PSet 4 Discussion

Problem 4-1: Wavefunction corrections from Schrieffer-Wolff transformations

(a)

From the lecture notes, the Schrieffer–Wolff (SW) transformation gives an approximate effective Hamiltonian of the form

\[ \hat H'=\hat H_0-\frac{\lambda^2 g^2}{\omega_c-\Omega}\left(2\hat J_z\,\hat a^\dagger\hat a+\hat J_+\hat J_-\right), \]

where \(\hat a,\hat a^\dagger\) are the cavity ladder operators, and \(\hat{\mathbf J}\) is the collective spin of the \(N\) qubits (so \(\hat J_z=\frac{1}{2}\sum_{i=1}^N \sigma_z^{(i)}\) and \(\hat J_\pm=\sum_{i=1}^N \sigma_\pm^{(i)}\)). The “free” Hamiltonian \(\hat H_0\) is a sum of a cavity term and a qubit term, i.e. it is built only from \(\hat a^\dagger\hat a\) and \(\hat J_z\) (up to irrelevant constants).

Define the cavity photon-number operator

\[ \hat n\equiv \hat a^\dagger\hat a. \]

Also note a key structural fact: cavity operators act on the cavity Hilbert space, while \(\hat{\mathbf J}\) acts on the qubit Hilbert space, hence they commute:

\[ [\hat n,\hat J_z]=[\hat n,\hat J_\pm]=0. \]

workout

We now show the two commutators.

1. Commutation with total photon number \(\hat n\)

   First, since \(\hat H_0\) is built from \(\hat n\) and \(\hat J_z\), we have \([\hat n,\hat H_0]=0\). For the SW correction, use linearity of the commutator and check each term:

   • For the dispersive term \(2\hat J_z \hat n\):
   \[ [\hat n,\hat J_z\hat n]=[\hat n,\hat J_z]\hat n+\hat J_z[\hat n,\hat n]=0\cdot \hat n+\hat J_z\cdot 0=0. \]
   • For the spin-only term \(\hat J_+\hat J_-\), since \(\hat n\) commutes with both \(\hat J_+\) and \(\hat J_-\):
   \[ [\hat n,\hat J_+\hat J_-]=0. \]

   Putting these pieces together gives

   \[ [\hat n,\hat H']=0. \]

   So the effective Hamiltonian conserves cavity photon number.

2. Commutation with \(\hat J_z\)

   Again, \([\hat J_z,\hat H_0]=0\) because \(\hat H_0\) is built from \(\hat J_z\) and \(\hat n\), and \([\hat J_z,\hat n]=0\).

   For the SW correction:

   • The term \(\hat J_z \hat n\) is immediate:
   \[ [\hat J_z,\hat J_z\hat n]=[\hat J_z,\hat J_z]\hat n+\hat J_z[\hat J_z,\hat n]=0\cdot \hat n+\hat J_z\cdot 0=0. \]
   • For \(\hat J_+\hat J_-\), use the standard angular momentum commutators \([\hat J_z,\hat J_\pm]=\pm \hat J_\pm\) and the Leibniz rule \([A,BC]=[A,B]C+B[A,C]\):
   \[ \begin{aligned} [\hat J_z,\hat J_+\hat J_-] &=[\hat J_z,\hat J_+]\hat J_-+\hat J_+[\hat J_z,\hat J_-] \\ &=(+\hat J_+)\hat J_-+\hat J_+(-\hat J_-) \\ &=0. \end{aligned} \]

   Therefore

   \[ [\hat J_z,\hat H']=0. \]

Conclusion: simultaneous quantum numbers

Since \([\hat H',\hat n]=0\) and \([\hat H',\hat J_z]=0\), we can choose energy eigenstates of \(\hat H'\) that are simultaneously eigenstates of \(\hat n\) and \(\hat J_z\). Denoting their eigenvalues by

\[ \hat n\,|n,m\rangle=n\,|n,m\rangle,\qquad \hat J_z\,|n,m\rangle=m\,|n,m\rangle, \]

we can label the eigenstates by the photon number \(n\) and the \(J_z\) value \(m\) (with the understanding that there may be additional degeneracy labels within a fixed \((n,m)\) sector).

(b)

From the lecture notes, the unperturbed Hamiltonian is

\[ \hat H_0=\omega_c\,\hat a^\dagger \hat a+\frac{\Omega}{2}\sum_{j=1}^N \hat\sigma_z^{(j)} =\omega_c\,\hat n+\Omega\,\hat J_z, \]

where \(\hat n=\hat a^\dagger \hat a\) and \(\hat J_z=\frac{1}{2}\sum_{j=1}^N \hat\sigma_z^{(j)}\).

The SW-transformed Hamiltonian to order \(\lambda^2\) is

\[ \hat H'=\hat H_0-\frac{\lambda^2 g^2}{\omega_c-\Omega}\left(2\hat J_z\,\hat n+\hat J_+\hat J_-\right). \]

Let \(|n,m\rangle\) be a simultaneous eigenstate of \(\hat n\) and \(\hat J_z\):

\[ \hat n\,|n,m\rangle=n\,|n,m\rangle,\qquad \hat J_z\,|n,m\rangle=m\,|n,m\rangle. \]

workout

Then the contribution from \(\hat H_0\) is immediate:

\[ \hat H_0|n,m\rangle=(\omega_c n+\Omega m)\,|n,m\rangle. \]

The dispersive interaction term is also immediate:

\[ (2\hat J_z\hat n)\,|n,m\rangle = 2mn\,|n,m\rangle. \]

It remains to evaluate the eigenvalue of \(\hat J_+\hat J_-\). Since \([\hat J_+\hat J_-,\hat J_z]=0\), we can choose a basis that also diagonalizes \(\hat J^2\), i.e. \(|j,m\rangle\) with

\[ \hat J^2|j,m\rangle=j(j+1)\,|j,m\rangle,\qquad \hat J_z|j,m\rangle=m\,|j,m\rangle. \]

Using the standard ladder-operator action,

\[ \hat J_-|j,m\rangle=\sqrt{j(j+1)-m(m-1)}\,|j,m-1\rangle, \]

workout

we get

\[ \hat J_+\hat J_-|j,m\rangle=\left(j(j+1)-m(m-1)\right)|j,m\rangle. \]

Therefore, for a state with photon number \(n\) and \(J_z\) value \(m\) (and total-spin label \(j\)),

\[ E'_{n,m;j} =\omega_c n+\Omega m-\frac{\lambda^2 g^2}{\omega_c-\Omega}\left(2mn+j(j+1)-m(m-1)\right). \]

