[NOTES/QCQI-03002] Density Matrix

1.Properties of density matrix

A mixed system is described by a density matrix $\rho$ having properties (i) $Tr \rho=1$ and (ii) $\rho$ is a positive matrix. A matrix $\rho$ represents a pure state if and only if $\rho^2=\rho$. We list some results about the density matrix.

1. The positivity property implies that the density matrix is hermitian.
2. Condition $\rho^2=\rho$ implies that $\rho$ has eigenvalues 0 and 1.
3. $\rho^2=\rho$ together with $Tr\rho=1$ implies that the eigenvalue 1 is nondegenerate. All other eigenvalues are zero. Hence spectral theorem implies that $\rho=\ket{\psi}\bra{\psi}$, where $\ket{\psi}$ is the eigenvector corresponding to the eigenvalue 1. In this case the state is describe by vector $\ket{\psi}$ and is a pure state.
4. $\rho$ being, a positive operator, is necessarily hermitian.
5. A necessary and sufficient condition that $\rho$ may represent a pure state is $Tr(\rho^2)=Tr(\rho^3)=Tr(\rho)$. Note that this is stated entirely in terms of traces of the density operator.

Question for you: Why do we need $Tr\rho^3=Tr(\rho^2)=1$. Give an example to show that the condition $Tr(\rho^2)=1$ is not sufficient to restrict $\rho$ to be a density operator. However $Tr\rho^3=1$ and $Tr\rho^2=1$ taken together are sufficient for $\rho$ to be a density operator for a pure state.

2.Density matrix and ensemble of pure states 

A mixed state can be thought of as an ensemble of pure states $\{\ket{\psi_k}\}$. If $p_k$ is the probability of system being in state $\ket{\psi_k}$, then the density operator can be written as a convex linear combination of projection operators $\ket{\psi_k}\bra{\psi_k}$ \begin{equation}\label{eq01} \rho = \sum_kp_k \ket{\psi_k}\bra{\psi_k}~,\qquad p_k>0. \end{equation} with $\sum p_k=1$. The representation (\ref{eq01}) of a density operator in terms of ensemble of pure state is not unique.\\ To see a proof of this statement let $\ket{\lambda_k}$ denote normalized eigenvectors of $\rho$ thus \begin{equation}\label{eq02} \rho = \sum_{k=1}^n \lambda_k\ket{\lambda_k}\bra{\lambda_k}~,\qquad \sum \lambda_k=1~, \end{equation} where $\lambda_k$ are eigenvalues of $\rho$. If we define \begin{equation}\label{eq03} \sqrt{p_m}\ket{\phi_m} = \sum U_{mn}\sqrt{\lambda n} \ket{\lambda_n}~. \end{equation} Then it is easy to verify that \begin{equation}\label{eq04} \sum p_m\ket{\phi_m}\bra{\phi_m} = \sum \lambda_k \ket{\lambda_k}\bra{\lambda_k} = \rho\,. \end{equation} Thus there are infinite ways of interpreting the system as an ensemble of pure state $\ket{\phi_i}$, are for each unitary matrix.

We quote a result [Subahash] about restrictions on the probability distributions $\{p_i\}$ consistent with a given $\rho$. If the eigenvalues $\lambda_i$ of $\rho$ and the probabilities $\{p_i\}$ are arranged in descending order to obtain ordered sequences $\{\lambda_\alpha\}_\downarrow$ and $\{p_\alpha\}_\downarrow$, then the former sequence majorized the latter; the partial sums of the first $r$ entries of the form is greater than or equal to the corresponding sums of the latter for any $r$. Thus result shows, in particular, that uniform distribution - all $p_r$'s equal is always admissible for any $\rho$. \noindent\textbf{Reference:} Subash Chaturvedi, ``Lecture Notes'' (2013).

