Partial Trace Exercises

These exercises practice the partial trace as a computational rule and as a physical rule for local statistics. All state vectors below are normalized.

Helpful background. Partial trace explains the index contraction; local measurement statistics connects it to probabilities.

Partial trace in an orthonormal basis

The main identity is

$$\operatorname{Tr}_B \bigl( \lvert i\rangle_A\langle k\rvert_A \otimes \lvert j\rangle_B\langle l\rvert_B \bigr) = \delta_{jl} \lvert i\rangle_A\langle k\rvert_A.$$

Equivalently, the traced subsystem contributes an inner product between the ket and bra labels that remain on that subsystem.

Exercises

1. Let

$$\lvert\Psi\rangle_{AB} = \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B.$$

Compute $\rho_A=\operatorname{Tr}_B\lvert\Psi\rangle\langle\Psi\rvert$.

Solution

The joint density operator is

$$\rho_{AB} = \lvert\psi\rangle\langle\psi\rvert_A \otimes \lvert\phi\rangle\langle\phi\rvert_B.$$

Taking the trace over $B$ gives

$$\rho_A = \lvert\psi\rangle\langle\psi\rvert_A \operatorname{Tr} \bigl( \lvert\phi\rangle\langle\phi\rvert_B \bigr).$$

Since $\lvert\phi\rangle$ is normalized,

$$\operatorname{Tr} \bigl( \lvert\phi\rangle\langle\phi\rvert \bigr) = 1.$$

Therefore

$$\rho_A = \lvert\psi\rangle\langle\psi\rvert_A.$$

A product pure state has pure reduced states.

2. Compute the one-qubit reduced state of

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr).$$

Solution

The density operator is

$$\rho_{AB} = \frac12 \bigl( \lvert00\rangle\langle00\rvert +\lvert00\rangle\langle11\rvert +\lvert11\rangle\langle00\rvert +\lvert11\rangle\langle11\rvert \bigr).$$

Trace over $B$. The two off-diagonal terms vanish because $\langle1\vert0\rangle=0$ and $\langle0\vert1\rangle=0$:

$$\rho_A = \frac12 \bigl( \lvert0\rangle\langle0\rvert +\lvert1\rangle\langle1\rvert \bigr) = \frac12 I.$$

The same calculation gives $\rho_B=I/2$.

3. The classically correlated state

$$\rho_{AB} = \frac12\lvert00\rangle\langle00\rvert +\frac12\lvert11\rangle\langle11\rvert$$

has the same one-qubit reduced states as $\lvert\Phi^+\rangle$. Verify this, and state one global difference between the two states.

Solution

Trace over $B$ term by term:

$$\operatorname{Tr}_B \bigl( \lvert00\rangle\langle00\rvert \bigr) = \lvert0\rangle\langle0\rvert,$$

and

$$\operatorname{Tr}_B \bigl( \lvert11\rangle\langle11\rvert \bigr) = \lvert1\rangle\langle1\rvert.$$

Therefore

$$\rho_A = \frac12 I, \qquad \rho_B = \frac12 I.$$

The global states are nevertheless different. The Bell state is a pure entangled state with coherence between $\lvert00\rangle$ and $\lvert11\rangle$. The classically correlated state is mixed and has no such off-diagonal coherence.

4. Let

$$\lvert\mathrm{GHZ}_3\rangle = \frac{1}{\sqrt2} \bigl( \lvert000\rangle+\lvert111\rangle \bigr).$$

Trace out qubit $C$ and compute $\rho_{AB}$.

Solution

The density operator contains four terms:

$$\rho_{ABC} = \frac12 \bigl( \lvert000\rangle\langle000\rvert +\lvert000\rangle\langle111\rvert +\lvert111\rangle\langle000\rvert +\lvert111\rangle\langle111\rvert \bigr).$$

Tracing over $C$ kills the cross terms because the $C$ labels are orthogonal:

$$\langle1\vert0\rangle=0, \qquad \langle0\vert1\rangle=0.$$

The remaining terms are

$$\rho_{AB} = \frac12 \lvert00\rangle\langle00\rvert +\frac12 \lvert11\rangle\langle11\rvert.$$

Thus the two-qubit reduction is a separable classical mixture, not a Bell state.

5. Let

$$\lvert W_3\rangle = \frac{1}{\sqrt3} \bigl( \lvert100\rangle+\lvert010\rangle+\lvert001\rangle \bigr).$$

Trace out qubit $C$ and compute $\rho_{AB}$.

Solution

Group the state by the value of qubit $C$:

$$\lvert W_3\rangle = \frac{1}{\sqrt3} \bigl( \lvert10\rangle_{AB} +\lvert01\rangle_{AB} \bigr) \lvert0\rangle_C +\frac{1}{\sqrt3} \lvert00\rangle_{AB}\lvert1\rangle_C.$$

The two $C$ states are orthogonal, so the trace over $C$ removes cross terms between the $C=0$ and $C=1$ branches. The $C=0$ branch has norm squared $2/3$ and normalized $AB$ state

$$\lvert\Psi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert01\rangle+\lvert10\rangle \bigr).$$

Therefore

$$\rho_{AB} = \frac23 \lvert\Psi^+\rangle\langle\Psi^+\rvert +\frac13 \lvert00\rangle\langle00\rvert.$$

This reduction remains partly coherent in the one-excitation subspace.

