Fidelity between one bipartite quantum state and another undergoing local unitary dynamics

Abstract

The fidelity and local unitary transformation are two widely useful notions in quantum physics. We study two constrained optimization problems in terms of the maximal and minimal fidelity between two bipartite quantum states undergoing local unitary dynamics. The problems are related to the geometric measure of entanglement and the distillability problem. We show that the problems can be reduced to semidefinite programming optimization problems. We give closed-form formulae of the fidelity when the two states are pure states, or a pure product state and the Werner state. We explain from the point of view of local unitary actions that why the entanglement in Werner states is hard to accessible. For general mixed states, we give upper and lower bounds of the fidelity using tools such as affine fidelity, channels, and relative entropy from information theory. We also investigate the power of local unitaries and the equivalence of the two optimization problems.

Appendix

1.1 Isotropic state

The isotropic state is the convex mixture of a maximally entangled state and the maximally mixed state:

where $\lambda \in [0,1]$ and $|\varPsi ^+\rangle =\frac{1}{\sqrt{d}}\sum ^d_{j=1}|jj\rangle$. By Theorem 2.1, we have

$$\begin{aligned} \max _{|u\rangle ,|v\rangle } \left|\langle uv | \varPsi ^+\rangle \right|^2 = \frac{1}{d}~\text {and}~\min _{|u\rangle ,|v\rangle } \left|\langle uv | \Psi ^+\rangle \right|^2 = 0. \end{aligned}$$

Thus,

$$\begin{aligned} \left\langle uv \left| \rho _{\text {iso}}(\lambda ) \right| uv \right\rangle = \frac{1-\lambda }{d^2-1} + \frac{d^2\lambda -1}{d^2-1}\left|\langle uv | \varPsi ^+\rangle \right|^2. \end{aligned}$$

(5.1)

To further characterize the maximum and minimum of this function, we discuss two subcases.

• (1) If $\frac{1}{d^2}\leqslant \lambda \leqslant 1$, then $\max _{|u\rangle ,|v\rangle }\left\langle uv \left| \rho _{\text {iso}}(\lambda ) \right| uv \right\rangle = \frac{d\lambda +1}{d(d+1)}$ and $\min _{|u\rangle ,|v\rangle }\left\langle uv \left| \rho _{\text {iso}}(\lambda ) \right| uv \right\rangle = \frac{1-\lambda }{d^2-1}$;

• (2) If $0\leqslant \lambda <\frac{1}{d^2}$, then $\min _{|u\rangle ,|v\rangle }\left\langle uv \left| \rho _{\text {iso}}(\lambda ) \right| uv \right\rangle = \frac{d\lambda +1}{d(d+1)}$ and $\max _{|u\rangle ,|v\rangle }\left\langle uv \left| \rho _{\text {iso}}(\lambda ) \right| uv \right\rangle = \frac{1-\lambda }{d^2-1}$. In summary, we have

  $$\begin{aligned} \mathrm {G}_{\max }(\rho _{\text {iso}}(\lambda ),| uv\rangle \langle uv |) = \max \left( \sqrt{\frac{d\lambda +1}{d(d+1)}},\sqrt{\frac{1-\lambda }{d^2-1}}\right) , \end{aligned}$$

  (5.2)

  and

  $$\begin{aligned} \mathrm {G}_{\min }(\rho _{\text {iso}}(\lambda ),| uv\rangle \langle uv |) = \min \left( \sqrt{\frac{d\lambda +1}{d(d+1)}},\sqrt{\frac{1-\lambda }{d^2-1}}\right) . \end{aligned}$$

  (5.3)

1.2 Proof of Proposition 3.1

For any semidefinite positive matrix X, the following inequality is easily derived via the spectral decomposition of X:

$$\begin{aligned} \sqrt{{\text {rank}}(X) \cdot {{\mathrm{Tr}}}_{}\left( X\right) } \geqslant {{\mathrm{Tr}}}_{}\left( \sqrt{X}\right) \geqslant \sqrt{{{\mathrm{Tr}}}_{}\left( X\right) }, \end{aligned}$$

(5.4)

where the first equality holds if and only if X has identical nonzero eigenvalues, and the second equality holds if and only if X has rank one. Let $X=A^{1/2}BA^{1/2}$ for any two semidefinite positive matrices A, B. Then, (5.4) implies

$$\begin{aligned} \sqrt{{\text {rank}}(A^{1/2}B^{1/2}) \cdot {{\mathrm{Tr}}}_{}(AB)}\geqslant \mathrm {F}(A,B)\geqslant \sqrt{{{\mathrm{Tr}}}_{}(AB)}, \end{aligned}$$

