Rotating wave approximation and conservation of energy

Question

So in my understanding, the RWA allows us to neglect terms that go as $e^{i(\omega+\omega_0)t}$ in an interaction Hamiltonian because if $\omega+\omega_0>>\omega_r$ where $\omega_r$ is the resonance frequency of the relevant interaction then the terms that oscillate with that high frequency tend to average out to zero over the relevant timespans. This makes sense to me.

However, I also tend to see the RWA invoked in going from an interaction Hamiltonian like $\hat{H}=\hbar\omega(\hat{a}^{\dagger}_i+\hat{a}_i)(\hat{a}^{\dagger}_j+\hat{a}_j)$ to the "approximate" interaction Hamiltonian $\hat{H}\approx\hbar\omega(\hat{a}^{\dagger}_j\hat{a}_i+\hat{a}^{\dagger}_i\hat{a}_j)$ where, for instance, the index $j$ corresponds to creation and annihilation operators for a cavity and the index $i$ corresponds to creation and annihilation operators for an atom (as in the Jaynes-Cummings model).

My confusion is that the terms that are getting approximated away from the "exact" Hamiltonian by the RWA like $\hat{a}_j^{\dagger}\hat{a}_i^{\dagger}$ seem to manifestly violate conservation of energy, so why would they ever be kept? And what does this approximation have to do with the version of the RWA that I'm familiar with pertaining to fast-oscillating terms?

Answer 1

You are being misled by thinking that the terms $a^{\dagger}_j a_i$ do conserve energy. What if the two systems $i$ and $j$ had different energy spacings? In the case of two oscillators, they can have different frequencies, and in the case of the light-matter interactions with the JCM the light can be off-resonance from the atomic transition. Then all of the terms seem to violate energy conservation!

This is all a misunderstanding: the energy that is conserved is of the entire system, including the interaction energy. This bears repeating: the interaction contributes to the total energy. Sure, on resonance it looks like $a^{\dagger}_j a_i$ transfers some unit of energy from system $i$ to system $j$, and that can be helpful for intuition, but one must always remember that one is mixing concepts.

In cases like these, there is an amount of energy contributed by each excitation in the free system $i$, an amount of energy contributed by each excitation in the free system $j$, and an amount of energy contributed by the interactions. Sometimes the energies contributed by the free systems are conserved but usually they are not, so terms like $a_j^\dagger a_i^\dagger$ are not problematic.

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Here is an additional intuition for how to think about why conservation of energy is related to what terms can be neglected or not. Think about the Heisenberg equations $$\frac{dA}{dt}=\frac{i}{\hbar}[H,A]$$ for the evolution of any operator $A$. In the interaction picture (rotating frame, not rotating wave approximation), the operator is actually $a_i(t)=a_i e^{-i\omega_i t}$, where we typically absorb the time dependence arising from the system Hamiltonians into the evolution of the creation/annihilation operators themselves such that they are "rotating" with frequencies $\omega_i$. For some operator $A$, we have $$\frac{dA}{dt}=\frac{i}{\hbar}[g(a_i a_j e^{-i(\omega_i+\omega_j) t}+a_i a_j^\dagger e^{-i(\omega_i-\omega_j) t}+a_i^\dagger a_j e^{-i(-\omega_i+\omega_j) t}+a_i^\dagger a_j^\dagger e^{-i(-\omega_i-\omega_j) t}),A],$$ where I use $g$ for the interaction strength to distinguish it from the bare system energy level spacings $\hbar\omega_i$. We can now get some intuition by thinking about what the equations of motion say classically and how they are affected by different terms in the Hamiltonian. If we consider classical dynamics and don't worry about the time dependence of $A$ or how it commutes with itself at different times, we would integrate the exponentials to find \begin{align}A(t)-A(0)&=\frac{i}{\hbar}\int_0^t dt [H,A]\\ &{``="} \frac{i}{\hbar}[g(a_i a_j \frac{e^{-i(\omega_i+\omega_j) t}-1}{-i(\omega_i+\omega_j)}+a_i a_j^\dagger \frac{e^{-i(\omega_i-\omega_j) t}-1}{-i(\omega_i-\omega_j)}+a_i^\dagger a_j \frac{e^{-i(-\omega_i+\omega_j) t}-1}{-i(-\omega_i+\omega_j) }+a_i^\dagger a_j^\dagger \frac{e^{-i(-\omega_i-\omega_j) t}-1}{-i(-\omega_i-\omega_j)}),A]. \end{align} When $\omega_i\approx \omega_j$, the contributions from the terms with $1/(\omega_i-\omega_j)\approx 1/0$ are much more significant than those with $1/(\omega_i+\omega_j)$. This is especially true in optics where the frequencies are large, so $1/(\omega_i+\omega_j)$ is very small anyway. The most relevant terms are those that conserve the most energy, with $\omega_i$ closest to $\omega_j$.

This classical handwaving is more exact for small evolutions $A(dt)-A(0)$ and is what aficionados directly understand from unitary operators with terms that oscillate very rapidly and others that oscillate slowly (really requiring the time-ordered exponential $U(t)=\mathcal{T}\exp(-i\int_0^t dt H(t)/\hbar)$). This is almost an identical calculation to what you see in the perturbation theory analysis and gives more of a dynamical intuition to why the fast-oscillating terms are those that don't conserve energy and thus are those that contribute less to the changes of any operator (including where $A$ is the quantum state itself).

