Determination of Qubit Entanglement in One-step Double Photoionization of Helium Atom

Abstract

Quantum entanglement is a unique phenomenon of quantum mechanics that explains how two subatomic particles are correlated even if they are separated by a vast distance. The phenomena of quantum entanglement are useful resources for quantum information. In this chapter, we will study the entanglement properties of bipartite states of two electronic qubits, without observing spin-orbit interaction (SOI), produced by single-step double photoionization in helium atom following the absorption of a single photon. In absence of SOI, Russell-Saunders coupling (L-S coupling) is applicable. We observe that the entanglement depends significantly on the direction of the ejection, as well as the spin quantization of photoelectrons.

1. Introduction

Quantum entanglement is the foremost prediction of quantum mechanics and one of the resources needed in quantum computing [1, 2]. Nowadays, it is an approach to solving time and processing power-consuming problems using quantum computers. A qubit is a quantum bit, the counterpart in quantum computing to the binary digit or bit of classical computing [3, 4, 5]. An important distinguishing feature between a qubit and a classical bit is that qubits can exhibit quantum entanglement. Quantum entanglement is a nonlocal property that allows a set of qubit to express a higher correlation that is not possible in classical systems. Qubits are employed in areas other than quantum computing, such as sensors. This helps in network and communication channels, improving the security of data stored in the quantum field and protecting it, as well as the speedier delivery of messages and encrypted networks for security-related information.

Einstein et al. began studying quantum entanglement in 1935 after presenting the EPR conundrum [6]. Bell [7] demonstrated the fault in EPR arguments by demonstrating that the notion of locality invoked in the EPR conundrum was incompatible with the hidden variables interpretation of the quantum theory. On the other hand, Bell’s theorem offers one of the potential ways to determine if two specific particles form an EPR pair or entangled state. The strong correlations of entangled particles are used as resources for quantum cryptography [8], quantum teleportation [9], and quantum computation [10] because entanglement allows multiple states of qubits to be acted on simultaneously unlike classical bits that can have one value at a time.

The bipartite entangled states of photons have been generated [11, 12, 13]. Being an excellent carrier of information, a photon is not suitable for long-term storage as it is immediately destroyed as soon as one tries to detect it. Some theories have been developed for studying bipartite states of electronic qubits produced by photoionization [14]. One of the simplest processes for producing EPR pairs of particles with non-zero rest mass is the simultaneous ejection of two electrons following the absorption of a single photon in an atom or molecule [15, 16]. This process is known as double photoionization (DPI). Here, entanglement is produced due to the process taking place inside an atom. We discuss DPI of an atom for generating bipartite states of two electronic qubits (say e₁ and e₂). In the absence of spin-orbit interaction (SOI), these particles are entangled with regard to their spin-angular momenta. In addition, the electrons in DPI are released in all directions and with all kinetic energies (subject to the conservation of total energy). Additionally, any direction can be quantized for its spins. Utilizing the density matrix (DM) of DPI concurrence [17, 18, 19] allows us to measure the level of entanglement.

We first establish some pertinent conventions in Section 2 and then quickly discuss the density operator (DO) and states for an atom’s DPI. This operator is appropriate when the target atom is in its ground state before DPI and the ionizing electromagnetic radiation is in a pure state of polarization. In the absence of SOI, this DO is used to develop an expression for the density matrix (DM) required to examine the quantum entanglement characteristics of the photoelectrons’ and photons’ spin states in Russell-Saunders coupling. In Section 3, we examine entanglement in a DPI system for qubits. Section 4 provides a quantitative application of the qubits for the DPI of the helium atom. The conclusion is given in Section 5, the last section.

2. Preliminaries

2.1 Density operator

Let’s use $e_1$ and $e_2$ to symbolize the two freely moving electrons whose entanglement characteristics we are interested in. The i(=1, 2)th electron’s propagation vector is ${\overrightarrow{k}}_i=\left(k_i,{\theta }_i,{\varphi }_i\right)$, and its kinetic energy is provided by ${\epsilon }_i={\mathrm{\hslash }}^2{k}_i^2/2m.$ These two electrons are a crucial component of atom A and are supposed to be simultaneously released from it after the absorption of a single photon. ${\mu }_i\left(=\pm 1/2\right)$ denotes the projection of the spin angular momentum of the ith electron along its spin quantization direction ${\widehat{u}}_i=\left({\alpha }_i,{\beta }_i\right)$. If A²⁺ denotes the residual dication, then our process can schematically be represented by

$$h{\nu }_r\left(\left|l_r\right|=1,m_r\right)+A\left|0\right.⟩\to A^{2+}\left|f\right.⟩+e_1\left({\overrightarrow{k}}_1;{\mu }_1{\widehat{u}}_1\right)+e_2\left({\overrightarrow{k}}_2;{\mu }_2{\widehat{u}}_2\right).$$E1

Here, in the electric dipole (E₁) approximation, E_r = $h{\nu }_r$ and $\left|l_r\right|=1$ represent, respectively, the energy and angular momentum of the photon absorbed by atom A. The parameter $m_r$ stands for the photon’s state of polarization, where $\left|0\right.⟩$ and $\left|f\right.⟩$, respectively, represent the bound electronic states of A with energy E₀ and the remaining doubly charged photoion A²⁺ with energy E_f. The direction of the electric vector of the linearly polarized ($m_r$ = 0) radiation involved in the process (1) defines the quantization axis of our space (or photon) frame of reference; if the ionizing radiation is circularly polarized ($m_r$ = $\pm 1$) or unpolarized, the direction of incidence then determines the polar axis of the photon frame.

