Information Constraints

A CNOT with its target in \(\ket0\) copies a computational label: \(\ket0\ket0\mapsto\ket{00}\) and \(\ket1\ket0\mapsto\ket{11}\). Does it copy every state? When we built a Bell pair, linearity forced \(\ket+\ket0\mapsto(\ket{00}+\ket{11})/\sqrt2\). Two copies of plus would be \(\ket+\ket+\), with four terms of amplitude one half. So CNOT copies the basis states and fails on their superposition.

Perhaps a cleverer machine, with an apparatus of its own, could do better. Suppose a unitary acts on the input, a blank qubit in \(\ket0\), and an apparatus in \(\ket e\), and copies two states \(\ket\psi\) and \(\ket\phi\):

\[\ket\psi\ket0\ket e\mapsto\ket\psi\ket\psi\ket{e_\psi},\qquad\ket\phi\ket0\ket e\mapsto\ket\phi\ket\phi\ket{e_\phi}.\]

A unitary keeps inner products. Before, the inner product of the two inputs is \(\braket\psi\phi\). After, it is \(\braket\psi\phi^2\braket{e_\psi}{e_\phi}\). Write \(s=|\braket\psi\phi|\) for the overlap. Since \(|\braket{e_\psi}{e_\phi}|\leq1\), the equation gives \(s\leq s^2\), which holds only for \(s=0\) or \(s=1\).

Put in numbers for zero and plus. The inputs have overlap \(1/\sqrt2\). Two exact copies would have overlap \(1/2\), and the apparatus could only make that smaller. A unitary cannot shrink an overlap, so no such machine exists. Because the apparatus can be anything, the argument covers every process built by the ancilla recipe. This is the no-cloning theorem (Wootters and Zurek, and Dieks, 1982): no device copies two distinct nonorthogonal pure states exactly and every time.

The theorem fits what we know about discrimination. Copying would raise the trace distance of zero and plus from \(1/\sqrt2\) to the two-copy value \(\sqrt3/2\). But no process can make states easier to tell apart.

For mixed states we can ask for less. Suppose we only require that each output, viewed on its own, has the input's density operator. The two outputs may be correlated. This weaker task is called broadcasting. Is it ever possible for an unknown mixed state?

Take the family \(\rho_q=q\ket0\bra0+(1-q)\ket1\bra1\), with \(q\) unknown. Measure Z and write the result into two output qubits. The joint output is \(q\ket{00}\bra{00}+(1-q)\ket{11}\bra{11}\). Each marginal is \(\rho_q\), so this broadcasts the whole family without learning \(q\).

It does not produce \(\rho_q\otimes\rho_q\). Its two bits always agree, while two independent samples of \(\rho_q\) sometimes disagree. Broadcasting gives correct marginals, not independent copies.

Why did this family allow it? All its members are diagonal in one basis, so measuring that basis loses nothing and a classical label can be copied freely. Matrices that are diagonal in a common basis commute: \(\rho\sigma=\sigma\rho\). Conversely, commuting Hermitian matrices always share a basis of eigenvectors, a linear-algebra fact we quote. So any commuting family can be broadcast by the same measure-and-copy trick.

What about zero and plus? Their projectors do not commute: \(\ket0\bra0\cdot\ket+\bra+=\tfrac1{\sqrt2}\ket0\bra+\), while the reverse product is \(\tfrac1{\sqrt2}\ket+\bra0\). We quote the theorem that settles the general case (Barnum, Caves, Fuchs, Jozsa and Schumacher, 1996): a family of density operators can be broadcast exactly when its members commute pairwise. For pure states this reduces to no-cloning. A noncommuting family cannot be treated as a classical label that we merely do not know.

Alice and Bob share a joint state \(\rho_{AB}\). Alice applies a channel to her system, with Kraus operators \(K_j\). A measurement whose record she keeps to herself is such a channel, as far as Bob is concerned. Then Bob measures, and outcome \(E\) is one of his effects. Can his probability of \(E\) depend on what Alice did?

