Entanglement of the quantum system with spin–spin coupling created by optical excitation

Abstract

In this paper, we investigate the quantum entanglement characteristics of the system consisting an intermediary molecule with an optically excited triplet and two bilateral spin-1/2 nucleons. The two nuclear spins both couple to the excitation state which is caused by a pulsed laser. We study the linear entropy and entangling power of the evolution operator acting on the product state of the system. We deduce the entangling power when the energy state has a uniform distribution, and we find that the entanglement of the system shows a certain stability. In this paper, several standard expressions are analyzed and calculated in detail, including the detailed solution for the quantum entropy as well as the calculation of the linear entropy and entangling power, which are based on this solution. In comparing the linear entropy and entangling power, we find that the latter is the average of the former. Subsequently, we present an alternative derivation of the evolution operator and find that the result is consistent with that of the traditional method. When the evolution operator acts on the average of the product states, the entangling power of the evolution operator presents a distinct changing trend.

Appendix

The eigenvalues and eigenstates of the system are calculated in detail. Moreover, a transformation matrix is introduced to derive the effective Hamiltonian, which can be used to analyze the quantum characteristics of the entire system and determine the quantum entanglement of the energy eigenstates. When the Zeeman splitting energy of the electron is far greater than those of the nucleons as well as the interaction between the intermedium and the nucleons, the eigenvalue problem of the system can be studied using degenerate perturbation theory. The Hamiltonian of the system has been expressed above in Eq. 2:

$$\begin{aligned} H_{sym}= & {} -\,\omega _{n}S_{z,n}+\omega _{e}S_{z,e}-\omega _{n}S_{z,n^{\prime }} \nonumber \\&+A\left( \mathbf S _{n}\cdot \mathbf S _{e}+\mathbf S _{n^{\prime }}\cdot \mathbf S _{e}\right) +DS_{z,e}^{2}. \end{aligned}$$

And, under the assumption of Eq. 3, the electronic Zeeman splitting is much than the interaction between the nucleon and the electron, the terms $A\left( \mathbf S _{n}\cdot \mathbf S _{e}+\mathbf S _{n^{\prime }}\cdot \mathbf S _{e}\right)$ can be treated as the perturbation term.

The eigenvalues and eigenstates of the Hamiltonian of the system can be obtained as follows:

$$\begin{aligned} E_{1}= & {} A+\omega _{e}-\omega _{n}+D, \nonumber \\ E_{2}= & {} \omega _{e}+D, \nonumber \\ E_{3}= & {} \omega _{e}+D+\frac{A^{2}}{\omega _{e}+\omega _{n}+D}, \nonumber \\ E_{4}= & {} \omega _{e}-A+\omega _{n}+D+\frac{A^{2}}{\omega _{e}+\omega _{n}+D}, \nonumber \\ E_{5}= & {} -\omega _{n}-\frac{A^{2}}{\omega _{e}+\omega _{n}+D}, \nonumber \\ E_{6}= & {} 0, \nonumber \\ E_{7}= & {} \frac{2A^{2}D}{\omega _{e}^{2}+2\omega _{e}\omega _{n}+\omega _{n}^{2} -D^{2}}, \nonumber \\ E_{8}= & {} \omega _{n}+\frac{A^{2}}{\omega _{e}+\omega _{n}-D}, \nonumber \\ E_{9}= & {} D-\omega _{e}-\omega _{n}-A-\frac{A^{2}}{\omega _{e}+\omega _{n}-D}, \nonumber \\ E_{10}= & {} D-\omega _{e}, \nonumber \\ E_{11}= & {} D-\omega _{e}-\frac{A^{2}}{\omega _{e}+\omega _{n}-D}, \nonumber \\ E_{12}= & {} A-\omega _{e}+\omega _{n}+D. \end{aligned}$$

(38)

