Six Properties of the Density Matrix

The density operator is the most general way to describe the state of a quantum system. Pure states (a definite quantum state) and mixtures (statistical ensembles of states) are both described by the same object, which is why it appears everywhere in open quantum systems, quantum statistical mechanics and quantum information. Below we go through six of its properties, and each one is proved explicitly (so no “it is easy to show”).

The density operator is defined as

\rho = \sum_i w_i \left|\alpha^{(i)}\right\rangle\!\left\langle\alpha^{(i)}\right|, \qquad \sum_i w_i = 1

where the states $|\alpha^{(i)}\rangle$ do not have to be orthogonal and the $w_i \geq 0$ are classical probabilities. A pure state has exactly one $w_i \neq 0$ , and anything else is a mixture. One thing to keep in mind from the start is that the $w_i$ are not quantum amplitudes (the $|\alpha^{(i)}\rangle$ can already be superpositions on their own). They are ordinary probabilities, i.e. they describe our classical ignorance of which state the system is actually in.

1. $\text{Tr}(\rho) = 1$ #

In some basis $|i\rangle$ , the matrix elements of $\rho$ are

\rho_{ij} = \langle i|\rho|j\rangle = \sum_k w_k \langle i|\alpha^{(k)}\rangle\langle\alpha^{(k)}|j\rangle

The trace is the sum of the diagonal elements,

\text{Tr}(\rho) = \sum_i \rho_{ii} = \sum_i \sum_k w_k \langle\alpha^{(k)}|i\rangle\langle i|\alpha^{(k)}\rangle

and completeness, $\sum_i |i\rangle\langle i| = \mathbf{1}$ , removes the sum over the basis,

\text{Tr}(\rho) = \sum_k w_k \langle\alpha^{(k)}|\alpha^{(k)}\rangle = \sum_k w_k = 1 \quad \blacksquare

2. Pure states: $\rho^2 = \rho$ and $\text{Tr}(\rho^2) = 1$ #

For a pure state there is a single normalized state $|i\rangle$ with $w_i = 1$ , so $\rho = |i\rangle\langle i|$ . Squaring it,

\rho^2 = |i\rangle\langle i|i\rangle\langle i| = |i\rangle\langle i| = \rho \quad \blacksquare

and taking the trace,

\text{Tr}(\rho^2) = \text{Tr}(\rho) = 1

For a mixture one finds $\text{Tr}(\rho^2) < 1$ instead, and this quantity is used as a measure of purity. The closer $\text{Tr}(\rho^2)$ is to 1, the closer the state is to a pure one.

3. Ensemble averages: $[A] = \text{Tr}(\rho A)$ #

The ensemble average of an operator $A$ is defined as the weighted average of its expectation values,

[A] = \sum_i w_i \langle\alpha^{(i)}|A|\alpha^{(i)}\rangle

We start from the other side, $\text{Tr}(\rho A)$ , and write it out using the matrix elements $A_{ji} = \langle j|A|i\rangle$ ,

\text{Tr}(\rho A) = \sum_i \sum_j \rho_{ij} A_{ji} = \sum_{i,j,k} w_k \langle\alpha^{(k)}|j\rangle\langle j|A|i\rangle\langle i|\alpha^{(k)}\rangle

Using completeness twice, once for the sum over $i$ and once for $j$ , gives

\text{Tr}(\rho A) = \sum_k w_k \langle\alpha^{(k)}|A|\alpha^{(k)}\rangle = [A] \quad \blacksquare

This is the reason the density matrix is so convenient in practice. Any expectation value, in any ensemble, is a single trace.

4. Cyclic property of trace, and basis independence#

Cyclic property: $\text{Tr}(AB) = \text{Tr}(BA)$ .

We insert a complete set $\sum_j |\phi_j\rangle\langle\phi_j| = \mathbf{1}$ between $A$ and $B$ ,

\text{Tr}(AB) = \sum_i \langle\psi_i|AB|\psi_i\rangle = \sum_{i,j}\langle\psi_i|A|\phi_j\rangle\langle\phi_j|B|\psi_i\rangle

The two matrix elements are just numbers, so we can swap their order and then use completeness for the $|\psi_i\rangle$ ,

= \sum_{i,j}\langle\phi_j|B|\psi_i\rangle\langle\psi_i|A|\phi_j\rangle = \sum_j \langle\phi_j|BA|\phi_j\rangle = \text{Tr}(BA) \quad \blacksquare

For longer products the same argument extends by induction, $\text{Tr}(ABC\cdots) = \text{Tr}(BC\cdots A) = \cdots$

Basis independence follows directly. If $U$ is a unitary change of basis, the trace in the new basis is $\text{Tr}(U\rho U^\dagger) = \text{Tr}(U^\dagger U \rho) = \text{Tr}(\rho)$ , by the cyclic property.

