Partial Trace & Reduced States

You hold one qubit of an entangled pair; your collaborator across the galaxy holds the other. What is your qubit's state? Not a ket — there isn't one. The operation that answers the question is the partial trace, and its output, the reduced density matrix, is the single most important construction in quantum information after the density operator itself. It is how a pure global state becomes a mixed local one, how entanglement is quantified, and (Term 4) how an open system "feels" its environment. It also delivers a clean punchline: the reduced state of a maximally entangled pair is the maximally mixed state — total local ignorance born from perfect global knowledge.

Learning Objectives

After this lesson you will be able to:

1. Define the partial trace $\rho_A = \operatorname{Tr}_B(\rho_{AB})$ operationally (on product operators) and by index contraction, and prove it yields a valid density operator.
2. Justify the partial trace as the unique operation reproducing local measurement statistics ($\operatorname{Tr}(\rho_A M_A) = \operatorname{Tr}(\rho_{AB}\,M_A\otimes I_B)$).
3. Prove the headline result: the reduced state of an entangled pure state is mixed, and compute $\rho_A = I/2$ for the Bell state $|\Phi^+\rangle$.
4. State and apply purification: every mixed state is a reduced state of a pure state on a larger space (Bell example).
5. Articulate no-signaling: a local operation on $B$ leaves $\rho_A$ unchanged.

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Intuition

A bipartite system $AB$ lives in $\mathcal H_A\otimes\mathcal H_B$ (0.1.7). Suppose you only ever touch subsystem $A$: every measurement you can do is of the form $M_A\otimes I_B$ — act on $A$, do nothing to $B$. The question "what is the state of $A$?" can only mean: what single operator $\rho_A$ on $\mathcal H_A$ reproduces all of these statistics? That demand — match $\operatorname{Tr}(\rho_{AB}, M_A\otimes I_B)$for every$M_A$ — has exactly one answer, and computing it is the partial trace.

The mechanical picture: the full $\operatorname{Tr}$ collapses both tensor factors to a number; the partial trace $\operatorname{Tr}_B$ collapses only the $B$ factor, leaving an operator on $A$. You are "summing $B$ out" — averaging over everything you cannot see. And here is the twist that makes quantum mechanics strange: even when the global $\rho_{AB} = |\Psi\rangle\langle\Psi|$ is pure (maximal knowledge), summing out an entangled partner injects genuine uncertainty into $A$. Knowledge of the whole does not imply knowledge of the parts. That is the formal signature of entanglement, and the rest of the lesson makes it precise and runs it in code.

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Theory

Definition (operational form)

Definition (partial trace). $\operatorname{Tr}_B : \mathcal L(\mathcal H_A\otimes\mathcal H_B) \to \mathcal L(\mathcal H_A)$ is the linear map defined on product operators by

$$\operatorname{Tr}_B\big(\,|a\rangle\langle a'|\otimes|b\rangle\langle b'|\,\big) = |a\rangle\langle a'|\;\langle b'|b\rangle ,$$

and extended linearly to all operators. The reduced density matrix of $A$ is $\rho_A = \operatorname{Tr}B(\rho{AB})$. Symmetrically,$\rho_B = \operatorname{Tr}A(\rho{AB})$.

In words: leave the $A$ part of each term alone, and replace the $B$ part $|b\rangle\langle b'|$ by the number $\langle b'|b\rangle = \operatorname{Tr}(|b\rangle\langle b'|)$ — you take the ordinary trace on the $B$ factor only. This matches the preview in 0.1.7. Since every operator on $\mathcal H_A\otimes\mathcal H_B$ is a linear combination of product operators $|a\rangle\langle a'|\otimes|b\rangle\langle b'|$ (these form a basis of $\mathcal L(\mathcal H_A\otimes \mathcal H_B)$), linear extension defines$\operatorname{Tr}_B$ uniquely on everything.