In the common collective-spin setting used in the notes (fully symmetric Dicke manifold), one has a fixed \(j=N/2\), so the energy can be written purely as a function of \((n,m)\):

\[ \boxed{ E'_{n,m} =\omega_c n+\Omega m-\frac{\lambda^2 g^2}{\omega_c-\Omega}\left(2mn+\frac{N}{2}\left(\frac{N}{2}+1\right)-m(m-1)\right) }. \]

(c)

Let the SW transformation be defined by a Hermitian generator \(\hat S=\hat S^\dagger\) via the unitary \(\hat U \equiv e^{-i\lambda \hat S}\), so that

\[ \hat H'=\hat U^\dagger \hat H \hat U = e^{i\lambda \hat S}\,\hat H\,e^{-i\lambda \hat S}. \]

workout

1. Since \(\hat H'\) is related to \(\hat H\) by a unitary similarity transformation, \(\hat H\) and \(\hat H'\) have exactly the same spectrum for any \(\lambda\).

   To see this explicitly: if \(\hat H\ket{\psi}=E\ket{\psi}\), then

   \[ \hat H'\left(\hat U^\dagger\ket{\psi}\right) =\hat U^\dagger \hat H \hat U\left(\hat U^\dagger\ket{\psi}\right) =\hat U^\dagger \hat H\ket{\psi} =E\left(\hat U^\dagger\ket{\psi}\right). \]

   Thus \(E\) is also an eigenvalue of \(\hat H'\).

2. In practical SW we do not keep \(\hat H'\) exactly; instead we expand in \(\lambda\) and truncate at \(O(\lambda^2)\), producing an approximate effective Hamiltonian \(\hat H'_{(2)}\) such that

   \[ \hat H'=\hat H'_{(2)}+\hat R,\qquad \hat R=O(\lambda^3). \]

   Now treat \(\hat R\) as a perturbation on \(\hat H'_{(2)}\). For an eigenstate \(\ket{\phi}\) of \(\hat H'_{(2)}\), the leading correction to its eigenvalue from \(\hat R\) is the usual first-order expectation value \(\bra{\phi}\hat R\ket{\phi}\), which is \(O(\lambda^3)\). Therefore the eigenvalues of \(\hat H'_{(2)}\) differ from the exact eigenvalues of \(\hat H'\) only at \(O(\lambda^3)\), and hence (since \(\hat H\) and exact \(\hat H'\) have identical spectra) we conclude:

   \[ E_{\hat H}=E_{\hat H'_{(2)}}+O(\lambda^3). \]

   Equivalently, \(\hat H\) and the truncated \(\hat H'\) have the same energy eigenvalues through order \(\lambda^2\).

(d)

Let \(\ket{n,m}\) be an eigenstate of the (truncated) effective Hamiltonian \(\hat H'_{(2)}\) with eigenvalue \(E\):

\[ \hat H'_{(2)}\ket{n,m}=E\ket{n,m}. \]

The defining SW relation implies (exactly)

\[ \hat H=\hat U\,\hat H'\,\hat U^\dagger \quad\text{with}\quad \hat U=e^{-i\lambda \hat S}. \]

If we ignore \(O(\lambda^3)\) differences between \(\hat H'\) and \(\hat H'_{(2)}\), the corresponding eigenstate of the original Hamiltonian \(\hat H\) is

\[ \ket{\psi_{n,m}} \equiv \hat U\,\ket{n,m}=e^{-i\lambda \hat S}\ket{n,m}, \]

because then

\[ \hat H\ket{\psi_{n,m}} =\hat U\,\hat H'\,\hat U^\dagger \hat U\ket{n,m} =\hat U\,\hat H'\ket{n,m} \simeq \hat U\,E\ket{n,m} =E\ket{\psi_{n,m}}. \]

We want \(\ket{\psi_{n,m}}\) to first order in \(\lambda\). Expanding the unitary:

\[ e^{-i\lambda \hat S}=1-i\lambda \hat S+O(\lambda^2), \]

workout

so

\[ \boxed{ \ket{\psi_{n,m}} =\left(1-i\lambda \hat S\right)\ket{n,m}+O(\lambda^2). } \]

To make this explicit as a superposition of \(\hat H_0\) eigenstates, we now insert the \(\hat S\) found from the SW condition. For the qubits+cavity dispersive setup, write

• \(\hat H=\hat H_0+\lambda \hat W\),
• \(\hat H_0=\omega_c\,\hat a^\dagger \hat a+\Omega\,\hat J_z\),
• \(\hat W=g\left(\hat J_+ \hat a+\hat J_- \hat a^\dagger\right)\),
• \(\Delta\equiv \omega_c-\Omega\).

The SW condition “no \(O(\lambda)\) term in \(\hat H'\)” is

\[ \hat W+[\hat H_0,-i\hat S]=0, \]

and the standard ansatz yields the Hermitian solution

\[ \hat S=\frac{i g}{\Delta}\left(\hat J_+\hat a-\hat J_- \hat a^\dagger\right). \]

Plugging this into \(\ket{\psi_{n,m}}=(1-i\lambda \hat S)\ket{n,m}\) gives

\[ \ket{\psi_{n,m}} =\left[1+\lambda\frac{g}{\Delta}\left(\hat J_+\hat a-\hat J_- \hat a^\dagger\right)\right]\ket{n,m} +O(\lambda^2). \]

Now expand in the \(\hat H_0\) eigenbasis using ladder-operator actions:

\[ \hat a\ket{n}=\sqrt{n}\ket{n-1},\qquad \hat a^\dagger\ket{n}=\sqrt{n+1}\ket{n+1}, \]

and (for a fixed total spin \(J\) manifold)

\[ \hat J_+\ket{J,m}=\sqrt{J(J+1)-m(m+1)},\ket{J,m+1}, \]

\[ \hat J_-\ket{J,m}=\sqrt{J(J+1)-m(m-1)},\ket{J,m-1}. \]

workout

Therefore

\[ \boxed{ \begin{aligned} \ket{\psi_{n,m}} &=\ket{n,m} +\lambda\frac{g}{\Delta} \Bigg[ \sqrt{n}\,\sqrt{J(J+1)-m(m+1)}\,\ket{n-1,m+1} \\ & -\sqrt{n+1}\,\sqrt{J(J+1)-m(m-1)}\,\ket{n+1,m-1} \Bigg] +O(\lambda^2). \end{aligned} } \]

This is already a superposition of \(\hat H_0\) eigenstates (each term has definite photon number and definite \(J_z=m\) value shifted by \(\pm 1\)). Normalization corrections only enter at \(O(\lambda^2)\), so the expression above is the correctly normalized state to first order in \(\lambda\) (up to an overall phase).

