Editing No-cloning theorem (section)

== Theorem and proof ==
Suppose we have two quantum systems ''A'' and ''B'' with a common Hilbert space <math>H = H_A = H_B</math>. Suppose we want to have a procedure to copy the state <math>|\phi\rangle_A</math> of quantum system ''A'', over the state <math>|e\rangle_B</math> of quantum system ''B,'' for any original state <math>|\phi\rangle_A</math> (see [[bra–ket notation]]). That is, beginning with the state <math>|\phi\rangle_A \otimes |e\rangle_B </math>, we want to end up with the state <math>|\phi\rangle_A \otimes |\phi\rangle_B </math>. To make a "copy" of the state ''A'', we combine it with system ''B'' in some unknown initial, or blank, state <math>|e\rangle_B</math> independent of <math>|\phi\rangle_A</math>, of which we have no prior knowledge.

The state of the initial composite system is then described by the following [[tensor product]]:
<math display="block">|\phi\rangle_A \otimes |e\rangle_B.</math>
(in the following we will omit the <math>\otimes</math> symbol and keep it implicit).

There are only two permissible [[quantum operation]]s with which we may manipulate the composite system:

* We can perform an [[Measurement in quantum mechanics|observation]], which irreversibly [[wave function collapse|collapses]] the system into some [[eigenstate]] of an [[observable]], corrupting the information contained in the [[qubit|qubit(s)]]. This is obviously not what we want.
* Alternatively, we could control the [[Hamiltonian (quantum mechanics)|Hamiltonian]] of the ''combined'' system, and thus the [[mathematical formulation of quantum mechanics|time-evolution operator]] ''U''(''t''), e.g. for a time-independent Hamiltonian, {{nowrap|<math>U(t) = e^{-iHt/\hbar}</math>.}} Evolving up to some fixed time <math>t_0</math> yields a [[unitary operator]] ''U'' on {{nowrap|<math>H \otimes H</math>,}} the Hilbert space of the combined system. However, no such unitary operator ''U'' can clone all states.

The no-cloning theorem answers the following question in the negative: Is it possible to construct a unitary operator ''U'', acting on <math>H_A \otimes H_B = H \otimes H</math>, under which the state the system B is in always evolves into the state the system A is in, ''regardless'' of the state system A is in?

{{math theorem | math_statement = There is no unitary operator ''U'' on <math>H \otimes H</math> such that for all normalised states <math>|\phi \rangle_A</math> and <math>|e\rangle_B</math> in <math>H</math>
<math display="block">U(|\phi\rangle_A |e\rangle_B) = e^{i \alpha(\phi,e)} |\phi\rangle_A |\phi\rangle_B</math>
for some real number <math>\alpha</math> depending on <math>\phi</math> and <math>e</math>.}}

The extra phase factor expresses the fact that a quantum-mechanical state defines a normalised vector in Hilbert space only up to a phase factor i.e. as an element of [[projective space|projectivised Hilbert space]].

To prove the theorem, we select an arbitrary pair of states <math>|\phi\rangle_A</math> and <math>|\psi\rangle_A</math> in the Hilbert space <math>H</math>. Because ''U'' is supposed to be unitary, we would have
<math display="block">
 \langle \phi| \psi\rangle \langle e | e \rangle \equiv
 \langle \phi|_A \langle e|_B |\psi\rangle_A |e\rangle_B =
 \langle \phi|_A \langle e|_B U^\dagger U |\psi\rangle_A |e\rangle_B =
 e^{-i(\alpha(\phi, e) - \alpha(\psi, e))} \langle \phi|_A \langle \phi|_B  |\psi\rangle_A |\psi\rangle_B \equiv
 e^{-i(\alpha(\phi, e) - \alpha(\psi, e))} \langle \phi |\psi\rangle^2.
</math>
Since the  quantum state <math>|e\rangle</math> is assumed to be normalized, we thus get
<math display="block"> |\langle \phi | \psi \rangle|^2 = |\langle \phi | \psi \rangle|.</math>

This implies that either <math>|\langle \phi | \psi \rangle| = 1</math> or <math>|\langle \phi | \psi \rangle| = 0</math>. Hence by the [[Cauchy–Schwarz inequality]] either <math>|\phi\rangle = e^{i\beta}|\psi\rangle</math> or <math>|\phi\rangle</math> is [[orthogonality|orthogonal]] to <math>|\psi\rangle</math>. However, this cannot be the case for two ''arbitrary'' states. Therefore, a single universal ''U'' cannot clone a ''general'' quantum state. This proves the no-cloning theorem.

Take a qubit for example. It can be represented by two [[complex number]]s, called [[probability amplitude]]s ([[Qubit#Qubit states|normalised to 1]]), that is three real numbers (two polar angles and one radius). Copying three numbers on a classical computer using any [[copy and paste]] operation is trivial (up to a finite precision) but the problem manifests if the qubit is unitarily transformed (e.g. by the [[Quantum logic gate#Hadamard gate|Hadamard quantum gate]]) to be polarised (which [[Unitary operator|unitary transformation]] is a [[Isometry#Introduction|surjective isometry]]). In such a case the qubit can be represented by just two real numbers (one polar angle and one radius equal to 1), while the value of the third can be arbitrary in such a representation. Yet a [[Qubit#Physical implementations|realisation]] of a qubit (polarisation-encoded photon, for example) is capable of storing the whole qubit information support within its "structure". Thus no single universal unitary evolution ''U'' can clone an arbitrary quantum state according to the no-cloning theorem. It would have to depend on the transformed qubit (initial) state and thus would not have been ''universal''.