Let us use ${\rho }_r=\left|m_r\right.⟩\left.⟨m_r\right|$ and ${\rho }_0=\left|0\right.⟩\left.⟨0\right|$ to represent the density operators of the ionizing radiation and the unpolarized atom A before DPI, respectively. Before the interaction between the two occurs, the incident photon and the atom are uncorrelated. This indicates that the direct product determines the density operator for the combined (atom + photon) system of Eq. (1).

$${\rho }_i={\rho }_0\otimes {\rho }_r$$E2

Let us denote by F_p the photoionization operator in the E₁ approximation. Then the density operator of the combined (A²⁺ + e₁ + e₂) system in Eq. (1) after DPI becomes

$${\rho }_f=K_p{F}_p{\rho }_i{F}_p^{+}.$$E3

The E₁ photoionization operator F_p has been defined in Appendix A. Here, $K_p=3\pi {\left(e^2/{\alpha }_0{E}_r\right)}^2$ with ${\alpha }_0$ the dimensionless fine structure constant [20].

2.2 Definitions of E₁ photoionization operator and concurrence

In the present case, E₁ photoionization operator ${\mathit{F}}_{\mathit{p}}$ for ${\mathit{n}}_e$-electron system can be defined as [20]:

$${\mathit{F}}_{\mathit{p}}=\sqrt{\frac{\mathrm{4}{\mathit{\pi \alpha }}^{\mathrm{3}}{\mathit{E}}_{\mathit{r}}^{\mathrm{3}}{\mathit{m}}_{ℯ}}{\mathrm{4}{ℯ}^{\mathrm{4}}{\mathrm{\hslash }}^{\mathrm{2}}}}\sum _{\mathit{i}=\mathrm{1}}^{{\mathit{n}}_e}{\widehat{\xi }}_{{\mathit{m}}_{\mathit{r}}}.{\overrightarrow{\mathit{r}}}_{\mathit{i}},$$E4

and

$${\mathit{F}}_{\mathit{p}}=\sqrt{\frac{\mathrm{4}{\mathit{\pi \alpha }}^{\mathrm{3}}{\mathit{E}}_{\mathit{r}}^{\mathrm{3}}{\mathit{a}}_{\mathrm{0}}^{\mathrm{2}}{\mathit{m}}_{ℯ}}{\mathrm{3}{\mathrm{\hslash }}^{\mathrm{2}}}}\sum _{\mathit{i}=\mathrm{1}}^{{\mathit{n}}_e}{\widehat{\xi }}_{{\mathit{m}}_{\mathit{r}}}.{\overrightarrow{\nabla }}_{\mathit{i}}.$$E5

The operators of Eqs. (4) and (5) represent the interaction of the atomic electrons (their number being ${\mathit{n}}_e$) with the incident electromagnetic radiation in the E₁ length and velocity approximations, respectively. In Eq. (4), ${\overrightarrow{\mathit{r}}}_{\mathit{i}}$ is the position vector and in Eq. (5), ${\overrightarrow{\nabla }}_{\mathit{i}}\left[={\left(-\mathrm{1}\right)}^{\mathrm{1}/\mathrm{2}}{\overrightarrow{\mathit{p}}}_{\mathit{i}}/\mathrm{\hslash }\right]$ is the linear momentum of ith bound atomic electron. Here ${\widehat{\xi }}_{{\mathit{m}}_{\mathit{r}}}$ is the spherical unit vector [21] in the direction of polarization of the incident.

The concurrence (C) is a very successful and widely used measure for quantifying quantum entanglement between two qubits. It is an additive and operational measure of entanglement. Additivity is a very desirable property that can reduce calculational complexity of entanglement [4, 17, 22]. For any state ${\left|\varphi \right.⟩}_{\mathit{AB}}$ in a $\mathit{d}\otimes {\mathit{d}}^{\text{'}}\left(\mathit{d}\le {\mathit{d}}^{\text{'}}\right)$ quantized system, it can be written as:

$$\mathrm{C}=\sqrt{2\left(1-\mathit{\text{Trace}}{\rho }_A^2\right)},$$E6

where ${\rho }_A$ is the reduced density matrix defined as ${\rho }_{\mathit{A}}={\mathit{\text{trace}}}_{\mathit{B}}\left({\left|\varphi \right.⟩}_{\mathit{AB}}\left.⟨\varphi \right|\right)$. When $0\le C\le 1$ always; C = 0 for a separable (unentangled) state; C > 0 for a nonseparable (entangled) state.

3. Entanglement between two electronic qubits for DPI

Here, we calculate the DM for the angle- and spin-resolved DPI of an atom without considering SOI into account in either the bound electronic states of A and A²⁺ or the continua of the two photoelectrons (e₁, e₂) ejected in the process (1). Here LS-coupling is applicable. Therefore, the total orbital angular momenta (${\overrightarrow{L}}_0,{\overrightarrow{L}}_f$) and the spin angular momenta (${\overrightarrow{S}}_0,{\overrightarrow{S}}_f$) of A and A²⁺ are conserved quantities. If the orbital angular momentum of the photoelectron e₁ is ${\overrightarrow{l}}_1$ and that of e₂ is ${\overrightarrow{l}}_2$ with their respective spin angular momenta ${\left(\frac{1}{2}\right)}_1$ and ${\left(\frac{1}{2}\right)}_2$, we then have

$${\overrightarrow{L}}_0+{\overrightarrow{l}}_r={\overrightarrow{L}}_f+\overrightarrow{l}\left(={\overrightarrow{l}}_1+{\overrightarrow{l}}_2\right)$$E7

and

$${\overrightarrow{S}}_0={\overrightarrow{S}}_f+{\overrightarrow{s}}_t\left(={\left(\frac{1}{2}\right)}_1+{\left(\frac{1}{2}\right)}_2\right).$$E8