(e)

Here we verify that the first-order SW wavefunction from part (d) agrees with standard first-order (non-degenerate) time-independent perturbation theory for

\[ \hat H=\hat H_0+\lambda \hat W,\qquad \hat W=g\left(\hat a^\dagger \hat J_-+\hat a\,\hat J_+\right), \qquad \Delta\equiv \omega_c-\Omega. \]

Take as the unperturbed eigenstate

\[ \ket{\phi^{(0)}}=\ket{n;j,m}, \qquad \hat H_0\ket{n;j,m}=E^{(0)}_{n,m}\ket{n;j,m}, \qquad E^{(0)}_{n,m}=\omega_c n+\Omega m. \]

Standard first-order perturbation theory gives the corrected (unnormalized) eigenstate

\[ \ket{\phi}=\ket{\phi^{(0)}}+\ket{\phi^{(1)}}+O(\lambda^2), \qquad \ket{\phi^{(1)}}=\lambda \sum_{\alpha\neq 0} \ket{\alpha}\,\frac{\bra{\alpha}\hat W\ket{\phi^{(0)}}}{E^{(0)}_{n,m}-E^{(0)}_{\alpha}}. \]

The operator \(\hat W\) changes photon number and \(m\) by one unit, so from \(\ket{n;j,m}\) it connects only to

\[ \ket{\alpha_+}=\ket{n+1;j,m-1} \quad\text{via }\hat a^\dagger \hat J_-, \qquad \ket{\alpha_-}=\ket{n-1;j,m+1} \quad\text{via }\hat a\,\hat J_+. \]

Compute the needed matrix elements using ladder-operator actions:

\[ \hat a^\dagger\ket{n}=\sqrt{n+1}\ket{n+1},\quad \hat a\ket{n}=\sqrt{n}\ket{n-1}, \]

\[ \hat J_-\ket{j,m}=\sqrt{j(j+1)-m(m-1)}\,\ket{j,m-1},\quad \hat J_+\ket{j,m}=\sqrt{j(j+1)-m(m+1)}\,\ket{j,m+1}. \]

Therefore,

\[ \bra{n+1;j,m-1}\hat W\ket{n;j,m} = g\,\sqrt{n+1}\,\sqrt{j(j+1)-m(m-1)}, \]

\[ \bra{n-1;j,m+1}\hat W\ket{n;j,m} = g\,\sqrt{n}\,\sqrt{j(j+1)-m(m+1)}. \]

Next compute the energy denominators:

\[ \begin{aligned} E^{(0)}_{n,m}-E^{(0)}_{n+1,m-1} &=(\omega_c n+\Omega m)-(\omega_c(n+1)+\Omega(m-1)) \\ &=-(\omega_c-\Omega)=-\Delta, \end{aligned} \]

\[ \begin{aligned} E^{(0)}_{n,m}-E^{(0)}_{n-1,m+1} &=(\omega_c n+\Omega m)-(\omega_c(n-1)+\Omega(m+1)) \\ &=+(\omega_c-\Omega)=+\Delta. \end{aligned} \]

workout

Plugging into the first-order correction formula yields

\[ \begin{aligned} \ket{\phi^{(1)}} &=\lambda\left[ \ket{n+1;j,m-1}\frac{g\sqrt{n+1}\sqrt{j(j+1)-m(m-1)}}{-\Delta} + \ket{n-1;j,m+1}\frac{g\sqrt{n}\sqrt{j(j+1)-m(m+1)}}{+\Delta} \right] \\ &= -\lambda\,\frac{g}{\Delta}\,\sqrt{n+1}\,\sqrt{j(j+1)-m(m-1)}\,\ket{n+1;j,m-1} +\lambda\,\frac{g}{\Delta}\,\sqrt{n}\,\sqrt{j(j+1)-m(m+1)}\,\ket{n-1;j,m+1}. \end{aligned} \]

Thus the perturbation-theory eigenstate to \(O(\lambda)\) is

\[ \boxed{ \begin{aligned} \ket{\phi} &=\ket{n;j,m} -\lambda\,\frac{g}{\Delta}\,\sqrt{n+1}\,\sqrt{j(j+1)-m(m-1)}\,\ket{n+1;j,m-1} \\ &\quad +\lambda\,\frac{g}{\Delta}\,\sqrt{n}\,\sqrt{j(j+1)-m(m+1)}\,\ket{n-1;j,m+1} +O(\lambda^2), \end{aligned} } \]

which matches exactly the SW result from part (d), i.e. \(\ket{\psi_{n,m}}=(1-\hat S)\ket{n;j,m}+O(\lambda^2)\) with

\[ \hat S=\lambda\,\frac{g}{\Delta}\left(\hat a^\dagger \hat J_- - \hat a\,\hat J_+\right). \]

(f)

workout

Take the unperturbed \(\hat H_0\) eigenstate with zero photons and \(N_e\) excited qubits. In the collective-spin notation, this is a state with

• photon number \(n=0\),
• \(\hat J_z\) eigenvalue \(m\) given by

\[ m=\frac{1}{2}(N_e-(N-N_e))=N_e-\frac{N}{2}. \]

From part (d), the corresponding eigenstate of \(\hat H\) to first order in \(\lambda\) is

\[ \ket{\psi} =\ket{0;j,m} -\lambda\,\frac{g}{\Delta}\,\sqrt{0+1}\,\sqrt{j(j+1)-m(m-1)}\,\ket{1;j,m-1} +O(\lambda^2), \]

where \(\Delta\equiv \omega_c-\Omega\). Since the unperturbed state has \(n=0\), the \((n-1,m+1)\) term from part (d) is absent.