Index form (the computational definition)

Fix orthonormal bases $\{|i\rangle_A\}$ and $\{|\mu\rangle_B\}$. Any bipartite operator has matrix elements $\langle i\mu|\,\rho_{AB}\,|j\nu\rangle =: (\rho_{AB})_{i\mu,\,j\nu}$. The partial trace sums over the repeated $B$ index:

$$\boxed{\;(\rho_A)_{ij} = \langle i|\rho_A|j\rangle = \sum_{\mu} \langle i\mu|\,\rho_{AB}\,|j\mu\rangle = \sum_\mu (\rho_{AB})_{i\mu,\,j\nu=\mu}\;}$$

Derivation from the operational form. Expand $\rho_{AB} = \sum_{i\mu,j\nu}(\rho_{AB}){i\mu,j\nu}, |i\rangle\langle j|\otimes|\mu\rangle\langle\nu|$. Apply$\operatorname{Tr}B$ termwise: $\operatorname{Tr}B(|i\rangle\langle j|\otimes|\mu\rangle\langle\nu|) = |i\rangle\langle j|, \langle\nu|\mu\rangle = \delta{\mu\nu},|i\rangle\langle j|$. The Kronecker delta sets$\nu = \mu$ and sums, giving $\rho_A = \sum{ij}\big(\sum\mu(\rho_{AB}){i\mu,j\mu}\big)|i\rangle\langle j|$, which is the boxed formula. ∎ Computationally this is a reshape-and-einsum: index $\rho{AB}$ as a 4-tensor $\rho[i,\mu,j,\nu]$ and contract $\mu = \nu$ — exactly the code below.

The defining property: partial trace reproduces local statistics

The operational definition is not arbitrary; it is forced by physics.

Theorem. $\rho_A = \operatorname{Tr}_B(\rho_{AB})$ is the unique operator on $\mathcal H_A$ such that, for every observable $M_A$ on $A$,

$$\operatorname{Tr}_A(\rho_A\,M_A) = \operatorname{Tr}_{AB}\big(\rho_{AB}\,(M_A\otimes I_B)\big).$$

This says: predicting any $A$-only measurement from the reduced state gives the same answer as predicting it from the full state with $I_B$ on the untouched side — as it must.

Proof. Existence. For a product operator $\rho_{AB} = \sigma\otimes\tau$ (with $\sigma\in\mathcal L(\mathcal H_A)$,$\tau\in\mathcal L(\mathcal H_B)$),$\operatorname{Tr}_B(\sigma\otimes\tau) = \sigma,\operatorname{Tr}(\tau)$ directly from the definition. Then

$$\operatorname{Tr}_{AB}\big((\sigma\otimes\tau)(M_A\otimes I_B)\big) = \operatorname{Tr}_{AB}\big((\sigma M_A)\otimes\tau\big) = \operatorname{Tr}_A(\sigma M_A)\,\operatorname{Tr}_B(\tau) = \operatorname{Tr}_A\big(\operatorname{Tr}_B(\sigma\otimes\tau)\,M_A\big),$$

using the mixed-product identity and $\operatorname{Tr}(A\otimes B) = \operatorname{Tr} A \operatorname{Tr} B$ (Appendix E §4–5). By linearity it holds for all $\rho_{AB}$. Uniqueness. If two operators $\rho_A,\rho_A'$ satisfied the identity for all $M_A$, then $\operatorname{Tr}_A((\rho_A - \rho_A')M_A) = 0$ for all $M_A$; taking $M_A = (\rho_A - \rho_A')^\dagger$ gives $\operatorname{Tr}((\rho_A-\rho_A')^\dagger(\rho_A-\rho_A')) = 0$, i.e. the Hilbert–Schmidt norm of the difference is zero, so $\rho_A = \rho_A'$. ∎

The reduced state is a valid density operator

Proposition. If $\rho_{AB}$ is a density operator, so is $\rho_A = \operatorname{Tr}_B(\rho_{AB})$.

Proof. Hermitian: taking $\dagger$ commutes with $\operatorname{Tr}_B$ (check on product operators: $(\operatorname{Tr}B(|a\rangle\langle a'|\otimes|b\rangle\langle b'|))^\dagger = (\langle b'|b\rangle|a\rangle\langle a'|)^\dagger = \overline{\langle b'|b\rangle}|a'\rangle\langle a| = \langle b|b'\rangle|a'\rangle\langle a| = \operatorname{Tr}B(|a'\rangle\langle a|\otimes|b'\rangle\langle b|)$), so $\rho_A^\dagger = \operatorname{Tr}_B(\rho_{AB}^\dagger) = \operatorname{Tr}_B(\rho_{AB}) = \rho_A$. Unit trace: $\operatorname{Tr}A(\rho_A) = \operatorname{Tr}{AB}(\rho{AB}) = 1$ (the iterated partial traces equal the full trace). Positive: for any $|\varphi\rangle_A$, set $M_A = |\varphi\rangle\langle\varphi|\succeq0$; by the defining identity$\langle\varphi|\rho_A|\varphi\rangle = \operatorname{Tr}A(\rho_A M_A) = \operatorname{Tr}{AB}(\rho{AB},M_A\otimes I_B) \ge 0$ because $\rho_{AB}\succeq0$ and $M_A\otimes I_B\succeq0$ (the trace of a product of two PSD operators is $\ge0$). ∎

