Quantumness-generating capability of quantum dynamics

Abstract

We study quantumness-generating capability of quantum dynamics, where quantumness refers to the noncommutativity between the initial state and the evolving state. In terms of the commutator of the square roots of the initial state and the evolving state, we define a measure to quantify the quantumness-generating capability of quantum dynamics with respect to initial states. Quantumness-generating capability is absent in classical dynamics and hence is a fundamental characteristic of quantum dynamics. For qubit systems, we present an analytical form for this measure, by virtue of which we analyze several prototypical dynamics such as unitary dynamics, phase damping dynamics, amplitude damping dynamics, and random unitary dynamics (Pauli channels). Necessary and sufficient conditions for the monotonicity of quantumness-generating capability are also identified. Finally, we compare these conditions for the monotonicity of quantumness-generating capability with those for various Markovianities and illustrate that quantumness-generating capability and quantum Markovianity are closely related, although they capture different aspects of quantum dynamics.

Appendix

1.1 A. Proof of Eq. (3)

Consider the quantum ensemble $\mathcal {E}=\{(p_1,\rho _1),(p_2,\rho _2)\}$ consisting of two-qubit states $\rho _i=\frac{1}{2}(\mathbf{1}+\mathbf{r}_i\cdot {\varvec{\sigma }})$ with fixed a prior probabilities $p_i,~i=1,2$, where $\mathbf{1}$ is the identity operator, $\mathbf{r}_i$ the Bloch vector of the state $\rho _i$. By the definition of the quantumness of ensembles Eq. (2), we get

$$\begin{aligned} Q(\mathcal {E})=-2\sqrt{p_1p_2} \mathrm{tr}\left[ \sqrt{\rho _1},\sqrt{\rho _2}\right] ^2=4\sqrt{p_1p_2}\left( \mathrm{tr}\rho _1\rho _2-\mathrm{tr}(\sqrt{\rho _1}\sqrt{\rho _2})^2\right) . \end{aligned}$$

By direct calculations, we obtain that $\mathrm{tr}\rho _1\rho _2=\frac{1}{2}(1+\mathbf{r}_1\cdot \mathbf{r}_2),$ and

$$\begin{aligned} \mathrm{tr}\left( \sqrt{\rho _1}\sqrt{\rho _2}\right) ^2=\frac{1}{2}(1+\mathbf{r}_1\cdot \mathbf{r}_2)-\left( 1-\sqrt{1-r_1^2}\right) \left( 1-\sqrt{1-r_2^2}\right) \frac{|\mathbf{r}_1\times \mathbf{r}_2|^2}{4 r_1^2 r_2^2}. \end{aligned}$$

Here ${r}_i=|\mathbf{r}_i|$. Combining these two expressions, we have

$$\begin{aligned} Q(\mathcal {E})= & {} \frac{\sqrt{p_1p_2}|\mathbf{r}_1 \times \mathbf{r}_2|^2}{\left( 1+\sqrt{1-r_1^2}\right) \left( 1+\sqrt{1-r_2^2}\right) }. \end{aligned}$$

Then Eq. (3) is derived from the above equation directly.

1.2 B. Proof of $\mathrm{d}Q(\varLambda _t|\rho )/\mathrm{d}|G_t|\le 0$ for amplitude damping dynamics

Let $\lambda _t=1-|G_t|^2$, then the statement of $\mathrm{d}Q(\varLambda _t|\rho )/\mathrm{d}|G_t|\le 0$ is equivalent to $\mathrm{d}Q(\varLambda _t|\rho )/\mathrm{d}\lambda _t\ge 0$. By numerical analysis, we know that $\mathrm{d}Q(\varLambda _t|\rho )/\mathrm{d}\lambda _t\ge 0$. Analytically, we will prove that

• (i) $\mathrm{d}Q(\varLambda _t|\rho )/\mathrm{d}\lambda _t\ge 0$ for any initial state when $1/2<\lambda _t<15/16$.

• (ii) $\mathrm{d}Q(\varLambda _t|\rho )/\mathrm{d}\lambda _t\ge 0$ holds for any pure initial state.