Here $M_{L_0},M_{S_0},M_{L_f},\text{and}{M}_{S_f}$ are the respective projections of ${\overrightarrow{L}}_0,{\overrightarrow{S}}_0,{\overrightarrow{L}}_f,\text{and}{\overrightarrow{S}}_f$ along the polar axis of the space frame, then, in Eq. (1), the bound electronic state of atom A is $\left|0\right.⟩\equiv \left|L_0{S}_0{M}_{L_0}{M}_{S_0}\right.⟩$ and that of the dication A²⁺ is $\left|f\right.⟩\equiv \left|L_f{S}_f{M}_{L_f}{M}_{S_f}\right.⟩$. The density operator (2) for the combined (atom + photon) system becomes

$${\rho }_i=\frac{1}{\left(2L_0+1\right)\left(2S_0+1\right)}\sum _{M_{L_0}{M}_{S_0}}\left|0;1m_r\right.⟩\left.⟨0;1m_r\right|,$$E9

here we have defined $\left|0;1m_r\right.⟩\equiv \left|0\right.⟩\left|1m_r\right.⟩.$

In order to calculate the DM for the (A²⁺+e₁+e₂) system in process (1), we now calculate the matrix elements ${\rho }_i$ and ${\rho }_f$. Following the procedures given in Ref. [15], the matrix elements in the present case are given by

$$\begin{array}{l}⟨f;{\overrightarrow{k}}_1,{\mu }_1{\widehat{u}}_{1;}{\overrightarrow{k}}_2,{\mu }_2{\widehat{u}}_2\left|{\rho }_f\right|f;{\overrightarrow{k}}_1,\mu {\text{'}}_1{\widehat{u}}_{1;}{\overrightarrow{k}}_2,\mu {\text{'}}_2{\widehat{u}}_2⟩=\frac{K_p}{\left(2L_0+1\right)\left(2S_0+1\right)}\\ \sum _{M_{L_0}{M}_{S_0}}⟨f;{\overrightarrow{k}}_1,{\mu }_1{\widehat{u}}_{1;}{\overrightarrow{k}}_2,{\mu }_2{\widehat{u}}_2\left|F_p\right|0;1{\mathrm{m}}_r⟩\mathrm{x}{⟨f;{\overrightarrow{k}}_1,\mu {\text{'}}_1{\widehat{u}}_{1;}{\overrightarrow{k}}_2,\mu {\text{'}}_2{\widehat{u}}_2\left|F_p\right|0;1{\mathrm{m}}_r⟩}^{\ast }.\end{array}$$E10

As the DM in Eq. (10) is Hermitian, we can write

$$\begin{array}{l}⟨f;{\overrightarrow{k}}_1,{\mu }_1{\hat{u}}_{1;}{\overrightarrow{k}}_2,{\mu }_2{\hat{u}}_2\left|{\rho }_f\right|f;{\overrightarrow{k}}_1,\mu {\text{'}}_1{\hat{u}}_{1;}{\overrightarrow{k}}_2,\mu {\text{'}}_2{\hat{u}}_2⟩\\ ={⟨f;{\overrightarrow{k}}_1,\mu {\text{'}}_1{\hat{u}}_{1;}{\overrightarrow{k}}_2,{\mu }_{\text{'}2}{\hat{u}}_2\left|{\rho }_f\right|f;{\overrightarrow{k}}_1,{\mu }_1{\hat{u}}_{1;}{\overrightarrow{k}}_2,{\mu }_2{\hat{u}}_2⟩}^{\ast }\end{array}$$E11

Next, we evaluate the matrix elements of E1 photoionization operator F_p occurring on the right-hand side of Eq. (10). To this end, we first introduce the coupling suggested by Eqs. (7) and (8) in each of the bras and kets used to calculate the matrix elements of F_p. We therefore can write

$$\left|\mathrm{0},\mathrm{1}{\mathit{m}}_{\mathit{r}}\right.⟩=\sum _{{\mathit{LM}}_{\mathit{L}}}{\left(-\mathrm{1}\right)}^{\mathrm{1}-{\mathit{L}}_{\mathrm{0}}-{\mathit{M}}_{\mathit{L}}}\sqrt{\mathrm{2}\mathit{L}+\mathrm{1}}\left(\begin{array}{l}{\mathit{L}}_{\mathrm{0}}\mathrm{1}\mathit{L}\\ {\mathit{M}}_{{\mathit{L}}_{\mathrm{0}}}{\mathit{m}}_{\mathit{r}}-{\mathit{M}}_{\mathit{L}}\end{array}\right)\left|\left({\mathit{L}}_{\mathrm{0}}\mathrm{1}\right){\mathit{LM}}_{\mathit{L}};{\mathit{S}}_{\mathrm{0}}{\mathit{M}}_{{\mathit{S}}_{\mathrm{0}}}\right.⟩$$E12

