2-Qubits

# Eigenenergies

(In this notes I will follow the schema of Rebolini’s thesis (Rebolini 2014-15). I will not use the “hat” ($$\hat{}$$) for the operators unless it is strictly necessary)

The Rabi Hamiltonian for 2 qubit is $\label{eqn:H2R} H_{2q}=\hbar \omega_0{a^{\dagger}}a + \hbar \frac{\omega_1}{2}\sigma^{(1)}_z + \hbar \frac{\omega_2}{2}\sigma^{(2)}_z + \hbar(\Gamma_1\sigma^{(1)}_x + \Gamma_2\sigma^{(2)}_x)({a^{\dagger}}+ a)$.

Using the following displacement operator properties $\begin{split} &{D^{\dagger}(\alpha)}aD(\alpha)=a + \alpha \\ &{D^{\dagger}(\alpha)}{a^{\dagger}}D(\alpha)={a^{\dagger}}+ \alpha^*\\ &{D^{\dagger}(\alpha)}D(\alpha)=D(\alpha){D^{\dagger}(\alpha)}={\mathbb{I}}\end{split}$

the Hamiltonian is easily rewritten in term of the displacement operator $\label{eqn:DHD} H_{2q}=\hbar\omega_0 D^{\dagger}(\omega_0^{-1}\hat{\gamma}){a^{\dagger}}a D(\omega_0^{-1}\hat{\gamma}) + \hbar \frac{\omega_1}{2}\sigma^{(1)}_z + \hbar \frac{\omega_2}{2}\sigma^{(2)}_z - \hbar\omega_0^{-1}\hat{\gamma}^2$

where $\hat{\gamma}=\Gamma_1\sigma^{(1)}_x + \Gamma_2\sigma^{(2)}_x$.

Where taking $$\Gamma_i\in\mathbb{R}$$ and remembering $$\sigma_x^{\dagger}=\sigma_x$$ one has $$\hat{\gamma}^{\dagger}=\hat{\gamma}$$.

Assuming the qubits totally-degenerate ($$\omega_1=0$$ and $$\omega_2=0$$), considering the displaced field operator \begin{align} b &= D^{\dagger}(\omega_0^{-1}\hat{\gamma}) a D(\omega_0^{-1}\hat{\gamma})\\ \nonumber b^{\dagger}&= D^{\dagger}(\omega_0^{-1}\hat{\gamma}){a^{\dagger}}D(\omega_0^{-1}\hat{\gamma}) \end{align}

and neglecting the constant terms, the Hamiltonian becomes

$H_{2q}=\hbar\omega_0 b^{\dagger}b - 2\hbar\frac{\Gamma_1\Gamma_2}{\omega_0}\sigma^{(1)}_x\otimes\sigma^{(2)}_x$

the eigenstates of this Hamiltonian are: $\label{eqn:basis} |N_{nm}nm>={D^{\dagger}(\omega_0^{-1}\gamma_{nm})}|N>|n>|m>$ with $$N\in\mathbb{N}$$, $$n,m=\{+,-\}$$ and $$\gamma_{nm}=n\Gamma_1+m\Gamma_2$$ $$\in\mathbb{R}$$ are the eigenvalues of the operator $$\hat{\gamma}$$ on the 2-qubits basis $$\{|nm>\}=\{|++>,|+->,|-+>,|-->\}$$.

Negletting the constant terms the Hamiltonian (\ref{eqn:DHD}) becomes

$H_{2q}=\hbar\omega_0 D^{\dagger}(\omega_0^{-1}\hat{\gamma}){a^{\dagger}}a D(\omega_0^{-1}\hat{\gamma}) + \hbar \frac{\omega_1}{2}\sigma^{(1)}_z + \hbar \frac{\omega_2}{2}\sigma^{(2)}_z - 2\hbar\frac{\Gamma_1\Gamma_2}{\omega_0}\sigma^{(1)}_x\otimes\sigma^{(2)}_x$

Calculate now the matrix’s elements in the basis of the eigenstates (\ref{eqn:basis}) in order to find later the eigenenergies of the system \begin{align}\label{eqn:Matrix_Elements} &<M_{st}st|H_1|N_{nm}nm>=\varepsilon_N\delta_{s,n}\delta_{t,m}<M_{st}|N_{nm}>\\ \nonumber &<M_{st}st|H_2|N_{nm}nm>=\varepsilon_1(1-\delta_{s,n})\delta_{t,m}<M_{st}|N_{\bar{n}m}>+\varepsilon_2\delta_{s,n}(1-\delta_{t,m})<M_{st}|N_{n\bar{m}}>\\ &<M_{st}st|H_3|N_{nm}nm>=-\varepsilon_{12}(nm)\delta_{s,n}\delta_{t,m}<M_{st}|N_{nm}> \nonumber \end{align} where $$\bar{n}$$ is the negation of $$n$$, $$H_1=\hbar\omega_0 D^{\dagger}(\omega_0^{-1}\hat{\gamma}){a^{\dagger}}a D(\omega_0^{-1}\hat{\gamma})$$, $$H_2=\hbar \frac{\omega_1}{2}\sigma^{(1)}_z + \hbar \frac{\omega_2}{2}\sigma^{(2)}_z$$, $$H_3=- 2\hbar\frac{\Gamma_1\Gamma_2}{\omega_0}\sigma^{(1)}_x\otimes\sigma^{(2)}_x$$, $$\varepsilon_N=\hbar\omega_0 N$$, $$\varepsilon_1=\hbar \frac{\omega_1}{2}$$, $$\varepsilon_2= \hbar \frac{\omega_2}{2}$$ and $$\varepsilon_{12}=2\hbar\frac{\Gamma_1\Gamma_2}{\omega_0}$$. One can easily identify two diagonal terms and an off-diagonal one.

