Lesson03

General quantum process¶

Constraint¶

\[ \text{Super Operator }\hat{\mathcal{E}}: \boxed{\text{physical state}} \xrightarrow{\text{Evolution}} \boxed{\text{physical state}} \]

Linearity: \(\hat{\mathcal{E}}(\lambda_1\rho_1+\lambda_2\rho_2) = \lambda_1\hat{\mathcal{E}}(\rho_1) + \lambda_2\hat{\mathcal{E}}(\rho_2)\)
Hermiticity: \(\hat{\mathcal{E}}(\rho) = \left(\hat{\mathcal{E}}(\rho)\right)^\dagger\)
Trace preserving: \(\text{Tr}[\hat{\mathcal{E}}(\rho)] = \text{Tr}[\rho]\)
Complete positivity: \(\left(\hat{\mathcal{E}}_A \otimes \hat I_B\right)(\rho_{AB}) \geq 0\) (all eigenvalues must be positive)

--> a Quantum process must be a CPTP map.

example: transpose is not a CP map¶

First, according to the combining rule on the indicies of two matrices:

\[ \left(\rho_A\right)_{i_A,\; j_A} \otimes \left(\rho_B\right)_{i_B,\; j_B} = \left(\rho_{AB}\right)_{\overline{i_Ai_B}^{(2)},\; \overline{j_Aj_B}^{(2)}} \]

For example, in the \(\overline{i_Ai_B}^{(2)}\) part, \(\overline{00}^{(2)}=0,\; \overline{01}^{(2)}=1,\; \overline{10}^{(2)}=2,\; \overline{11}^{(2)}=3\).
Now if we perform a partial transpose on system A,

\[ \left(\rho_{AB}\right)_{\overline{i_Ai_B}^{(2)},\; \overline{j_Aj_B}^{(2)}} \xrightarrow{\text{transpose only on }A} \left(\rho_{AB}\right)_{\overline{j_Ai_B}^{(2)},\; \overline{i_Aj_B}^{(2)}} \]

Therefore, if our initial state is \(\ket\psi_{AB} = \frac{1}{\sqrt2}(\ket{00}+\ket{11})\), then its density matrix undergoes:

\[ \frac12\begin{bmatrix} 1&0&0&1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 1&0&0&1 \end{bmatrix} \xrightarrow{\text{transpose only on }A} \frac12\begin{bmatrix} 1&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{bmatrix} \]

(basically the bottom-left and the top-right sub-block (2x2-sized) would exchange their position).

But the eigenvalues under transpose become:

\[ \text{eigenvalues: }(1,0,0,0) \xrightarrow{\text{transpose only on }A} \text{eigenvalues: }(\frac12,\frac12,\boxed{-\frac12},\frac12) \]

This \(\boxed{-\frac12}\) eigenvalue (associated eigenvector \(\frac{\ket{01}-\ket{10}}{\sqrt2}\)), indicates a "negative probability", which is unphysical! (Or mathematically speaking, it violates CP map.) Hence, transpose is not a CP map, therefore cannot be a quantum process.

Kraus Representation Theorem¶

(c.f. Preskill Notes. Chapt 3)

Any superoperator \(\hat{\mathcal{E}}\) satisfying the above 4 properties (i.e., Linearity, Hermiticity, Trace preserving, and Complete positivity) has an operator representation: \(\exists\{K_\mu\}\), such that

\[ \boxed{\hat{\mathcal{E}}(\rho) = \sum_\mu K_\mu\rho K_\mu^\dagger} \\ \text{ with }\textcolor{red}{\sum_\mu K_\mu^\dagger K_\mu = I} \]

The constraint \(\textcolor{red}{\sum_\mu K_\mu^\dagger K_\mu = I}\), can ensure \(\rho\)'s trace is preserved after \(\hat{\mathcal{E}}\)

\[ \begin{align*} \text{Tr}[\mathcal{E}(\rho)] &= \sum_\mu \text{Tr}[K_\mu\rho K_\mu^\dagger] \\ &= \sum_\mu \text{Tr}[K_\mu^\dagger K_\mu\rho] \\ &= \text{Tr}\left[\sum_\mu K_\mu^\dagger K_\mu\rho\right] \\ &= \text{Tr}[I\rho] \end{align*} \]

We can also check on its:
- positivity: \(\sum_\mu\bra\psi K_\mu\rho K_\mu^\dagger\ket\psi \geq0\), and
- complete positivity: \(\sum_\mu\bra{\psi_{AB}} (K_\mu\otimes I)\rho_{AB} (K_\mu\otimes I)^\dagger\ket{\psi_{AB}} \geq0\).

