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Derivation

Reduced States and the Partial Trace

D-403 Home PU-404 Threads chance · matter Depends on born-rule, tensor-product-of-vector-spaces
Statement

Given a joint state \(\rho_{AB}\) on a tensor-product Hilbert space \(\mathcal{H}_A \otimes \mathcal{H}_B\), there exists a unique operator \(\rho_A\) on \(\mathcal{H}_A\) such that \(\operatorname{Tr}_A[M_A\,\rho_A] = \operatorname{Tr}_{AB}\!\big[(M_A \otimes I_B)\,\rho_{AB}\big]\) for every local observable \(M_A\). That operator is the partial trace \(\rho_A = \operatorname{Tr}_B[\rho_{AB}]\), and it reproduces the Born-rule statistics of every measurement performed on subsystem \(A\) alone.

Why it matters

Once two systems are entangled, no wavefunction of \(A\) by itself exists — yet an experimenter with access only to \(A\) must still get definite probabilities for every measurement they can do. The partial trace is exactly the object that packages those local probabilities. It is the bridge between the global pure-state description and the local, generally mixed, description that any real, spatially bounded laboratory actually sees.

It is also the engine of decoherence and of entanglement entropy: tracing out an unobserved environment turns coherent superpositions into apparent classical mixtures, and the entropy of \(\rho_A\) measures how entangled the two halves are. Almost every statement about open quantum systems begins here.