The orthonormalized energy eigenstates corresponding to these energy eigenvalues are as shown by the following expressions:

$$\begin{aligned} \left| 1\right\rangle= & {} g_{1}, \nonumber \\ \left| 2\right\rangle= & {} \frac{1}{\sqrt{2}}\left( g_{2}-g_{3}\right) , \nonumber \\ \left| 3\right\rangle= & {} c_{1}\left( \frac{1}{\sqrt{2}}\left( g_{2}+g_{3}\right) +\frac{A}{\omega _{e}+\omega _{n}+D}g_{5}\right) , \nonumber \\ \left| 4\right\rangle= & {} c_{1}\left( g_{4}+\frac{1}{\sqrt{2}}\frac{A}{ \omega _{e}+\omega _{n}+D}\left( g_{6}+g_{7}\right) \right) , \nonumber \\ \left| 5\right\rangle= & {} c_{1}\left( g_{5}-\frac{1}{\sqrt{2}}\frac{A}{ \omega _{e}+\omega _{n}+D}\left( g_{2}+g_{3}\right) \right) , \nonumber \\ \left| 6\right\rangle= & {} \frac{1}{\sqrt{2}}\left( g_{6}-g_{7}\right) , \nonumber \\ \left| 7\right\rangle= & {} c_{2}\left( \frac{1}{\sqrt{2}}\left( g_{6}+g_{7}\right) -\frac{A}{\omega _{e}+\omega _{n}+D}g_{4}+\frac{A}{\omega _{e} +\omega _{n}-D}g_{9}\right) , \nonumber \\ \left| 8\right\rangle= & {} c_{3}\left( g_{8}+\frac{1}{\sqrt{2}}\frac{A}{ \omega _{e}+\omega _{n}-D}\left( g_{10}+g_{11}\right) \right) , \nonumber \\ \left| 9\right\rangle= & {} c_{3}\left( g_{9}-\frac{1}{\sqrt{2}}\frac{A}{ \omega _{e}+\omega _{n}-D}\left( g_{6}+g_{7}\right) \right) , \nonumber \\ \left| 10\right\rangle= & {} \frac{1}{\sqrt{2}}\left( g_{10}-g_{11}\right) , \nonumber \\ \left| 11\right\rangle= & {} c_{3}\left( \frac{1}{\sqrt{2}}\left( g_{10}+g_{11}\right) -\frac{A}{\omega _{e}+\omega _{n}-D}g_{8}\right) , \nonumber \\ \left| 12\right\rangle= & {} g_{12}. \end{aligned}$$

(39)

Here, the ${g_{i},i=1,2,\ldots ,12,}$ form the natural basis (defined as $\left| T_{j}n_{1}n_{2}\right\rangle$, where $j=1/0/+, T_{-}=\left( \begin{array}{c} 1 \\ 0 \\ 0 \\ \end{array} \right) , T_{0}=\left( \begin{array}{c} 0 \\ 1 \\ 0 \\ \end{array} \right) , T_{+}=\left( \begin{array}{c} 0 \\ 0 \\ 1 \\ \end{array} \right) , n_{1/2}= \left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right) , \left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)$) of the action space of the Hamiltonian operators. These eigenstates are orthonormal. And

$$\begin{aligned} c_{1}= & {} \frac{\omega _{e}+\omega _{n}+D}{\sqrt{\left( \omega _{e}+\omega _{n}+D \right) ^{2}+A^{2}}}, \nonumber \\ c_{2}= & {} \frac{\left( \omega _{e}+\omega _{n}-D\right) \left( \omega _{e}+\omega _{n}+D\right) }{\sqrt{A^{2}\left( \left( \omega _{e}+\omega _{n}-D\right) ^{2}+\left( \omega _{e}+\omega _{n}+D\right) ^{2}\right) +\left( \omega _{e}+\omega _{n}-D\right) ^{2}\left( \omega _{e}+\omega _{n}+D\right) ^{2}}}, \nonumber \\ c_{3}= & {} \frac{\omega _{e}+\omega _{n}-D}{\sqrt{\left( \omega _{e}+\omega _{n}-D\right) ^{2}+A^{2}}}. \end{aligned}$$

(40)

The Hamiltonian of this system is complex. We set an transformation matrix ${\varvec{V}}=(V_{i,j}), i,j=1,2,\ldots ,12$ to obtain a simple and regular form for the Hamiltonian, where the matrix elements $V_{ij}$ are as follows:

$$\begin{aligned} V_{1,1}= & {} V_{12,12}=1, \end{aligned}$$

(41)

$$\begin{aligned} V_{2,2}= & {} -V_{3,2}=V_{10,10}=-V_{11,10}=V_{6,6}=-V_{7,6}=\frac{1}{\sqrt{2}}, \end{aligned}$$