The overlap between two displaced number states $$|N_{nm}>$$ is $\label{eqn:Lag} <M_{st}|N_{nm}>=<M|D(\alpha_{stnm})|N>=e^{-2\alpha_{stnm}^2}(2\alpha_{stnm})^{(M-N)}\sqrt{\frac{N!}{M!}}L^{(M-N)}_N(4\alpha_{stnm}^2)\;\;\;\;\;(M\geq N)$ where $$\alpha_{stnm}=\omega_0^{-1}\left(\gamma_{st}-\gamma_{nm}\right)$$. Therefore, one can easily see that for $$s=n$$ and $$t=m$$ the matrix element becomes $$<M|N>=\delta_{MN}$$ (since $$D(0)={\mathbb{I}}$$).

Perform now the adiabatic approximation that consist in discarding the far-from diagonal matrix elements, i.e. those therms for which $$M\neq N$$. For the remaining terms from (\ref{eqn:Lag}) one can easily see that $$<N_{st}|N_{nm}>=<N_{nm}|N_{st}>$$, or for simplicity of notation $$N_{st}N_{nm}=N_{nm}N_{st}$$, due to the quadratic dependence from $$\alpha_{stnm}$$,.

A deeper study of $$\alpha_{stnm}$$ gives the following result $\begin{split} \alpha_{stnm}&=\omega_0^{-1}\left(\gamma_{st}-\gamma_{nm}\right)\\ &=\omega_0^{-1}\left(s\Gamma_1+t\Gamma_2-n\Gamma_1-m\Gamma_2\right)\\ &=\omega_0^{-1}\left[(s-n)\Gamma_1+(t-m)\Gamma_2\right]\\ &=\omega_0^{-1}\left[2s(1-\delta_{s,n})\Gamma_1+2t(1-\delta_{t,m})\Gamma_2\right] \end{split}$

where $$n,m,s,t=\pm1$$.

Combining this result with (\ref{eqn:Matrix_Elements}) and keeping in mind the adiabatic approximation, one obtain a diagonal-block matrix in which every block is a $$4\times4$$ matrix like

$(H_{2q})_{ad}= \begin{bmatrix} & & |N_{--}> & |N_{-+}> & |N_{+-}> & |N_{++}> & \\ & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ <N_{--}| & \cdots & \varepsilon_N - \varepsilon_{12} & \varepsilon_2\; G(\Gamma_2) & \varepsilon_1\; G(\Gamma_1)& 0 & \cdots \\ <N_{-+}| & \cdots & \varepsilon_2\; G(\Gamma_2) & \varepsilon_N + \varepsilon_{12} & 0 & \varepsilon_1\; G(\Gamma_1) & \cdots \\ <N_{+-}| & \cdots & \varepsilon_1\; G(\Gamma_1) & 0 & \varepsilon_N + \varepsilon_{12} & \varepsilon_2\; G(\Gamma_2) & \cdots \\ <N_{++}| & \cdots & 0 & \varepsilon_1\; G(\Gamma_1) & \varepsilon_2\; G(\Gamma_2) & \varepsilon_N - \varepsilon_{12} & \cdots \\ & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ \end{bmatrix}$

where $G(\Gamma_i)=e^{-2\Gamma_i^2}L_N(4\Gamma_i^2)$

Each block can be easily diagonalized and one gets eigentates and eigenenergies of the total Hamiltonian \begin{align} &E^{1/2}_N=\varepsilon_N\mp \sqrt{\Big(\varepsilon_{12}\Big)^2+\Big(\varepsilon_2\; e^{-2\Gamma_2^2}L_N(4\Gamma_2^2)-\varepsilon_1\; e^{-2\Gamma_1^2}L_N(4\Gamma_1^2)\Big)^2}\\\nonumber &E^{3/4}_N=\varepsilon_N\mp \sqrt{\Big(\varepsilon_{12}\Big)^2+\Big(\varepsilon_2\; e^{-2\Gamma_2^2}L_N(4\Gamma_2^2)+\varepsilon_1\; e^{-2\Gamma_1^2}L_N(4\Gamma_1^2)\Big)^2}\\ \end{align}

It is easy to see that if one qubit is “turned-off”, i.e. it is not coupled with the resonator anymore, the eigenenergies (and eigenvectors ????) are the same found by Rebolini for one qubit.

### References

1. Andrea Rebolini. OPEN SYSTEM DYNAMICS OF A QUANTUM OSCILLATOR STRONGLY COUPLED TO A TWO-LEVEL SYSTEM. (2014-15).