(5.5)

where the first equality holds if and only if $A^{1/2}BA^{1/2}$ has identical nonzero eigenvalues, and the second equality holds if and only if $A^{1/2}BA^{1/2}$ has rank one. To prove the first inequality in (3.1), let $W_1\otimes W_2 = \arg \mathrm {G}_{\max }(\rho ,\sigma )$. We have

$$\begin{aligned}&\mathrm {G}_{\max }(\rho ,\sigma )^2 + (d_1d_2-1)\mathrm {G}_{\min }(\rho ,\sigma ')^2 \nonumber \\&\quad \leqslant \mathrm {F}(\rho ,(W_1\otimes W_2)\sigma (W_1\otimes W_2)^\dagger )^2\nonumber \\&\qquad +\, (d_1d_2-1) \mathrm {F}(\rho ,(W_1\otimes W_2)\sigma '(W_1\otimes W_2)^\dagger )^2 \nonumber \\&\quad \leqslant {\text {rank}}(\rho ^{1/2} (W_1\otimes W_2)\sigma ^{1/2}(W_1\otimes W_2)^\dagger )\nonumber \\&\qquad \times {{\mathrm{Tr}}}_{}\left( \rho (W_1\otimes W_2)\sigma (W_1\otimes W_2)^\dagger \right) \nonumber \\&\qquad + {\text {rank}}(\rho ^{1/2}(W_1\otimes W_2)(\sigma ')^{1/2}(W_1\otimes W_2)^\dagger )(d_1d_2-1)\nonumber \\&\qquad \times {{\mathrm{Tr}}}_{}\left( \rho (W_1\otimes W_2)\sigma '(W_1\otimes W_2)^\dagger \right) \nonumber \\&\quad \leqslant {\text {rank}}(\rho ) \left[ {{\mathrm{Tr}}}_{}\left( \rho (W_1\otimes W_2)\sigma (W_1\otimes W_2)^\dagger \right) \right. \nonumber \\&\left. \qquad +\, (d_1d_2-1) {{\mathrm{Tr}}}_{}(\rho (W_1\otimes W_2)\sigma '(W_1\otimes W_2)^\dagger )\right] \nonumber \\&\quad ={\text {rank}}(\rho ), \end{aligned}$$

(5.6)

where the first inequality follows from the definition of $\mathrm {G}_{\max }$ and $\mathrm {G}_{\min }$, and its equality is equivalent to condition (i). The second inequality in (5.6) follows from the first inequality in (5.5) by assuming $A=\rho$, $B=(W_1\otimes W_2)\sigma (W_1\otimes W_2)^\dagger$ and $(W_1\otimes W_2)\sigma '(W_1\otimes W_2)^\dagger$, respectively. Its equality is equivalent to condition (i) by the first inequality in (5.5). The third inequality in (5.6) follows from the fact ${\text {rank}}(A) \geqslant {\text {rank}}(A^{1/2} B^{1/2})$. Its equality holds if condition (iii) holds. So we have proved the first inequality and the three conditions by which the equality holds in (3.1).

To prove the second inequality in (3.1), let $V_1\otimes V_2 = \arg \mathrm {G}_{\min }(\rho ,\sigma ')$. We have

$$\begin{aligned}&\mathrm {G}_{\max }(\rho ,\sigma )^2 + (d_1d_2-1)\mathrm {G}_{\min }(\rho ,\sigma ')^2 \nonumber \\&\quad \geqslant \mathrm {F}(\rho ,(V_1\otimes V_2)\sigma (V_1\otimes V_2)^\dagger )^2 + (d_1d_2-1)\mathrm {F}(\rho ,(V_1\otimes V_2)\sigma '(V_1\otimes V_2)^\dagger )^2 \nonumber \\&\quad \geqslant {{\mathrm{Tr}}}_{}(\rho (V_1\otimes V_2)\sigma (V_1\otimes V_2)^\dagger ) + (d_1d_2-1) {{\mathrm{Tr}}}_{}(\rho (V_1\otimes V_2)\sigma '(V_1\otimes V_2)^\dagger ) \nonumber \\&\quad =1, \end{aligned}$$

(5.7)

where the second inequality follows from (5.5) by assuming $A=\rho$, $B=(V_1\otimes V_2)\sigma (V_1\otimes V_2)^\dagger$ and $(V_1\otimes V_2)\sigma '(V_1\otimes V_2)^\dagger$, respectively. So we have proved the second inequality in (3.1). The equality in (3.1) holds if and only if the first two equalities in (5.7) hold. The first equality is equivalent to condition (iv) by the definition of $\mathrm {G}_{\max }$ and $\mathrm {G}_{\min }$, and the second equality is equivalent to conditions (v) and (vi) by (5.5). This completes the proof.