(42)

$$\begin{aligned} V_{2,3}= & {} V_{3,3}=\frac{1}{\sqrt{2}}V_{4,4}=\frac{1}{\sqrt{2}}V_{5,5}=\frac{ \omega _{e}+\omega _{n}+D}{\sqrt{2}\sqrt{\left( \omega _{e}+\omega _{n}+D\right) ^ {2}+A^{2}}}, \end{aligned}$$

(43)

$$\begin{aligned} V_{6,4}= & {} V_{7,4}=-V_{2,5}=-V_{3,5}=\frac{1}{\sqrt{2}}V_{5,3}=\frac{A}{\sqrt{2} \sqrt{\left( \omega _{e}+\omega _{n}+D\right) ^{2}+A^{2}}}, \end{aligned}$$

(44)

$$\begin{aligned} V_{4,7}= & {} -\frac{A\left( \omega _{e}+\omega _{n}-D\right) }{\sqrt{A^{2}\left( \left( \omega _{e}+\omega _{n}+D\right) ^{2}+\left( \omega _{e}+\omega _{n}-D\right) ^{2} \right) +\left( \omega _{e}+\omega _{n}-D\right) ^{2}\left( \omega _{e}+\omega _{n}+D\right) ^{2}}}, \end{aligned}$$

(45)

$$\begin{aligned} V_{6,7}= & {} V_{7,7}=\frac{\left( \omega _{e}+\omega _{n}-D\right) \left( \omega _{e}+\omega _{n}+D\right) }{\sqrt{2}\sqrt{A^{2}\left( \left( \omega _{e}+\omega _{n}+D\right) ^{2}+\left( \omega _{e}+\omega _{n}-D\right) ^{2}\right) +\left( \omega _{e}+\omega _{n}-D\right) ^{2}\left( \omega _{e}+\omega _{n}+D\right) ^{2}}},\nonumber \\ \end{aligned}$$

(46)

$$\begin{aligned} V_{9,7}= & {} \frac{A\left( \omega _{e}+\omega _{n}+D\right) }{\sqrt{A^{2}\left( \left( \omega _{e}+\omega _{n}+D\right) ^{2}+\left( \omega _{e}+\omega _{n}-D\right) ^{2} \right) +\left( \omega _{e}+\omega _{n}-D\right) ^{2}\left( \omega _{e}+\omega _{n}+D\right) ^{2}}}, \end{aligned}$$

(47)

$$\begin{aligned} V_{10,8}= & {} V_{11,8}=-V_{6,9}=-V_{7,9}=-\frac{1}{\sqrt{2}}V_{8,11}=\frac{A}{ \sqrt{2}\sqrt{\left( \omega _{e}+\omega _{n}-D\right) ^{2}+A^{2}}}, \end{aligned}$$

(48)

$$\begin{aligned} V_{10,11}= & {} V_{11,11}=\frac{1}{\sqrt{2}}V_{8,8}=\frac{1}{\sqrt{2}}V_{9,9}= \frac{\omega _{e}+\omega _{n}-D}{\sqrt{2}\sqrt{\left( \omega _{e}+\omega _{n}-D\right) ^{2}+A^{2}}}. \end{aligned}$$

(49)

Aside from the matrix elements given above, all other matrix elements are zero. The effective Hamiltonian of the system can be written as

$$\begin{aligned} {\varvec{H}}_{sym, eff,-}= & {} \left( \begin{array}{cccc} H_{11,-} &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad H_{22,-} &{}\quad H_{23,-} &{}\quad 0 \\ 0 &{}\quad H_{32,-} &{}\quad H_{33,-} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{} \quad H_{44,-} \end{array} \right) , \end{aligned}$$

(50)

$$\begin{aligned} {\varvec{H}}_{sym, eff,0}= & {} \left( \begin{array}{cccc} H_{11,0} &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad H_{22,0} &{}\quad H_{23,0} &{}\quad 0 \\ 0 &{}\quad H_{32,0} &{}\quad H_{33,0} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad H_{44,0} \end{array} \right) , \end{aligned}$$

