for a single particle state, its wave function can only be a superposition of eigrnstates:

\left| \psi \right> = \sum_n \left| n \right> c_n

where |c_n|^2 is the probability of the corresponding eignstate. the eigen equation is:

Q = \sum |q \left> q \right< q |

\sum |q \left> \right< q| = 1

Notice that the state need not be the eigen state. The expectation of operator Q for a pure state is straight forward:

\left<\psi|Q|\psi\right>=\sum c_n*c_m\left<n|Q|m\right>

However, for many particles state, the wave function is a sum of individual wave function of each particle.

\left|\Psi\right>=\frac{1}{\sqrt{N}}\sum \left|\psi_i \right>

where N is the total number of state, such that the individual wave functions are independent.

\left<\psi_i|\psi_j\right> =\delta_{ij}

the expectation for operator Q will be:

\left<\Psi|Q|\Psi\right> = \frac{1}{N} \sum \left<\psi_i|Q|\psi_j\right>

due to the in dependency of each particle wave function, the double sum reduce to a single sum.

\left<\Psi|Q|\Psi\right>=\frac{1}{N} \sum \left<\psi_i|Q|\psi_i\right>

by using the identity,

\sum_i \left<\psi_i|Q|\psi_i\right> = \sum_{q,i} \left<\psi_i|q\right>\left<q|Q|\psi_i\right>

rearrange by taking out the eigen ket out:

\sum_q \left<q| \left( Q \sum_i |\psi_i\right>\left<\psi_i| \right) |q\right>

therefore, we can write it as the trace of matrix under eigen ket bais:

Tr[ Q \rho ]_q

such that the density matrix:

\rho =\frac{1}{N} \sum_i \left| \psi_i \right> \left< \psi_i \right|

Under rotation, like Rabi π-pulse, if we using the same ket as basis, the expectation value can be found by apply the rotation on the density matrix.


Pure state:

\left| \psi \right> = \frac{1}{\sqrt{2}} ( \left|\uparrow\right> + \left| \downarrow \right> )

\left< \psi | J_z | \psi \right> = 0

if we use the eigen basis and donate:

\left|\uparrow\right> = \begin {pmatrix} 1 \\ 0 \end {pmatrix}

\left|\downarrow\right> = \begin {pmatrix} 0 \\ 1 \end {pmatrix}

\rho = \left| \psi \right>\left<\psi \right| = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}

J_z=\begin{pmatrix}1&0 \\ 0 &-1\end{pmatrix}

J_z \rho = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}

Thus the trace is 0.

Mixed state:

\left| \psi_1 \right> = \frac{1}{\sqrt{2}} ( \left|\uparrow_1\right> + \left| \downarrow_1 \right> )

\left| \psi_2 \right> = \frac{1}{\sqrt{2}} ( \left|\uparrow_2\right> - \left| \downarrow_2 \right> )

notices that the eigen state from each particle cannot be added.

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

For a general 2 states system:

\rho = \sum_k p_k\left| \psi_k\right>\left< \psi_k\right|

the p_k is population of state k. and k is either up-state, down-state, transverse-state or random state. since the random state will cancel itself on both 3 axis, thus, it make no contribution.

\rho = \begin{pmatrix} p_{\uparrow} & p_1 \\ p_2 & p_{\downarrow} \end{pmatrix}

the meaning of the off-diagonal value can be understood by finding the expectation of J_x and J_y . If we we have some mixed state with 40% up-state , 20% down-state and 40 % in random state.

\rho= \begin{pmatrix}0.4&0 \\ 0 & 0.2 \end{pmatrix}