The headline result: entangled ⇒ mixed reduced state

Now the payoff. Take a pure global state $\rho_{AB} = |\Psi\rangle\langle\Psi|$. When is $\rho_A$ pure, and when is it mixed?

Theorem. For a bipartite pure state $|\Psi\rangle\in\mathcal H_A\otimes\mathcal H_B$, the reduced state $\rho_A$ is pure iff $|\Psi\rangle$ is a product state, and mixed iff $|\Psi\rangle$ is entangled. Moreover $\rho_A$ and $\rho_B$ have the same nonzero eigenvalues.

Proof (via the Schmidt decomposition). By the Schmidt decomposition (1.4.4, built on the SVD of 0.1.6), write

$$|\Psi\rangle = \sum_{k=1}^{r}\sqrt{\lambda_k}\,|u_k\rangle_A\otimes|w_k\rangle_B, \qquad \lambda_k > 0,\ \sum_k\lambda_k = 1,$$

with $\{|u_k\rangle\}$, $\{|w_k\rangle\}$ orthonormal and $r$ the Schmidt rank. Then

$$\rho_{AB} = \sum_{k,l}\sqrt{\lambda_k\lambda_l}\,|u_k\rangle\langle u_l|\otimes|w_k\rangle\langle w_l|,$$

and tracing out $B$ (using $\operatorname{Tr}B(|w_k\rangle\langle w_l|) = \langle w_l|w_k\rangle = \delta{kl}$):

$$\boxed{\;\rho_A = \operatorname{Tr}_B(\rho_{AB}) = \sum_{k=1}^{r}\lambda_k\,|u_k\rangle\langle u_k|\;}$$

So the eigenvalues of $\rho_A$ are exactly the squared Schmidt coefficients $\lambda_k$. By the symmetric computation, $\rho_B = \sum_k\lambda_k|w_k\rangle\langle w_l|$ has the same eigenvalues.

If $|\Psi\rangle$ is a product state, $r = 1$, one $\lambda_1 = 1$: $\rho_A = |u_1\rangle\langle u_1|$ is pure. If $|\Psi\rangle$ is entangled, $r \ge 2$, so at least two $\lambda_k\in(0,1)$: $\operatorname{Tr}(\rho_A^2) = \sum_k\lambda_k^2 < (\sum_k\lambda_k)^2 = 1$, hence mixed (Lesson 1's purity bound, with strict inequality whenever two weights are nonzero). ∎

This is striking: a pure $|\Psi\rangle$ carries all the information about $AB$, yet $\rho_A$ — your complete local description — is mixed. The "missing" information is not in $A$ or in $B$ separately; it lives in the correlations between them. The purity of $\rho_A$ becomes a measure of entanglement (the basis of the entropy of entanglement $S(\rho_A) = -\sum_k\lambda_k\log\lambda_k$, Term 4 and beyond).

Worked engine: $\rho_A = I/2$ for the Bell state

Take the Bell state (1.4.2) $|\Phi^+\rangle = \tfrac1{\sqrt2}(|00\rangle + |11\rangle)$. Its density operator is

$$\rho_{AB} = |\Phi^+\rangle\langle\Phi^+| = \tfrac12\big(|00\rangle + |11\rangle\big)\big(\langle00| + \langle11|\big) = \tfrac12\big(|00\rangle\langle00| + |00\rangle\langle11| + |11\rangle\langle00| + |11\rangle\langle11|\big).$$