(51)

$$\begin{aligned} {\varvec{H}}_{sym, eff,+}= & {} \left( \begin{array}{cccc} H_{11,+} &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad H_{22,+} &{}\quad H_{23,+} &{}\quad 0 \\ 0 &{}\quad H_{32,+} &{}v H_{33,+} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad H_{44,+} \end{array} \right) . \end{aligned}$$

(52)

where the matrix elements that appear in the matrices above are as follows:

$$\begin{aligned} H_{11,-}= & {} A+\omega _{e}-\omega _{n}+D, \nonumber \\ H_{22,-}= & {} H_{33,-}=\omega _{e}+D+\frac{1}{2}\frac{A^{2}}{ \omega _{e}+\omega _{n}+D}, \nonumber \\ H_{23,-}= & {} H_{32,-}=-\frac{1}{2}\frac{A^{2}}{\omega _{e}+\omega _{n}+D}, \nonumber \\ H_{44,-}= & {} \omega _{e}-A+\omega _{n}+D+\frac{A^{2}}{\omega _{e}+\omega _{n}+D}; \nonumber \\ H_{11,0}= & {} -\omega _{n}-\frac{A^{2}}{\omega _{e}+\omega _{n}+D}, \nonumber \\ H_{22,0}= & {} H_{33,0}=A^{2}\frac{D}{\omega _{e}^{2}+2\omega _{e}\omega _{n}+ \omega _{n}^{2}-D^{2}}, \nonumber \\ H_{23,0}= & {} H_{32,0}=-A^{2}\frac{D}{\omega _{e}^{2}+2\omega _{e}\omega _{n} +\omega _{n}^{2}-D^{2}}, \nonumber \\ H_{44,0}= & {} \omega _{n}+\frac{A^{2}}{\omega _{e}+\omega _{n}-D}; \nonumber \\ H_{11,+}= & {} D-\omega _{e}-\omega _{n}-A-\frac{A^{2}}{\omega _{e}+\omega _{n}-D}, \nonumber \\ H_{22,+}= & {} H_{33,+}=D-\omega _{e}-\frac{1}{2}\frac{A^{2}}{ \omega _{e}+\omega _{n}-D}, \nonumber \\ H_{23,+}= & {} H_{32,+}=\frac{1}{2}\frac{A^{2}}{\omega _{e}+\omega _{n}-D}, \nonumber \\ H_{44,+}= & {} A-\omega _{e}+\omega _{n}+D. \end{aligned}$$

(53)

Where, we use the subscript $-,0,+$ to describe the three blocks that the effective Hamiltonian is divided into. And using the effective Hamiltonian (Eq. 53), we obtain the corresponding eigenvalus and eigenstates. The eigenvectors obtained are as follows,

$$\begin{aligned} \left| \psi _{1,-/0/+}\right\rangle= & {} \overrightarrow{p_{1}}, \nonumber \\ \left| \psi _{2,-/0/+}\right\rangle= & {} \frac{1}{\sqrt{2}}(\overrightarrow{p_{2}} -\overrightarrow{p_{3}}), \nonumber \\ \left| \psi _{3,-/0/+}\right\rangle= & {} \frac{1}{\sqrt{2}}(\overrightarrow{p_{2}} +\overrightarrow{p_{3}}), \nonumber \\ \left| \psi _{4,-/0/+}\right\rangle= & {} \overrightarrow{p_{4}}. \end{aligned}$$

(54)

Where, the computation bases are $\overrightarrow{p_1}=\left( \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ \end{array} \right) , \overrightarrow{p_2}=\left( \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ \end{array} \right) , \overrightarrow{p_3}=\left( \begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \\ \end{array} \right) , \overrightarrow{p_4}=\left( \begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \\ \end{array} \right)$. By using the block processing to solve the eigenvalue and eigenstate problem, we can obtain the eigenspectrum of each block. Each block is with a $4\times 4$ dimension, and there are four eigenvalues and four eigenstates corresponding to each block of the effective Hamiltonian. In the above calculations, we retain the term relating to $\frac{A}{\omega _{e}+\omega _{n}\pm D}$. If we ignore the terms, that is $\frac{A}{\omega _{e}+\omega _{n}\pm D}=0$, then the transformation matrix and the effective Hamiltonian are greatly simplified. These effective Hamiltonians expressions can be transformed by a local rotation as follows,

$$\begin{aligned} {\varvec{H}}_{sym, eff,j}^{\prime }={\varvec{U}}_{j}{\varvec{H}}_{sym, eff,j}{\varvec{U}}_{j}^{\dagger }+i{\varvec{U}}_{j}\frac{d{\varvec{U}}_{j}^{\dagger }}{dt}\quad ,\end{aligned}$$

