3.6: Density Operators

We now introduce the density operator, which helps to streamline many calculations in multi-particle quantum mechanics.

Consider a quantum system with a $d$-dimensional Hilbert space $\mathscr{H}$. Given an arbitrary state $|\psi\rangle \in \mathscr{H}$, define

$$\hat{\rho} = |\psi\rangle\, \langle\psi|. \label{rho_pure}$$

This is just the projection operator for $|\psi\rangle$, but in this context we call it a “density operator”. Some other authors call it a density matrix, based on the fact that linear operators can be represented as matrices. It has the following noteworthy features:

1. It is Hermitian.

2. Suppose $\hat{Q}$ is an observable with eigenvalues $\{q_\mu\}$ and eigenstates $\{|\mu\rangle\}$ (where $\mu$ is some label that enumerates the eigenstates. If we do a $\hat{Q}$ measurement on $|\psi\rangle$, the probability of obtaining $q_\mu$ is

   $$P_\mu = \big|\langle \mu | \psi\rangle\big|^2 = \langle \mu |\, \hat{\rho}\, | \mu \rangle. \label{Pi_rho}$$

3. Moreover, the expectation value of the observable is

   $$\langle Q\rangle = \sum_\mu q_\mu P_\mu = \sum_\mu q_\mu \langle \mu | \hat{\rho}| \mu \rangle = \mathrm{Tr}\big[\,\hat{Q} \, \hat{\rho}\,\big]. \label{Qexpt}$$

   In the last equality, $\mathrm{Tr}[\cdots]$ denotes the trace, which is the sum of the diagonal elements of the matrix representation of the operator. The value of the trace is basis-independent.

Now consider, once again, a composite system consisting of two subsystems $A$ and $B$, with Hilbert spaces $\mathscr{H}_A$ and $\mathscr{H}_B$. Let’s say we are interested in the physical behavior of $A$, that is to say the outcome probabilities and expectation values of any measurements performed on $A$. These can be calculated from $|\psi\rangle$, the state of the combined system; however, $|\psi\rangle$ also carries information about $B$, which is not relevant to us as we only care about $A$.

There is a more economical way to encode just the information about $A$. We can define the density operator for subsystem $A$ (sometimes called the reduced density operator):

$$\hat{\rho}_A = \mathrm{Tr}_B \,\big[\,\hat{\rho}\,\big]. \label{rhoa_def}$$

Here, $\mathrm{Tr}_B[\cdots]$ refers to a partial trace. This means tracing over the $\mathscr{H}_B$ part of the Hilbert space $\mathscr{H} = \mathscr{H}_A \otimes \mathscr{H}_B$, which yields an operator acting on $\mathscr{H}_A$.

To better understand Equation $\eqref{rhoa_def}$, let us go to an explicit basis. Let $\hat{Q}_A$ be an observable for $\mathscr{H}_A$ with eigenbasis $\{|\mu\rangle\}$, and let $\hat{Q}_B$ be an observable for $\mathscr{H}_B$ with eigenbasis $\{|\nu\rangle\}$. If the density operator of the combined system is $\hat{\rho} = |\psi\rangle\langle \psi|$, then

$$\hat{\rho}_A = \sum_\nu \Big( \hat{I}\otimes \langle \nu| \Big) \; |\psi\rangle \langle \psi | \; \Big( \hat{I}\otimes | \nu\rangle \Big). \label{rhoa_explicit}$$

This is a Hermitian operator acting on the $\mathscr{H}_A$ space. In the $\{|\mu\rangle\}$ basis, its diagonal matrix elements are

$$\begin{align} \begin{aligned} \langle \mu | \hat{\rho}_A | \mu \rangle &= \sum_\nu \Big( \langle \mu| \langle \nu| \Big) \, |\psi\rangle \langle \psi | \, \Big( |\mu\rangle | \nu\rangle \Big) \\ &= \langle \psi | \, \left[ |\mu\rangle \langle \mu| \otimes \left(\sum_\nu | \nu\rangle \langle \nu|\right) \right] |\psi\rangle \\ &= \langle \psi | \, \Big( |\mu\rangle \langle \mu| \otimes \hat{I}_B\Big) |\psi\rangle. \end{aligned}\end{align}$$

According to the rules of partial measurements discussed in Section 3.2, this is precisely the probability of obtaining $q_\mu$ when measuring $\hat{Q}_A$ on subsystem $A$:

$$P_\mu = \langle \mu | \hat{\rho}_A | \mu \rangle. \label{rho_prob}$$

It follows that the expectation value for observable $\hat{M}$ is

$$\langle Q_A \rangle = \sum_\mu q_\mu \langle \mu | \hat{\rho}_A | \mu \rangle = \mathrm{Tr}\Big[\hat{Q}_A \, \hat{\rho}_A \Big]. \label{rho_expect}$$

These results hold for any choice of basis. Hence, knowing the density operator for $A$, we can determine the outcome probabilities of any partial measurement performed on $A$.