and

$$\begin{array}{c}\left|\mathit{f};{\hat{\mathit{k}}}_{\mathrm{1}},{\mu }_{\mathrm{1}}{\hat{\mathit{u}}}_{\mathrm{1};}{\hat{\mathit{k}}}_{\mathrm{2}},{\mu }_{\mathrm{2}}{\hat{\mathit{u}}}_{\mathrm{2}}\right.⟩={\left(-\mathrm{1}\right)}^{-{\mathit{L}}_{\mathit{f}}-{\mathit{S}}_{\mathit{f}}}\sum _{\begin{array}{l}{\ell }_{\mathrm{1}}{\ell }_{\mathrm{2}}\ell \\ {\mathit{n}}_{\mathrm{1}}{\mathit{n}}_{\mathrm{2}}\mathit{N}\end{array}}\sum _{\begin{array}{l}{\nu }_{\mathrm{1}}{\nu }_{\mathrm{2}}\\ {\mathit{s}}_{\mathit{t}}\nu \end{array}}\sum _{\begin{array}{l}{\mathit{L}}_{\mathit{T}}{\mathit{M}}_{{\mathit{L}}_{\mathit{T}}}\\ {\mathit{SM}}_{\mathit{S}}\end{array}}{\mathit{i}}^{{\ell }_{\mathrm{1}}+{\ell }_{\mathrm{2}}}{ℯ}^{-\mathit{i}\left({\sigma }_{{\ell }_{\mathrm{1}}}+{\sigma }_{{\ell }_{\mathrm{2}}}\right)}{\left(-\mathrm{1}\right)}^{-{\ell }_{\mathrm{1}}+{\ell }_{\mathrm{2}}+\ell +{\mathit{s}}_{\mathit{t}}-\mathit{N}-\nu -{\mathit{M}}_{{\mathit{L}}_{\mathit{T}}}-{\mathit{M}}_{\mathit{S}}}\\ \mathit{x}\sqrt{\left(\mathrm{2}\ell +\mathrm{1}\right)\left(\mathrm{2}{\mathit{s}}_{\mathit{t}}+\mathrm{1}\right)\left(\mathrm{2}{\mathit{L}}_{\mathit{T}}+\mathrm{1}\right)\left(\mathrm{2}\mathit{S}+\mathrm{1}\right)}\left(\begin{array}{l}{\ell }_{\mathrm{1}}{\ell }_{\mathrm{2}}\ell \\ {\mathit{m}}_{\mathrm{1}}{\mathit{m}}_{\mathrm{2}}-\mathit{m}\end{array}\right)\left(\begin{array}{l}\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{1}}{\mathrm{2}}{\mathit{s}}_{\mathit{t}}\\ {\nu }_{\mathrm{1}}{\nu }_{\mathrm{2}}-\nu \end{array}\right)\left(\begin{array}{l}{\mathit{L}}_{\mathit{f}}\ell {\mathit{L}}_{\mathit{T}}\\ {\mathit{M}}_{{\mathit{L}}_{\mathit{f}}}\mathit{N}-{\mathit{M}}_{{\mathit{L}}_{\mathit{T}}}\end{array}\right)\\ \mathit{x}\left(\begin{array}{l}{\mathit{S}}_{\mathit{f}}{\mathit{s}}_{\mathit{t}}\mathit{S}\\ {\mathit{M}}_{{\mathit{S}}_{\mathit{f}}}\nu -{\mathit{M}}_{\mathit{S}}\end{array}\right){\mathit{Y}}_{{\ell }_{\mathrm{1}}}^{{\mathit{n}}_{\mathrm{1}}^{\ast }}\left({\hat{\mathit{k}}}_{\mathrm{1}}\right){\mathit{Y}}_{{\ell }_{\mathrm{2}}}^{{\mathit{n}}_{\mathrm{2}}^{\ast }}\left({\hat{\mathit{k}}}_{\mathrm{1}}\right){\mathit{D}}_{{\mu }_{\mathrm{1}}{\nu }_{\mathrm{1}}}^{{\frac{\mathrm{1}}{\mathrm{2}}}^{\ast }}\left({\omega }_{\mathrm{1}}\right){\mathit{D}}_{{\mu }_{\mathrm{2}}{\nu }_{\mathrm{2}}}^{{\frac{\mathrm{1}}{\mathrm{2}}}^{\ast }}\left({\omega }_{\mathrm{2}}\right)\left|\left({\mathit{L}}_{\mathit{f}}\ell \right){\mathit{L}}_{\mathit{T}}{\mathit{M}}_{{\mathit{L}}_{\mathit{T}}};\left({\mathit{S}}_{\mathit{f}}{\mathit{s}}_{\mathit{t}}\right){\mathit{SM}}_{\mathit{S}}\right.⟩.\end{array}$$E13

Here, ${\sigma }_{{\ell }_{\mathrm{1}}}$ and ${\sigma }_{{\ell }_{\mathrm{2}}}$ are the Coulomb phases for l₁th and l₂th partial waves of the photoelectrons, respectively; Ds are the well-known rotational harmonics [23] with ${\omega }_{\mathrm{1}}\left({\alpha }_{\mathrm{1}},{\beta }_{\mathrm{1}},\mathrm{0}\right)$ and ${\omega }_{\mathrm{2}}\left({\alpha }_{\mathrm{2}},{\beta }_{\mathrm{2}},\mathrm{0}\right)$, the Euler angles that rotate the axis of the space-frame into the spin-polarization directions ${\widehat{u}}_1$ and ${\widehat{u}}_2$ (Figure 1), respectively. Furthermore, in (13), the properly antisymmetrized and asymptotically normalized [24] ket $\left|\left({\mathit{L}}_{\mathit{f}}\ell \right){\mathit{L}}_{\mathit{T}}{\mathit{M}}_{{\mathit{L}}_{\mathit{T}}};\left({\mathit{S}}_{\mathit{f}}{\mathit{s}}_{\mathit{y}}\right){\mathit{SM}}_{\mathit{S}}\right.⟩$ represents the photoion A²⁺ in its electronic state and the two photoelectrons with their total orbitaland spin angular momenta coupled according to the scheme expressed in Eqs. (7) and (8).

Figure 1.

Two electrons are emitted simultaneously from helium after photoabsorption.

Now we substitute in Eq. (9) the normalized condition of reference [25]

$$\begin{array}{l}\left.⟨\left({\mathit{L}}_{\mathit{f}}\ell \right){\mathit{L}}_{\mathit{T}}{\mathit{M}}_{{\mathit{L}}_{\mathit{T}}};\left({\mathit{S}}_{\mathit{f}}{\mathit{s}}_{\mathit{t}}\right){\mathit{SM}}_{\mathit{S}}\right|{\mathit{F}}_{\mathit{p}}\left|\left({\mathit{L}}_{\mathrm{0}}\mathrm{1}\right){\mathit{LM}}_{\mathit{L}};{\mathit{S}}_{\mathrm{0}}{\mathit{M}}_{{\mathit{S}}_{\mathrm{0}}}\right.⟩\\ ={\delta }_{{\mathit{LL}}_{\mathit{T}}}{\delta }_{{\mathit{M}}_{\mathit{L}}{\mathit{M}}_{{\mathit{L}}_{\mathit{T}}}}{\delta }_{{\mathit{S}}_{\mathrm{0}}\mathit{S}}{\delta }_{{\mathit{M}}_{{\mathit{S}}_{\mathrm{0}}}{\mathit{M}}_{\mathit{S}}}\left.⟨\left({\mathit{L}}_{\mathit{f}}\ell \right)\mathit{L}\right|{\mathit{F}}_{\mathit{p}}\left|\left({\mathit{L}}_{\mathrm{0}}\mathrm{1}\right)\mathit{L}\right.⟩.\end{array}$$

This result arises from the conservation circumstances in (7) and (8).

