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:
-
It is Hermitian.
-
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
\[P_\mu = \big|\langle \mu | \psi\rangle\big|^2 = \langle \mu |\, \hat{\rho}\, | \mu \rangle. \label{Pi_rho}\]
-
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:
-
\(\hat{\rho}_A\)
is Hermitian.
-
\(\langle\varphi|\hat{\rho}_A|\varphi\rangle \ge 0\)
for any
\(|\varphi\rangle \in \mathscr{H}_A\)
(i.e., the operator is positive semidefinite).
-
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 \(\eqref{prop3}\).
-
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 \(\hat{Q} = \hat{I}_A\) .