Identifying quantum correlations using explicit SO(3) to SU(2) maps

Abstract

Quantum state manipulation of two-qubits on the local systems by special unitaries induces special orthogonal rotations on the Bloch spheres. An exact formula is given for determining the local unitaries for some given rotation on the Bloch sphere. The solution allows for easy manipulation of two-qubit quantum states with a single definition that is programmable. With this explicit formula, modifications to the correlation matrix are made simple. Using our solution, it is possible to diagonalize the correlation matrix without solving for the parameters in SU(2) that define the local unitary that induces the special orthogonal rotation in SO(3). Since diagonalization of the correlation matrix is equivalent to diagonalization of the interaction Hamiltonian, manipulating the correlation matrix is important in time-optimal control of a two-qubit state. The relationship between orthogonality conditions on SU(2) and SO(3) is given and manipulating the correlation matrix when only one qubit can be accessed is discussed.

Appendix

1.1 A.1 Correlation matrices for the Bell states.

To determine the correlation matrix of any two-qubit density operator, simply perform the calculations $\text {tr}({\hat{\sigma }}_i \otimes \hat{\sigma _j} \cdot \rho ) = \{{\mathcal {T}}_{\rho }\}_{ij}$. Using this formula, we can directly determine the correlation matrices for the maximally entangled Bell operators:

$$\begin{aligned} {\mathcal {T}}_{\Phi ^+}&=\ \left( \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} -1 &{} 0 \\ 0 &{} 0 &{} 1 \end{array} \right) \quad {\mathcal {T}}_{\Phi ^-} =\ \left( \begin{array}{ccc} -1 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 \end{array} \right) \end{aligned}$$

(44)

$$\begin{aligned} {\mathcal {T}}_{\Psi ^+}&=\ \left( \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} -1 \end{array} \right) \quad {\mathcal {T}}_{\Psi ^-} =\ \left( \begin{array}{ccc} -1 &{} 0 &{} 0 \\ 0 &{} -1 &{} 0 \\ 0 &{} 0 &{} -1 \end{array} \right) . \end{aligned}$$

(45)

1.2 A.2 Proof of equation (30)

Let us now calculate each part of Eq. (30) case by case. Note that for all these cases $\alpha _1=0$ as seen from Eq. (18).

Case 1: $\text {sgn}(\alpha _2)\text {sgn}(\beta _1)\text {sgn}(\beta _2) \ne 0$

$$\begin{aligned} W({\mathcal {O}},|L_1|,|L_2|,|L_3|) = \eta _1\left( \begin{array}{cc} i \alpha _2 &{} \beta _1+i \beta _2 \\ -\beta _1 + i \beta _2 &{} -i \alpha _2 \end{array} \right) \end{aligned}$$

(46)

where $\eta _1 = \text {sgn}(\alpha _2)\text {sgn}(\beta _1)\text {sgn}(\beta _2)$.

Case 2: $\text {sgn}(\alpha _2) = 0 \; \text {and} \; \text {sgn}(\beta _1)\text {sgn}(\beta _2) \ne 0$

$$\begin{aligned} W({\mathcal {O}},{\mathbb {I}},{\mathbb {I}},-|L_1|) = \eta _2\left( \begin{array}{cc} -i|\alpha _2|\eta _2 &{} \beta _1 + i \beta _2 \\ -\beta _1 + i \beta _2 &{} i|\alpha _2|\eta _2 \end{array} \right) \end{aligned}$$

(47)

where $\eta _2 = -\text {sgn}(\beta _1)$. Since $\text {sgn}(\alpha _2) = 0$ implies that $\alpha _2 = 0$, this form is correct. The $\eta _2$ only adds a ± global phase. We also see that

$$\begin{aligned} (1-\gamma _1)\gamma _2\gamma _3 = (1-\text {sgn}(\alpha _2 \beta _1)^2)(1-\text {sgn}(\alpha _2 \beta _2)^2)\text {sgn}(\beta _1 \beta _2)^2 \end{aligned}$$

(48)

which can be expressed in cases as

$$\begin{aligned} (1-\gamma _1)\gamma _2\gamma _3 = {\left\{ \begin{array}{ll} 1 &{}\text { if } \text {sgn}(\alpha _2) = 0 \; \text {and} \; \text {sgn}(\beta _1)\text {sgn}(\beta _2) \ne 0 \\ 0 &{}\text { otherwise} \end{array}\right. }. \end{aligned}$$

(49)

Case 3: $\text {sgn}(\beta _1) = 0 \; \& \; \text {sgn}(\alpha _2)\text {sgn}(\beta _2) \ne 0$

$$\begin{aligned} W({\mathcal {O}},-|L_2|,{\mathbb {I}},{\mathbb {I}}) = \eta _3\left( \begin{array}{cc} i\alpha _2 &{} -|\beta _1| \eta _3 + i \beta _2 \\ |\beta _1| \eta _3 + i \beta _2 &{} -i\alpha _2 \end{array} \right) \end{aligned}$$