Let us begin with rewriting $Q(\varLambda _t|{\rho })$ as

$$\begin{aligned} Q(\varLambda _t|{\rho })=\frac{ r^2 \sin ^2 \theta \left( \lambda _t+(1-\lambda _t)r\cos \theta -\sqrt{1-\lambda _t}r\cos \theta \right) ^2}{2\left( 1+\sqrt{1-r^2}\right) \left( 1+\sqrt{1-r_t^2}\right) }, \end{aligned}$$

where $1-r_t^2=(1-\lambda _t)(1-r^2)+\lambda _t(1-\lambda _t)(1-\cos \theta )^2$. Its derivative with respect to $\lambda _t$ is

$$\begin{aligned} \frac{\mathrm{d}Q(\varLambda _t|{\rho })}{\mathrm{d}\lambda _t} =\frac{r^2\sin ^2\theta }{2(1+\sqrt{1-r^2})}\frac{\mathrm{d}f(\lambda _t)}{\mathrm{d} \lambda _t} \end{aligned}$$

where

$$\begin{aligned} f(\lambda _t)=\frac{\left( \lambda _t+(1-\lambda _t)r\cos \theta -\sqrt{1-\lambda _t}r\cos \theta \right) ^2}{1+\sqrt{1-r_t^2}}, \end{aligned}$$

from which we have

$$\begin{aligned} \frac{d f(\lambda _t)}{d \lambda _t}=\frac{ \lambda _t+(1-\lambda _t)r\cos \theta -\sqrt{1-\lambda _t} r\cos \theta }{\left( 1+\sqrt{1-r_t^2}\right) ^2}g(\lambda _t). \end{aligned}$$

Here

$$\begin{aligned} g(\lambda _t)= & {} 2\left( 1+\left( \frac{1}{2\sqrt{1-\lambda _t}}-1\right) r\cos \theta \right) \left( \sqrt{1-r_t^2}+1\right) \\&+\,\frac{(1-r^2)-(1-2\lambda _t)(1-\cos \theta )^2}{2\sqrt{1-r_t^2}}\left( \lambda _t+(1-\lambda _t)r\cos \theta -r\cos \theta \sqrt{1-\lambda _t}\right) . \end{aligned}$$

From the above equation, it is easy to see that when

$$\begin{aligned} -1<\frac{1}{2\sqrt{1-\lambda _t}}-1<1 \text { and } 1-2\lambda _t<0, \end{aligned}$$

which is equivalent to $1/2<\lambda _t<15/16$, $\mathrm{d} Q(\varLambda _t|\rho )/\mathrm{d}\lambda _t \ge 0$. This completes the proof of (i).

Next, if the initial state is pure, let $x=1-\cos \theta$, then $x \in [0,2]$ and

$$\begin{aligned} Q(\varLambda _t|{\rho })=\frac{ x(2-x)\left( 1-\sqrt{1-\lambda _t}+\left( \sqrt{1-\lambda _t}-1+\lambda _t\right) x\right) ^2}{2\left( 1+\sqrt{\lambda _t(1-\lambda _t)}x\right) } \end{aligned}$$

The derivative can be evaluated as

$$\begin{aligned} \frac{\mathrm{d}Q(\varLambda _t|{\rho })}{\mathrm{d}\lambda _t}= & {} \frac{(2-x)\left( 1-\sqrt{1-\lambda _t}+(\sqrt{1-\lambda _t}-1+\lambda _t)x\right) }{4\sqrt{\lambda _t}(1-\lambda _t)\left( 1+\sqrt{\lambda _t(1-\lambda _t)}x\right) ^2} h(x) \end{aligned}$$

where $h(x)=a x^3 + b x^2 + c x,$ with

$$\begin{aligned} a&=(1-\lambda _t)\left( \sqrt{1-\lambda _t}+2\lambda _t\sqrt{1-\lambda _t}-1\right) ,\\ b&=\sqrt{1-\lambda _t}\left( \sqrt{1-\lambda _t}-1+2\lambda _t-2\sqrt{\lambda _t}+4\sqrt{\lambda _t(1-\lambda _t)}\right) ,\\ c&=2\sqrt{\lambda _t(1-\lambda _t)}. \end{aligned}$$

After tedious calculations, we have $h(x)\ge h(0)=0$ when $x \in [0,2]$, which leads directly to $\mathrm{d} Q(\varLambda _t|{\rho })/\mathrm{d}\lambda _t\ge 0$.

