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Emilio Pisanty
Emilio Pisanty
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Physically, a CP map is represents evolution processes even in the presence of entanglement.

After all positive maps sends a state to a state

This is false when the system is entangled with something else. The classic example is the transpose on the system of interest. (Technically, a transpose, since it's basis-dependent.) Since positivity can be characterized using determinants, transposing preserves positivity. However, if you only transpose a tensor half of a big matrix then this fails. Again, the standard example is a pair of qubits entangled in the $|\phi^+\rangle={1\over\sqrt{2}}\left(|00\rangle+|11\rangle\right)$ Bell state. Then the total density matrix is

$$\rho=\frac{1}{2}\begin{pmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{pmatrix}$$

and when transposed over the second qubit (done transposing each of the four submatrices) it goes to

$$\rho^{T_\textrm{B}}=\frac{1}{2}\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix},$$

which is not positive ($(0,1,-1,0)$ has eigenvalue -1).

That's the math. The physics it implies is that while you can ensure lone-system positivity of a map with purely local conditions, it does not imply that it works as a physical evolution for a larger, entangled system.

Physically, a CP map is represents evolution processes even in the presence of entanglement.

After all positive maps sends a state to a state

This is false when the system is entangled with something else. The classic example is the transpose on the system of interest. (Technically, a transpose, since it's basis-dependent.) Since positivity can be characterized using determinants, transposing preserves positivity. However, if you only transpose a tensor half of a big matrix then this fails. Again, the standard example is a pair of qubits entangled in the $|\phi^+\rangle={1\over\sqrt{2}}\left(|00\rangle+|11\rangle\right)$ Bell state. Then the total density matrix is

$$\rho=\frac{1}{2}\begin{pmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{pmatrix}$$

and when transposed over the second qubit (done transposing each of the four submatrices) it goes to

$$\rho^{T_\textrm{B}}=\frac{1}{2}\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix},$$

which is not positive ($(0,1,-1,0)$ has eigenvalue -1).

That's the math. The physics it implies is that while you can ensure lone-system positivity of a map with purely local conditions, it does not imply that it works as a physical evolution for a larger, entangled system.

Physically, a CP map represents evolution processes even in the presence of entanglement.

After all positive maps sends a state to a state

This is false when the system is entangled with something else. The classic example is the transpose on the system of interest. (Technically, a transpose, since it's basis-dependent.) Since positivity can be characterized using determinants, transposing preserves positivity. However, if you only transpose a tensor half of a big matrix then this fails. Again, the standard example is a pair of qubits entangled in the $|\phi^+\rangle={1\over\sqrt{2}}\left(|00\rangle+|11\rangle\right)$ Bell state. Then the total density matrix is

$$\rho=\frac{1}{2}\begin{pmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{pmatrix}$$

and when transposed over the second qubit (done transposing each of the four submatrices) it goes to

$$\rho^{T_\textrm{B}}=\frac{1}{2}\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix},$$

which is not positive ($(0,1,-1,0)$ has eigenvalue -1).

That's the math. The physics it implies is that while you can ensure lone-system positivity of a map with purely local conditions, it does not imply that it works as a physical evolution for a larger, entangled system.

Source Link
Emilio Pisanty
Emilio Pisanty
  • 138.3k
  • 34
  • 366
  • 699

Physically, a CP map is represents evolution processes even in the presence of entanglement.

After all positive maps sends a state to a state

This is false when the system is entangled with something else. The classic example is the transpose on the system of interest. (Technically, a transpose, since it's basis-dependent.) Since positivity can be characterized using determinants, transposing preserves positivity. However, if you only transpose a tensor half of a big matrix then this fails. Again, the standard example is a pair of qubits entangled in the $|\phi^+\rangle={1\over\sqrt{2}}\left(|00\rangle+|11\rangle\right)$ Bell state. Then the total density matrix is

$$\rho=\frac{1}{2}\begin{pmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{pmatrix}$$

and when transposed over the second qubit (done transposing each of the four submatrices) it goes to

$$\rho^{T_\textrm{B}}=\frac{1}{2}\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix},$$

which is not positive ($(0,1,-1,0)$ has eigenvalue -1).

That's the math. The physics it implies is that while you can ensure lone-system positivity of a map with purely local conditions, it does not imply that it works as a physical evolution for a larger, entangled system.