Rewrite each term as $|a\rangle\langle a'|_A\otimes|b\rangle\langle b'|_B$ and apply $\operatorname{Tr}_B$, which replaces $|b\rangle\langle b'|$ by $\langle b'|b\rangle$:

$$\begin{aligned} \operatorname{Tr}_B\big(|00\rangle\langle00|\big) &= |0\rangle\langle0|_A\,\langle0|0\rangle_B = |0\rangle\langle0|, \\ \operatorname{Tr}_B\big(|00\rangle\langle11|\big) &= |0\rangle\langle1|_A\,\langle1|0\rangle_B = 0, \\ \operatorname{Tr}_B\big(|11\rangle\langle00|\big) &= |1\rangle\langle0|_A\,\langle0|1\rangle_B = 0, \\ \operatorname{Tr}_B\big(|11\rangle\langle11|\big) &= |1\rangle\langle1|_A\,\langle1|1\rangle_B = |1\rangle\langle1|. \end{aligned}$$

The two off-diagonal (coherence) terms are killed by the orthogonality $\langle 1|0\rangle = 0$ — this is the crux. Summing the survivors:

$$\boxed{\;\rho_A = \tfrac12\big(|0\rangle\langle0| + |1\rangle\langle1|\big) = \tfrac12 I\;}$$

The reduced state of a maximally entangled pair is maximally mixed: purity $\operatorname{Tr}(\rho_A^2) = \tfrac12$, the smallest a qubit allows. Globally you know everything (a single pure ket); locally you know nothing (a fair coin in every basis). The Schmidt coefficients are $\lambda_1 = \lambda_2 = \tfrac12$, confirming maximal entanglement.

Purification

The headline result has a powerful converse.

Theorem (purification). Every density operator $\rho_A$ on $\mathcal H_A$ is the reduced state of some pure state $|\Psi\rangle_{AB}$ on $\mathcal H_A\otimes\mathcal H_B$ with $\dim\mathcal H_B \ge \operatorname{rank}(\rho_A)$. We call$|\Psi\rangle$a **purification** of$\rho_A$.

Proof (construction). Diagonalize $\rho_A = \sum_k\lambda_k|u_k\rangle\langle u_k|$ (Lesson 1). Pick an orthonormal set $\{|w_k\rangle\}$ in a reference system $\mathcal H_B$ and define

$$|\Psi\rangle = \sum_k\sqrt{\lambda_k}\,|u_k\rangle_A\otimes|w_k\rangle_B .$$

This is normalized ($\langle\Psi|\Psi\rangle = \sum_k\lambda_k = 1$) and pure, and by the Schmidt computation above $\operatorname{Tr}_B(|\Psi\rangle\langle\Psi|) = \sum_k\lambda_k|u_k\rangle\langle u_k| = \rho_A$. ∎

Purification is the formal statement that every mixed state can be seen as part of a larger pure state — classical uncertainty is "really" entanglement with a system you are ignoring (a reference, or, physically, the environment). This "go to the church of the larger Hilbert space" move underpins quantum channels and the Stinespring dilation in Term 4.

Bell example. For $\rho_A = \tfrac12 I$ we have $\lambda_1 = \lambda_2 = \tfrac12$, $|u_k\rangle = |k\rangle$. Choosing $|w_k\rangle = |k\rangle$ rebuilds $|\Psi\rangle = \tfrac1{\sqrt2}(|00\rangle + |11\rangle) = |\Phi^+\rangle$ — the Bell state is a purification of the maximally mixed qubit. (Purifications are not unique: any $|w_k'\rangle = V|w_k\rangle$ for unitary $V$ on $B$ gives another, e.g. $\tfrac1{\sqrt2}(|0+\rangle + |1-\rangle)$ — the freedom matching Lesson 1's ensemble ambiguity.)

No-signaling

Suppose your distant collaborator performs any physical operation on $B$ alone — a unitary $U_B$, a measurement, even discarding it. Can that change your $\rho_A$? No.

No-signaling (unitary case). If $\rho_{AB}\to (I_A\otimes U_B)\rho_{AB}(I_A\otimes U_B)^\dagger$, then $\rho_A$ is unchanged.