Let \(\hat\Pi_{n>0}\) be the projector onto the subspace with nonzero photon number. Using only the first-order state, the probability to have nonzero photon population is

\[ P(n>0)=\bra{\psi}\hat\Pi_{n>0}\ket{\psi}. \]

workout

To order \(\lambda^2\), only the \(n=1\) component contributes (normalization corrections are also \(O(\lambda^2)\) and do not change the leading result). Therefore,

\[ P(n>0) = \left|\,-\lambda\,\frac{g}{\Delta}\,\sqrt{j(j+1)-m(m-1)}\,\right|^2 +O(\lambda^4) = \left(\lambda\,\frac{g}{\Delta}\right)^2\left[j(j+1)-m(m-1)\right] +O(\lambda^4). \]

In the fully symmetric Dicke manifold relevant for collective coupling, one has \(j=N/2\), and with \(m=N_e-N/2\) which then simplifies to

\[ j(j+1)-m(m-1)=N_e\,(N-N_e+1), \]

so

\[ \boxed{ P(n>0) =\left(\lambda\,\frac{g}{\Delta}\right)^2\,N_e\,(N-N_e+1) +O(\lambda^4). } \]

Finally, in the regime \(\omega_c\gg \Omega\), we have \(\Delta=\omega_c-\Omega\simeq \omega_c\), giving the useful estimate

\[ \boxed{ P(n>0)\simeq \left(\lambda\,\frac{g}{\omega_c}\right)^2\,N_e\,(N-N_e+1). } \]

(g)

From part (f), starting in a zero-photon unperturbed state with \(N_e\) excited qubits, the first-order corrected state has a one-photon mixture with probability

\[ P(n>0)\approx \left(\lambda\,\frac{g}{\Delta}\right)^2\,N_e\,(N-N_e+1), \qquad \Delta\equiv \omega_c-\Omega, \]

keeping only the leading contribution in \(\lambda\).

A basic consistency requirement for perturbation theory (and for truncating the wavefunction at first order) is that the correction remains small, i.e. the total weight outside the unperturbed subspace should satisfy

\[ P(n>0)\ll 1. \]

However, even if the single-particle dispersive parameter is small,

\[ \frac{g}{|\Delta|}\ll 1, \]

the collective enhancement factor \(N_e(N-N_e+1)\) can grow with system size. For a generic many-excitation state, \(N_e\) scales with \(N\) (e.g. \(N_e\sim cN\) with \(0<c<1\)), so

\[ N_e\,(N-N_e+1)=O(N^2). \]

Then

\[ P(n>0)\sim \left(\lambda\,\frac{g}{\Delta}\right)^2\,O(N^2). \]

workout

Thus, for sufficiently large \(N\) the probability becomes order unity:

\[ P(n>0)\gtrsim 1 \quad\Longleftrightarrow\quad N_e\,(N-N_e+1)\gtrsim \left(\frac{\Delta}{\lambda g}\right)^2. \]

In particular, in the “worst-case” scaling (e.g. \(N_e\approx N/2\)),

\[ N_e\,(N-N_e+1)\approx \frac{N}{2}\left(\frac{N}{2}+1\right)\sim \frac{N^2}{4}, \]

so breakdown occurs once

\[ \frac{N^2}{4}\left(\lambda\,\frac{g}{\Delta}\right)^2\sim 1 \quad\Longrightarrow\quad N\sim \frac{2|\Delta|}{\lambda g}. \]

Therefore, even with \(g\ll|\Delta|\), perturbation theory (and the SW truncation) fails when \(N\) is large enough that the collectively enhanced mixing amplitude is no longer small. Equivalently, the true small parameter is not \(g/|\Delta|\) alone but the collective combination

\[ \boxed{ \lambda\,\frac{g}{|\Delta|}\,\sqrt{N_e\,(N-N_e+1)}\ll 1, } \]

and this condition can be violated at large \(N\).

Problem 4-2: Using SW transformations to understand a two-qubit gate

(a)

Set \(\Omega=0\). Then the Hamiltonian is naturally split as

\[ \hat H = \hat H_0 + \lambda \hat V, \qquad \hat H_0=\frac{\Delta_1}{2}\hat\sigma_z^{(1)}, \qquad \hat V=g\left(\hat\sigma_+^{(2)}\hat\sigma_-^{(1)}+\hat\sigma_+^{(1)}\hat\sigma_-^{(2)}\right). \]

Intuitively, the interaction is extremely weak in the regime \(\Delta_1\gg g\) for the same reason the qubit–cavity coupling becomes weak in the dispersive limit: the interaction term tries to induce excitation exchange, but that exchange is far off-resonant because it changes the energy set by \(\hat H_0\).

A clean way to see this without calculation is to track what the coupling does in the \(\hat H_0\) eigenbasis.

• \(\hat H_0\) depends only on qubit 1, so its eigenstates are product states with qubit 1 in \(\ket{g_1}\) or \(\ket{e_1}\), split by an energy \(\Delta_1\).
• The interaction \(\hat V\) is a “flip-flop” term: it swaps an excitation between the qubits. In particular, within the single-excitation manifold it couples \(\ket{e_1 g_2}\) to \(\ket{g_1 e_2}\).
• But these two states differ in the state of qubit 1, so they differ in unperturbed energy by exactly \(\Delta_1\). Hence the flip-flop process is off-resonant by detuning \(\Delta_1\) and cannot efficiently produce real population transfer when \(\Delta_1\gg g\); it mainly produces small, rapidly oscillating admixtures that average away.

workout

What survives to order \(\lambda^2\) is therefore not a real exchange interaction, but a dispersive, “virtual-process” interaction (again exactly like the qubit–cavity case):

• At order \(\lambda\), \(\hat V\) changes which \(\hat H_0\) eigenstate we are in (it flips qubit 1), so it cannot contribute a static correction within a given unperturbed eigenspace in the large-detuning limit.
• At order \(\lambda^2\), we can have a virtual process that starts in an unperturbed eigenstate, briefly visits a detuned intermediate state via one application of \(\hat V\), and then returns via a second application of \(\hat V\). By symmetry/structure, such a second-order effect must commute with \(\hat H_0\) (it cannot cause real transitions if we have eliminated them to first order), so it must be diagonal in the \(\hat\sigma_z^{(1)}\) basis.

Given the conserved quantities:

• The flip-flop interaction conserves the total excitation number (it only moves excitations around), so any effective interaction generated perturbatively also conserves total excitation number.
• The second-order effective Hamiltonian in the dispersive limit must commute with \(\hat\sigma_z^{(1)}\) (equivalently commute with \(\hat H_0\) up to constants), because real flips of qubit 1 are suppressed and removed by the SW procedure.