1.3 C. Proof of $\mathrm{d}Q(\varLambda _t|\rho )/\mathrm{d}p_{0t}\le 0$ for Case 2 of random unitary dynamics

Let $\lambda _t=1-p_{0t}$, then the quantumness-generating capability $Q(\varLambda _t|\rho )$ can be rewritten as

$$\begin{aligned} Q(\varLambda _t|\rho )=\frac{2\left( 3\alpha -1\right) ^2\left( r_x^2+r_y^2\right) r_z^2\lambda _t^2}{\left( 1+\sqrt{1-r^2}\right) \left( 1+\sqrt{1-r_t^2}\right) }, \end{aligned}$$

in which $r_t^2=(1-2(1-\alpha )\lambda _t)^2(r_x^2+r_y^2)+(1-4\alpha \lambda _t)^2 r_z^2$. The derivative of $Q(\varLambda _t|\rho )$ with respect to $\lambda _t$ is

$$\begin{aligned} \frac{\mathrm{d}Q(\varLambda _t|{\rho })}{\mathrm{d}\lambda _t}=\frac{4\lambda _t (3\alpha -1)^2(r_x^2+r_y^2)r_z^2}{\left( 1+\sqrt{1-r^2}\right) \left( 1+\sqrt{1-r_t^2}\right) ^2\sqrt{1-r_t^2}}f(\lambda _t), \end{aligned}$$

where

$$\begin{aligned} f(\lambda _t)= & {} 1-r_t^2+\sqrt{1-r_t^2}-\lambda _t\left( 1-\alpha \right) \left( 1-2(1-\alpha )\lambda _t\right) \left( r_x^2+r_y^2\right) \\&-2\alpha \lambda _t(1-4\alpha \lambda _t)r_z^2\\\ge & {} 2-2r^2\left( 1-4\alpha \lambda _t\right) \left( 1-3\alpha \lambda _t\right) +\lambda _t \left( r_x^2+r_y^2\right) \left( 1-3\alpha \right) \left( 7-6 \lambda _t(\alpha +1)\right) . \end{aligned}$$

When $\alpha \in [0,1/2]$, we have $f(\lambda _t)\ge 0$, which is equivalent to $\mathrm{d}Q(\varLambda _t|{\rho })/ \mathrm{d}\lambda _t>0$, and $\mathrm{d}Q(\varLambda _t|\rho )/\mathrm{d}p_{0t}\le 0$ follows directly by the chain rule.

1.4 D. Proof of $\mathrm{d}Q(\varLambda _t|\rho )/\mathrm{d}t\ge 0$ for Case 3 of random unitary dynamics

For the initial state $\rho$ with Bloch vector $\mathbf{r}=(r_x,r_y,r_z)$ and the evolving state $\rho _t$ with the Bloch vector $\mathbf{r}_t=(r_x,r_y,e^{-4t}r_z)$, we know

$$\begin{aligned} Q(\varLambda _t|\rho )= & {} \frac{ \left( r_x^2+r_y^2\right) r_z^2\left( 1-e^{-4t}\right) ^2}{2\left( 1+\sqrt{1-r^2}\right) \left( 1+\sqrt{1-r_t^2}\right) }. \end{aligned}$$

Now we prove that

$$\begin{aligned} \frac{\mathrm{d}Q(\varLambda _t|{\rho })}{\mathrm{d}t}\ge 0 \ \mathrm{for\ any}\ t\ge 0. \end{aligned}$$

Let $f_t=(1-e^{-4t})^2/(1+\sqrt{1-r_t^2})$, then

$$\begin{aligned} \frac{\mathrm{d}Q(\varLambda _t|{\rho })}{\mathrm{d}t}=\frac{(r_x^2+r_y^2)r_z^2}{2(1+\sqrt{1-r^2})}\frac{\mathrm{d}f_t}{\mathrm{d}t} \end{aligned}$$

where

$$\begin{aligned} \frac{\mathrm{d}f_t}{\mathrm{d}t}= & {} \frac{4(1-e^{-4t})}{\sqrt{1-r_t^2} \left( 1+\sqrt{1-r_t^2}\right) ^2}\left( 2e^{-4t}\left( 1-r_t^2+\sqrt{1-r_t^2}\right) -e^{-8t}(1-e^{-4t})r_z^2\right) \\\ge & {} \frac{4(1-e^{-4t})}{\sqrt{1-r_t^2}\left( 1+\sqrt{1-r_t^2}\right) ^2}\left( 4e^{-4t}(1-r_t^2)-e^{-8t}(1-e^{-4t})r_z^2\right) \\= & {} \frac{4(1-e^{-4t})}{\sqrt{1-r_t^2}\left( 1+\sqrt{1-r_t^2}\right) ^2}\left( 4e^{-4t}(1-r^2)+e^{-4t}(3e^{-4t}+4)(1-e^{-4t})r_z^2\right) \end{aligned}$$

which is positive when $t\ge 0$. This completes the proof.