No-Broadcasting Theorem

The no-broadcasting theorem gives an exact operational characterization of commutativity. In finite-dimensional quantum mechanics, one physical channel can place every state in a promised family into two output systems with both marginals unchanged if and only if all states in that family commute pairwise.

Broadcasting is weaker than cloning: the two correct marginals may belong to a correlated joint state. The theorem therefore reaches mixed-state families that the usual pure-state no-cloning argument does not describe.

Required background. Density Operators supplies the state space; Partial Trace supplies the marginal operation; and Quantum Operations supplies completely positive trace-preserving maps and Kraus operators.

Helpful background. Entangled States clarifies why a joint output need not be a product. The pure-state no-cloning boundary is recovered below without being assumed.

Broadcasting a family of states

Let $\mathcal H$ be finite dimensional, and identify each output space $\mathcal H_A$ and $\mathcal H_B$ with $\mathcal H$. A channel

$$\mathcal B: \mathcal L(\mathcal H) \longrightarrow \mathcal L(\mathcal H_A\otimes\mathcal H_B)$$

broadcasts a family $\mathsf S=\{\rho_i\}$ when, for every $i$,

$$\tau_i^{AB}=\mathcal B(\rho_i)$$

has marginals

$$\operatorname{Tr}_B\tau_i^{AB}=\rho_i, \qquad \operatorname{Tr}_A\tau_i^{AB}=\rho_i.$$

The same channel must work without being told which member of the family was supplied.

No-broadcasting theorem. A family of density operators on a finite-dimensional Hilbert space is exactly broadcastable by one CPTP map if and only if

$$[\rho_i,\rho_j]=0$$

for every pair $i,j$.

Pairwise commuting finite-dimensional Hermitian matrices are simultaneously diagonalizable. The theorem can therefore be read as follows: exactly the state families encoded in one common classical basis can be broadcast.

Constructing the channel for commuting states

Suppose the family commutes. Choose a common orthonormal eigenbasis $\{|k\rangle\}_{k=1}^d$ so that

$$\rho_i = \sum_{k=1}^d p_i(k)|k\rangle\langle k|.$$

Define Kraus operators from the input to the two outputs by

$$K_k = |k\rangle_A|k\rangle_B\langle k|.$$

They satisfy

$$\sum_k K_k^\dagger K_k = \sum_k |k\rangle\langle k| = I,$$

so

$$\mathcal B(X) = \sum_k K_kXK_k^\dagger$$

is completely positive and trace preserving. On every promised input,

$$\mathcal B(\rho_i) = \sum_k p_i(k) |kk\rangle\langle kk|.$$

Tracing out either output gives $\rho_i$. Operationally, this channel dephases in the common eigenbasis and writes the surviving classical label into both systems. Dephasing does not alter any state in the promised commuting family.

The output is generally not two independent copies:

$$\sum_k p_i(k)|kk\rangle\langle kk| \ne \rho_i\otimes\rho_i.$$

Its two records are classically correlated. That distinction is why broadcasting can succeed for overlapping commuting mixed states even when cloning cannot.

Why noncommuting states cannot be broadcast

The difficult direction is necessity. A useful proof follows the fidelity argument of Barnum, Caves, Fuchs, Jozsa, and Schumacher. Define the root fidelity

$$f(\rho,\sigma) = \left\|\sqrt\rho\sqrt\sigma\right\|_1 = \operatorname{Tr} \sqrt{\sqrt\rho\,\sigma\sqrt\rho}.$$

This convention is the square root of the quantity that some texts call fidelity. It is monotone under every CPTP map:

$$f(\Phi(\rho),\Phi(\sigma)) \geq f(\rho,\sigma).$$

A channel cannot make two inputs more distinguishable. Partial trace is a CPTP map and obeys the same inequality.

Assume one channel broadcasts two states $\rho$ and $\sigma$, with outputs $\tau_\rho$ and $\tau_\sigma$. Channel monotonicity and the $A$ marginal give

$$f(\rho,\sigma) \leq f(\tau_\rho,\tau_\sigma) \leq f(\operatorname{Tr}_B\tau_\rho, \operatorname{Tr}_B\tau_\sigma) = f(\rho,\sigma).$$

The $B$ marginal gives the same chain. Every inequality is therefore an equality:

$$f(\rho,\sigma) = f(\tau_\rho,\tau_\sigma) = f(\operatorname{Tr}_B\tau_\rho, \operatorname{Tr}_B\tau_\sigma) = f(\operatorname{Tr}_A\tau_\rho, \operatorname{Tr}_A\tau_\sigma).$$

This equality chain is necessary, but writing it down is not yet a proof of commutativity. A channel can preserve fidelity for a pair without broadcasting it. The nontrivial step is to impose the equality conditions for both output marginals.

