-T/2 \leq \tau_1 \leq \tau_2 \leq T/2}{\int d\tau_1 d\tau_2} \sum_{k} V_{fk}V_{ki}\,\, e^{i(E_f - E_k) \tau_2 / \hbar} \, e^{i(E_k - E_i)\tau_1 / \hbar}$

The sum represents sum over intermediate states. To eliminate the ordering $\tau_1 \leq \tau_2$, we use the integral representation of Heaviside theta function:

$e^{-i E_k (\tau_2 - \tau_1) /\hbar} \theta (\tau_2 - \tau_1) = \underset{\eta \to 0_{+}}{\lim} -\frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{e^{-i E (\tau_2 - \tau_1) / \hbar} }{E + i\eta - E_k} dE$

Using this expression, we can re-write $\mathcal{P}_{fi}^{(2)}$ as

$\mathcal{P}_{fi}^{(2)} = \left(\frac{1}{i\hbar}\right)^2 \left(\frac{-1}{2\pi i}\right) \int_{-T/2}^{T/2} d\tau_2 \int_{-T/2}^{T/2} d\tau_1 \int_{-\infty}^{\infty} dE \,\, e^{i(E_f - E)\tau_2 / \hbar} e^{-i(E_i - E)\tau_1 / \hbar} \underset{\eta \to 0_{+}}{\lim} \sum_{k} \frac{V_{fk}V_{ki}}{E+i\eta - E_k}$

Now we need to generalize this to the situation when the interaction $\hat{V}$ has some time-dependence. The matrix element $V_{fk}$ will be written as $V_{fk}(\tau_2)$ and $V_{ki}$ as $V_{ki}(\tau_1)$. In the problem we consider, the explicit time dependence comes from the envelop $v(\tau)$. So $V_{fk}(\tau_1) = V_{fk}v(\tau_1)$ and so on.

For a sinusoidal drive $v(\tau) = A_1\sin (\omega_{d_1} \tau) + A_2 \sin (\omega_{d_2} \tau)$ we obtain sinc functions after we carry out the time integrals. As Emilio said, we can ignore the counter-rotating terms ($e^{+i\omega\tau}$). And I will also focus on contribution from both $\omega_{d_1}$ and $\omega_{d_2}$, assuming that both of them and their harmonics are detuned from resonance. With this approximation we get

$\left(\frac{2 \pi \hbar}{i\hbar}\right)^2 \left(\frac{-1}{2\pi i} \right) \int dE \,\delta^{(T)} (E-E_i - \hbar\omega_{d_1}) \delta^{(T)}(E+\hbar\omega_{d_2}-E_f)\underset{\eta \to 0_{+}}{\lim} \sum_{k} A_1 A_2\frac{V_{fk}V_{ki}}{E+i\eta - E_k} $

where the function $\delta^{(T)} (E_a - E_b)$ being

$\delta^{(T)} (E_a - E_b) = \frac{1}{2\pi \hbar} \int_{-T/2}^{T/2} d\tau \, e^{i(E_a - E_b) \tau / \hbar} = \frac{1}{\pi} \frac{\sin(E_f - E_i) T/ 2 \hbar}{E_f - E_i}$

obviously $\delta^{(T)}$ approximates a Dirac delta function at $T \to \infty$ and hence the notation.

Here we take a break from re-writing expressions and make some physical interpretations...The strength of the transition will be determined by the magnitude of the matrix elements and the energy widths of $W_{fi}(E) = \underset{\eta \to 0_{+}}{\lim} \sum_{k} A_1 A_2\frac{V_{fk}V_{ki}}{E+i\eta - E_k}$ and $\delta^{(T)}(E)$'s. Let's assume that there is no "special" intermediate state with large matrix element, so that we focus on the effects of the energy widths of those aforementioned functions. The $\delta^{(T)}(E)$'s are practically zero when $E > \hbar / T$. So we see that we get major contribution to the energy integral when $E$ lies in the range such that $E_i + \hbar\omega_{d_1} - E $ and $E_f-\hbar\omega_{d_2}-E$ are within $\hbar / T$ and $W_{fi} (E)$ varies slowly in the same range. We see that the more dissimilar $\omega_{d_1}$ and $\omega_{d_2}$ are, the harder it is to satisfy the overlap problem.

To complete the analysis, once you assume $W_{fi}(E)$ is slowly varying, you approximate it as $W_{fi}(E_m)$, where $E_m$ gives you the strongest overlap. If $\omega_{d_1} = \omega_{d_2}$, then $E_m$ is simply $\hbar\omega_d$. The integral of the product of delta function gives you another delta function that enforces the conservation of energy for two -photon transition (basically reduces to what Emilio wrote at the end).