(50)

where $\eta _3 = -\text {sgn}(\beta _2)$. Since $\text {sgn}(\beta _1) = 0$ implies that $\beta _1 = 0$, this form is correct. The $\eta _3$ only adds a ± global phase. We also see that

$$\begin{aligned} \gamma _1(1-\gamma _2)\gamma _3 = (1-\text {sgn}(\alpha _2 \beta _1)^2)(1-\text {sgn}(\beta _1 \beta _2)^2)\text {sgn}(\alpha _2 \beta _2)^2 \end{aligned}$$

(51)

which can be expressed in cases as

$$\begin{aligned} \gamma _1(1-\gamma _2)\gamma _3 = {\left\{ \begin{array}{ll} 1 &{}\text { if } \text {sgn}(\beta _1) = 0 \; \text {and} \; \text {sgn}(\alpha _2)\text {sgn}(\beta _2) \ne 0 \\ 0 &{}\text { otherwise} \end{array}\right. }. \end{aligned}$$

(52)

Case 4: $\text {sgn}(\beta _2) = 0 \; \& \; \text {sgn}(\alpha _2)\text {sgn}(\beta _1) \ne 0$

$$\begin{aligned} W({\mathcal {O}},{\mathbb {I}},-|L_3|,{\mathbb {I}}) = \eta _4\left( \begin{array}{cc} i\alpha _2 &{} \beta _1 - i |\beta _2| \eta _4 \\ -\beta _1 - i |\beta _2| \eta _4 &{} -i\alpha _2 \end{array} \right) \end{aligned}$$

(53)

where $\eta _4 = -\text {sgn}(\alpha _2)$. Since $\text {sgn}(\beta _2) = 0$ implies that $\beta _2 = 0$, this form is correct. The $\eta _4$ only adds a ± global phase. We also see that

$$\begin{aligned} \gamma _1 \gamma _2 (1-\gamma _3) = (1-\text {sgn}(\alpha _2 \beta _2)^2)(1-\text {sgn}(\beta _1 \beta _2)^2)\text {sgn}(\alpha _2 \beta _1)^2 \end{aligned}$$

(54)

which can be expressed in cases as

$$ \begin{aligned} \gamma _1 \gamma _2 (1-\gamma _3) = {\left\{ \begin{array}{ll} 1 &{}\text { if } \text {sgn}(\beta _2) = 0 \; \& \; \text {sgn}(\alpha _2)\text {sgn}(\beta _1) \ne 0 \\ 0 &{}\text { otherwise} \end{array}\right. }. \end{aligned}$$

(55)

$$ \begin{aligned} {\textbf {Case 5: }} \text {sgn}(\alpha _2) \ne 0 \; \& \; \text {sgn}(\beta _1),\text {sgn}(\beta _2) = 0 \nonumber \\ {\textbf {Case 6: }} \text {sgn}(\beta _1) \ne 0 \; \& \; \text {sgn}(\alpha _2),\text {sgn}(\beta _2) = 0 \nonumber \\ {\textbf {Case 7: }} \text {sgn}(\beta _2) \ne 0 \; \& \; \text {sgn}(\alpha _2),\text {sgn}(\beta _1) = 0 \end{aligned}$$

(56)

$$\begin{aligned} W({\mathcal {O}},-{\mathbb {I}},-{\mathbb {I}},-{\mathbb {I}}) = \left( \begin{array}{cc} i|\alpha _2| &{} |\beta _1| + i |\beta _2| \\ -|\beta _1| + i |\beta _2| &{} -i|\alpha _2| \end{array} \right) . \end{aligned}$$

(57)

Since only one of the elements of $\{\alpha _2, \beta _1, \beta _2\}$ are nonzero, this form is correct. The missing sign is only a ± global phase. We also see that

$$\begin{aligned} \gamma _1\gamma _2\gamma _3 = (1-\text {sgn}(\alpha _2 \beta _1)^2)(1-\text {sgn}(\alpha _2 \beta _2)^2)(1-\text {sgn}(\beta _1 \beta _2)^2) \end{aligned}$$

(58)

which can be expressed in cases as

$$ \begin{aligned} \gamma _1\gamma _2\gamma _3 = {\left\{ \begin{array}{ll} 1 &{}\text { if } \text {sgn}(\alpha _2),\text {sgn}(\beta _1),\text {sgn}(\beta _2) = 0 \\ &{}\text { if } \text {sgn}(\alpha _2),\text {sgn}(\beta _1) = 0 \; \& \; \text {sgn}(\beta _2) \ne 0 \\ &{}\text { if } \text {sgn}(\alpha _2),\text {sgn}(\beta _2) = 0 \; \& \; \text {sgn}(\beta _1) \ne 0 \\ &{}\text { if } \text {sgn}(\beta _1),\text {sgn}(\beta _2) = 0 \; \& \; \text {sgn}(\alpha _2) \ne 0 \\ 0 &{}\text { otherwise} \end{array}\right. }, \end{aligned}$$

(59)

which completes the rest of the cases involved when $\text {tr}({\mathcal {O}}) = -1$. The $\gamma$ functions ensure that there are no repeats of any solutions in Eq. (30). Now we can safely say that all of the 8 possible cases described by Eq. (32) have been proven. Case 8 is when $\text {tr}({\mathcal {O}}) \ne -1$ and it has been proven in Eq. (20).