Chapter 06

Why an unknown state cannot be copied

A machine copies $|0\rangle$ and $|1\rangle$ perfectly. Feed it $|+\rangle$: what comes out? And could any machine, however it is built, copy every state it is given?

Before you start: Inner products (Chapter 1); unitaries and the controlled-NOT (Chapter 4). One section uses effects (Chapter 2) and one problem uses reduced states (Chapter 3). Notation guide.

Specify what copying would mean

A copier receives a system in a state $|\psi\rangle$ and a blank register in a fixed state $|b\rangle$. It must produce $|\psi\rangle|\psi\rangle$ for every allowed input without being told which input it received. The requirement that one device handle every input, without being told which, is what makes the problem hard: if a classical description of $|\psi\rangle$ is supplied, a preparation procedure can simply be run twice.

Try the controlled-NOT from Chapter 4, with the second qubit initially $|0\rangle$. It passes two tests: input $|0\rangle$ gives $|00\rangle$, and input $|1\rangle$ gives $|11\rangle$. In each case the original survives and the blank becomes a second copy.

Now feed it $|+\rangle$. The two-register input is $(|00\rangle+|10\rangle)/\sqrt2$. CNOT leaves the first term alone and changes the second to $|11\rangle$, giving the Bell pair $(|00\rangle+|11\rangle)/\sqrt2$. Compare that with the required output:

$$|+\rangle|+\rangle=\frac{|00\rangle+|01\rangle+|10\rangle+|11\rangle}{2}.$$

These make different predictions. Measure both registers in the computational basis. The machine's output always gives matching results, 00 or 11. Two independent copies give all four results with equal probability, including mismatches. Matching the two registers to each other was not enough to copy their input state.

For a general input $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, the same calculation gives

$$\operatorname{CNOT}(|\psi\rangle|0\rangle)=\alpha|00\rangle+\beta|11\rangle,$$

whereas expanding the two copies gives

$$|\psi\rangle^{\otimes2}=\alpha^2|00\rangle+\alpha\beta|01\rangle+\alpha\beta|10\rangle+\beta^2|11\rangle.$$

The first output follows by applying the machine to each input term and retaining its amplitude. The requested output instead multiplies amplitudes together. This is the obstruction from linearity: choosing how a linear device handles the basis states already fixes how it handles their superpositions.

CNOT is only one example. Specify a copier as any unitary $U$ on the two registers alone that sends $|00\rangle$ to $|00\rangle$ and $|10\rangle$ to $|11\rangle$, with no extra phases and no apparatus. Linearity then fixes its output for $|+\rangle$: $U(|+\rangle|0\rangle)=(U|00\rangle+U|10\rangle)/\sqrt2=(|00\rangle+|11\rangle)/\sqrt2$. The figure sets that forced output beside what two copies would be.

The inner-product proof

The first argument fixed how the device copied $|0\rangle$ and $|1\rangle$. To rule out every universal copier, we can ask a smaller question: could the same device copy both $|0\rangle$ and $|+\rangle$? Those two inputs overlap. A reversible evolution must preserve that overlap, while making two copies would reduce it.

The overlap proof uses one property of composite systems that is worth writing out. If the two registers are independent product states, their inner product factorises:

$$(\langle a|\otimes\langle b|)(|c\rangle\otimes|d\rangle)=\langle a|c\rangle\,\langle b|d\rangle.$$

Why the inner product of product states factorises

To derive it, expand each local state in an orthonormal basis. The joint inner product is $\sum_{i,j}a_i^*b_j^*c_id_j$. The $i$ and $j$ sums are independent, so it equals $(\sum_i a_i^*c_i)(\sum_jb_j^*d_j)$. We will apply this rule to the proposed product outputs, not to an arbitrary entangled output.

Next, unitarity gives $\langle Uv|Uw\rangle=\langle v|U^\dagger U|w\rangle=\langle v|w\rangle$. Thus a unitary cannot change the overlap of two possible inputs. The blank adds no distinguishability because it is the same normalised state in both cases: its overlap with itself is one. Two copies would multiply the input overlap by itself. For an overlap strictly between zero and one in magnitude, squaring makes it smaller, contradicting preservation by the unitary.