Properties¶

maximum number of Kraus operators¶

If \(\hat{\mathcal{E}}\) is a valid, physical quantum process, it will always have an operator representation. If it has an operator representation, then it can have at most \(N^2\) Kraus operators. (\(N\) is the \(\text{dim.}\) of the system, e.g., for a 4-qubit system then \(N=16\).)

ambiguity¶

Kraus representation \((K_\mu)\) is not unique, i.e., we can also use \((M_a)\) that is some linear (unitary-ish) combination of \(K_\mu\):

\[ M_a = \sum_\mu \boxed{(\mathcal{U})_{a\mu}} K_\mu \\ \text{with }\sum_a M_a^\dagger M_a = I \text{ and } \boxed{(\mathcal{U})_{a\mu}} \text{ is the element of a unitary matrix }\mathcal{U}. \]

proof for non-uniqueness

\[ \begin{align*} \hat{\mathcal{E}}_M(\rho) &= \sum_a M_a\rho M_a^\dagger \\ &= \sum_{\textcolor{red}{a}, \mu, \mu'} \textcolor{red}{(\mathcal{U})_{a\mu}} K_\mu \rho K_{\mu'}^\dagger \textcolor{red}{(\mathcal{U}^\dagger)_{\mu'a}} \\ &= \sum_{\mu, \mu'} \textcolor{red}{\delta_{\mu'\mu}} K_\mu \rho K_{\mu'}^\dagger \\ &= \sum_{\mu} K_\mu \rho K_{\mu}^\dagger \\ &= \hat{\mathcal{E}}_K(\rho) \end{align*} \]

System-Ancilla model¶

ANY evolution of a density operator \(\hat{\mathcal{E}}(\rho_A) = \sum_\mu K_\mu\rho_A K_\mu^\dagger\) can be implemented as a unitary evolution \(\mathcal{U}_{AB}\) on some extended (or we can say "bigger") system (with Hilbert space \(H_A\otimes H_B\)).

How to construct the unitary?¶

For any state \(\ket{\psi_A}\in\mathcal{H}_d\), \(\mathcal{U}\) needs to satisfy

\[ \mathcal{U}\left(\ket{\psi_A}\otimes\ket{0_B}\right) = \sum_{\mu=0}^{d^2-1} \left(K_\mu\ket{\psi_A}\otimes\ket{0_B}\right) \]

Note that for arbitrary states \(\ket{\psi_A}\) and \(\ket{\phi_A}\), from completeness condition we have

\[ \begin{align*} \bra{\phi_A} \bra{0_B} \mathcal{U}^\dagger \mathcal{U} \ket{\psi_A} \ket{0_B} &= \sum_\mu \bra{\phi_A} K_\mu^\dagger K_\mu \ket{\psi_A} \\ &= \bra{\phi_A} \ket{\psi_A} \end{align*} \]

Thus \(\mathcal U\) can be extended to a unitary operator acting on the entire space of the joint system (see N&C page 366, Box 8.1). We can think \(\mathcal U_{d^3\times d^3}\) matrix as:

\[ \mathcal{U} = \begin{bmatrix} K_0 & * & \cdots & * \\ K_1 & * & \dots & * \\ K_2 & * & \dots & * \\ \vdots & \vdots & \ddots \\ K_{d^2-1} & * &&* \end{bmatrix}_{d^2\times d^2} \]

The \(*\) elements can be artificially choosen to make \(\mathcal{U}^\dagger\mathcal{U} = \mathcal{U}\mathcal{U}^\dagger = I\).