The two-marginal equality step

First suppose $\rho$ and $\sigma$ are invertible. Fidelity also has the measurement representation

$$f(\rho,\sigma) = \min_{\{E_k\}} \sum_k \sqrt{\operatorname{Tr}(\rho E_k)} \sqrt{\operatorname{Tr}(\sigma E_k)},$$

where the minimum is over POVMs. An optimal projective POVM may be chosen in an eigenbasis of the positive operator

$$M = \sigma^{-1/2} \sqrt{\sigma^{1/2}\rho\sigma^{1/2}} \sigma^{-1/2},$$

which is the unique positive solution of

$$\rho=M\sigma M.$$

Let $E_k=|k\rangle\langle k|$ be its spectral projections, with $M|k\rangle=m_k|k\rangle$. Because broadcasting preserves the input probabilities on each marginal, the two lifted POVMs

$$\{E_k\otimes I\}, \qquad \{I\otimes E_k\}$$

have the same classical overlap as the input-optimal POVM. The fidelity equalities force both lifted POVMs to be optimal for the output pair.

Here is the equality lemma needed for the remaining implication. It is the step that the fidelity sandwich by itself does not provide.

Two-marginal fidelity lemma. Let $\alpha$ and $\beta$ be states on $A\otimes B$ whose marginals satisfy

$$\operatorname{Tr}_A\alpha=\operatorname{Tr}_B\alpha=\rho, \qquad \operatorname{Tr}_A\beta=\operatorname{Tr}_B\beta=\sigma.$$

Suppose $\{E_k\}$ is a projective POVM that is optimal for $\rho,\sigma$, and that both lifted POVMs $\{E_k\otimes I\}$ and $\{I\otimes E_k\}$ are optimal for $\alpha,\beta$. Then there are nonnegative numbers $g_k,h_k$ and polar unitaries $U,V$ such that

$$\begin{aligned} U\sqrt\alpha\,(E_k\otimes I) &= g_k\sqrt\beta\,(E_k\otimes I),\\ V\sqrt\alpha\,(I\otimes E_k) &= h_k\sqrt\beta\,(I\otimes E_k). \end{aligned}$$

To verify the lemma for either lifted projective POVM $\{P_k\}$, choose a unitary extension of the polar partial isometry such that

$$U\sqrt\alpha\sqrt\beta = \sqrt{\sqrt\beta\,\alpha\sqrt\beta}.$$

Then

$$\begin{aligned} f(\alpha,\beta) &= \sum_k \operatorname{Tr}(\sqrt\beta\,U\sqrt\alpha P_k)\\ &\le \sum_k \left| \operatorname{Tr} \left[ (\sqrt\beta P_k)^\dagger (U\sqrt\alpha P_k) \right] \right|\\ &\le \sum_k \|\sqrt\beta P_k\|_2 \|U\sqrt\alpha P_k\|_2\\ &= \sum_k \sqrt{\operatorname{Tr}(\beta P_k)} \sqrt{\operatorname{Tr}(\alpha P_k)}. \end{aligned}$$

For an optimal measurement the endpoints agree, so both intermediate inequalities are equalities. Because the marginals $\rho,\sigma$ are invertible, every nonzero $E_k$ has positive probability. Equality in each Hilbert–Schmidt Cauchy–Schwarz term therefore gives $U\sqrt\alpha P_k=c_k\sqrt\beta P_k$, with the common phase absorbed into $U$ and

$$c_k = \sqrt{ \frac{\operatorname{Tr}(\alpha P_k)} {\operatorname{Tr}(\beta P_k)} } \ge0.$$

Applying this argument to the two lifted PVMs gives the two displayed relations. Their polar unitaries may differ on the null spaces of the output states, which is why they were denoted $U$ and $V$.