Proof. Using $\operatorname{Tr}_B((I_A\otimes U_B)\,M\,(I_A\otimes U_B^\dagger))$ and cyclicity within the $B$ factor: on a product operator $\sigma\otimes\tau$,

$$\operatorname{Tr}_B\big((I\otimes U_B)(\sigma\otimes\tau)(I\otimes U_B^\dagger)\big) = \operatorname{Tr}_B\big(\sigma\otimes U_B\tau U_B^\dagger\big) = \sigma\,\operatorname{Tr}(U_B\tau U_B^\dagger) = \sigma\,\operatorname{Tr}(\tau) = \operatorname{Tr}_B(\sigma\otimes\tau),$$

and linear extension gives $\operatorname{Tr}B((I\otimes U_B)\rho{AB}(I\otimes U_B^\dagger)) = \operatorname{Tr}B(\rho{AB}) = \rho_A$. ∎ (For a measurement on$B$, the same holds after averaging over $B$'s outcomes — Term 1.3.)

This is why entanglement cannot transmit information: Alice's local description is immune to anything Bob does. Bob can correlate his outcomes with Alice's (that is what entanglement is), but he cannot steer her marginal $\rho_A$ — consistent with relativity, and the reason the Bell-state correlations of Term 1.4 coexist peacefully with no faster-than-light signaling.

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Worked Examples

Example 1 — A product state has a pure reduced state

Take the separable $|\Psi\rangle = |+\rangle_A\otimes|0\rangle_B$. Then $\rho_{AB} = |+\rangle\langle+|\otimes|0\rangle\langle0|$ is a product operator, so

$$\rho_A = \operatorname{Tr}_B\big(|+\rangle\langle+|\otimes|0\rangle\langle0|\big) = |+\rangle\langle+|\,\operatorname{Tr}(|0\rangle\langle0|) = |+\rangle\langle+|\cdot 1 = |+\rangle\langle+| .$$

The reduced state is pure ($\operatorname{Tr}(\rho_A^2) = 1$) — exactly as the theorem predicts for a product state (Schmidt rank $1$). Contrast this with the Bell state, where the same qubit $A$ comes out maximally mixed. The difference is entanglement, full stop.

Example 2 — A partially entangled state: tunable purity

Consider $|\Psi_\theta\rangle = \cos\theta\,|00\rangle + \sin\theta\,|11\rangle$ (already in Schmidt form: $\sqrt{\lambda_1} = \cos\theta$, $\sqrt{\lambda_2} = \sin\theta$, orthonormal local bases $\{|0\rangle,|1\rangle\}$). Then

$$\rho_A = \cos^2\theta\,|0\rangle\langle0| + \sin^2\theta\,|1\rangle\langle1| = \begin{pmatrix}\cos^2\theta & 0\\ 0 & \sin^2\theta\end{pmatrix},$$

with purity

$$\operatorname{Tr}(\rho_A^2) = \cos^4\theta + \sin^4\theta = 1 - 2\sin^2\theta\cos^2\theta = 1 - \tfrac12\sin^2(2\theta).$$

This interpolates continuously between the extremes: at $\theta = 0$ the state is the product $|00\rangle$ and purity $= 1$ (pure $\rho_A$); at $\theta = \tfrac\pi4$ it is $|\Phi^+\rangle$ and purity $= 1 - \tfrac12 = \tfrac12$ (maximally mixed). So the reduced purity is a dial for the amount of entanglement — minimized exactly at the maximally entangled point. We will see in Lesson 3 that this corresponds to the Bloch vector shrinking from the sphere ($|\vec r| = 1$) to the center ($|\vec r| = 0$) as $\theta: 0\to\tfrac\pi4$.

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Hands-on (Python)

We implement the partial trace via reshape/einsum, confirm $\rho_A = I/2$ for $|\Phi^+\rangle$, watch the purity drop, and check no-signaling numerically.

import numpy as np

# Big-endian: qubit A = index 0 (leftmost), qubit B = index 1.
ket0 = np.array([1, 0], dtype=complex)
ket1 = np.array([0, 1], dtype=complex)

def partial_trace_B(rho_AB, dimA=2, dimB=2):
    """
    Tr_B(ρ_AB): contract the B factor of a bipartite density matrix.
    Reshape the (dA·dB)×(dA·dB) matrix into a 4-tensor ρ[i, μ, j, ν]
    (row = (i,μ), col = (j,ν)), then sum the diagonal over the B indices μ=ν.
    """
    t = rho_AB.reshape(dimA, dimB, dimA, dimB)   # ρ[i, μ, j, ν]
    return np.einsum('imjm->ij', t)              # Σ_μ ρ[i, μ, j, μ]