(55)

where $j=-,0,+$, and

$$\begin{aligned} {\varvec{U}}_{j}(t)=e^{-iS_{z,n}\phi _{j}t}\otimes e^{-iS_{z,n^{\prime }}\phi _{j}t}, \end{aligned}$$

(56)

Here, $\phi _{-}$, $\phi _{0}$, and $\phi _{+}$ are,

$$\begin{aligned} \phi _{-}= & {} -\,\omega _{n}+A-\xi _{-}, \nonumber \\ \phi _{0}= & {} -\,\omega _{n}-\xi _{-}-\xi _{+}, \nonumber \\ \phi _{+}= & {} -\,\omega _{n}-A-\xi _{+}. \end{aligned}$$

(57)

Where, $\xi _{-}=\frac{1}{2}\frac{A^{2}}{\omega _{e}+\omega _{n}+D}$, $\xi _{+}=\frac{1}{2}\frac{A^{2}}{\omega _{e}+\omega _{n}-D}$, and $\xi _{0}=\xi _{+}-\xi _{-}$. In this way, the effective Hamiltonian of the system can be expressed as

$$\begin{aligned} {\varvec{H}}_{sym, eff,j}^{\prime }=2k\left( j\right) \xi _{j}\left( S_{x,n}S_{x,n^{\prime }}+S_{y,n}S_{y,n^{\prime }}\right) \end{aligned}$$

(58)

Where, $j=-,0,+$, $k\left( -\right) =-1$, $k\left( 0\right) =-1$, $k\left( +\right) =1$. $S_{x,n/n^{\prime }/e}$ and $S_{y,n/n^{\prime }/e}$ denote the x and y components of the Pauli spin operators, respectively. Therefore, the effective Hamiltonian that is acting on the subspace of the system indicates an XY-type interaction between the nucleon spins. Overall, the effective Hamiltonian can enable block processing and the interaction between the nucleons is the direct simple Heisenberg XY interaction.

Given the above, firstly, under the assumption of $\left| \omega _{n}\right| ,\left| D\right| ,A\ll \omega _{e}$, the terms $A\left( \mathbf S _{n}\cdot \mathbf S _{e}+\mathbf S _{n^{\prime }}\cdot \mathbf S _{e}\right)$ can be treated as the perturbation term, and the degenerate perturbation theory can be employed to determine the eigenvalues and eigenstates of the Hamiltonian of the system. Secondly, we can obtain an effective Hamiltonian ${\varvec{H}}_{sym, eff}$ which can be divided into three subspace blocks ${\varvec{H}}_{sym, eff,j}(j=-,0,+)$ with all matrix elements between the different subspaces neglected for the electronic Zeeman splitting $\omega _{e}$ is much larger than other parameters. Therefore, the dynamics in each of the subspace is closed and can be described by a $4 \times 4$ effective Hamiltonian ${\varvec{H}}_{sym, eff,j}(j=-,0,+)$. That is to say, in each block there is a direct Heisenberg XY interaction between the nucleons and the dynamics in each block is closed. For each closed subspace, no loss occurs during the coupling of the two nucleons. Furthermore, nuclear spins have low decoherence rates and long coherence time [27,28,29] which is much longer than the time that the electron is in its optically excited state. The transient optically excited state can be used to handle the fast and controllable generation of the nuclear spin–spin entanglement. Finally, it is worth noting that this method of producing entanglement must be modified when the electronic Zeeman splitting energy is not sufficiently high. When the Zeeman splitting energy is not very high, an optical excitation not only can induce entanglement but also may modulate or interfere with the entanglement, thereby leading to much more complex operation. However, that is worthy of further study in the future.