6. For $\lvert\Phi^+\rangle$, use the reduced state to compute the probabilities for measuring $Z$ on subsystem $A$. Then compare with measuring the parity observable $Z\otimes Z$.

Solution

From Exercise 2,

$$\rho_A=\frac12 I.$$

The projectors for measuring $Z$ are

$$P_0=\lvert0\rangle\langle0\rvert, \qquad P_1=\lvert1\rangle\langle1\rvert.$$

Thus

$$p(0) = \operatorname{Tr}(\rho_A P_0) = \frac12, \qquad p(1) = \operatorname{Tr}(\rho_A P_1) = \frac12.$$

The local outcome is random. But the joint observable has perfect correlation:

$$(Z\otimes Z)\lvert\Phi^+\rangle = \lvert\Phi^+\rangle.$$

Parity is $+1$ with certainty, so $\langle Z\otimes Z\rangle=1$. Measuring each qubit’s $Z$ separately gives the finer outcomes $00$ and $11$, each with probability $1/2$. An ideal Lüders parity measurement preserves this Bell state, whereas separately measuring both qubits and forgetting their outcomes gives the mixture in Exercise 3. The observable’s expectation alone does not specify a measurement’s disturbance; see local and global observables.

7. Construct a purification of

$$\rho_A = p\lvert0\rangle\langle0\rvert +(1-p)\lvert1\rangle\langle1\rvert, \qquad 0\le p\le1,$$

and verify that tracing out the reference system gives $\rho_A$.

Solution

Introduce a reference qubit $R$ and define

$$\lvert\Psi\rangle_{AR} = \sqrt p\,\lvert0\rangle_A\lvert0\rangle_R +\sqrt{1-p}\, \lvert1\rangle_A\lvert1\rangle_R.$$

The projector is

$$\lvert\Psi\rangle\langle\Psi\rvert = p\lvert00\rangle\langle00\rvert +\sqrt{p(1-p)}\, \lvert00\rangle\langle11\rvert +\sqrt{p(1-p)}\, \lvert11\rangle\langle00\rvert +(1-p)\lvert11\rangle\langle11\rvert.$$

Tracing over $R$ removes the cross terms because the reference states are orthogonal. The result is

$$\operatorname{Tr}_R \bigl( \lvert\Psi\rangle\langle\Psi\rvert \bigr) = p\lvert0\rangle\langle0\rvert +(1-p)\lvert1\rangle\langle1\rvert = \rho_A.$$

8. Let

$$\rho_{AB} = \frac14 \bigl( I\otimes I +r\,Z\otimes I +s\,I\otimes Z +t\,Z\otimes Z \bigr).$$

For real $r,s,t$, find the conditions for this matrix to be positive. Compute $\rho_A$ and identify which coefficients are invisible to measurements on $A$ alone.

Solution

The matrix is diagonal. Its four eigenvalues are

$$\lambda_{ab}=\frac{1+r z_a+s z_b+t z_a z_b}{4}, \qquad z_0=1,\quad z_1=-1.$$

All four must be nonnegative; the trace is already one. These inequalities restrict $r,s,t$ jointly, not just one parameter at a time. For example, $r=s=1,t=0$ gives a negative eigenvalue and is not a state. Next use

$$\operatorname{Tr}(I)=2, \qquad \operatorname{Tr}(Z)=0.$$

Then

$$\begin{aligned} \rho_A &= \operatorname{Tr}_B\rho_{AB}\\ &= \frac14 \bigl( 2I +2rZ +sI\,\operatorname{Tr}Z +tZ\,\operatorname{Tr}Z \bigr)\\ &= \frac12 \bigl( I+rZ \bigr). \end{aligned}$$

The coefficients $s$ and $t$ do not appear in $\rho_A$. Similarly, $\rho_B=(I+sZ)/2$. The joint state is a product exactly when $t=rs$; other allowed values of $t$ describe classical correlations in this diagonal family.

9. In the ordered qubit–qutrit basis, reduce the complex state

$$|\psi\rangle=\frac{|00\rangle+i|11\rangle+|12\rangle}{\sqrt3}.$$

Compute both local purities and the expectation of $Y_{12}=-i|1\rangle\langle2|+i|2\rangle\langle1|$ on $B$.

Solution

The coefficient array is $C=\begin{pmatrix}1&0&0\\0&i&1\end{pmatrix}/\sqrt3$. Direct contraction gives

$$\rho_A=\frac13\begin{pmatrix}1&0\\0&2\end{pmatrix}, \qquad \rho_B=\frac13\begin{pmatrix}1&0&0\\0&1&i\\0&-i&1\end{pmatrix}.$$

Both have nonzero eigenvalues $1/3,2/3$, hence purity $5/9$. The local expectation is

$$\operatorname{Tr}(\rho_B Y_{12})=-\frac23.$$

In this coefficient convention $\rho_B=C^{\mathsf T}C^*$, not $C^\dagger C$. The latter would reverse the imaginary entries and the sign of this measured expectation, while leaving the eigenvalues and purity unchanged. This is why a complex example tests more than normalization alone.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press, 2002.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.