Apply the lemma with $\alpha=\tau_\rho$, $\beta=\tau_\sigma$, and $E_k=|k\rangle\langle k|$. Define

$$G=\sum_k g_kE_k, \qquad H=\sum_k h_kE_k.$$

Summing the equality relations over $k$ gives

$$\begin{aligned} U\sqrt{\tau_\rho} &= \sqrt{\tau_\sigma}(G\otimes I),\\ V\sqrt{\tau_\rho} &= \sqrt{\tau_\sigma}(I\otimes H). \end{aligned}$$

Taking adjoint products and then the appropriate partial trace uses the broadcasting marginals to give

$$\rho=G\sigma G, \qquad \rho=H\sigma H.$$

The positive solution $X$ of $\rho=X\sigma X$ is unique: multiplying by $\sigma^{1/2}$ on both sides shows that

$$\sigma^{1/2}X\sigma^{1/2} = \sqrt{\sigma^{1/2}\rho\sigma^{1/2}}.$$

Hence $G=H=M$. Because $M$ is invertible, eliminating $\sqrt{\tau_\sigma}$ between the two summed relations yields, for the unitary $W=V^\dagger U$,

$$W\sqrt{\tau_\rho} = \sqrt{\tau_\rho} (M\otimes M^{-1}).$$

Act on the product eigenvector $|k\rangle_A|\ell\rangle_B$ and compare norms. The left side has norm $\|\sqrt{\tau_\rho}|k\ell\rangle\|$, while the right side has norm $(m_k/m_\ell)\|\sqrt{\tau_\rho}|k\ell\rangle\|$. Positivity of the eigenvalues therefore implies

$$\sqrt{\tau_\rho}|k\ell\rangle=0 \qquad \text{whenever }m_k\ne m_\ell.$$

The $M$-basis matrix elements of the $A$ marginal are

$$\langle k|\rho|k'\rangle = \sum_\ell \left\langle \sqrt{\tau_\rho}|k\ell\rangle, \sqrt{\tau_\rho}|k'\ell\rangle \right\rangle.$$

If $m_k\ne m_{k'}$, the two vectors in every summand cannot both be nonzero: the first requires $m_k=m_\ell$ and the second requires $m_{k'}=m_\ell$. Thus the matrix element vanishes and $[M,\rho]=0$. Finally,

$$\sigma=M^{-1}\rho M^{-1} = M^{-2}\rho = \rho M^{-2},$$

so $[\rho,\sigma]=0$. The argument has now used both marginal equalities, the equality condition itself, and the uniqueness of $M$ explicitly.

For states that are not invertible, let $K=\operatorname{supp}(\rho+\sigma)$ and restrict to $K$. This restriction also contains the output supports. If $P_K$ projects onto $K$ and a positive output $\tau$ has its $A$ marginal supported in $K$, then

$$\operatorname{Tr} \left[ ((I-P_K)\otimes I)\tau \right] =0.$$

Positivity forces $((I-P_K)\otimes I)\tau=0$ on the support of $\tau$; the other marginal gives the analogous statement for $B$. Hence every broadcast output is supported in $K\otimes K$.

Define the broadcast state

$$\omega = \frac{\rho+\sigma}{2}.$$

Linearity means the channel also broadcasts every convex combination of $\rho$ and $\sigma$. For $0<\varepsilon<1$, the two states

$$\rho_\varepsilon =(1-\varepsilon)\rho+\varepsilon\omega, \qquad \sigma_\varepsilon =(1-\varepsilon)\sigma+\varepsilon\omega$$

are invertible on $K$ and are broadcast by the same channel. The invertible argument gives $[\rho_\varepsilon,\sigma_\varepsilon]=0$, while direct expansion gives

$$[\rho_\varepsilon,\sigma_\varepsilon] =(1-\varepsilon)[\rho,\sigma].$$

Hence $[\rho,\sigma]=0$. Applying this pairwise to a broadcast family proves necessity for the whole family.

Broadcasting versus cloning

Cloning requires the stronger output condition

$$\mathcal C(\rho_i) = \rho_i\otimes\rho_i.$$

Fidelity is multiplicative on tensor products. If one channel cloned two states exactly, partial trace would recover each input from its output. The same two-sided monotonicity argument would therefore require

$$f(\rho,\sigma) = f(\rho\otimes\rho, \sigma\otimes\sigma) = f(\rho,\sigma)^2.$$

Thus $f(\rho,\sigma)$ must be $0$ or $1$: the states have orthogonal supports or are identical. Overlapping commuting mixed states can therefore be broadcast but cannot be cloned into product copies.

For a pure input, broadcasting already reduces to cloning. If a bipartite state has the pure marginal $|\psi\rangle\langle\psi|$, positivity forces the joint state to factor as

$$|\psi\rangle\langle\psi|\otimes\eta.$$

If the other marginal is the same pure state, then $\eta=|\psi\rangle\langle\psi|$. Rank-one projectors commute only when they represent the same ray or orthogonal rays, recovering the exact deterministic no-cloning theorem.