By using Racah algebra to analytically evaluate as many of the sums that are contained there, the following subsequent equation is made simpler. It necessitates, for instance, the application of (a) the addition theorems (i.e., Eqs. (4.3.2) and (4.6.5) from [23] for rotational and spherical harmonics, (b) Eq. (6.2.5) [23] for transforming a single sum of the product of three 3-j symbols into a product of one 3-j and one 6-j symbols, (c) identity (5) given on page 453 in Ref. [26] for converting a product of two 3-j symbols and one 6-j symbol summed over two variables into a product of two 3-j and one 6-j symbols, (d) Eq. (14.42) from [27] that transforms a quadruple sum of the product of four 3-j symbols into a double sum containing two 3-j and one 9-j symbols, (e) Eq. (3.7.9) [23] to convert a phase factor into a 3-j symbol, (f) the orthogonality of 3-j symbols (3.7.7) [23], (g) relation (6.4.14) [23] to write a 9-j symbol in terms of a 6-j symbol, and (h) relation (6.4.14) [23] to turn the sum of the products of two 6-j symbols into one 6-j symbol. Due to these and other simplifications, the DM (10) is written as follows:

$$\begin{array}{l}⟨f;{\overrightarrow{k}}_1,{\mu }_1{\hat{u}}_{1;}{\overrightarrow{k}}_2,{\mu }_2{\hat{u}}_2\left|{\rho }_f\right|f;{\overrightarrow{k}}_1,\mu {\text{'}}_1{\hat{u}}_{1;}{\overrightarrow{k}}_2,\mu {\text{'}}_2{\hat{u}}_2⟩=\\ =\begin{array}{l}\frac{K_p}{4\pi \left(2L_0+1\right)}\sum _{\begin{array}{l}{l}_1{l}_{2}l\\ l{\text{'}}_{1}l{\text{'}}_2{l}^{\text{'}}\end{array}}\sum _{\begin{array}{l}{L}_1{L}_{2}M\\ {\mathit{LL}}^{\text{'}}{L}_r\end{array}}\sum _{{\mathit{sQm}}_1{m}_{2}n}{\left(-1\right)}^{l{\text{'}}_1+l{\text{'}}_2+l\text{'}+\mathrm{Q}-\mathit{n}+{\mu }_{\text{'}1}+\mu {\text{'}}_2+M{\text{'}}_{S_f}}\left(2S_f+1\right)\left(2{\mathrm{L}}_r+1\right)\left(2s+1\right)\\ \mathrm{x}\left(2Q+1\right)\sqrt{\left(2L_1+1\right)\left(2L_2+1\right)}\left(\begin{array}{l}{l}_{1}l{\text{'}}_1{L}_1\\ 000\end{array}\right)\left(\begin{array}{l}{l}_{2}l{\text{'}}_2{L}_2\\ 000\end{array}\right)\left(\begin{array}{l}11L_r\\ m_r-m_{r}0\end{array}\right)\left(\begin{array}{l}{L}_1{L}_2{L}_r\\ M-M0\\ \end{array}\right)\\ \mathrm{x}\left\{\begin{array}{l}11L_r\\ {\mathrm{L}}^{\text{'}}\mathrm{L}{\mathrm{L}}_0\end{array}\right\}\left\{\begin{array}{l}l{l}^{\text{'}}{L}_r\\ L^{\text{'}}L{L}_f\end{array}\right\}\left\{\begin{array}{l}{l}_1{l}_{2}l\\ l{\text{'}}_{1}l{\text{'}}_2{l}^{\text{'}}\\ L_1{L}_2{L}_r\end{array}\right\}\left(\begin{array}{l}\frac{1}{2}\frac{1}{2}s\\ {\mu }_1-\mu {\text{'}}_1{m}_1\end{array}\right)\left(\begin{array}{l}\frac{1}{2}\frac{1}{2}s\\ {\mu }_2-\mu {\text{'}}_2{m}_2\end{array}\right)\left(\begin{array}{l}{S}_f{S}_{f}Q\\ M_{S_f}-M{\text{'}}_{S_f}n\end{array}\right)\\ \mathrm{x}\left(\begin{array}{l}ssQ\\ m_1{m}_{2}n\end{array}\right)\left\{\begin{array}{l}\frac{1}{2}\frac{1}{2}s\\ S_f{S}_{f}Q\end{array}\right\}{Y}_{L_1}^M\left({\hat{k}}_1\right)Y_{L_2}^{-M}\left({\hat{k}}_2\right)D_{m_{1}n}^{s^{\ast }}\left({\omega }_1\right)D_{m_2-n}^{s^{\ast }}\left({\omega }_2\right)\\ \mathrm{x}⟨L_{f}l\left|F\left(L\right)\right|L_{0}1⟩{⟨L_f{l}^{\text{'}}\left|F\left(L^{\text{'}}\right)\right|L_{0}1⟩}^{\ast }\end{array}\end{array}$$E14

with

$$⟨L_{f}l\left|F\left(\mathrm{L}\right)\right|L_{0}1⟩={\left(-i\right)}^{l_1+l_2}{e}^{-i\left({\sigma }_{l_1}+{\sigma }_{l_2}\right)}\left(2L+1\right)\sqrt{\left(2l_1+1\right)\left(2l_2+1\right)\left(2l+1\right)}\left.⟨\left(L_{f}l\right)\mathrm{L}\right|F_p\left|\left(L_{0}1\right)L\right.⟩.$$

