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)

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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

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2018.03.14 Wednesday (c) All rights reserved by ACHK