For that pair, the overlap before copying is $\langle0|+\rangle=1/\sqrt2$. Attaching the same blank does not change it. After successful copying, it would be $\langle00|++\rangle=(1/\sqrt2)^2=1/2$. A unitary cannot make both of those statements true. This already excludes a device that copies every state.

Writing the same argument for any pair tells us exactly which pairs escape it. Suppose one unitary copies two states $|\psi\rangle$ and $|\phi\rangle$, with the same normalised blank. Unitarity preserves their input overlap. Writing $s=\langle\psi|\phi\rangle$ gives

$$\langle\psi|\phi\rangle\langle b|b\rangle=\langle\psi|\phi\rangle\langle\psi|\phi\rangle,\qquad s=s^2.$$

For fixed vector representatives this requires $s=0$ or $s=1$. Allowing physically irrelevant output phases gives the phase-independent conclusion $|s|=|s|^2$: the input physical states must be orthogonal or identical. A ray is the set of nonzero vectors that differ only by an overall scalar; after normalisation, its representatives differ only by a common phase. Quantum pure states are rays in this sense. Distinct nonorthogonal states cannot both be copied exactly by that unitary. Since a universal copier would have to copy $|0\rangle$ and $|+\rangle$, whose overlap magnitude is $1/\sqrt2$, it cannot exist.

An extra apparatus does not evade the result

A real copier can have internal parts as well as the two output registers. Perhaps those parts retain a record of the input and make copying possible. Include that record in the calculation: it contributes one more overlap factor. For normalised record states that factor can never have magnitude greater than one, so it cannot compensate for the reduction caused by the two proposed copies.

Allow the copier to retain a record, $U|\psi,b,a\rangle=|\psi,\psi\rangle|a_\psi\rangle$. The two output copies are pure, so any pure global output with those exact copies must factor from the remaining apparatus. Preservation of overlap now requires

$$s=s^2\langle a_\psi|a_\phi\rangle.$$

For $0<|s|<1$, taking magnitudes would require $|\langle a_\psi|a_\phi\rangle|=1/|s|>1$, which is impossible for normalised apparatus states. General deterministic physical processes (channels, Chapter 13) can be represented by such a unitary with an environment and a discarded record. Thus adding an environment or forgetting an outcome does not create a deterministic exact copier.

Why measuring first does not solve the problem

Another approach is to measure the input, learn which state it was, and use that description to prepare two copies. The difficulty is the first step. A detector that reads 0 cannot say whether it received $|0\rangle$ or $|+\rangle$: either input can give that reading. Changing the detector cannot make these two nonorthogonal states perfectly distinguishable either.

Chapter 2 showed what is possible if failure is allowed: some readings identify the state without error, and other runs are explicitly inconclusive. An exact copier that must succeed on every run cannot use that escape. The proof below shows why certainty for one input and impossibility for the other require orthogonal states.

The proof that certainty on one input forces orthogonality

Let effect $E$ identify $|\psi\rangle$ with certainty and never occur on $|\phi\rangle$. Then $\langle\psi|E|\psi\rangle=1$ and $\langle\phi|E|\phi\rangle=0$. Here $0\le E\le I$ means both $E$ and $I-E$ are positive: all their expectations are nonnegative.

Positivity implies $(I-E)|\psi\rangle=0$: a positive operator $P$ with $\langle\psi|P|\psi\rangle=0$ satisfies $\|\sqrt P|\psi\rangle\|^2=0$, hence $\sqrt P|\psi\rangle=0$ and $P|\psi\rangle=0$. Applying the same argument to $E$ on $|\phi\rangle$ gives $E|\phi\rangle=0$. Thus $E|\psi\rangle=|\psi\rangle$ while $E|\phi\rangle=0$. Therefore $\langle\psi|\phi\rangle=\langle\psi|E|\phi\rangle=0$. Certainty for one and impossibility for the other requires orthogonality. Chapter 2's discrimination scheme avoids this obstruction by explicitly allowing an inconclusive outcome.