From Eq. (12), the DM for the angle- and spin-resolved DPI process (1) in the absence of SOI can be written in the following form:

$$⟨f;{\overrightarrow{k}}_1,{\mu }_1{\widehat{u}}_{1;}{\overrightarrow{k}}_2,{\mu }_2{\widehat{u}}_2\left|{\rho }_f\right|f;{\overrightarrow{k}}_1,\mu {\text{'}}_1{\widehat{u}}_{1;}{\overrightarrow{k}}_2,\mu {\text{'}}_2{\widehat{u}}_2⟩=\frac{d^3\sigma \left(m_r\right)}{d{\epsilon }_{1}d{\widehat{k}}_{1}d{\widehat{k}}_2}\sigma {\left(S_0;S_f;{\widehat{u}}_1,{\widehat{u}}_2\right)}_{M_{S_f}{\mu }_1{\mu }_2;M{\text{'}}_{S_f}\mu {\text{'}}_1\mu {\text{'}}_2}.$$E15

The first term, that is, the triple differential cross section (TDCS, i.e., $d^3\sigma \left(m_r\right)/d{\epsilon }_{1}d{\widehat{k}}_{1}d{\widehat{k}}_2$) on the right-hand side of (15) depends upon the orbital angular momenta of A and A²⁺; phase shifts, energies (${\epsilon }_1,{\epsilon }_2$), and the directions (${\widehat{k}}_1,{\widehat{k}}_2$) of the emitted electrons (e₁, e₂); the state of polarization (m_r) of the ionizing radiation; and the photoionization dynamics. It does not include spins of the photoelectrons or the target atom or the residual dication. Thus, $d^3\sigma \left(m_r\right)/d{\epsilon }_{1}d{\widehat{k}}_{1}d{\widehat{k}}_2$ in the DM (15) describes purely the angular correlation between the photoelectrons in the L-S coupling scheme for the angular momenta of the particles involved in DPI (1). Its value is always positive. Here,

$$\begin{array}{l}\frac{d^3\sigma \left(m_r\right)}{d{\epsilon }_{1}d{\widehat{k}}_{1}d{\widehat{k}}_2}={\left(-1\right)}^{m_r+L_0+L_f}\frac{K_p}{4\pi \left(2L_0+1\right)}\sum _{\begin{array}{l}{l}_1{l}_{2}l\\ l{\text{'}}_{1}l{\text{'}}_2{l}^{\text{'}}\end{array}}\sum _{\begin{array}{l}{L}_1{L}_{2}M\\ {\mathit{LL}}^{\text{'}}{L}_r\end{array}}{\left(-1\right)}^{l{\text{'}}_1+l{\text{'}}_2+l\text{'}}\left(2{\mathrm{L}}_r+1\right)\\ \mathrm{x}\sqrt{\left(2L_1+1\right)\left(2L_2+1\right)}\left(\begin{array}{l}{l}_{1}l{\text{'}}_1{L}_1\\ 000\end{array}\right)\left(\begin{array}{l}{l}_{2}l{\text{'}}_2{L}_2\\ 000\end{array}\right)\left(\begin{array}{l}11L_r\\ m_r-m_{r}0\end{array}\right)\left(\begin{array}{l}{L}_1{L}_2{L}_r\\ M-M0\end{array}\right)\left\{\begin{array}{l}11L_r\\ {\mathrm{L}}^{\text{'}}\mathrm{L}{\mathrm{L}}_0\end{array}\right\}\\ \mathrm{x}\left\{\begin{array}{l}l{l}^{\text{'}}{L}_r\\ L^{\text{'}}L{L}_f\end{array}\right\}\left\{\begin{array}{l}{l}_1{l}_{2}l\\ l{\text{'}}_{1}l{\text{'}}_2{l}^{\text{'}}\\ L_1{L}_2{L}_r\end{array}\right\}{Y}_{L_1}^M\left({\widehat{k}}_1\right)Y_{L_2}^{-M}\left({\widehat{k}}_2\right)⟨L_{f}l\left|F\left(L\right)\right|L_{0}1⟩{⟨L_f{l}^{\text{'}}\left|F\left({\mathrm{L}}^{\text{'}}\right)\right|L_{0}1⟩}^{\ast }.\end{array}$$E16

The second term (i.e., $\sigma {\left(S_0;S_f;\widehat{u}\right)}_{M_{S_f}\mu ;M{\text{'}}_{S_f}\mu \text{'}}$) present on the right-hand side of Eq. (15) is the spin-correlation density matrix (SCDM), which completely determines the entanglement properties among electronic qubit (e_p) and ionic qudit $\left(M^{+}\right)$, can be written as:

$$\begin{array}{l}{\sigma }_{{{}_{e_1}}_{e_2}}={\left(-1\right)}^{{\mu }_{\text{'}1}+\mu {\text{'}}_2}\sum _{{\mathit{ss}}_t{m}_1{m}_{2}n}{\left(-1\right)}^{s_t-n}\left(2s+1\right)\\ \mathrm{x}\left(\begin{array}{l}\frac{1}{2}\frac{1}{2}s\\ {\mu }_1-\mu {\text{'}}_1{m}_1\end{array}\right)\left(\begin{array}{l}\frac{1}{2}\frac{1}{2}s\\ {\mu }_2-\mu {\text{'}}_2{m}_2\end{array}\right)\left\{\begin{array}{l}\frac{1}{2}\frac{1}{2}s\\ \frac{1}{2}\frac{1}{2}{s}_t\end{array}\right\}{D}_{m_{1}n}^{s^{\ast }}\left({\omega }_1\right)D_{m_2-n}^{s^{\ast }}\left({\omega }_2\right).\end{array}$$E17

Considering the condition given by Eq. (8), we see that $\left|S_f-S_0\right|$ can take only two values: 0 and 1. Let us consider the entanglement between two electrons e₁ and e₂ in both cases. In order to calculate the SCDMs, we have to consider the real part of the outgoing wave function [28]. Before writing the DM, we first consider

$$\alpha ={\alpha }_1-{\alpha }_2,c_1=cos{\beta }_1,s_1=sin{\beta }_1,c_2=cos{\beta }_2,s_2=sin{\beta }_2,\text{and}c=cos\alpha$$