Assumptions
The total Hilbert space factorises as \(\mathcal{H}_A \otimes \mathcal{H}_B\).Without a clean tensor factorisation there is no well-defined "subsystem \(A\)" to reduce to, and the partial trace is undefined (see fermionic modes and superselection below). The state \(\rho_{AB}\) is a valid density operator: Hermitian, positive, unit trace.If positivity fails, \(\rho_A\) can have negative eigenvalues and no longer yields a probability distribution for any local measurement. Observables of interest are local, of the form \(M_A \otimes I_B\).If one asks about joint observables \(M_A \otimes N_B\), the single-system \(\rho_A\) is insufficient — the correlations it discards are exactly what such observables probe. Operators are trace-class (automatic in finite dimension).In infinite dimensions the traces need not converge; the reduction and its uniqueness argument require trace-class \(\rho_{AB}\).
Derivation
1
\[ \langle M_A \otimes I_B \rangle \;=\; \operatorname{Tr}_{AB}\!\big[(M_A \otimes I_B)\,\rho_{AB}\big] \]
Born rule (prior result) for the expectation of the local observable \(M_A \otimes I_B\) in the joint state \(\rho_{AB}\). This is the quantity a lab restricted to \(A\) can measure. A
2
\[ \text{Require } \rho_A \text{ on } \mathcal{H}_A \text{ with } \operatorname{Tr}_A[M_A\,\rho_A] = \operatorname{Tr}_{AB}\!\big[(M_A \otimes I_B)\,\rho_{AB}\big]\ \ \forall\, M_A \]
We demand a self-contained local description: a single operator whose ordinary expectation values on \(A\) match the joint statistics for all local observables. The derivation must (i) exhibit such a \(\rho_A\) and (ii) show it is unique. A
3
\[ \operatorname{Tr}_{AB}[X] = \sum_{i}\sum_{j}\big(\langle i|_A \otimes \langle j|_B\big)\,X\,\big(|i\rangle_A \otimes |j\rangle_B\big) \]
Expand the joint trace in a product orthonormal basis \(\{|i\rangle_A\}\), \(\{|j\rangle_B\}\) of \(\mathcal{H}_A \otimes \mathcal{H}_B\). The trace is basis-independent, so any product basis will do. A
4
\[ \operatorname{Tr}_{AB}\!\big[(M_A \otimes I_B)\rho_{AB}\big] = \sum_{i,j}\sum_{i'} \langle i|M_A|i'\rangle\,\langle i',j|\,\rho_{AB}\,|i,j\rangle \]
Insert the resolution \(\sum_{i'}|i'\rangle\langle i'|_A\) after \(M_A\); since \(M_A\) acts only on \(A\) and \(I_B\) leaves the \(B\) label untouched, the \(j\) index passes straight through to \(\rho_{AB}\). B
5
\[ \langle i'|\,\rho_A\,|i\rangle \;:=\; \sum_{j} \langle i',j|\,\rho_{AB}\,|i,j\rangle \;=\; \sum_j \big(I_A \otimes \langle j|_B\big)\,\rho_{AB}\,\big(I_A \otimes |j\rangle_B\big) \]
Define \(\rho_A\) by contracting only the \(B\) indices of \(\rho_{AB}\). This is precisely the partial trace \(\rho_A = \operatorname{Tr}_B[\rho_{AB}]\); on product operators it acts as \(\operatorname{Tr}_B\big[\,|a\rangle\langle a'|\otimes|b\rangle\langle b'|\,\big]=\langle b'|b\rangle\,|a\rangle\langle a'|\). B
6
\[ \sum_{i,i'}\langle i|M_A|i'\rangle\,\langle i'|\rho_A|i\rangle = \sum_i \langle i|\,M_A\,\rho_A\,|i\rangle = \operatorname{Tr}_A[M_A\,\rho_A] \]
Substituting the definition of Step 5 into Step 4 collapses the double \(A\)-sum into a single trace over \(A\). Thus \(\rho_A=\operatorname{Tr}_B[\rho_{AB}]\) satisfies the requirement of Step 2 for every \(M_A\): existence is proved. B
7
\[ \operatorname{Tr}_A\!\big[M_A(\rho_A-\rho_A')\big]=0\ \ \forall M_A \;\Longrightarrow\; \big\| \rho_A-\rho_A' \big\|_{\mathrm{HS}}^2 = 0 \]
Uniqueness: suppose \(\rho_A'\) also works. Their difference \(\Delta=\rho_A-\rho_A'\) obeys \(\operatorname{Tr}[M_A\Delta]=0\) for all \(M_A\); choosing \(M_A=\Delta^{\dagger}\) gives \(\operatorname{Tr}[\Delta^{\dagger}\Delta]=0\). The Hilbert–Schmidt inner product \(\langle X,Y\rangle=\operatorname{Tr}[X^{\dagger}Y]\) is positive-definite, so \(\Delta=0\). The partial trace is the unique such map. C
8
\[ \operatorname{Tr}_A[\rho_A] = \operatorname{Tr}_{AB}[\rho_{AB}] = 1,\qquad \rho_A = \rho_A^{\dagger} \ge 0 \]
Setting \(M_A=I_A\) in Step 6 gives unit trace. Hermiticity is manifest from Step 5, and positivity follows because \(\langle\phi|\rho_A|\phi\rangle=\sum_j\langle\phi,j|\rho_{AB}|\phi,j\rangle\ge0\) for any \(|\phi\rangle_A\), since each term is a diagonal matrix element of the positive operator \(\rho_{AB}\). So \(\rho_A\) is itself a legitimate density operator. C
Result
\[ \rho_A = \operatorname{Tr}_B[\rho_{AB}], \qquad \langle i'|\rho_A|i\rangle = \sum_j \big(\langle i'|\otimes\langle j|\big)\,\rho_{AB}\,\big(|i\rangle\otimes|j\rangle\big) \]

Reading. The reduced density matrix is obtained by "summing over" — tracing out — everything belonging to \(B\). It is the one and only operator on \(\mathcal{H}_A\) that reproduces the Born-rule statistics of every local observable \(M_A\otimes I_B\). For an entangled \(\rho_{AB}\) it is generically a mixed state even when \(\rho_{AB}\) is pure: entanglement, viewed locally, looks like classical uncertainty.

