The density matrix

Intuition

One object carries everything an ensemble can predict: the density matrix $\hat{\rho}=\sum p_{k}\ket{\psi_{k}}\bra{\psi_{k}}$ . Every probability is $\bra{a}\hat{\rho}\ket{a}$ and every mean value is $\operatorname{tr}(\hat{\rho}\hat{A})$ . It is Hermitian, has no negative eigenvalues and has trace one, and every matrix with those three properties is the density matrix of some ensemble. Different ensembles can have the same density matrix — half $\ket{\uparrow}$ and half $\ket{\downarrow}$ gives the same one as half $\ket{+x}$ and half $\ket{-x}$ — and then no experiment can tell them apart: the density matrix is the state. Its diagonal elements are populations, its off-diagonal elements coherences.

Two recipes can make the same cake, and once it is baked nothing tells which was used. The density matrix is the cake: everything measurable about an ensemble is in it, whichever way it was prepared.

A cut through the Bloch sphere. The even mixture of $\ket{\uparrow}$ and $\ket{\downarrow}$ and the even mixture of $\ket{+x}$ and $\ket{-x}$ both sit at the centre: the same density matrix , and no measurement can tell which was prepared.

For the ensemble $\{p_{k},\ket{\psi_{k}}\}$ :

\begin{aligned}\hat{\rho}&=\sum_{k}p_{k}\ket{\psi_{k}}\bra{\psi_{k}}\\ P(a)&=\bra{a}\hat{\rho}\ket{a},\qquad \langle\hat{A}\rangle=\operatorname{tr}\bigl(\hat{\rho}\hat{A}\bigr)\end{aligned}

Properties

$\hat{\rho}^{\dagger}=\hat{\rho}$ , no eigenvalue of $\hat{\rho}$ is negative, and ; every such matrix is a density matrix, of the ensemble of its eigenvectors weighted by its eigenvalues.

Mean values from the density matrix

\sum_{k}p_{k}\bra{\psi_{k}}\hat{A}\ket{\psi_{k}}=\operatorname{tr}\bigl(\hat{\rho}\hat{A}\bigr)

Each member of the ensemble has its own mean value, and the ensemble averages them. Inserting a complete basis turns each mean value into the trace of the member’s projector times the operator; the trace is linear, so the weighted sum becomes the trace of the density matrix times the operator.

Proof steps

\langle\hat{A}\rangle=\sum_{k}p_{k}\bra{\psi_{k}}\hat{A}\ket{\psi_{k}}

Average the mean values of the members.

\bra{\psi}\hat{A}\ket{\psi}=\sum_{n}\braket{n|\psi}\bra{\psi}\hat{A}\ket{n}

Insert $\sum_{n}\ket{n}\bra{n}=\hat{1}$ after $\overset{}{A}$ and reorder the two numbers.

\bra{\psi}\hat{A}\ket{\psi}=\operatorname{tr}\bigl(\ket{\psi}\bra{\psi}\hat{A}\bigr)

$\sum_{n}\bra{n}\hat{X}\ket{n}$ is the trace, here of $\hat{X} = ∣ ψ ⟩$ .

\langle\hat{A}\rangle=\operatorname{tr}\Bigl(\sum_{k}p_{k}\ket{\psi_{k}}\bra{\psi_{k}}\hat{A}\Bigr)=\operatorname{tr}\bigl(\hat{\rho}\hat{A}\bigr)

The trace is linear.

Applications

Practice

Populations

In the basis $\ket{\uparrow}$ , $\ket{\downarrow}$ the diagonal elements of the density matrix are the probabilities of up and down along $z$ .

P(a)=\bra{a}\hat{\rho}\ket{a}

Try it

For $\hat{\rho}=\begin{pmatrix}0.7&0.2\\0.2&0.3\end{pmatrix}$ in the basis $\ket{\uparrow},\ket{\downarrow}$ , which number is the probability of spin up along $z$ ?