[NOTES/QCQI-03003] Density Matrix of Mixed State

Example
Given mixed a state \(\ket{\psi}\), the corresponding density matrix is uniquely determined as \(\rho = \ket{\psi}\bra{\psi}\). However the same is not true for a mixed state. We shall demonstrate that same averages can be obtained by two different density matrices. Consider a density matrix for spin half system given by \begin{equation} \rho = \frac{3}{4}\ket{0}\bra{0} + \frac{1}{4} \ket{1}\bra{1} \end{equation} Let \(\ket{a},\ket{b}\) be two orthogonal vectors defined as follows. \begin{eqnarray} \ket{a} = \sqrt{\frac{3}{4}}\ket{0} + \sqrt{\frac{1}{4}}\ket{1} \\ \ket{b} = \sqrt{\frac{3}{4}}\ket{0} + \sqrt{\frac{1}{4}}\ket{1} \end{eqnarray} So that \begin{equation} \ket{0}=\frac{1}{\sqrt{3}}(\ket{a}+\ket{b}),\qquad \ket{1} =(\ket{a}-\ket{b}) \end{equation} The given density matrix takes the form \begin{eqnarray} \rho &=& \frac{3}{4}\ket{0}\bra{0} + \frac{1}{4} \ket{1}\bra{1} \nonumber\\ &=& \frac{3}{4}\big( \frac{1}{3} (\ket{a}+\ket{b})( \bra{a} + \bra{b})\big) + \frac{1}{4}\big((\ket{a}-\ket{b})(\bra{a} - \bra{b})\big)\\ &=& \frac{1}{2}\ket{a}\bra{a} + \frac{1}{2} \ket{b}\bra{b} \end{eqnarray} Thus same averages can be obtained from two different ensembles. In general if we are given a density matrix \begin{equation} \rho = \sum_n p_n \ket{\psi_n}\bra{\psi_n} \end{equation} where \(\{\ket{\psi_n}\}\) are a set of orthogonal vectors. Let \(U\) be a unitary matrix. Define \(\ket{\phi_n}\) by \begin{equation} \ket{\phi_n} = \sum_n U_{nm}\ket{\psi_m}. \end{equation} Then the set of vectors \(\{\ket{\phi_n}\}\) is an orthogonal set. The density matrix \(\tilde{\rho}\) defined as \begin{equation} \tilde{\rho} = \sum_n p_n \ket{\phi_n}\bra{\phi_n} \end{equation} gives the same average for every observable \(X\). To prove this statement let us define the matrix element \begin{equation} (X)_{mn}= \matrixelement{\psi_m}{\widehat{X}}{\psi_n} \end{equation} Then the average of \(\widehat{X}\) is given by \begin{eqnarray} \widehat{X}_\rho &=& \text{tr} {\rho \widehat{X}}\\ &=& \sum_{mn} \rho_{nm} (X)_{mn}\\ &=& \sum_{mn}\matrixelement{n}{\rho}{m}\cdot \matrixelement{m}{\widehat{X}}{n}\\ &=& \sum_{mn} \sum_j p_j\innerproduct{n}{\psi_j} \innerproduct{\psi_j}{m}\cdot \matrixelement{m}{\widehat{X}}{n}\\ &=& \sum_j p_j \average {\psi_j}{\widehat{X}}\\ &=& \sum_j p_j \text{tr} \big( \widehat{X} \ket{\psi_j}\bra{\psi_j}\big). \end{eqnarray} Since \(U\) is a unitary transformation \Flagged[This document is flagged for completion] \noindent

Bloch representation A general density matrix for mixed state of a two level system is \[\rho= \frac{1}{2}\Big(I+ \vec{r}\cdot\vec{\sigma}\Big)\]
Problem Show thatthe above density matrix represents a pure state if and only if vector \(\vec{r}\) is a unit vector, \(|\vec{r}|=1\).
Problem Give explicit expressions of two different density matrices for a two level system which will give rise to the same average values.