Justification for Dropping Cross Terms in Transmon Hamiltonian

I am working through the "Introduction to Transmon Physics" page on Qiskit, and there is a part of the transmon Hamiltonian derivation that I am stuck on.

The initial form of the transmon Hamiltonian used in the document is

$$\hat H_{tr} = 4E_c \hat n^2-E_j \cos\hat \phi.$$

Next, ladder operators are introduced, and the Hamiltonian is approximated as

$$\hat H_{tr} \approx \sqrt{8E_cE_J}\left(\hat c^\dagger \hat c + \frac{1}{2}\right)-E_j-\frac{E_c}{12}\left(\hat c^\dagger + \hat c\right)^4$$

The next step is where I am having trouble. It is suggested that the terms with an uneven number of $$\hat c$$ and $$\hat c^\dagger$$ are "fast-rotating," and are therefore neglected. I can't figure out why terms like $$E_c/12(\hat c^\dagger\hat c \hat c^\dagger\hat c + h.c.)$$ (the terms that are kept) should be treated differently than terms like $$(\hat c \hat c \hat c \hat c + h.c)$$ (terms that are dropped). Can anybody explain why the cross terms can be dropped in the derivation?

• This is one of the most frustratingly poorly explained things in the field of quantum superconducting devices. It's ridiculous how often people wave their hands about this and don't explain it. I'm writing an answer. Commented Dec 18, 2022 at 2:15
• I can't help myself: Here is a paper showing that dropping those terms is actually a bad idea if you want to understand real experiments. Commented Dec 18, 2022 at 2:20
• @DanielSank I'm glad I'm not crazy. Thank you for the paper, I'm looking forward to going through it! Commented Dec 18, 2022 at 2:22
• Have you studied the "rotating frame" or "interaction picture" in quantum mechanics? Commented Dec 18, 2022 at 2:30
• @DanielSank Yes I have, although it's been a little while. Commented Dec 18, 2022 at 2:39

There are two aspects of this question:

1. Why can we drop terms with unequal numbers of $$\hat{c}$$ and $$\hat{c}^\dagger$$?
2. Why does the tutorial you're reading, and many other references, say that those terms are "fast rotating"?

Why you can drop the terms

Perturbation series

Let's write the Hamiltonian as $$H = H_0 + V$$ where $$H_0 = \sqrt{8 E_C E_J} \, c^\dagger c \qquad V = - \frac{E_C}{12} \left(c + c^\dagger\right)^4$$ and we've dropped the constants. Because the transmon is precisely the case where $$E_C \ll E_J$$, we can treat $$V$$ as a perturbation. Note that we're taking the energies and states of $$H_0$$ as the unperturbed, so we're perturbing away from a harmonic oscillator. The first order energy shift is $$\delta E_n = \left \langle \Psi_n | V | \Psi_n \right \rangle \tag{1}$$ where $$\Psi_n$$ are the unperturbed states. The unperturbed states of the harmonic oscillator have the property $$\left \langle \Psi_m | c | \Psi_n \right \rangle = \delta_{m,n-1} \sqrt{n} \qquad \left \langle \Psi_m | c^\dagger | \Psi_n \right \rangle = \delta_{m,n+1} \sqrt{n+1} \,.$$ Because of those deltas, the only terms in (1) that can be nonzero are those where there are an equal number of $$c$$ and $$c^\dagger$$. Therefore, to first order in perturbation theory, we can drop all terms in the perturbing Hamiltonian with unequal numbers of $$c$$ and $$c^\dagger$$.

Gershgorin Circle theorem

But have we really gained any intuition or even a convincing mathematical argument? Let's see if we can understand, from a mathematical point of view, why the perturbation series has only equal numbers of $$c$$ and $$c^\dagger$$ to first order.

Let's think about what the matrix $$H$$ looks like... but to write a matrix we have to choose a basis. Let's choose the harmonic oscillator basis, i.e. the basis of eigenvectors of $$c$$ and $$c^\dagger$$ (the same basis we used as the start of our perturbation series). Because we're working in that basis, the first term in the Hamiltonian has the usual Harmonic oscillator form (again dropping constants): \begin{align} \sqrt{8 E_J E_C} \, c^\dagger c &= \begin{pmatrix} 1 & 0 & 0 & \cdots \\ 0 & 2 & 0 & \\ 0 & 0 & 3 \\ \vdots & & &\ddots \end{pmatrix}\end{align} i.e. $$c^\dagger c$$ is just the number operator $$n$$. Now what about that second term $$-\frac{E_C}{12} \left( c + c^\dagger \right)^4 \, ?$$ Expanding this out, we get three kinds of terms:

1. Constants,
2. Terms with equal numbers (powers) of $$c$$ and $$c^\dagger$$
3. Terms with unequal numbers of $$c$$ and $$c^\dagger$$.

It's obvious that terms 1 contribute to the diagonal of the matrix, and just a little thought reveals that terms 2 also contribute only to the diagonal. One way to see this is like this: given a string with an equal number of $$c$$ and $$c^\dagger$$ we can commute the operators around to form them into $$c^\dagger c = n$$ pairs at the cost of introducing constants, because the commutator is $$[c, c^\dagger] = 1$$. Another way to see this is to just observe that a term with equal numbers of $$c$$ and $$c^\dagger$$ pushes the state up as many times as it pushes the state down, so we wind up where we started, which in a matrix representation corresponds to a diagonal element [1].

Then we have terms 3, which you now reason create off-diagonal elements. And now, crucially, if the off diagonal elements of a matrix are relatively small compared to the diagonals, it is a mathematical theorem that they don't affect the eigenvalues much [2]. A transmon is exactly a circuit where $$E_C \ll E_J$$, so we are indeed in a case where the off-diagonal terms are smaller than the diagonals, and so we can expect the eigenvalues (energies) and eigenvectors (states) of the transmon are mostly determined by the terms on the diagonal of this matrix representation of the Hamiltonian, which we've proven come from the Hamiltonian terms with equal numbers of $$c$$ and $$c^\dagger$$.

I'm going to post this as-is for now and come back later to add a section on why we call terms with unequal numbers of $$c$$ and $$c^\dagger$$ "fast rotating".

[1] The $$i^\text{th}$$ column of a matrix tells you what comes out when you act that matrix on the $$i^\text{th}$$ eigenvector.

• Thank you for the detailed explanation. I am much more satisfied with the dropping of terms when viewing it as a perturbation theory. Looking forward to the extension of the answer! Commented Dec 20, 2022 at 0:07