The only nontrivial two-qubit interaction consistent with these symmetries (diagonal in the \(\hat\sigma_z\) basis, conserving excitation number) is a longitudinal “cross-Kerr / ZZ” type term, along with single-qubit energy shifts (Lamb/Stark shifts). Schematically, to order \(\lambda^2\) one expects

\[ \hat H_{\mathrm{eff}} = \frac{\Delta_1}{2}\hat\sigma_z^{(1)} +\lambda^2\left( \delta_1\,\hat\sigma_z^{(1)} +\delta_2\,\hat\sigma_z^{(2)} +\chi\,\hat\sigma_z^{(1)}\hat\sigma_z^{(2)} \right), \]

where the coefficients are set by the strength of the virtual processes.

Finally, regarding the scaling, each virtual process uses the coupling twice and pays one energy denominator set by the detuning (recall the second-order perturbation energy formula we just derived during midterm exam), so the characteristic second-order scale is

workout

\[ (\delta_1,\ \delta_2,\ \chi) \sim \frac{g^2}{\Delta_1}. \]

Restoring the perturbative bookkeeping, these are order \(\lambda^2\):

\[ \boxed{ \text{effective interaction strength (rate)} \;\;\sim\;\; \lambda^2\,\frac{g^2}{\Delta_1}. } \]

So even though there is a direct exchange term at order \(\lambda g\), it is far off-resonant and averages out; the leading static interaction in the dispersive regime is a much weaker second-order dispersive (ZZ-type) coupling scaling like \(\lambda^2 g^2/\Delta_1\).

(b)

Set \(\Omega=0\) and split the Hamiltonian as

\[ \hat H=\hat H_0+\lambda \hat V, \qquad \hat H_0=\frac{\Delta_1}{2}\hat\sigma_z^{(1)}, \qquad \hat V=g\left(\hat\sigma_+^{(2)}\hat\sigma_-^{(1)}+\hat\sigma_+^{(1)}\hat\sigma_-^{(2)}\right). \]

We use the SW unitary \(\hat U=e^{-i\lambda \hat S}\) (with \(\hat S\) Hermitian), so

\[ \hat H'=\hat U^\dagger \hat H \hat U = e^{i\lambda \hat S}\hat H e^{-i\lambda \hat S}. \]

Expanding to first order in \(\lambda\) (BCH expansion),

\[ \hat H' =\hat H+i\lambda[\hat S,\hat H]+O(\lambda^2) =\hat H_0+\lambda \hat V+i\lambda[\hat S,\hat H_0]+O(\lambda^2), \]

where in the last step we used \([\hat S,\lambda \hat V]=O(\lambda)\), hence \(i\lambda[\hat S,\lambda \hat V]=O(\lambda^2)\).

To eliminate the qubit–qubit coupling to first order, we demand that the entire \(O(\lambda)\) term vanishes:

\[ \hat V+i[\hat S,\hat H_0]=0 \qquad\Longleftrightarrow\qquad i[\hat S,\hat H_0]=-\hat V. \]

workout

Now use the Pauli commutators on qubit 1:

\[ [\hat\sigma_z^{(1)},\hat\sigma_+^{(1)}]=2\hat\sigma_+^{(1)}, \qquad [\hat\sigma_z^{(1)},\hat\sigma_-^{(1)}]=-2\hat\sigma_-^{(1)}. \]

Since operators on different qubits commute, \(\hat\sigma_\pm^{(2)}\) commute with \(\hat H_0\). Hence

\[ [\hat H_0,\hat\sigma_+^{(2)}\hat\sigma_-^{(1)}] =\hat\sigma_+^{(2)}[\hat H_0,\hat\sigma_-^{(1)}] =\hat\sigma_+^{(2)}\left(\frac{\Delta_1}{2}\right)[\hat\sigma_z^{(1)},\hat\sigma_-^{(1)}] =-\Delta_1\,\hat\sigma_+^{(2)}\hat\sigma_-^{(1)}, \]

and similarly

\[ [\hat H_0,\hat\sigma_+^{(1)}\hat\sigma_-^{(2)}] =[\hat H_0,\hat\sigma_+^{(1)}]\hat\sigma_-^{(2)} =\Delta_1\,\hat\sigma_+^{(1)}\hat\sigma_-^{(2)}. \]

Define the anti-Hermitian combination

\[ \hat A \equiv \hat\sigma_+^{(2)}\hat\sigma_-^{(1)}-\hat\sigma_+^{(1)}\hat\sigma_-^{(2)}. \]

Using the two commutators above,

\[ [\hat H_0,\hat A]=-\Delta_1\left(\hat\sigma_+^{(2)}\hat\sigma_-^{(1)}+\hat\sigma_+^{(1)}\hat\sigma_-^{(2)}\right) =-\frac{\Delta_1}{g}\,\hat V, \]

so

\[ [\hat A,\hat H_0]=+\frac{\Delta_1}{g}\,\hat V. \]

Now choose \(\hat S = \alpha \hat A\). Then

\[ i[\hat S,\hat H_0]=i\alpha[\hat A,\hat H_0]=i\alpha\frac{\Delta_1}{g}\hat V. \]

Imposing \(i[\hat S,\hat H_0]=-\hat V\) gives \(i\alpha(\Delta_1/g)=-1\), i.e. \(\alpha=i g/\Delta_1\). Therefore,

\[ \boxed{ \hat S = i\,\frac{g}{\Delta_1}\left(\hat\sigma_+^{(2)}\hat\sigma_-^{(1)}-\hat\sigma_+^{(1)}\hat\sigma_-^{(2)}\right) } \]

which is Hermitian (since the bracketed operator is anti-Hermitian). With this choice, the transformed Hamiltonian has no \(O(\lambda)\) exchange term.

(c)

Now include the drive term in the “unperturbed” Hamiltonian used in the BCH expansion, but keep the same \(\hat S\) found in part (b) (i.e., \(\hat S\) is computed assuming \(\Omega=0\)):

\[ \hat H = \hat H_0 + \lambda \hat V, \qquad \hat H_0=\frac{\Delta_1}{2}\hat\sigma_z^{(1)}+\frac{\Omega}{2}\hat\sigma_x^{(1)}, \qquad \hat V=g\left(\hat\sigma_+^{(2)}\hat\sigma_-^{(1)}+\hat\sigma_+^{(1)}\hat\sigma_-^{(2)}\right). \]

We perform the SW rotation with \(\hat U=e^{-i\lambda \hat S}\), so

\[ \hat H' = e^{i\lambda \hat S}\hat H e^{-i\lambda \hat S}. \]

Expand to first order in \(\lambda\):

\[ \hat H'=\hat H+i\lambda[\hat S,\hat H]+O(\lambda^2) =\hat H_0+\lambda \hat V+i\lambda[\hat S,\hat H_0]+O(\lambda^2), \]

where we used \(i\lambda[\hat S,\lambda \hat V]=O(\lambda^2)\).

