Action of tensor product of operators on entangled state

Question

Let $S$ be a system described by the density operator $\rho_S$. Consider the operator

$$\mathcal{L_t}\left[ \rho_S \right] = \gamma (t) \left[ \sigma_z \rho_S \sigma_z - \rho_S\right] $$

where $\sigma_z$ is the $z$ Pauli matrix and $\gamma(t)$ a time dependent coefficient.

Now consider an additional system A. The overall system $S + A$ is in a maximally entangled state

$$\rho_{SA} = |\Psi \rangle \langle \Psi |.$$

In this case, the density matrix of $S$ is obtained by tracing out the ancillary system $A$ $$\rho_S = Tr_A \left[ \rho_{SA} \right]$$

How to explicit the action of the operator

$$\mathcal{L}_t \otimes Id_A $$

where $Id_A$ is the identity of tha ancillary system $A$, on the maximally entangled state $\rho_{SA}$?

In particular, prove that if $\gamma (t) < 0$ then

$$lim_{\epsilon \rightarrow 0} \frac {|| \left[ Id_{SA} + \left[ \mathcal{L}_t \otimes Id_A \right]\epsilon \right] |\Psi \rangle \langle \Psi |||_1 -1}{\epsilon} = -2 \gamma (t)$$

where $|| X ||_1 = Tr\left[\sqrt{ X X^{\dagger} } \right]$ is the trace norm.

Answer 1

We will evaluate this limit in several steps. First of all one readily verifies that $\mathcal L_t$ acts like $$ \mathcal L_t\begin{pmatrix}x_{11}&x_{12}\\x_{21}&x_{22}\end{pmatrix}=\begin{pmatrix}0&-2\gamma(t)x_{12}\\-2\gamma(t)x_{21}&0\end{pmatrix} $$ which lets us compute${}^1$ \begin{align*} ({\rm id}\otimes\mathcal L_t)(|\Psi\rangle\langle\Psi|)&=\frac12\begin{pmatrix}\mathcal L_t\begin{pmatrix}1&0\\0&0\end{pmatrix}&\mathcal L_t\begin{pmatrix}0&1\\0&0\end{pmatrix}\\\mathcal L_t\begin{pmatrix}0&0\\1&0\end{pmatrix}&\mathcal L_t\begin{pmatrix}0&0\\0&1\end{pmatrix}\end{pmatrix}\\ &=\frac12\begin{pmatrix}0&0&0&-2\gamma(t)\\0&0&0&0\\0&0&0&0\\-2\gamma(t)&0&0&0\end{pmatrix}\,. \end{align*} This, in turn, yields $$ |\Psi\rangle\langle\Psi|+\varepsilon ({\rm id}\otimes\mathcal L_t)(|\Psi\rangle\langle\Psi|)=\frac12\begin{pmatrix} 1&0&0&1-2\varepsilon\gamma(t)\\0&0&0&0\\0&0&0&0\\1-2\varepsilon\gamma(t)&0&0&1 \end{pmatrix}\,. $$ Now a general matrix $$ \frac12\begin{pmatrix}1&1-x\\1-x&1\end{pmatrix}$$ has eigenvalues $1-\frac x2,\, \frac x2$ for all $x\in\mathbb R$ meaning its trace norm equals $|1-\frac x2|+\frac{|x|}2$. For us this—assuming $\gamma(t)\leq 0$—means that \begin{align*} \lim_{\varepsilon\to 0^+}\frac{\|\,|\Psi\rangle\langle\Psi|+\varepsilon ({\rm id}\otimes\mathcal L_t)(|\Psi\rangle\langle\Psi|)\|_1-1}\varepsilon&= \lim_{\varepsilon\to 0^+}\frac{|1-\varepsilon\gamma(t)|+|\varepsilon\gamma(t)|-1}\varepsilon\\ &=\lim_{\varepsilon\to 0^+}\frac{1-\varepsilon\gamma(t)-\varepsilon\gamma(t)-1}\varepsilon\\ &=\lim_{\varepsilon\to 0^+}{-2\varepsilon\gamma(t)}\varepsilon=-2\gamma(t)\,, \end{align*} as was to be shown.

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_{1: I choose to apply $\mathcal L_t$ to the second and not the first component which does make the calculations more instructive but does not change the result: generally, this swap just differs by a global permutation which does not affect the trace norm and, in this special case, the matrix itself is the same regardless of what subsystem $\mathcal L_t$ acts on}