To better understand the properties of $\hat{\rho}_A$, let us write $|\psi\rangle$ explicitly as

$$|\psi\rangle = \sum_{\mu\nu} \psi_{\mu\nu} |\mu\rangle |\nu\rangle,$$

where $\sum_{\mu\nu} |\psi_{\mu\nu}|^2 = 1$. Then

$$\begin{align} \begin{aligned} \hat{\rho} &= \sum_{\mu\mu'\nu\nu'} \psi_{\mu\nu} \psi_{\mu'\nu'}^* \; |\mu\rangle |\nu\rangle \, \langle\mu'|\langle \nu'| \\ \hat{\rho}_A &= \sum_{\mu\mu'\nu} \psi_{\mu\nu}\psi_{\mu'\nu}^* |\mu\rangle \langle\mu'| \\ &= \sum_\nu \left(\sum_\mu \psi_{\mu\nu} |\mu\rangle\right) \left(\sum_{\mu'} \psi_{\mu'\nu}^*\langle\mu'|\right) \\ &= \sum_\nu |\varphi_\nu\rangle \langle \varphi_\nu|, \;\;\;\mathrm{where}\;\;\; |\varphi_\nu\rangle = \sum_\mu \psi_{\mu\nu} |\mu\rangle. \end{aligned}\end{align}$$

But $|\varphi_\nu\rangle$ is not necessarily normalized to unity: $\langle \varphi_\nu | \varphi_\nu\rangle = \sum_{\mu}|\psi_{\mu\nu}|^2 \le 1$. Let us define

$$|\tilde{\varphi}_\nu\rangle = \frac{1}{\sqrt{P_\nu}} |\varphi_\nu\rangle, \;\;\;\mathrm{where} \;\; P_\nu = \sum_{\mu}|\psi_{\mu\nu}|^2.$$

Note that each $P_\nu$ is a non-negative real number in the range $[0,1]$. Then

$$\hat{\rho}_A = \sum_\nu P_\nu\, |\tilde{\varphi}_\nu\rangle \langle \tilde{\varphi}_\nu|, \;\;\;\mathrm{where}\;\; \begin{cases} \;\;\textrm{each $P_\nu$ is a real number in $[0,1]$, and} \\ \;\;\textrm{each}\; |\tilde{\varphi}_\nu\rangle \in \mathscr{H}_A, \;\;\mathrm{with} \;\;\langle\tilde{\varphi}_\nu|\tilde{\varphi}_\nu\rangle = 1. \end{cases} \label{rhoform}$$

In general, we can define a density operator as any operator that has the form of Equation $\eqref{rhoform}$, regardless of whether or not it was formally derived via a partial trace. We can interpret it as describing a ensemble of quantum states weighted by a set of classical probabilities. Each term in the sum consists of (i) a weighting coefficient $P_\nu$ which can be regarded as a probability (the coefficients are all real numbers in the range $[0,1]$, and sum to 1), and (ii) a projection operator associated with some normalized state vector $|\tilde{\varphi}_\nu\rangle$. Note that the states in the ensemble do not have to be orthogonal to each other.

From this point of view, a density operator of the form $|\psi\rangle\langle\psi|$ corresponds to the special case of an ensemble containing only one quantum state $|\psi\rangle$. Such an ensemble is called a pure state, and describes a quantum system that is not entangled with any other system. If an ensemble is not a pure state, we call it a mixed state; it describes a system that is entangled with some other system.

We can show that any linear operator $\hat{\rho}_A$ obeying Equation $\eqref{rhoform}$ has the following properties:

1. $\hat{\rho}_A$ is Hermitian.

2. $\langle\varphi|\hat{\rho}_A|\varphi\rangle \ge 0$ for any $|\varphi\rangle \in \mathscr{H}_A$ (i.e., the operator is positive semidefinite).

3. For any observable $\hat{Q}_A$ acting on $\mathscr{H}_A$,

   $$\begin{align} \begin{aligned} \langle Q_A \rangle &\equiv \sum_\nu P_\nu \langle \tilde{\varphi}_\nu|\hat{Q}_A|\tilde{\varphi}_\nu\rangle \\ &= \sum_{\mu\nu} P_\nu\, \langle \tilde{\varphi}_\nu|\mu\rangle \, \langle\mu|\hat{Q}|\tilde{\varphi}_\nu\rangle \;\;\;\big(\textrm{using some basis} \;\{|\mu\rangle\}\big) \\ &= \sum_\mu \langle\mu|\hat{Q} \left(\sum_\nu |\tilde{\varphi}_\nu\rangle \langle \tilde{\varphi}_\nu|\right) |\mu\rangle \\ &= \mathrm{Tr}\left[\,\hat{Q} \,\hat{\rho}_A\,\right]. \end{aligned} \label{prop3} \end{align}$$

   This property can be used to deduce the probability of obtaining any measurement outcome: if $|\mu\rangle$ is the eigenstate associated with the outcome, the outcome probability is $\langle\mu|\hat\rho_A|\mu\rangle$, consistent with Equation $\eqref{rho_prob}$. To see this, take $\hat{Q} = |\mu\rangle \langle \mu|$ in Equation $\eqref{prop3}$.

4. The eigenvalues of $\hat{\rho}_A$, denoted by $\{p_1, p_2, \dots, p_{d_A}\}$, satisfy

   $$p_j \in \mathbb{R} \;\;\;\mathrm{and}\;\; 0 \le p_j \le 1 \;\; \mathrm{for}\;\; j = 1,\dots,d_A, \quad\mathrm{with}\;\; \sum_{j=1}^{d_A} p_j = 1. \label{trrho_reduced}$$

   In other words, the eigenvalues can be interpreted as probabilities. This also implies that $\mathrm{Tr}[\hat\rho_A] = 1$.

   This property follows from Property 3 by taking $\hat{Q} = |\varphi\rangle\langle\varphi|$, where $|\varphi\rangle$ is any eigenvector of $\hat\rho_A$, and then taking $\hat{Q} = \hat{I}_A$.