def partial_trace_A(rho_AB, dimA=2, dimB=2):
    """Tr_A(ρ_AB): contract the A factor instead (sum the A index k=k)."""
    t = rho_AB.reshape(dimA, dimB, dimA, dimB)   # ρ[k, μ, k, ν]
    return np.einsum('kmkn->mn', t)              # Σ_k ρ[k, μ, k, ν] -> operator on B

def purity(rho):
    return np.trace(rho @ rho).real

# Build the Bell state |Φ+⟩ = (|00⟩ + |11⟩)/√2 and its density matrix (big-endian).
ket00 = np.kron(ket0, ket0)
ket11 = np.kron(ket1, ket1)
phi_plus = (ket00 + ket11) / np.sqrt(2)
rho_AB = np.outer(phi_plus, phi_plus.conj())

print("Global purity Tr(ρ_AB²):", round(purity(rho_AB), 6))   # 1.0  (pure global state)

rho_A = partial_trace_B(rho_AB)
print("ρ_A =\n", np.round(rho_A, 6))
print("Is ρ_A = I/2 ?", np.allclose(rho_A, np.eye(2) / 2))     # True
print("Local purity Tr(ρ_A²):", round(purity(rho_A), 6))       # 0.5  -> maximally mixed!

# Entangled pure global state (purity 1) -> mixed local state (purity 1/2).

# Tunable entanglement: |Ψθ⟩ = cosθ|00⟩ + sinθ|11⟩ (Worked Example 2).
for theta in [0.0, np.pi/8, np.pi/4]:
    psi = np.cos(theta) * ket00 + np.sin(theta) * ket11
    rho = np.outer(psi, psi.conj())
    rA = partial_trace_B(rho)
    print(f"θ={theta:.4f}  Tr(ρ_A²)={purity(rA):.4f}  "
          f"(predicted 1 - ½sin²2θ = {1 - 0.5*np.sin(2*theta)**2:.4f})")
# θ=0     -> 1.0000 (product, pure ρ_A)
# θ=π/4   -> 0.5000 (Bell, maximally mixed) -- minimum purity = max entanglement

# No-signaling: any local unitary on B leaves ρ_A untouched.
H = np.array([[1, 1], [1, -1]], dtype=complex) / np.sqrt(2)
S = np.array([[1, 0], [0, 1j]], dtype=complex)
U_B = S @ H                                   # an arbitrary unitary on B
op = np.kron(np.eye(2), U_B)                  # I_A ⊗ U_B
rho_AB_after = op @ rho_AB @ op.conj().T
rho_A_after = partial_trace_B(rho_AB_after)
print("ρ_A unchanged by Bob's operation?", np.allclose(rho_A_after, rho_A))  # True

# Optional cross-check on Braket's LocalSimulator (free): density-matrix simulator
# reproduces ρ_A = I/2 for the Bell state. Deep Braket intro is Term 2 — basic pattern only.
try:
    from braket.circuits import Circuit
    from braket.devices import LocalSimulator

    bell = Circuit().h(0).cnot(0, 1)          # prepares (|00⟩+|11⟩)/√2, big-endian
    dm_sim = LocalSimulator("braket_dm")      # density-matrix local simulator
    full_rho = dm_sim.run(bell.density_matrix(), shots=0).result().values[0]
    rho_A_braket = partial_trace_B(np.asarray(full_rho))
    print("Braket ρ_A = I/2 ?", np.allclose(rho_A_braket, np.eye(2) / 2))
except Exception as e:                        # SDK not installed / offline — NumPy result already proved it
    print("Braket optional check skipped:", type(e).__name__)

Why einsum? Indexing $\rho_{AB}$ as the 4-tensor $\rho[i,\mu,j,\nu]$ makes the index definition literal: 'imjm->ij' says "fix free indices $i,j$ on $A$, sum the repeated $B$ index $\mu$." For $n$ qubits, reshape to $2n$ axes and contract the traced-out subset — the same idea scales (with the usual exponential memory cost of dense simulation, 0.1.7).

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Exercises

E1 (easy). Compute $\rho_B = \operatorname{Tr}_A(|\Phi^+\rangle\langle\Phi^+|)$ and confirm it equals $\rho_A = \tfrac12 I$. What does the equality of eigenvalues say in general?