1. Let us first consider the case $\left|S_f-S_0\right|$ = 0. We obtain the following SCDM from (17):

   $${\sigma }_{{{}_{e_1}}_{e_2}}=$$

   We calculate from Table 1 that the det (${\sigma }_{{{}_{e_1}}_{e_2}}$) = $-\frac{1}{16}{cos}^2\alpha$. In order to quantifying entanglement, we have obtained concurrence, using SCDM of Table 1, according to the definition given in Eq. (6), which yields

   $$C=\sqrt{1-\frac{1+{\left(cos{\alpha }_{1}cos{\alpha }_2+cos\alpha sin{\alpha }_{1}sin{\alpha }_2\right)}^2}{16384}}.$$E18

   Eq. (18) means [4, 17, 22] that the values of C are positive and so the spin state $\left|S_f-S_0\right|$ = 0 is entangled, depending on ${\alpha }_1$ and ${\alpha }_2.$ This variation of concurrence is shown in Figure 2.

   We can see from Figure 2 that the states are entangled (the values of concurrence are positive) and separable (for the zero value of concurrence) depending on spin quantization directions.

2. Considering the case of $\left|S_f-S_0\right|$ = 1, we procure the following SCDM from Eq. (17):

$\mu {\text{'}}_1\mu {\text{'}}_2$; ${\mu }_1{\mu }_2$	$\frac{1}{2}\frac{1}{2}$	$\frac{1}{2}-\frac{1}{2}$	$-\frac{1}{2}\frac{1}{2}$	$-\frac{1}{2}-\frac{1}{2}$
$\frac{1}{2}\frac{1}{2}$	$0.25-\frac{1}{4}\left(c_1{c}_2+s_1{s}_{2}c\right)$	$\frac{1}{4}\left(c_1{s}_2-s_1{c}_{2}c\right)$	$\frac{1}{4}\left(s_1{c}_2-c_1{s}_{2}c\right)$	$\frac{1}{4}\left(c-s_1{s}_2-c_1{c}_{2}c\right)$
$\frac{1}{2}-\frac{1}{2}$	$\frac{1}{4}\left(c_1{s}_2-s_1{c}_{2}c\right)$	$0.25+\frac{1}{4}\left(c_1{c}_2+s_1{s}_{2}c\right)$	$-\frac{1}{4}\left(c+s_1{s}_2+c_1{c}_{2}c\right)$	$-\frac{1}{4}\left(s_1{c}_2-c_1{s}_{2}c\right)$
$-\frac{1}{2}\frac{1}{2}$	$\frac{1}{4}\left(s_1{c}_2-c_1{s}_{2}c\right)$	$-\frac{1}{4}\left(c+s_1{s}_2+c_1{c}_{2}c\right)$	$0.25+\frac{1}{4}\left(c_1{c}_2+s_1{s}_{2}c\right)$	$-\frac{1}{4}\left(c_1{s}_2-s_1{c}_{2}c\right)$
$-\frac{1}{2}-\frac{1}{2}$	$\frac{1}{4}\left(c-s_1{s}_2-c_1{c}_{2}c\right)$	$-\frac{1}{4}\left(s_1{c}_2-c_1{s}_{2}c\right)$	$-\frac{1}{4}\left(c_1{s}_2-s_1{c}_{2}c\right)$	$0.25-\frac{1}{4}\left(c_1{c}_2+s_1{s}_{2}c\right)$

Table 1.

SCDM of qubit-qubit system for the case $\left|S_f-S_0\right|$ = 0.

$${\sigma }_{{{}_{e_1}}_{e_2}}=$$

Figure 2.

Variation of concurrence for the case $\left|S_f-S_0\right|$ = 0 with respect to spin quantization directions α₁ and α₂.

According to Table 2, we thus obtain

$\mu {\text{'}}_1\mu {\text{'}}_2$; ${\mu }_1{\mu }_2$	$\frac{1}{2}\frac{1}{2}$	$\frac{1}{2}-\frac{1}{2}$	$-\frac{1}{2}\frac{1}{2}$	$-\frac{1}{2}-\frac{1}{2}$
$\frac{1}{2}\frac{1}{2}$	$0.25+\frac{1}{12}\left(c_1{c}_2+s_1{s}_{2}c\right)$	$-\frac{1}{12}\left(c_1{s}_2-s_1{c}_{2}c\right)$	$-\frac{1}{12}\left(s_1{c}_2-c_1{s}_{2}c\right)$	$-\frac{1}{12}\left(c-s_1{s}_2-c_1{c}_{2}c\right)$
$\frac{1}{2}-\frac{1}{2}$	$-\frac{1}{12}\left(c_1{s}_2-s_1{c}_{2}c\right)$	$0.25-\frac{1}{12}\left(c_1{c}_2+s_1{s}_{2}c\right)$	$\frac{1}{12}\left(c+s_1{s}_2+c_1{c}_{2}c\right)$	$\frac{1}{12}\left(s_1{c}_2-c_1{s}_{2}c\right)$
$-\frac{1}{2}\frac{1}{2}$	$-\frac{1}{12}\left(s_1{c}_2-c_1{s}_{2}c\right)$	$\frac{1}{12}\left(c+s_1{s}_2+c_1{c}_{2}c\right)$	$0.25-\frac{1}{4}\left(c_1{c}_2+s_1{s}_{2}c\right)$	$\frac{1}{12}\left(c_1{s}_2-s_1{c}_{2}c\right)$
$-\frac{1}{2}-\frac{1}{2}$	$-\frac{1}{12}\left(c-s_1{s}_2-c_1{c}_{2}c\right)$	$\frac{1}{12}\left(s_1{c}_2-c_1{s}_{2}c\right)$	$\frac{1}{12}\left(c_1{s}_2-s_1{c}_{2}c\right)$	$0.25+\frac{1}{4}\left(c_1{c}_2+s_1{s}_{2}c\right)$

Table 2.