Split \(\hat H_0\) into the detuning part and the drive part,

\[ \hat H_0=\hat H_{0z}+\hat H_{0\Omega}, \qquad \hat H_{0z}=\frac{\Delta_1}{2}\hat\sigma_z^{(1)}, \qquad \hat H_{0\Omega}=\frac{\Omega}{2}\hat\sigma_x^{(1)}. \]

By construction in part (b), \(\hat S\) satisfies

\[ \hat V+i[\hat S,\hat H_{0z}]=0, \]

so the \(O(\lambda)\) terms simplify to

\[ \lambda \hat V+i\lambda[\hat S,\hat H_0] =\lambda \hat V+i\lambda[\hat S,\hat H_{0z}]+i\lambda[\hat S,\hat H_{0\Omega}] =i\lambda[\hat S,\hat H_{0\Omega}]. \]

Thus, to order \(\lambda^1\),

\[ \hat H'=\hat H_0+i\lambda[\hat S,\hat H_{0\Omega}]+O(\lambda^2). \]

workout

Now compute the remaining commutator using the \(\hat S\) from part (b):

\[ \hat S = i\,\frac{g}{\Delta_1}\left(\hat\sigma_+^{(2)}\hat\sigma_-^{(1)}-\hat\sigma_+^{(1)}\hat\sigma_-^{(2)}\right). \]

Since \(\hat H_{0\Omega}=(\Omega/2)\hat\sigma_x^{(1)}\), we need \([\hat S,\hat\sigma_x^{(1)}]\). Write \(\hat\sigma_x^{(1)}=\hat\sigma_+^{(1)}+\hat\sigma_-^{(1)}\) and use (on qubit 1)

\[ [\hat\sigma_-^{(1)},\hat\sigma_x^{(1)}]=[\hat\sigma_-^{(1)},\hat\sigma_+^{(1)}]=-\hat\sigma_z^{(1)}, \qquad [\hat\sigma_+^{(1)},\hat\sigma_x^{(1)}]=[\hat\sigma_+^{(1)},\hat\sigma_-^{(1)}]=+\hat\sigma_z^{(1)}. \]

Also, operators on different qubits commute. Therefore,

\[ \begin{aligned} \left[\hat\sigma_+^{(2)}\hat\sigma_-^{(1)},\hat\sigma_x^{(1)}\right] &=\hat\sigma_+^{(2)}\left[\hat\sigma_-^{(1)},\hat\sigma_x^{(1)}\right] =-\hat\sigma_+^{(2)}\hat\sigma_z^{(1)}, \\ \left[\hat\sigma_+^{(1)}\hat\sigma_-^{(2)},\hat\sigma_x^{(1)}\right] &=\left[\hat\sigma_+^{(1)},\hat\sigma_x^{(1)}\right]\hat\sigma_-^{(2)} =\hat\sigma_z^{(1)}\hat\sigma_-^{(2)}. \end{aligned} \]

Hence

\[ \begin{aligned} [\hat S,\hat\sigma_x^{(1)}] &= i\,\frac{g}{\Delta_1}\left( \left[\hat\sigma_+^{(2)}\hat\sigma_-^{(1)},\hat\sigma_x^{(1)}\right] -\left[\hat\sigma_+^{(1)}\hat\sigma_-^{(2)},\hat\sigma_x^{(1)}\right] \right) \\ &= i\,\frac{g}{\Delta_1}\left( -\hat\sigma_+^{(2)}\hat\sigma_z^{(1)}-\hat\sigma_z^{(1)}\hat\sigma_-^{(2)} \right) \\ &= -i\,\frac{g}{\Delta_1}\,\hat\sigma_z^{(1)}\left(\hat\sigma_+^{(2)}+\hat\sigma_-^{(2)}\right) = -i\,\frac{g}{\Delta_1}\,\hat\sigma_z^{(1)}\hat\sigma_x^{(2)}. \end{aligned} \]

Therefore,

\[ i\lambda[\hat S,\hat H_{0\Omega}] =i\lambda\frac{\Omega}{2}[\hat S,\hat\sigma_x^{(1)}] =i\lambda\frac{\Omega}{2}\left(-i\,\frac{g}{\Delta_1}\hat\sigma_z^{(1)}\hat\sigma_x^{(2)}\right) =\lambda\,\frac{\Omega g}{2\Delta_1}\,\hat\sigma_z^{(1)}\hat\sigma_x^{(2)}. \]

Putting everything together, the transformed Hamiltonian to order \(\lambda^1\) is

\[ \boxed{ \hat H' = \frac{\Delta_1}{2}\hat\sigma_z^{(1)} +\frac{\Omega}{2}\hat\sigma_x^{(1)} +\lambda\,\frac{\Omega g}{2\Delta_1}\,\hat\sigma_z^{(1)}\hat\sigma_x^{(2)} +O(\lambda^2) }. \]

(d)

workout

From part (c), to order \(\lambda^1\) we found

\[ \hat H' = \frac{\Delta_1}{2}\hat\sigma_z^{(1)} +\frac{\Omega}{2}\hat\sigma_x^{(1)} +\lambda\,\frac{\Omega g}{2\Delta_1}\,\hat\sigma_z^{(1)}\hat\sigma_x^{(2)} +O(\lambda^2). \]

Each term has a clear physical meaning.

1. The detuning term on qubit 1

   \[ \frac{\Delta_1}{2}\hat\sigma_z^{(1)}. \]

   This is the energy splitting of qubit 1 in the chosen rotating frame. It sets the large “fast” precession of qubit 1 about its \(z\) axis, and it is the dominant scale in the regime of interest (\(|\Delta_1|\) is the largest frequency).

2. The drive term on qubit 1

   \[ \frac{\Omega}{2}\hat\sigma_x^{(1)}. \]

   This is the coherent drive applied to qubit 1 (in the rotating frame), causing Rabi oscillations of qubit 1 about the \(x\) axis at rate \(\Omega\). If qubit 1 were isolated, this would simply rotate its Bloch vector.