Compute it. After Alice's channel the joint state is \(\sum_j(K_j\otimes I)\rho_{AB}(K_j^\dagger\otimes I)\). Bob's outcome probability is the trace of this with \(I\otimes E\). Moving \(K_j^\dagger\otimes I\) around the trace, and using that it commutes with \(I\otimes E\),

\[p(E)=\mathrm{tr}\Bigl[\Bigl(\sum_jK_j^\dagger K_j\otimes E\Bigr)\rho_{AB}\Bigr]=\mathrm{tr}\bigl[(I\otimes E)\rho_{AB}\bigr].\]

The last step uses trace preservation, \(\sum_jK_j^\dagger K_j=I\). The result is Bob's probability with Alice doing nothing. So no operation on Alice's side, entangled or not, changes any of Bob's outcome probabilities. Equivalently, Bob's reduced state is unchanged. This is the no-signalling property. Shared entanglement alone cannot carry a message; communication still needs something to be sent.

Yet Alice's outcome does change Bob's state, given that outcome. How can both be true?

Take the Bell pair. If Alice measures Z and sees 0, Bob's qubit is \(\ket0\). If she measures X and sees minus, Bob's qubit is \(\ket-\). So her choice of basis decides which ensemble Bob is in. Now average over her outcomes, as Bob must, since he has not seen them:

\[\tfrac12\ket0\bra0+\tfrac12\ket1\bra1=\tfrac12\ket+\bra++\tfrac12\ket-\bra-=I/2.\]

Both choices leave Bob with \(I/2\). Every effect he could use gives \(\mathrm{tr}(E\,I/2)\) either way, so no measurement, however clever, reveals Alice's basis. That is stronger than checking his Z counts.

Conditioning selects one branch; Bob's own prediction sums all of them. To use the conditional state he must know which branch each trial belongs to, and that requires Alice's record. When the record arrives, Bob can sort his earlier data and find strong correlations. Nothing reached him before the record did.

The overlap argument allowed \(s=0\) and \(s=1\). Which sets of pure states can one device copy exactly?

Apply the argument to each pair in the set. Every pair must be orthogonal or identical. Conversely, a set of mutually orthogonal states can be copied: measure in a basis containing them and prepare two copies of the result. So a set of pure states can be cloned exactly by one device exactly when its distinct members are mutually orthogonal. The basis states \(\ket0,\ket1\) qualify; the set \(\{\ket0,\ket1,\ket+\}\) does not.

Two further qualifications matter. A known state is no problem at all: whoever knows its description can prepare it again as often as resources allow. And the theorem concerns one device acting on an unknown member of the set, exactly and on every trial. It says nothing against devices that only approximate, or that sometimes refuse to answer.

The overlap argument needed exact copies. Does "almost copying" escape it? To answer we need a measure of "almost". For a pure target \(\ket\psi\) and an output \(\sigma\), use the probability that \(\sigma\) passes the test for \(\psi\), \(F=\bra\psi\sigma\ket\psi\). This number is the fidelity of the output with the target; it is 1 for a perfect copy.

A crude copier measures Z and prepares two copies of the result. On \(\ket0\) both outputs have fidelity 1. On \(\ket+\) each output is \(\ket0\) or \(\ket1\), which passes the plus test with probability \(1/2\). So its fidelity depends strongly on the input. Can one device do equally well on every input? We quote the answer: the best such universal copier for qubits gives each output fidelity \(5/6\) for every pure input (Bužek and Hillery, 1996; optimality shown by Gisin and Massar, 1997). Approximate copies are possible, with a fidelity strictly below one.

A different relaxation keeps the copies exact but allows the device to fail on some trials and say so. For linearly independent sets of states, such probabilistic exact cloning is possible (Duan and Guo, 1998). The failures are declared, so they do not contradict the deterministic theorem.

So every claim of impossibility must say which inputs, which resources and which success criterion it concerns. A security argument built on no-cloning is only as good as the match between those assumptions and the real protocol.

Copying and signalling are ruled out. What can a shared Bell pair accomplish, then, when it is combined with ordinary communication?

Sources and further study: Barnum et al., Noncommuting mixed states cannot be broadcast.