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  • 2-Qubits

    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).