Units check. Density operators and the trace are dimensionless (\(\operatorname{Tr}\rho=1\), a pure number). Contracting the \(B\) indices removes the \(\dim\mathcal{H}_B\) factors of the joint operator, leaving an operator on \(\mathcal{H}_A\) with \(\operatorname{Tr}_A\rho_A=1\) — dimensionally and normalisation-wise a valid state. Both sides of \(\operatorname{Tr}_A[M_A\rho_A]=\operatorname{Tr}_{AB}[(M_A\otimes I_B)\rho_{AB}]\) carry the units of \(M_A\).

Limiting cases
  • Product state \(\rho_{AB}=\rho_A\otimes\rho_B\): \(\operatorname{Tr}_B[\rho_A\otimes\rho_B]=\rho_A\operatorname{Tr}[\rho_B]=\rho_A\). No information is lost by reduction.
  • Pure product \(|\psi\rangle_A\otimes|\chi\rangle_B\): \(\rho_A=|\psi\rangle\langle\psi|\) stays pure — zero entanglement.
  • Maximally entangled (Bell) state: \(\rho_A=I_A/d\), maximally mixed; local measurements are completely random.
  • Trivial \(B\) (\(\dim\mathcal{H}_B=1\)): \(\rho_A=\rho_{AB}\); the partial trace acts as the identity.
  • Schmidt form \(|\psi\rangle=\sum_k\sqrt{p_k}\,|a_k\rangle|b_k\rangle\): \(\rho_A=\sum_k p_k|a_k\rangle\langle a_k|\), a mixture with weights equal to the squared Schmidt coefficients.
Breaks when
  • Non-local observables. \(\rho_A\) reproduces \(\langle M_A\otimes I_B\rangle\) but says nothing about correlators \(\langle M_A\otimes N_B\rangle\). Two very different joint states can share the same \(\rho_A\); the partial trace deliberately throws away all cross-system correlation.
  • No tensor factorisation. Identical particles (symmetrised/antisymmetrised states), fermionic modes, or systems with superselection rules do not split as \(\mathcal{H}_A\otimes\mathcal{H}_B\). Without a chosen factorisation "\(\operatorname{Tr}_B\)" is ambiguous or ill-defined.
  • Non-trace-class operators. In infinite-dimensional \(\mathcal{H}_B\) (e.g. a field mode) the sum \(\sum_j\langle j|\rho_{AB}|j\rangle\) may diverge; the construction and its uniqueness proof presuppose a convergent trace.
Failure modes
  • "Pure stays pure." Assuming \(\rho_{AB}\) pure forces \(\rho_A\) pure — false: entangled pure states have mixed reductions.
  • Full trace instead of partial. Writing \(\operatorname{Tr}[\rho_{AB}]=1\) (a number) where an operator \(\operatorname{Tr}_B[\rho_{AB}]\) on \(\mathcal{H}_A\) is required.
  • Contracting the wrong index. Sandwiching \(\langle j|_B\) but letting it act on the \(A\) factor, or tracing out \(A\) when \(B\) was intended.
  • Basis anxiety. Believing \(\rho_A\) depends on the chosen \(\{|j\rangle_B\}\); it does not — cyclicity makes the partial trace basis-independent.
  • Correlation amnesia in reverse. Treating \(\rho_A\) as the whole story and trying to predict \(A\)–\(B\) correlations from it alone.
  • Improper normalisation. Forgetting the cross terms \(\langle j'|j\rangle=\delta_{j'j}\) and mis-summing off-diagonal \(B\) elements.
Discussion

The deepest content of this result is operational: it is the statement that the local physics of \(A\) is completely encoded in one Hermitian operator. Any two global states indistinguishable by local measurements on \(A\) are mapped to the same \(\rho_A\); conversely, everything a local observer can ever learn is a functional of \(\rho_A\). This is why the reduced density matrix, not the global wavefunction, is the correct notion of "the state of a subsystem".

