Partial trace and reduced states · Quantum Mechanics

Intuition

When the whole is in $\hat{\rho}$ and only part $A$ is looked at, everything $A$ ’s measurements can show is in the reduced density matrix $\hat{\rho}_{A}=\operatorname{tr}_{B}\hat{\rho}$ , found by tracing out $B$ . For a product state it is $A$ ’s own state. For an entangled pure state it is mixed: each spin of the singlet, looked at alone, is completely mixed, $\tfrac12\cdot1$ , up or down at random along every axis. So a mixed state need not come from ignorance of a preparation: it can come from entanglement with something else. This is one way into entanglement: a part whose reduced state is not pure.

Listen to only one track of a duet recorded on two. You hear that singer fully, but whatever the two voices did together — the harmony — is lost to you. Tracing out $B$ keeps $A$ ’s track and discards the rest.

The purity of qubit $A$ alone when the pair is in $\cos\theta\ket{00}+\sin\theta\ket{11}$ , against $\theta$ . At and the pair is a product and is pure; at it is completely mixed, purity , although the pair as a whole is pure throughout.

The reduced density matrix

For a state $\hat{\rho}$ of the whole, with bases $\ket{a}$ of $A$ and $\ket{b}$ of :

\begin{aligned}\hat{\rho}_{A}&=\operatorname{tr}_{B}\hat{\rho}\\ \bra{a}\hat{\rho}_{A}\ket{a'}&=\sum_{b}\bra{a\,b}\hat{\rho}\ket{a'\,b}\\ \langle\hat{A}\otimes\hat{1}\rangle&=\operatorname{tr}_{A}\bigl(\hat{\rho}_{A}\hat{A}\bigr)\end{aligned}

Properties

$\hat{\rho}_{A}$ is a density matrix: Hermitian, no negative eigenvalues, trace one.
For $\hat{\rho}=\hat{\rho}_{A}\otimes\hat{\rho}_{B}$ , tracing out gives back .

The reduced state predicts every measurement on A

\operatorname{tr}\bigl(\hat{\rho}\,(\hat{A}\otimes\hat{1})\bigr)=\operatorname{tr}_{A}\bigl(\hat{\rho}_{A}\hat{A}\bigr)

Write the mean value of $\hat{A}\otimes\hat{1}$ as a trace in the product basis and insert the product basis between the two operators. The identity on $B$ forces both $B$ labels to be the same, and the sum over that label is exactly the definition of the reduced density matrix.

Proof steps

\langle\hat{A}\otimes\hat{1}\rangle=\sum_{a,b}\bra{a\,b}\hat{\rho}\,(\hat{A}\otimes\hat{1})\ket{a\,b}

The trace in the product basis.

\langle\hat{A}\otimes\hat{1}\rangle=\sum_{a,a',b}\bra{a\,b}\hat{\rho}\ket{a'\,b}\bra{a'}\hat{A}\ket{a}

Insert $\sum\ket{a'b'}\bra{a'b'}$ ; the identity on makes .

\bra{a}\hat{\rho}_{A}\ket{a'}=\sum_{b}\bra{a\,b}\hat{\rho}\ket{a'\,b}

Collect the sum over $b$ : the reduced density matrix.

\langle\hat{A}\otimes\hat{1}\rangle=\sum_{a,a'}\bra{a}\hat{\rho}_{A}\ket{a'}\bra{a'}\hat{A}\ket{a}=\operatorname{tr}_{A}\bigl(\hat{\rho}_{A}\hat{A}\bigr)

The remaining double sum is a trace over $A$ .

Applications

Practice

Tracing Out

The partial trace over $B$ sums the density matrix over the states of $B$ , keeping $A$ ’s labels. What remains predicts every measurement made on $A$ alone.

\bra{a}\hat{\rho}_{A}\ket{a'}=\sum_{b}\bra{a\,b}\hat{\rho}\ket{a'\,b}

Try it

What does the partial trace over $B$ keep?