Qubit examples

Consider a commuting $z$-diagonal family

$$\rho_r = \frac{I+r\sigma_z}{2}, \qquad -1\leq r\leq1.$$

The Kraus channel with $K_0=|00\rangle\langle0|$ and $K_1=|11\rangle\langle1|$ broadcasts every member at once:

$$\mathcal B(\rho_r) = \frac{1+r}{2}|00\rangle\langle00| + \frac{1-r}{2}|11\rangle\langle11|.$$

Both marginals equal $\rho_r$.

By contrast, let

$$\rho = \frac{I+r\sigma_z}{2}, \qquad \sigma = \frac{I+s\sigma_x}{2}.$$

Their commutator is

$$[\rho,\sigma] = \frac{irs}{2}\sigma_y.$$

If $r$ and $s$ are both nonzero, no exact channel broadcasts the pair. This includes noisy, genuinely mixed examples; purity is not the obstruction.

Scope and edge cases

Singleton and identical families. A singleton commutes trivially and can be broadcast by a constant preparation channel. Repeated copies of the same density operator do not make a nontrivial family.

Orthogonal supports. Orthogonal states commute and can be perfectly identified, after which a channel may prepare two records. They can even be cloned exactly.

Degenerate spectra. Pairwise commutativity, not nondegeneracy, is the criterion. Any common eigenbasis within degenerate subspaces gives a valid constructive channel.

Correlated output. Broadcasting constrains only the two marginals. The joint output may contain classical or quantum correlations; it need not be a product state.

External classical label. If a device is told which state was prepared, it can prepare as many copies as desired. The theorem assumes that the same channel receives only the unknown quantum input.

Exact finite-dimensional setting. The theorem here is deterministic and exact, with one input copy and two faithful marginals. Approximate, probabilistic, asymmetric, many-input, and infinite-dimensional broadcasting require separate statements and error measures.

Common pitfalls

Calling the theorem universal no-copying. A known singleton or a commuting family is broadcastable. The obstruction is a promised family containing noncommuting alternatives.

Replacing broadcasting by cloning. Correct marginals need not be independent product copies. Commuting mixed states expose the difference.

Claiming that measurement and preparation always work. Measuring the common eigenbasis works for a commuting family because that dephasing leaves every promised state invariant. No single measurement has this property for a noncommuting family.

Stopping at fidelity preservation. Equality under the channel and one partial trace is not enough. Necessity uses the equality conditions for both marginals and the uniqueness of the positive operator $M$.

Extending the theorem to infinite dimension without hypotheses. Support, continuity, normality, and operator-algebraic issues must be controlled in an infinite-dimensional version.

Exercises

1. Verify the constructive channel

For $K_k=|kk\rangle\langle k|$, prove that the Kraus map is trace preserving and that it broadcasts every density operator diagonal in the basis $\{|k\rangle\}$.

Solution

Trace preservation follows from

$$\sum_kK_k^\dagger K_k = \sum_k|k\rangle\langle k| =I.$$

For $\rho=\sum_kp_k|k\rangle\langle k|$,

$$\mathcal B(\rho) = \sum_kp_k|kk\rangle\langle kk|.$$

Using $\operatorname{Tr}_B(|kk\rangle\langle kk|)=|k\rangle\langle k|$ and the analogous identity for $A$, both marginals are $\rho$.

2. Test a noncommuting qubit pair

For $\rho=(I+r\sigma_z)/2$ and $\sigma=(I+s\sigma_x)/2$, calculate the commutator and determine exactly when the pair is broadcastable.

Solution

Since $[\sigma_z,\sigma_x]=2i\sigma_y$,

$$[\rho,\sigma] = \frac{rs}{4}[\sigma_z,\sigma_x] = \frac{irs}{2}\sigma_y.$$

The commutator vanishes exactly when $r=0$ or $s=0$. In either case one state is maximally mixed and commutes with the other, so the pair is broadcastable. If $rs\ne0$, the no-broadcasting theorem forbids an exact broadcasting channel.

3. Derive the pure-state corollary

Show that two rank-one projectors commute if and only if they are identical or orthogonal. Conclude that two distinct nonorthogonal pure states cannot be broadcast.