3. The induced interaction term between qubits

   \[ \lambda\,\frac{\Omega g}{2\Delta_1}\,\hat\sigma_z^{(1)}\hat\sigma_x^{(2)}. \]
   workout

   This is the key SW-generated effect at order \(\lambda^1\) after eliminating the direct flip-flop exchange. Physically:

   • The bare coupling \(g(\hat\sigma_+^{(2)}\hat\sigma_-^{(1)}+\mathrm{h.c.})\) cannot efficiently swap excitations because it is off-resonant by \(\Delta_1\).
   • Nevertheless, the coupling allows qubit 1 to “virtually” exchange an excitation with qubit 2.
   • When qubit 1 is additionally being driven (\(\Omega\neq 0\)), these virtual processes convert the drive on qubit 1 into an effective drive on qubit 2, but the effective drive depends on the state of qubit 1.

   More concretely, since \(\hat\sigma_z^{(1)}\) has eigenvalues \(\pm 1\), the term acts like a conditional \(x\)-drive on qubit 2:

   • If qubit 1 is in \(\hat\sigma_z^{(1)}=+1\) (its “excited” eigenstate of \(\sigma_z\)), qubit 2 feels an \(x\)-drive of amplitude \(+\lambda(\Omega g/2\Delta_1)\).
   • If qubit 1 is in \(\hat\sigma_z^{(1)}=-1\) (its “ground” eigenstate of \(\sigma_z\)), qubit 2 feels an \(x\)-drive of amplitude \(-\lambda(\Omega g/2\Delta_1)\).

   So the drive on qubit 1 has been converted into a state-dependent transverse field acting on qubit 2. This is the hallmark of a controlled interaction: qubit 1 controls the sign (phase) of the rotation applied to qubit 2.

4. What is missing at this order

   To order \(\lambda^1\), there is no residual exchange term (by construction), and there are no dispersive energy shifts (which would appear at order \(\lambda^2\)). Those \(O(\lambda^2)\) terms would include Stark/Lamb shifts and a static entangling \(ZZ\)-type coupling, but they are beyond our scope of order here.

   In summary: \(\hat H'\) describes (i) fast \(z\)-precession of qubit 1 from detuning, (ii) direct Rabi driving of qubit 1, and (iii) an induced conditional \(x\)-drive on qubit 2 of strength \(\sim \lambda\,\Omega g/\Delta_1\) arising from virtual exchange combined with the applied drive.

(e)

To order \(\lambda^1\), the transformed Hamiltonian contains the interaction

\[ \hat H_{\mathrm{int}} =\lambda\,\frac{\Omega g}{2\Delta_1}\,\hat\sigma_z^{(1)}\hat\sigma_x^{(2)} \equiv \frac{\kappa}{2}\,\hat\sigma_z^{(1)}\hat\sigma_x^{(2)}, \qquad \kappa\equiv \lambda\,\frac{\Omega g}{\Delta_1}. \]

This term generates a unitary evolution

\[ \hat U_{\mathrm{int}}(t)=\exp\!\left(-i t\,\frac{\kappa}{2}\,\hat\sigma_z^{(1)}\hat\sigma_x^{(2)}\right). \]

Key point: \(\hat\sigma_z^{(1)}\) has eigenvalues \(\pm 1\), so \(\hat U_{\mathrm{int}}(t)\) applies different rotations to qubit 2 depending on the state of qubit 1. This conditional action is what produces entanglement.

workout

A concrete demonstration:

1. Prepare an initial product state with qubit 1 in a superposition of \(\hat\sigma_z\) eigenstates, and qubit 2 in an eigenstate of \(\hat\sigma_z\) (or any state not an eigenstate of \(\hat\sigma_x\)). For example,

   \[ \ket{\psi(0)}=\ket{+_x}_1\otimes\ket{g}_2 =\frac{1}{\sqrt{2}}\left(\ket{g}_1+\ket{e}_1\right)\otimes\ket{g}_2, \]

   where \(\hat\sigma_z\ket{e}=+\ket{e}\) and \(\hat\sigma_z\ket{g}=-\ket{g}\).

2. Apply \(\hat U_{\mathrm{int}}(t)\) to this state. Since \(\hat\sigma_z^{(1)}\) acts as \(\pm 1\) on \(\ket{e}_1,\ket{g}_1\), we get

   \[ \hat U_{\mathrm{int}}(t)\ket{e}_1\otimes\ket{\phi}_2 =\ket{e}_1\otimes e^{-i(\kappa t/2)\hat\sigma_x^{(2)}}\ket{\phi}_2, \]
   \[ \hat U_{\mathrm{int}}(t)\ket{g}_1\otimes\ket{\phi}_2 =\ket{g}_1\otimes e^{+i(\kappa t/2)\hat\sigma_x^{(2)}}\ket{\phi}_2, \]

   because the sign flips when \(\sigma_z^{(1)}=-1\).

   Taking \(\ket{\phi}_2=\ket{g}_2\) and combining,

   \[ \ket{\psi(t)} =\frac{1}{\sqrt{2}} \left( \ket{g}_1\otimes e^{+i(\kappa t/2)\hat\sigma_x^{(2)}}\ket{g}_2 + \ket{e}_1\otimes e^{-i(\kappa t/2)\hat\sigma_x^{(2)}}\ket{g}_2 \right). \]

3. This state is entangled for generic times because the two conditional states of qubit 2 are different. In other words, qubit 2 becomes correlated with the computational-basis value of qubit 1.

   We can make this especially explicit by choosing a time such that the two conditional rotations map \(\ket{g}_2\) to orthogonal states. For a rotation about \(x\),

   \[ e^{-i\theta \hat\sigma_x}\ket{g} =\cos\theta\,\ket{g}-i\sin\theta\,\ket{e}, \qquad e^{+i\theta \hat\sigma_x}\ket{g} =\cos\theta\,\ket{g}+i\sin\theta\,\ket{e}. \]

   Here \(\theta=\kappa t/2\). Then

   \[ \ket{\psi(t)} =\frac{1}{\sqrt{2}}\left[ \ket{g}_1\otimes\left(\cos\theta\,\ket{g}_2+i\sin\theta\,\ket{e}_2\right) + \ket{e}_1\otimes\left(\cos\theta\,\ket{g}_2-i\sin\theta\,\ket{e}_2\right) \right]. \]

   At \(\theta=\pi/4\) (i.e. \(\kappa t=\pi/2\)), the conditional states of qubit 2 are

   \[ \frac{1}{\sqrt{2}}\left(\ket{g}_2\pm i\ket{e}_2\right), \]

   which are orthogonal. The joint state becomes a maximally entangled Bell-like state (up to local phases). Indeed, at \(\theta=\pi/4\),

   \[ \ket{\psi(t)} =\frac{1}{2}\left[ \left(\ket{g}_1+\ket{e}_1\right)\ket{g}_2 +i\left(\ket{g}_1-\ket{e}_1\right)\ket{e}_2 \right], \]

   which cannot be factorized into a product of single-qubit states.