Answer 2

Les us look at it from the point of view of time-independent perturbation theory: $$ H = H_0+H_1,\\ H_0=\sum_i\hbar\Omega_ia_i^\dagger a_i,\\ H_1=\sum_{i,j, j<i}\hbar\omega_{ij}(a_i^\dagger + a_i)(a_j^\dagger+a_j)= \sum_{i,j, j<i}\hbar\omega_{ij}(a_i^\dagger a_j^\dagger + a_i a_j + a_i^\dagger a_j + a_j^\dagger a_i) $$ The first order preturbative corrections in $H_1$ are zero, whereas the second order perturbative corrections are of the order of

• for terms $a_i^\dagger a_j, a_j^\dagger a_i$: $$ \frac{\hbar^2\omega_{ij}^2}{\hbar\Omega_i-\hbar\Omega_j} $$
• for terms $a_i^\dagger a_j^\dagger + a_i a_j$ $$ \frac{\hbar^2\omega_{ij}^2}{\hbar\Omega_i+\hbar\Omega_j}. $$ It is then clear that for frequencies $\Omega_i\approx\Omega_j$ (i.e., for small detuning) and for perturbation such that $\omega_{ij}/\Omega_i\ll 1$ the corrections due to resonance terms are much bigger than those due to the two-photon creation/annihilation. This is the same that we see in usual time-dependent perturbation theory.

Remarks: As has been pointed out in the other answer, energy conservation is helpful here for intuitive reasoning, but less obvious when applied rigorously - it must be applied to a process as a whole, not for separate terms in the approximation series. Moreover, in time-dependent problem the energy is manifestly not conserved - it is pumped into the system.

Answer 3

I'm going to add a long comment in the form of an answer for future learners spelling out a bit @RogerV. excellent answer.

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We start with the energy perturbation series up to second-order $$E_n(\lambda) = E_n^{(0)} + 0 + \lambda^2 E_n^{(2)} + \mathcal{O}(\lambda^3),$$ where the first order correction it's zero.

Next, we work on the second-order corrections due to the $a_i^\dagger a_j^\dagger$ terms $$ E_n^{(2)} = \sum_{n\neq m} \frac{|\langle n_0, n_1, \cdots | a^\dagger_i a_j^\dagger | m_0, m_1, \cdots \rangle|^2}{E_{m_0, m_1, \cdots}-E_{ n_0, n_1, \cdots}} = \sum_{n\neq m} \frac{(m_j+1)(m_i+1)|\langle n_0, n_1, \cdots | m_0, m_1, \cdots, m_i + 1, \cdots, m_j + 1, \cdots \rangle|^2}{E_{m_0, m_1, \cdots}-E_{ n_0, n_1, \cdots}} $$ we proceed with the algebra $$ E_n^{(2)} = \sum_{n\neq m} \frac{(m_j+1)(m_i+1)\delta_{n_0,m_0}\cdots\delta_{n_i, m_i + 1} \delta_{n_j, m_j + 1}\cdots}{(E_{m_0, m_1, \cdots}-E_{ n_0, n_1, \cdots})} = \frac{n_jn_i}{E_{n_0, n_1, \cdots, n_i,\cdots, n_j, \cdots}-E_{n_0, n_1, \cdots, n_i-1,\cdots, n_j-1, \cdots}} $$ and we spot that $$ E_n^{(2)} = \frac{n_jn_i}{n_0\omega_0+\cdots-n_0\omega_0-\cdots-(n_i-1)\omega_i-\cdots} = \frac{n_jn_i}{\omega_i + \omega_j}. $$ We conclude that, up to the second-order, the corrections due to the $a_i^\dagger a_j^\dagger$ terms are.

$$ E_n(\lambda) = \sum_j \omega_j n_j + \sum_{ij}\frac{\lambda^2}{\omega_i + \omega_j}n_jn_i + \mathcal{O}(\lambda^3) $$

On the other hand, it is now easy to see that the second-order corrections due to the $a_i^\dagger a_j$ terms are $$ E_n^{(2)} = \sum_{n\neq m} \frac{|\langle n_0, n_1, \cdots | a^\dagger_i a_j| m_0, m_1, \cdots \rangle|^2}{E_{m_0, m_1, \cdots}-E_{ n_0, n_1, \cdots}} = \sum_{n\neq m} \frac{(m_i+1)(m_j)|\langle n_0, n_1, \cdots | m_0, m_1, \cdots, m_i + 1, \cdots, m_j - 1, \cdots \rangle|^2}{E_{m_0, m_1, \cdots}-E_{ n_0, n_1, \cdots}} $$ we proceed with the algebra $$ E_n^{(2)} = \sum_{n\neq m} \frac{(m_i+1)(m_j)\delta_{n_0,m_0}\cdots\delta_{n_i, m_i + 1} \delta_{n_j, m_j - 1}\cdots}{(E_{m_0, m_1, \cdots}-E_{ n_0, n_1, \cdots})} = \frac{n_i(n_j+1)}{E_{n_0, n_1, \cdots, n_i,\cdots, n_j, \cdots}-E_{n_0, n_1, \cdots, n_i+1,\cdots, n_j-1, \cdots}} $$ and we spot that $$ E_n^{(2)} = \frac{n_j(n_i+1)}{n_0\omega_0+\cdots-n_0\omega_0-\cdots-(n_j+1)\omega_j-\cdots} = \frac{n_j(n_i+1)}{\omega_i-\omega_j}. $$ We conclude that, up to the second-order, the corrections due to the $a_i^\dagger a_j$ terms are

$$ E_n(\lambda) = \sum_j \omega_j n_j + \sum_{ij}\frac{\lambda^2}{\omega_i - \omega_j}n_j(n_i+1) + \mathcal{O}(\lambda^3). $$

As Roger V. pointed out, when $\omega \gg \lambda$ the corrections due to the $a_i^\dagger a_j$ and $a_i a_j^\dagger$ terms are large compared to the corrections due to the $a_i^\dagger a_j^\dagger$ and $a_i a_j$ terms and can be ignored.