Density matrix, 4

Consider a system that is in a mixed state. The system has 0.3 of probability in a pure state $|\psi_1 \rangle$ and 0.7 of probability in another pure state $|\psi_2 \rangle$ . Then the density matrix $\rho$ is

$0.3 | \psi_1 \rangle \langle \psi_1 | + 0.7 | \psi_2 \rangle \langle \psi_2 |$

In the most general cases, neither $|\psi_1 \rangle$ nor $|\psi_2 \rangle$ is an eigenstate. So we cannot expect that $\rho$ is diagonal.

For example, if each of the pure state $|\psi_1 \rangle$ and $|\psi_2 \rangle$ is a superposition of two eigenstates $(|\phi_1\rangle, |\phi_2\rangle)$ , then

$| \psi_1 \rangle = \frac{1}{\sqrt 2} |\phi_1 \rangle + \frac{1}{\sqrt 2} |\phi_2 \rangle$

$| \psi_2 \rangle = \frac{1}{\sqrt 3} |\phi_1 \rangle + \sqrt{\frac{2}{3}} |\phi_2 \rangle$

and

$\rho$

$= 0.3 | \psi_1 \rangle \langle \psi_1 | + 0.7 | \psi_2 \rangle \langle \psi_2 |$

$= 0.3 \left( \frac{1}{\sqrt 2} |\phi_1 \rangle + \frac{1}{\sqrt 2} |\phi_2 \rangle \right) \left( \frac{1}{\sqrt 2} \langle \phi_1 | + \frac{1}{\sqrt 2} \langle \phi_2 | \right)$

$+ 0.7 \left( \frac{1}{\sqrt 3} |\phi_1 \rangle + \sqrt{\frac{2}{3}} |\phi_2 \rangle \right) \left( \frac{1}{\sqrt 3} \langle \phi_1 | + \sqrt{\frac{2}{3}} \langle \phi_2 | \right)$

For simplicity, assume that the eigenstates $\{ |\phi_1\rangle, |\phi_2\rangle \}$ form a complete orthonormal set.

If we use $\{ | \phi_1 \rangle, |\phi_2 \rangle \}$ as basis,

$\rho$

$= 0.3 \begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} \end{bmatrix} + 0.7 \begin{bmatrix} \frac{1}{\sqrt 3} \\ \sqrt{\frac{2}{3}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt 3} & \sqrt{\frac{2}{3}} \end{bmatrix}$

$= \frac{0.3}{2} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} + \frac{0.7}{3} \begin{bmatrix} 1 & \sqrt{2} \\ \sqrt{2} & 2 \end{bmatrix}$

$=\cdots$

— Me@2018.03.12 11:51 AM

Physics Town

continues

Density matrix, 4

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