Operationally, this interaction is a controlled-rotation gate:

• Qubit 1 is the control (its \(\hat\sigma_z\) eigenvalue selects the sign of the rotation),
• Qubit 2 is the target (it undergoes an \(x\)-rotation),
• The entanglement arises when the control is in a superposition, so the target undergoes a superposition of two different rotations.

(f)

Introduce a second small parameter by writing the drive as \(\Omega\to \lambda_2\Omega\), and split

\[ \hat H=\hat H_{0z}+\lambda_2 \hat H_{0\Omega}+\lambda \hat V, \qquad \hat H_{0z}=\frac{\Delta_1}{2}\hat\sigma_z^{(1)}, \qquad \hat H_{0\Omega}=\frac{\Omega}{2}\hat\sigma_x^{(1)}, \qquad \hat V=g\left(\hat\sigma_+^{(2)}\hat\sigma_-^{(1)}+\hat\sigma_+^{(1)}\hat\sigma_-^{(2)}\right). \]

We want an SW generator \(\hat S\) (Hermitian) such that, after \(\hat U=e^{-i\lambda \hat S}\), the transformed Hamiltonian has no “exchange” term at order \(\lambda\). The first-order condition is

\[ \hat V+i[\hat S,\hat H_{0z}+\lambda_2 \hat H_{0\Omega}]=0 \quad\text{(to the order we choose to enforce).} \]

The justification for dropping \(\lambda_2\hat H_{0\Omega}\) when solving for \(\hat S\) is a controlled power-counting statement:

• Assume the hierarchy of scales

\[ |\Delta_1|\gg |\lambda g|,\qquad |\Delta_1|\gg |\lambda_2 \Omega|. \]

• Expand \(\hat S\) itself in \(\lambda_2\):

\[ \hat S=\hat S^{(0)}+\lambda_2 \hat S^{(1)}+O(\lambda_2^2), \]

where \(\hat S^{(0)}\) is exactly what we computed in part (b) by solving

\[ \hat V+i[\hat S^{(0)},\hat H_{0z}]=0. \]

This yields \(\hat S^{(0)}\sim g/\Delta_1\).

workout

Now look at what we miss by ignoring \(\lambda_2\)-dependent corrections in \(\hat S\).

If we plug the expansion into the full condition and collect powers of \(\lambda_2\), the next equation would be

\[ i[\hat S^{(1)},\hat H_{0z}]= -\,i[\hat S^{(0)},\hat H_{0\Omega}], \]

so parametrically

\[ \hat S^{(1)} \sim \frac{1}{\Delta_1}\,[\hat S^{(0)},\hat H_{0\Omega}] \sim \frac{1}{\Delta_1}\left(\frac{g}{\Delta_1}\right)\Omega \sim \frac{g\,\Omega}{\Delta_1^2}. \]

Thus the correction we neglected in \(\hat S\) has size

\[ \delta \hat S \equiv \lambda_2 \hat S^{(1)} \sim \lambda_2\,\frac{g\,\Omega}{\Delta_1^2}, \]

which is smaller than \(\hat S^{(0)}\) by a factor \(\sim \lambda_2\Omega/\Delta_1\ll 1\).

Now check how this affects \(\hat H'\) at the order we kept in part (c). To first order in \(\lambda\),

\[ \hat H'=\hat H_0+\lambda \hat V+i\lambda[\hat S,\hat H_0]+O(\lambda^2), \qquad \hat H_0=\hat H_{0z}+\lambda_2\hat H_{0\Omega}. \]

Using only \(\hat S^{(0)}\) cancels \(\lambda\hat V\) against \(i\lambda[\hat S^{(0)},\hat H_{0z}]\), leaving the residual term

\[ i\lambda[\hat S^{(0)},\lambda_2\hat H_{0\Omega}] \sim \lambda\lambda_2\,\frac{g\,\Omega}{\Delta_1}, \]

which is exactly the \(\hat\sigma_z^{(1)}\hat\sigma_x^{(2)}\) interaction found in part (c).

If instead we included \(\delta \hat S=\lambda_2\hat S^{(1)}\), its contribution at this same BCH order would enter through \(i\lambda[\delta \hat S,\hat H_{0z}]\), which scales as

\[ i\lambda[\lambda_2\hat S^{(1)},\hat H_{0z}] \sim \lambda\lambda_2\,\Delta_1\left(\frac{g\,\Omega}{\Delta_1^2}\right) \sim \lambda\lambda_2\,\frac{g\,\Omega}{\Delta_1}. \]

So including \(\hat S^{(1)}\) would modify terms of the same parametric order \(\lambda\lambda_2\); this is why one must be clear about what is being “eliminated.”

The point is:

• In parts (c)–(e) we are not trying to eliminate all effects proportional to \(\lambda\lambda_2\).
• We purposely keep the leading drive-assisted interaction (the term \(\propto \hat\sigma_z^{(1)}\hat\sigma_x^{(2)}\)) because it is the physically useful gate mechanism.

workout

Dropping \(\Omega\) in the construction of \(\hat S\) is justified when the drive is a slow perturbation compared to the large detuning, i.e.

\[ \boxed{ \frac{\lambda_2\Omega}{|\Delta_1|}\ll 1, \qquad \frac{\lambda g}{|\Delta_1|}\ll 1, } \]

so that the dominant “fast” off-resonant exchange physics is controlled by \(\hat H_{0z}\), and the drive can be treated as an additional weak term whose leading effect is captured by the commutator \(i\lambda[\hat S^{(0)},\lambda_2\hat H_{0\Omega}]\).

Equivalently: we may ignore \(\Omega\) when finding \(\hat S\) if the drive does not significantly change the energy denominators that suppress exchange. If \(\Omega\) were not small compared to \(|\Delta_1|\) (or if the drive strongly mixes the \(\hat\sigma_z^{(1)}\) eigenstates on the same timescale as \(\Delta_1^{-1}\)), then computing \(\hat S\) with \(\Omega=0\) would not be a controlled approximation.