Entanglement makes reduction irreversible. For a pure entangled \(|\psi\rangle_{AB}\), the reduced state is mixed and its von Neumann entropy \(S(\rho_A)=-\operatorname{Tr}[\rho_A\log\rho_A]\) — the entanglement entropy — quantifies the correlation with \(B\). It equals \(S(\rho_B)\) by the symmetry of the Schmidt decomposition, and vanishes exactly when the state factorises. Decoherence is this mechanism in motion: coupling to an unobserved environment \(B\) and tracing it out suppresses the off-diagonal (coherence) elements of \(\rho_A\), producing the appearance of a classical statistical mixture.

A subtler consequence is no-signalling. Any operation Bob performs on \(B\) — a unitary, or even a measurement whose outcome is not communicated — leaves \(\rho_A=\operatorname{Tr}_B[\rho_{AB}]\) unchanged, because the partial trace is invariant under \(I_A\otimes U_B\) conjugation and under coarse-graining over measurement outcomes. Hence no local statistic on \(A\) can reveal what was done at \(B\), and entanglement cannot transmit information faster than light. The partial trace is thus not a computational convenience but the formal guarantor of relativistic causality in quantum theory. It also sits inside the general framework of completely positive trace-preserving maps: \(\operatorname{Tr}_B\) is a CPTP channel, and every such channel arises as a unitary on a larger space followed by a partial trace (Stinespring dilation), which is the reverse operation to purification.

Common misconceptions. A "proper" mixture (a classical ensemble of pure states Alice actually prepared) and an "improper" mixture (the reduced state of an entangled pair) are described by the same \(\rho_A\) and are locally indistinguishable — but they are not the same physical situation, since the improper mixture retains correlations with \(B\) that no experiment on \(A\) alone can access. Reading \(\rho_A\) as "\(A\) is secretly in one definite pure state we just don't know" is the classic error.

Worked examples

Example 1 — reduced state of a Bell pair.

1
\[ |\Phi^{+}\rangle=\tfrac{1}{\sqrt{2}}\big(|00\rangle+|11\rangle\big),\qquad \rho_{AB}=|\Phi^{+}\rangle\langle\Phi^{+}| \]
Write the joint pure state and its density operator. A
2
\[ \rho_{AB}=\tfrac{1}{2}\big(|00\rangle\langle00|+|00\rangle\langle11|+|11\rangle\langle00|+|11\rangle\langle11|\big) \]
Expand the outer product term by term. A
3
\[ \operatorname{Tr}_B\big[|ij\rangle\langle kl|\big]=\langle l|j\rangle\,|i\rangle\langle k| \;\Rightarrow\; \operatorname{Tr}_B|00\rangle\langle11|=\langle1|0\rangle\,|0\rangle\langle1|=0 \]
Apply the partial-trace rule; the off-diagonal (coherence) terms carry \(\langle1|0\rangle=0\) and vanish. B
4
\[ \rho_A=\tfrac{1}{2}\big(|0\rangle\langle0|+|1\rangle\langle1|\big)=\tfrac{1}{2}I_2 \]
Only the diagonal terms survive, giving the maximally mixed qubit. B
\[ \rho_A=\begin{pmatrix}\tfrac12&0\\[2pt]0&\tfrac12\end{pmatrix},\quad \operatorname{Tr}\rho_A^2=\tfrac12,\quad S(\rho_A)=\log_2 2 = 1\ \text{bit} \]

Reading. Locally, Alice's qubit is completely random: every measurement direction gives \(50/50\). The purity \(\tfrac12<1\) and the one bit of entropy signal maximal entanglement, even though the global state is perfectly pure.

Units check. Dimensionless; \(\operatorname{Tr}\rho_A=\tfrac12+\tfrac12=1\), and \(0\le\operatorname{Tr}\rho_A^2\le1\) as required for a density matrix.

Example 2 — a partially entangled state and a cross-check.