Solution

Let $P=|\psi\rangle\langle\psi|$ and $Q=|\phi\rangle\langle\phi|$. Their products are

$$PQ = \langle\psi|\phi\rangle |\psi\rangle\langle\phi|, \qquad QP = \langle\phi|\psi\rangle |\phi\rangle\langle\psi|.$$

If the overlap is zero, both products vanish. If it is nonzero and $PQ=QP$, the one-dimensional ranges of the two products coincide, so $|\psi\rangle$ and $|\phi\rangle$ define the same ray. The theorem therefore permits broadcasting only for identical or orthogonal pure states.

4. Check the rank-deficient reduction

With $\omega=(\rho+\sigma)/2$, verify

$$[\rho_\varepsilon,\sigma_\varepsilon] =(1-\varepsilon)[\rho,\sigma]$$

for the regularized states used in the necessity proof.

Solution

The regularized states are

$$\rho_\varepsilon = \left(1-\frac\varepsilon2\right)\rho + \frac\varepsilon2\sigma, \qquad \sigma_\varepsilon = \frac\varepsilon2\rho + \left(1-\frac\varepsilon2\right)\sigma.$$

Using bilinearity and antisymmetry of the commutator, the coefficient of $[\rho,\sigma]$ is

$$\left(1-\frac\varepsilon2\right)^2 - \left(\frac\varepsilon2\right)^2 = 1-\varepsilon.$$

All self-commutators vanish. Thus commutativity of the regularized pair for any $0<\varepsilon<1$ implies commutativity of the original pair.

5. Use fidelity to rule out cloning

Assume a channel maps both $\rho$ and $\sigma$ to two product copies. Use fidelity monotonicity and multiplicativity to show that the two states must be identical or have orthogonal supports.

Solution

Monotonicity gives

$$f(\rho,\sigma) \leq f(\rho\otimes\rho, \sigma\otimes\sigma) = f(\rho,\sigma)^2.$$

Because $0\leq f\leq1$, one also has $f^2\leq f$. Both inequalities can hold only when $f=0$ or $f=1$. Root fidelity is zero exactly for orthogonal supports and one exactly for identical states.

6. Complete the sector argument

Let $M|k\rangle=m_k|k\rangle$ with every $m_k>0$, and suppose a state $\tau$ and a unitary $W$ obey

$$W\sqrt\tau = \sqrt\tau(M\otimes M^{-1}).$$

Show first that $\sqrt\tau|k\ell\rangle=0$ whenever $m_k\ne m_\ell$. Then prove that the $A$ marginal of $\tau$ commutes with $M$.

Solution

Taking norms after acting on $|k\ell\rangle$ gives

$$\left\|\sqrt\tau|k\ell\rangle\right\| = \frac{m_k}{m_\ell} \left\|\sqrt\tau|k\ell\rangle\right\|.$$

If $m_k\ne m_\ell$, positivity makes the ratio different from one, so the norm must vanish. For $\tau_A=\operatorname{Tr}_B\tau$,

$$\langle k|\tau_A|k'\rangle = \sum_\ell \left\langle \sqrt\tau|k\ell\rangle, \sqrt\tau|k'\ell\rangle \right\rangle.$$

When $m_k\ne m_{k'}$, a nonzero first vector requires $m_k=m_\ell$, whereas a nonzero second vector requires $m_{k'}=m_\ell$. These conditions cannot both hold, so every summand vanishes. Thus $\tau_A$ is block diagonal in the eigenspaces of $M$, which is equivalent to $[\tau_A,M]=0$.

References

• H. Barnum, C. M. Caves, C. A. Fuchs, R. Jozsa, and B. Schumacher, “Noncommuting mixed states cannot be broadcast,” Physical Review Letters 76, 2818–2821, 1996, doi:10.1103/PhysRevLett.76.2818.
• D. Dieks, “Communication by EPR devices,” Physics Letters A 92, 271–272, 1982, doi:10.1016/0375-9601(82)90084-6.
• C. A. Fuchs and C. M. Caves, “Mathematical techniques for quantum communication theory,” Open Systems & Information Dynamics 3, 345–356, 1995, doi:10.1007/BF02228997.
• R. Jozsa, “Fidelity for mixed quantum states,” Journal of Modern Optics 41, 2315–2323, 1994, doi:10.1080/09500349414552171.
• W. K. Wootters and W. H. Zurek, “A single quantum cannot be cloned,” Nature 299, 802–803, 1982, doi:10.1038/299802a0.

Further reading

• M. M. Wilde, Quantum Information Theory, 2nd ed., Cambridge University Press, 2017, sections on fidelity, recoverability, and no-broadcasting.