Solution

By the symmetric calculation (trace out $A$ instead, using $\langle 1|0\rangle_A = 0$ to kill the coherences), $\rho_B = \tfrac12(|0\rangle\langle0| + |1\rangle\langle1|) = \tfrac12 I = \rho_A$. In general $\rho_A$ and $\rho_B$ share the same nonzero eigenvalues (the squared Schmidt coefficients), so they have equal purity and equal entropy — the entanglement is a symmetric property of the pair.

E2 (easy). For the product state $|\Psi\rangle = |0\rangle_A\otimes|+\rangle_B$, compute $\rho_A$ and $\rho_B$ and state their purities.

Solution

$\rho_{AB} = |0\rangle\langle0|\otimes|+\rangle\langle+|$ is a product operator, so $\rho_A = |0\rangle\langle0|,\operatorname{Tr}(|+\rangle\langle+|) = |0\rangle\langle0|$and$\rho_B = |+\rangle\langle+|,\operatorname{Tr}(|0\rangle\langle0|) = |+\rangle\langle+|$. Both pure: purity$= 1$. A product state has pure reduced states on both sides (Schmidt rank 1).

E3 (medium). Show that $\operatorname{Tr}_B$ followed by $\operatorname{Tr}_A$ equals the full trace: $\operatorname{Tr}_A(\operatorname{Tr}_B(\rho_{AB})) = \operatorname{Tr}_{AB}(\rho_{AB})$.

Solution

On a product operator $|a\rangle\langle a'|\otimes|b\rangle\langle b'|$: $\operatorname{Tr}_B(\cdot) = \langle b'|b\rangle\,|a\rangle\langle a'|$, then $\operatorname{Tr}A(\cdot) = \langle b'|b\rangle\langle a'|a\rangle$. Meanwhile$\operatorname{Tr}{AB}(|a\rangle\langle a'|\otimes |b\rangle\langle b'|) = \operatorname{Tr}(|a\rangle\langle a'|)\operatorname{Tr}(|b\rangle\langle b'|) = \langle a'|a\rangle\langle b'|b\rangle$— the same number. Linearity extends to all$\rho_{AB}$. ∎ (In particular $\operatorname{Tr}\rho_A = \operatorname{Tr}\rho_{AB} = 1$, reconfirming $\rho_A$ is unit-trace.)

E4 (medium). Let $\rho_{AB} = \tfrac12|00\rangle\langle00| + \tfrac12|11\rangle\langle11|$ — the classically correlated (separable, mixed) cousin of the Bell state. Compute $\rho_A$. Compare with the Bell-state reduced state and explain why they coincide despite very different global states.

Solution

$\operatorname{Tr}_B(|00\rangle\langle00|) = |0\rangle\langle0|$ and $\operatorname{Tr}_B(|11\rangle\langle11|) = |1\rangle\langle1|$(no cross terms exist here), so$\rho_A = \tfrac12(|0\rangle\langle0| + |1\rangle\langle1|) = \tfrac12 I$— identical to the Bell-state$\rho_A$. The reduced state cannot tell a quantum-entangled global state from a classically correlated one: both look maximally mixed locally. The distinction lives in correlations and only shows up in joint measurements (e.g. CHSH, Term 1.4) — the local marginal throws that information away. This is exactly the ensemble ambiguity of Lesson 1, viewed through the partial trace.

E5 (hard). Prove that $\operatorname{Tr}_B(\rho_{AB})$ is independent of the orthonormal basis $\{|\mu\rangle_B\}$ used in the index definition $\sum_\mu\langle i\mu|\rho_{AB}|j\mu\rangle$.

Solution

Let $\{|\mu'\rangle_B\}$ be another ONB, related by a unitary $U$ on $B$: $|\mu'\rangle = \sum_\nu U_{\nu\mu}|\nu\rangle$. Then$\sum_\mu\langle i\mu'|\rho_{AB}|j\mu'\rangle = \sum_{\mu,\nu,\sigma} \overline{U_{\nu\mu}}U_{\sigma\mu}\langle i\nu|\rho_{AB}|j\sigma\rangle = \sum_{\nu,\sigma}\big(\sum_\mu U_{\sigma\mu}\overline{U_{\nu\mu}}\big)\langle i\nu|\rho_{AB}|j\sigma\rangle$. The inner sum is $(UU^\dagger)_{\sigma\nu} = \delta_{\sigma\nu}$, collapsing to $\sum_\nu\langle i\nu|\rho_{AB}|j\nu\rangle$ — the original expression. So $\operatorname{Tr}_B$ is basis-independent (as it must be, being defined operationally without reference to a basis). ∎

E6 (hard). Use the Schmidt decomposition to prove that any purification of a given $\rho_A$ is related to the canonical one $\sum_k\sqrt{\lambda_k}|u_k\rangle_A|w_k\rangle_B$ by an isometry on the purifying system $B$ (assuming $\dim\mathcal H_B$ is large enough). Conclude purifications are unique up to such an isometry.