3.Partial Trace

Given a density matrix $\rho$ for a composite system, having subsystems $A$ and $B$, we ask what happens if we restrict our attention to one of the subsystems, say $A$ (or $B$). The subsystems are described by taking partial trace of the density matrix $\rho$. The process of taking partial trace will be described now. Let $\rho_A, \rho_B$ and $\rho$ denote the density operators describing systems $A,B$ and the composite system $AB$ respectively. Let $\{\ket{A_i}\}$ and $\{\ket{B_r}\}$ denote orthonormal bases on $\mathcal{H}_A$ and $\mathcal{H}_B$ respectively. \noindent Recall that the trace of an operator $\hat{X}$ in a Hilbert space can be defined by writing its matrix representation $(\hat{X})_{mn}$ w.r.t. an orthonormal basis $\{\ket{n}\}$ and summing over diagonal elements \begin{eqnarray}\label{eq05} tr(\hat{X})&=& \sum_{m,n=1}^N \delta_{mn}(\hat{X})_{mn}\nonumber\\ &=& \sum_m(\hat{X})_{mn} \end{eqnarray} In case of an operator an tensor product Hilbert space $\mathcal{H}_A\otimes\mathcal{H}_B$, we first form orthonormal basis in $\mathcal{H}_A\otimes\mathcal{H}_B$ as \begin{equation}\label{eq06} \ket{aA,\nu B} = \ket{aA}\otimes\ket{\nu B} \end{equation} where $\{\ket{aA}|\nu=1,N\}$ and $\{\ket{\nu B}|\nu=1\cdots N\}$ are orthonormal bases in $\mathcal{H}_A$ and $\mathcal{H}_B$ respectively. The partial traces of an operator $\mathscr X$ are defined as \begin{eqnarray} \text{partial~trace~of}~ \mathscr X\!\!\!\!\!\!-~\text{over}~B = \sum_{\nu=1}^N \bra{aA,\nu B}\hat{\mathscr X}\ket{bA,\sigma B}\delta_{\nu\sigma}\equiv (\mathscr X_A)_{ab}\label{eq07}\\ \text{partial~trace~of}~ \mathscr X\!\!\!\!\!\!-~\text{over}~A = \sum_{a=1}^N \bra{aA,\nu B}\hat{\mathscr X}\ket{bA,\sigma B}\delta_{ab}\equiv (\mathscr X_B)_{\nu\sigma}\label{eq08} \end{eqnarray} These lead to matrices $(\mathscr X_A)_{ab}$ and $(\mathscr X_B)_{\nu\sigma}$ which represent operators in $\mathcal{A}$ and $\mathcal{H}_B$ respectively.\\ Thus we have $$ (\rho_A)_{ab} = \sum_{\nu=1}^N \bra{aA,\nu B}\rho\ket{bA,\nu B} $$ and the partial trace of $\rho$ over $B$ can expressed in an operator form as \begin{eqnarray*} \rho_A &=& \sum_{a,b}(\rho_A)_{ab}\ket{a}\bra{b}\\ &=& \sum_{a,b}\sum_\nu \bra{aA,\nu B}\rho\ket{bA,\nu B}\bra{a}\ket{b} \end{eqnarray*} with similar expressions for partial trace of density operator $\rho$ over $A$. It is obvious that $Tr(\rho)=Tr(\rho_A)=Tr(\rho_B)$, partial trace operation does not change the trace of the operators. However if $\rho$ is a positive operator, the act of taking partial trace does not always lead to a positive operator in the subspace.


4. An example

The density operator for composite system in pure state in $\mathcal{H}_A\otimes\mathcal{H}_B$, after partial trace leads to density matrix which, in general, does not correspond to a pure state.\\ \noindent\textbf{Example:} Consider Bell state ${\ket{00}+\ket{11}\over \sqrt{2}}$ having density matrix \begin{eqnarray*} \rho&=& {1\over2}(\ket{00}+\ket{11})(\bra{00}+\bra{11})\\ &=& {1\over2} \ket{00}>\bra{001}+{1\over2}\ket{00}\bra{11}+{1\over2}\ket{11}\bra{00} + {1\over2}\ket{11}\bra{11} \end{eqnarray*} $\rho_A$ is then given by $$ \rho_A = {1\over2}\ket{0}\bra{0} + {1\over2}\ket{1}\bra{1} $$ and we compute $$ \rho^2_A = {1\over4} \ket{0}\bra{0}+{1\over4} \ket{1}\bra{1}\ne \rho_A $$ $\therefore$ ~The density operator $\rho_A$ for the subsystem $A$ does not correspond to a pure state.