The boundaries of the theorem

No-cloning forbids deterministic, exact copying of arbitrary unknown states. It allows copying a known orthogonal alphabet, preparing a known state repeatedly, approximate copying with imperfect outputs, and some probabilistic procedures with reported failure. It does not forbid transferring a state: teleportation consumes the original encoding and requires classical communication.

It also does not forbid encoding one qubit into many physical qubits. The map $\alpha|0\rangle+\beta|1\rangle\mapsto\alpha|000\rangle+\beta|111\rangle$ is linear. Its output stores amplitudes jointly; the individual qubits do not each contain $|\psi\rangle$. Chapter 14 uses precisely this distinction.

Where this leaves us

Fed $|+\rangle$, the copying unitary specified above produces $(|00\rangle+|11\rangle)/\sqrt2$, not $|+\rangle|+\rangle$, because a linear map must act on the two terms separately. A copier that adds input-dependent phases or keeps an apparatus record gives a different output, but not two copies either. No machine copies every state: copying two states requires their overlap $s$ to satisfy $|s|=|s|^2$, so they must be identical or orthogonal. An apparatus that keeps a record, or a measurement made first, does not escape this. What remains allowed is copying a known orthogonal set, approximate or probabilistic copying, and moving a state from one place to another.

The output $(|00\rangle+|11\rangle)/\sqrt2$ is not two copies of anything. It is a two-qubit state that cannot be written as a product at all. The next chapter turns “cannot be written as a product” into a calculation, and measures how far a state is from one.

Problems

1. For the CNOT output $\alpha|00\rangle+\beta|11\rangle$, find the state of the second qubit. When does it equal the original pure input?

Worked answer

Form the projector of the output $|\Psi\rangle=\alpha|00\rangle+\beta|11\rangle$. Its four terms have coefficients $|\alpha|^2,\alpha\beta^*,\beta\alpha^*,|\beta|^2$. When tracing the first qubit, the cross terms vanish because their first-qubit overlaps are zero. Therefore

$$\rho_2=\begin{pmatrix}|\alpha|^2&0\\0&|\beta|^2\end{pmatrix},\qquad\rho_{\rm input}=\begin{pmatrix}|\alpha|^2&\alpha\beta^*\\\alpha^*\beta&|\beta|^2\end{pmatrix}.$$

Equality requires $\alpha\beta^*=0$. Since these are complex numbers, at least one amplitude must vanish. Normalisation then makes the input $|0\rangle$ or $|1\rangle$ up to a common phase. A general superposition is not copied into the second register.

The tempting wrong turn is to compare only the diagonals. They agree, so every computational-basis measurement on the second qubit gives the input’s statistics. The missing off-diagonal entries show up in any measurement that combines the two amplitudes: for input $|+\rangle$, the second qubit gives $+$ with probability $\tfrac12$ instead of $1$.

2. Can one apparatus copy both $|+\rangle$ and $|-\rangle$? Give a circuit.

Worked answer

Convert the known orthogonal alphabet into the computational basis, copy that basis, then convert both outputs back. With the second register initially zero, the two paths are

$$\begin{aligned}|+0\rangle&\xrightarrow{H\otimes I}|00\rangle\xrightarrow{\mathrm{CNOT}}|00\rangle\xrightarrow{H\otimes H}|++\rangle,\\|-0\rangle&\xrightarrow{H\otimes I}|10\rangle\xrightarrow{\mathrm{CNOT}}|11\rangle\xrightarrow{H\otimes H}|--\rangle.\end{aligned}$$

The same circuit works for both promised inputs because they are orthogonal. Its success on this two-state alphabet does not extend to arbitrary superpositions of its members; linearity would again produce correlated outputs instead of two independent copies.

References and further reading

Nielsen & Chuang, §1.3.5 (copying circuit), §2.2.4 (distinguishability), and Chapter 12 (quantum information).