1
\[ |\psi\rangle=\cos\theta\,|00\rangle+\sin\theta\,|11\rangle,\qquad \theta=\tfrac{\pi}{6}\ \ (\cos\theta=\tfrac{\sqrt3}{2},\ \sin\theta=\tfrac12) \]
A Schmidt-form state with unequal weights; \(\langle\psi|\psi\rangle=\cos^2\theta+\sin^2\theta=1\). A
2
\[ \rho_A=\operatorname{Tr}_B|\psi\rangle\langle\psi|=\cos^2\theta\,|0\rangle\langle0|+\sin^2\theta\,|1\rangle\langle1| \]
Cross terms \(\propto\langle0|1\rangle_B=0\) drop out, leaving the diagonal Schmidt mixture. B
3
\[ \rho_A=\tfrac34\,|0\rangle\langle0|+\tfrac14\,|1\rangle\langle1|,\qquad \operatorname{Tr}\rho_A^2=\tfrac{9}{16}+\tfrac{1}{16}=\tfrac58 \]
Insert numbers: eigenvalues \(\tfrac34,\tfrac14\); purity between \(\tfrac12\) and \(1\), so partial entanglement. A
4
\[ \langle\sigma_z\otimes I\rangle \overset{\rho_A}{=}\operatorname{Tr}[\sigma_z\rho_A]=\tfrac34-\tfrac14=\tfrac12,\qquad \langle\psi|\sigma_z\otimes I|\psi\rangle=\cos2\theta=\cos\tfrac{\pi}{3}=\tfrac12 \]
Cross-check: the reduced state reproduces the local expectation computed directly from the joint state — the defining property in action. B
\[ \rho_A=\begin{pmatrix}\tfrac34&0\\[2pt]0&\tfrac14\end{pmatrix},\quad S(\rho_A)=-\tfrac34\log_2\tfrac34-\tfrac14\log_2\tfrac14\approx0.811\ \text{bit} \]

Reading. Weaker entanglement than the Bell pair: purity \(\tfrac58\) and \(0.811\) bits of entropy, both strictly between the product-state and maximally-entangled limits. The local expectation \(\langle\sigma_z\rangle=\tfrac12\) matches the direct joint computation, confirming \(\rho_A\) carries all local statistics.

Units check. Dimensionless; \(\operatorname{Tr}\rho_A=\tfrac34+\tfrac14=1\); \(S\) in bits (\(\log_2\)), bounded by \(0\le S\le\log_2 2=1\).