5. Superposition $\neq$ mixture: an explicit example#

If there is one property in this list to remember, it is this one. A 50/50 mixture and a 50/50 superposition sound like the same thing, but they are physically different, and we can show it with a two-line calculation. Take a spin-1/2 particle with $|\pm\rangle \equiv |S_z = \pm\tfrac{\hbar}{2}\rangle$ .

The mixture $\rho_\text{mix} = \tfrac{1}{2}(|+\rangle\langle+| + |-\rangle\langle-|)$ is

\rho_\text{mix} = \frac{1}{2}\begin{pmatrix}1&0\\0&0\end{pmatrix} + \frac{1}{2}\begin{pmatrix}0&0\\0&1\end{pmatrix} = \frac{1}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix} = \frac{\mathbf{1}}{2}

With $S_x = \frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}$ , the average spin along $x$ is

[S_x]_\text{mix} = \text{Tr}(\rho_\text{mix} S_x) = \frac{\hbar}{4}\text{Tr}\begin{pmatrix}0&1\\1&0\end{pmatrix} = 0

The superposition $|\psi_\text{sup}\rangle = \frac{1}{\sqrt{2}}(|+\rangle + |-\rangle)$ instead gives

\rho_\text{sup} = |\psi_\text{sup}\rangle\langle\psi_\text{sup}| = \frac{1}{2}\begin{pmatrix}1\\1\end{pmatrix}\begin{pmatrix}1&1\end{pmatrix} = \frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix}

[S_x]_\text{sup} = \text{Tr}(\rho_\text{sup} S_x) = \frac{\hbar}{4}\text{Tr}\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \frac{\hbar}{4}\text{Tr}\begin{pmatrix}1&1\\1&1\end{pmatrix} = \frac{\hbar}{2}

So $[S_x]_\text{mix} = 0$ while $[S_x]_\text{sup} = \hbar/2$ , and the two states make different measurable predictions. The superposition has a definite spin along $x$ and the mixture does not. In the density matrix, the difference is entirely in the off-diagonal elements (the coherences). They carry the quantum interference, and in the mixture they are simply zero.

6. Equation of motion: the von Neumann equation#

Finally, the time evolution. We differentiate $\rho = \sum_i w_i(t)|\alpha^{(i)}(t)\rangle\langle\alpha^{(i)}(t)|$ with the product rule,

\frac{d\rho}{dt} = \sum_i \frac{dw_i}{dt}|\alpha^{(i)}\rangle\langle\alpha^{(i)}| + \sum_i w_i\left(\frac{d}{dt}|\alpha^{(i)}\rangle\langle\alpha^{(i)}| + |\alpha^{(i)}\rangle\frac{d}{dt}\langle\alpha^{(i)}|\right)

The time-dependent Schrödinger equation $i\hbar\frac{d}{dt}|\alpha^{(i)}\rangle = H|\alpha^{(i)}\rangle$ and its conjugate give

\frac{d}{dt}|\alpha^{(i)}\rangle = -\frac{i}{\hbar}H|\alpha^{(i)}\rangle, \qquad \frac{d}{dt}\langle\alpha^{(i)}| = +\frac{i}{\hbar}\langle\alpha^{(i)}|H

and substituting these into the second sum gives

\frac{d\rho}{dt} = \sum_i \frac{dw_i}{dt}|\alpha^{(i)}\rangle\langle\alpha^{(i)}| + \left(-\frac{i}{\hbar}H\rho + \frac{i}{\hbar}\rho H\right)

\boxed{\frac{d\rho}{dt} = \sum_i \frac{dw_i}{dt}|\alpha^{(i)}\rangle\langle\alpha^{(i)}| - \frac{i}{\hbar}[H,\,\rho]}

If the weights $w_i$ do not change in time (a closed system), the first term drops out and we are left with the von Neumann equation,

\frac{d\rho}{dt} = -\frac{i}{\hbar}[H, \rho]

This is the quantum version of the Liouville equation from classical statistical mechanics. Watch the sign though. It is $-\frac{i}{\hbar}[H,\rho]$ , which is the opposite sign to the Heisenberg equation for operators, $\frac{dA}{dt} = +\frac{i}{\hbar}[H,A]$ .

1. Tr(ρ)=1\text{Tr}(\rho) = 1Tr(ρ)=1#

2. Pure states: ρ2=ρ\rho^2 = \rhoρ2=ρ and Tr(ρ2)=1\text{Tr}(\rho^2) = 1Tr(ρ2)=1#

3. Ensemble averages: [A]=Tr(ρA)[A] = \text{Tr}(\rho A)[A]=Tr(ρA)#