Science-2020-K. C. Miao, et al.

Universal coherence protection in a solid-state spin qubit

Divacancy Ground State Hamiltonian

we start from the Hamiltonian of divacancy ground state

$\begin{gathered} H / h=D\left(\hat{S}_{z}^{2}-\frac{S(S+1)}{3}\right)+E\left(\hat{S}_{+}^{2}+\hat{S}_{-}^{2}\right)+ \\ g \mu_{\mathrm{B}} \mathbf{B} \cdot \hat{\mathbf{S}}+\sum_{i} \hat{\mathbf{S}} \cdot \mathbf{A}_{i} \cdot \hat{\mathbf{I}}_{i} \end{gathered}$

ZEFOZ eigen basis

Operating at the ZEFOZ point $\mathrm{B}_{\mathrm{eff}}=0$, and neglect the coupling with nuclear spins, the Hamiltonian can be expressed in the $S_{z}=\{|+1\rangle,|0\rangle,|-1\rangle\}$ basis

$H_{\mathrm{ZFFOZ}} / h=\left[\begin{array}{ccc}D & 0 & E \\ 0 & 0 & 0 \\ E & 0 & D\end{array}\right]$

the eigenstate of the Hamiltonian is $\{|+\rangle,|0\rangle,|-\rangle\}$, and its relationship to the $S_{z}=\{|+1\rangle_z,|0\rangle_z,|-1\rangle_z\}$ basis is

$\begin{pmatrix} |+\rangle\\ |0\rangle\\ |-\rangle \end{pmatrix} = \frac{1}{\sqrt{2}}\left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & \sqrt{2} & 0 \\ 1 & 0 & -1 \end{array}\right] \begin{pmatrix} |+1\rangle_z\\ |0\rangle_z\\ |-1\rangle_z \end{pmatrix}\\ \ \\ U=\frac{1}{\sqrt{2}}\left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & \sqrt{2} & 0 \\ -1 & 0 & 1 \end{array}\right]$

The hamiltonian is diagonalized in this basis

$\begin{gathered} H_{\text {ZEFOZ }}^{\prime} / h=U^{*} H_{\mathrm{ZEFOZ}} U^{T} / h=\left[\begin{array}{ccc} D+E & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & D-E \end{array}\right] \end{gathered}$

For basal divacancies in 4H-SiC, E is not neglectable, so there exhibits a transverse anisotropy spin splitting (E) in addition to D in zero field, which results in two zero-field resonance at frequencies defined by ($D\pm E$)

The image(Nature-2011-W. F. Koehl, et al.) above is an ODMR result of 4H-SiC ensemble, from the linewidth of the dip we can infer that the splitting is quite uniform in 4H-SiC. The splitting is 18.6 and 82 MHz for PL4 and PL3 respectively.

Question: why there is a peak?

Couple to Magnetic Field

Spin-1matrices in $\{|+\rangle,|0\rangle,|-\rangle\}$ basis can be expressed as

$S_{x}^{\prime}=\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] S_{y}^{\prime}=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{array}\right] S_{z}^{\prime}=\left[\begin{array}{lll} 0 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{array}\right]$

Switch from eigenstate to dress state

Dressing state basis $\{|+1\rangle,|0\rangle,|-1\rangle\}$

$\begin{pmatrix} |+1\rangle\\ |0\rangle\\ |-1\rangle \end{pmatrix} = \frac{1}{\sqrt{2}}\left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & \sqrt{2} & 0 \\ 1 & 0 & -1 \end{array}\right] \begin{pmatrix} exp(i(D+E)t) & 0 & 0 \\ 0 & exp(iDt) & 0 \\ 0 & 0 & exp(i(D-E)t) \end{pmatrix} \begin{pmatrix} |+\rangle\\ |0\rangle\\ |-\rangle \end{pmatrix}$

$S_x''= \left( \begin{array}{ccc} 0 & \frac{e^{-i \text{E} t}}{\sqrt{2}} & 0 \\ \frac{e^{i \text{E} t}}{\sqrt{2}} & 0 & \frac{e^{i \text{E} t}}{\sqrt{2}} \\ 0 & \frac{e^{-i \text{E} t}}{\sqrt{2}} & 0 \\ \end{array} \right) S_y''= \left( \begin{array}{ccc} 0 & -\frac{i e^{i \text{E} t}}{\sqrt{2}} & 0 \\ \frac{i e^{-i \text{E} t}}{\sqrt{2}} & 0 & -\frac{i e^{-i \text{E} t}}{\sqrt{2}} \\ 0 & \frac{i e^{i \text{E} t}}{\sqrt{2}} & 0 \\ \end{array} \right)\\ \ \\ S_z''= \left( \begin{array}{ccc} \frac{1}{2} e^{-2 i \text{E} t} \left(1+e^{4 i \text{E} t}\right) & 0 & \frac{1}{2} e^{-2 i \text{E} t} \left(-1+e^{4 i \text{E} t}\right) \\ 0 & 0 & 0 \\ -\frac{1}{2} e^{-2 i \text{E} t} \left(-1+e^{4 i \text{E} t}\right) & 0 & -\frac{1}{2} e^{-2 i \text{E} t} \left(1+e^{4 i \text{E} t}\right) \\ \end{array} \right)$

Dress state decoherence

Question: Why $\left|\psi_{0}\right\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|+\rangle)$( red ) much shorter than $\left|\psi_{0}\right\rangle=\frac{1}{\sqrt{2}}(|+1\rangle+|-1\rangle)$ ?

Energy dispersion of dress state

怎么把这个实验拉到室温？

系综有什么好处？

E测量会比我们平常的测量更准吗？

测应力的灵敏度怎么样？