Solution

Let $|\Psi\rangle = \sum_k\sqrt{\lambda_k}|u_k\rangle_A|w_k\rangle_B$ and $|\Psi'\rangle = \sum_k\sqrt{\lambda_k}|u_k\rangle_A|w_k'\rangle_{B'}$both purify$\rho_A = \sum_k\lambda_k|u_k\rangle\langle u_k|$ (same $A$-Schmidt vectors and coefficients, since these are fixed by $\rho_A$'s spectral decomposition). Define $W:\mathcal H_B\to\mathcal H_{B'}$ on the relevant subspace by $W|w_k\rangle = |w_k'\rangle$; since both${|w_k\rangle}$and${|w_k'\rangle}$are orthonormal,$W$ is an isometry ($W^\dagger W = I$ on its support). Then $(I_A\otimes W)|\Psi\rangle = \sum_k\sqrt{\lambda_k}|u_k\rangle W|w_k\rangle = |\Psi'\rangle$. Hence any two purifications differ by an isometry on$B$ (a unitary when $\dim\mathcal H_B = \dim\mathcal H_{B'}$). ∎ This is the operator-level statement of "the freedom of purification," and it is the source of Lesson 1's ensemble ambiguity (HJW): different bases $\{|w_k\rangle\}$ for the reference correspond to different ensembles realizing the same $\rho_A$.

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Checkpoint

1. Define the partial trace on product operators and in index form. Which $B$-index gets summed?
2. What property forces the partial trace to be exactly this map (and not some other "discard $B$")?
3. State and justify in one line: the reduced state of an entangled pure state is mixed.
4. Compute $\rho_A$ for $|\Phi^+\rangle$ and give its purity. What does global-pure / local-mixed mean?
5. State purification and no-signaling, and say what each is "for."

Answers

1. $\operatorname{Tr}B(|a\rangle\langle a'|\otimes|b\rangle\langle b'|) = \langle b'|b\rangle, |a\rangle\langle a'|$; in indices$(\rho_A){ij} = \sum_\mu\langle i\mu|\rho_{AB}|j\mu\rangle$ — the repeated $B$ index $\mu$ is summed (set $\nu = \mu$).
2. It is the unique map with $\operatorname{Tr}A(\rho_A M_A) = \operatorname{Tr}{AB}(\rho_{AB}, M_A\otimes I_B)$for all$M_A$, i.e. it reproduces all local measurement statistics.
3. Schmidt: $\rho_A = \sum_k\lambda_k|u_k\rangle\langle u_k|$ with eigenvalues $=$ squared Schmidt coefficients; entangled means $\ge2$ nonzero $\lambda_k$, so $\operatorname{Tr}\rho_A^2 < 1$ — mixed.
4. $\rho_A = \tfrac12 I$, purity $\tfrac12$ (maximally mixed). Global-pure/local-mixed: you know the whole pair perfectly yet know nothing about either qubit alone — the information is in the correlations.
5. Purification: every mixed $\rho_A$ is $\operatorname{Tr}_B$ of a pure $|\Psi\rangle_{AB}$ — for treating mixedness as entanglement with a reference/environment (channels, Term 4). No-signaling: local operations on $B$ leave $\rho_A$ fixed — for ruling out faster-than-light signaling via entanglement.

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Further Reading

• [NC] Nielsen & Chuang, §2.4.3 (reduced density operator & partial trace), §2.5 (Schmidt decomposition & purification).
• [Pre] Preskill, Ph219, Ch. 2 — partial trace, the "church of the larger Hilbert space," and no-signaling.
• [Wat] Watrous, Theory of Quantum Information, Ch. 2 — partial trace as a linear map and its characterization.
• [Wil] Wilde, Quantum Information Theory, Ch. 5 — purification and the reduced-state formalism.
• Appendix E §4–5 — trace and Kronecker identities used in the proofs.

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