SCDM of qubit-qubit system for the case $\left|S_f-S_0\right|$ = 1.

$$\begin{array}{l}\mathit{det}\left({\sigma }_{{{}_{e_1}}_{e_2}}\right)=\frac{1}{576}(140+8cos2{\alpha }_1-4cos4{\alpha }_1+cos\left({\alpha }_1-5{\alpha }_2\right)-8cos2\left({\alpha }_1-2{\alpha }_2\right)\\ +8cos\left(4{\alpha }_1-2{\alpha }_2\right)+cos\left(5{\alpha }_1-{\alpha }_2\right)-8cos\alpha -120cos2\alpha -7cos3\alpha -4cos4\alpha -cos5\alpha \\ -8cos2{\alpha }_2-4cos4{\alpha }_2-8cos2\left({\alpha }_1+{\alpha }_2\right)+7cos\left(3{\alpha }_1+{\alpha }_2\right)+7cos\left({\alpha }_1+3{\alpha }_2\right)\end{array}$$

In order to quantifying entanglement, we obtain the concurrence, using SCDM of Table 2. According to the definition given in Eq. (6), which yields

$$C=\sqrt{1-\frac{9+{\left(cos{\alpha }_{1}cos{\alpha }_2+cos\alpha sin{\alpha }_{1}sin{\alpha }_2\right)}^2}{5184}}.$$E19

Eq. (19) indicates [4, 17, 22] that the values of C are positive and so the spin state $\left|S_f-S_0\right|$ = 1 is entangled, depending on ${\alpha }_1$and ${\alpha }_2.$ This variation of concurrence is given in Figure 3.

Figure 3.

Variation of concurrence for the case $\left|S_f-S_0\right|$ = 1 with respect to spin quantization directions α₁ and α₂.

We find from Figure 3 that the states for $\left|S_f-S_0\right|$ = 1 are entangled (the values of concurrence are positive) and separable at valley points (for the zero value of concurrence) depending on spin quantization directions.

4. Example for Entanglement in DPI for He

As an application qubit-qubit entanglement, we consider helium atom where DPI in its ground electronic state can be represented by

Figure 4.

Variation of concurrence for the case $\left|S_f-S_0\right|$ = 0 with respect to directions of ejection of photoelectrons.

Figure 5.

Variation of concurrences for the case $\left|S_f-S_0\right|$ = 0 with respect to spin quantization directions and directions of ejection of photoelectron.

$$h{\nu }_r+\mathit{He}\left(1s^2{{}^{1}S}_0\right)\to {\mathit{He}}^{2+}\left(1s^0{{}^{1}S}_0\right)+e_1\left({\overrightarrow{k}}_1;{\mu }_1{\widehat{u}}_1\right)+e_2\left({\overrightarrow{k}}_2;{\mu }_2{\widehat{u}}_2\right).$$E20

We use the values of TDCS of helium given in Ref. [29] for photon energy 99 eV along with linear polarization of photon and for equal energy sharing between the two electrons. The TDCS is

$${\left.\frac{d^3\sigma }{d{\epsilon }_{1}d{\widehat{k}}_{1}d{\widehat{k}}_2}\right|}_{{\epsilon }_1={\epsilon }_2}=a{\left(cos{\theta }_1+cos{\theta }_2\right)}^{2}exp\left\{-\frac{1}{2}{\left[\left({\theta }_{12}-{180}^0\right)/\gamma \right]}^2\right\},$$E21

where ${\theta }_{12}$ is the angle between two electrons, $\gamma$ named Gaussian half-width of value $90.2\pm 2^0$. The value of normalization factor $a=107\pm 16b{\mathit{eV}}^{-1}{\mathit{sr}}^{-2}$.

1. In case of $\left|S_f-S_0\right|$ = 0, the variations of concurrence with respect to the direction of ejection and spin polarization of the photoelectrons are shown in Figures 4 and 5.

   From Figures 4 and 5, we see that the values of concurrence are either zero or positive depending on the directions of ejection and spin quantization of e₁ and e₂. So we can conclude that depending on the values of ${\theta }_1,{\theta }_2,{\alpha }_1$, and ${\alpha }_2,$most of the states for $\left|S_f-S_0\right|$ = 0 of helium atom are entangled.

2. In case of $\left|S_f-S_0\right|$ = 1, the variations of concurrence with respect to the direction of ejection and spin polarization of the photoelectrons are shown in Figures 6 and 7.

Figure 6.

Variation of concurrence for the case $\left|S_f-S_0\right|$ = 1 with respect to directions of ejection of photoelectrons.

Figure 7.

Variation of concurrences for the case $\left|S_f-S_0\right|$ = 1 with respect to spin quantization directions and directions of ejection of photoelectron.

From Figures 6 and 7, we see that for the states for $\left|S_f-S_0\right|$ = 1 of a helium atom, the nature of variations as well as magnitudes of concurrence (i.e., entanglement) depend on the directions of ejection of photoelectrons along with their spin polarization.

5. Conclusion

For concurrently creating two electrons in continuum in a single step, DPI is the most natural approach. It is the most obvious example of electron-electron correlation in an atom because if the independent particle model were true, only one photon would have been absorbed before two electrons would have emerged at the same time. Thus, the presence of correlation effects between them could lead to the simultaneous ejection of two electrons from an atom following the absorption of a single photon. In this article, we have tried to demonstrate how effective the DPI method is for creating different types of entanglement between two qubits. A quantitative application for this case is studied for DPI in helium atom. For helium atom, we have studied the states for $\left|S_f-S_0\right|$ = 0 and $\left|S_f-S_0\right|$ = 1, we have shown that depending on the direction of ejection, as well as spin polarization of the ejected photoelectrons, the states are totally entangled, partially entangled, and separate.