Problems
  1. Compute the reduced state of Alice's qubit for the singlet \(|\Psi^{-}\rangle=\tfrac{1}{\sqrt2}(|01\rangle-|10\rangle)\).
    Solution \(\rho_{AB}=\tfrac12(|01\rangle-|10\rangle)(\langle01|-\langle10|)\). Tracing out \(B\): \(\operatorname{Tr}_B|01\rangle\langle01|=|0\rangle\langle0|\), \(\operatorname{Tr}_B|10\rangle\langle10|=|1\rangle\langle1|\), and the cross terms \(\operatorname{Tr}_B|01\rangle\langle10|=\langle0|1\rangle|0\rangle\langle1|=0\). Hence \(\rho_A=\tfrac12(|0\rangle\langle0|+|1\rangle\langle1|)=\tfrac12 I\). Maximally mixed, exactly as for \(|\Phi^{+}\rangle\): all four Bell states have the same maximally-mixed reductions.
  2. For the mixed joint state \(\rho_{AB}=p\,|\Phi^{+}\rangle\langle\Phi^{+}|+(1-p)\,|00\rangle\langle00|\), find \(\rho_A\) as a function of \(p\).
    Solution By linearity of the partial trace, \(\rho_A=p\,\operatorname{Tr}_B|\Phi^{+}\rangle\langle\Phi^{+}|+(1-p)\,\operatorname{Tr}_B|00\rangle\langle00|\). From Example 1 the first term gives \(p\cdot\tfrac12 I=\tfrac{p}{2}(|0\rangle\langle0|+|1\rangle\langle1|)\); the second gives \((1-p)\,|0\rangle\langle0|\). Adding, \(\rho_A=\big(1-\tfrac{p}{2}\big)|0\rangle\langle0|+\tfrac{p}{2}\,|1\rangle\langle1|\). Trace \(=1-\tfrac{p}{2}+\tfrac{p}{2}=1\). At \(p=0\), \(\rho_A=|0\rangle\langle0|\) (pure); at \(p=1\), \(\rho_A=\tfrac12 I\) (maximally mixed).
  3. Show directly from the definition that \(\operatorname{Tr}_B[\rho_A\otimes\rho_B]=\rho_A\) for any \(\rho_B\) with \(\operatorname{Tr}\rho_B=1\).
    Solution Using an orthonormal basis \(\{|j\rangle_B\}\): \(\operatorname{Tr}_B[\rho_A\otimes\rho_B]=\sum_j(I_A\otimes\langle j|)(\rho_A\otimes\rho_B)(I_A\otimes|j\rangle)=\rho_A\sum_j\langle j|\rho_B|j\rangle=\rho_A\,\operatorname{Tr}[\rho_B]=\rho_A\). The \(A\)-factor \(\rho_A\) passes through untouched because \(\langle j|_B\) acts only on the \(B\)-factor, and \(\sum_j\langle j|\rho_B|j\rangle\) is just the definition of \(\operatorname{Tr}\rho_B=1\).
  4. For the GHZ state \(|\mathrm{GHZ}\rangle=\tfrac{1}{\sqrt2}(|000\rangle+|111\rangle)\) on qubits \(A,B,C\), find the single-qubit reduced state \(\rho_A=\operatorname{Tr}_{BC}|\mathrm{GHZ}\rangle\langle\mathrm{GHZ}|\).
    Solution \(\rho_{ABC}=\tfrac12(|000\rangle+|111\rangle)(\langle000|+\langle111|)\). Tracing out \(C\) first: \(\operatorname{Tr}_C|000\rangle\langle111|=\langle1|0\rangle_C|00\rangle\langle11|=0\), so the coherences die and \(\operatorname{Tr}_C\rho_{ABC}=\tfrac12(|00\rangle\langle00|+|11\rangle\langle11|)\). Tracing out \(B\): \(\operatorname{Tr}_B|00\rangle\langle00|=|0\rangle\langle0|\), \(\operatorname{Tr}_B|11\rangle\langle11|=|1\rangle\langle1|\), giving \(\rho_A=\tfrac12(|0\rangle\langle0|+|1\rangle\langle1|)=\tfrac12 I\). Each qubit of a GHZ state looks maximally mixed on its own.
  5. Prove the no-signalling property: for any unitary \(U\) on \(B\), the reduced state \(\rho_A'=\operatorname{Tr}_B[(I_A\otimes U)\rho_{AB}(I_A\otimes U^{\dagger})]\) equals \(\rho_A=\operatorname{Tr}_B[\rho_{AB}]\).
    Solution Write the \(B\)-trace in the basis \(\{|j\rangle_B\}\): \(\rho_A'=\sum_j(I_A\otimes\langle j|)(I_A\otimes U)\rho_{AB}(I_A\otimes U^{\dagger})(I_A\otimes|j\rangle)=\sum_j(I_A\otimes\langle j|U)\rho_{AB}(I_A\otimes U^{\dagger}|j\rangle)\). Since \(U\) is unitary, \(\{U^{\dagger}|j\rangle\}\equiv\{|k\rangle\}\) is another orthonormal basis of \(\mathcal{H}_B\), and \(\langle j|U=\langle k|\) with the sum over \(j\) equal to the sum over \(k\). Therefore \(\rho_A'=\sum_k(I_A\otimes\langle k|)\rho_{AB}(I_A\otimes|k\rangle)=\operatorname{Tr}_B[\rho_{AB}]=\rho_A\). The partial trace is invariant under any local operation on \(B\), so nothing Bob does to \(B\) can change Alice's local statistics — the formal statement of no-signalling.