Quantum logic gate

The Hadamard gate acts on a single qubit. It maps the basis state ${\displaystyle |0\rangle }$ to ${\displaystyle {\frac {|0\rangle +|1\rangle }{\sqrt {2}}}}$ and ${\displaystyle |1\rangle }$ to ${\displaystyle {\frac {|0\rangle -|1\rangle }{\sqrt {2}}}}$, which means that a measurement will have equal probabilities to become 1 or 0 (i.e. creates a superposition). It represents a rotation of ${\displaystyle \pi }$ about the axis ${\displaystyle ({\hat {x}}+{\hat {z}})/{\sqrt {2}}}$ at the Bloch sphere. Equivalently, it is the combination of two rotations, ${\displaystyle \pi }$ about the Z-axis followed by ${\displaystyle \pi /2}$ about the Y-axis. It is represented by the Hadamard matrix:

Circuit representation of Hadamard gate

${\displaystyle H={\frac {1}{\sqrt {2}}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}}$.

The Hadamard gate is the one-qubit version of the quantum Fourier transform.

Since ${\displaystyle HH^{\dagger }=I}$ where I is the identity matrix, H is a unitary matrix (like all other quantum logical gates). Also, it is its own unitary inverse, ${\displaystyle H=H^{\dagger }}$.

Quantum circuit diagram of a NOT-gate

The Pauli-X gate acts on a single qubit. It is the quantum equivalent of the NOT gate for classical computers (with respect to the standard basis ${\displaystyle |0\rangle }$, ${\displaystyle |1\rangle }$, which distinguishes the Z-direction; in the sense that a measurement of the eigenvalue +1 corresponds to classical 1/true and -1 to 0/false). It equates to a rotation around the X-axis of the Bloch sphere by ${\displaystyle \pi }$ radians. It maps ${\displaystyle |0\rangle }$ to ${\displaystyle |1\rangle }$ and ${\displaystyle |1\rangle }$ to ${\displaystyle |0\rangle }$. Due to this nature, it is sometimes called bit-flip. It is represented by the Pauli X matrix:

${\displaystyle X={\begin{bmatrix}0&1\\1&0\end{bmatrix}}}$.

The Pauli-Y gate acts on a single qubit. It equates to a rotation around the Y-axis of the Bloch sphere by ${\displaystyle \pi }$ radians. It maps ${\displaystyle |0\rangle }$ to ${\displaystyle i|1\rangle }$ and ${\displaystyle |1\rangle }$ to ${\displaystyle -i|0\rangle }$. It is represented by the Pauli Y matrix:

${\displaystyle Y={\begin{bmatrix}0&-i\\i&0\end{bmatrix}}}$.

The Pauli-Z gate acts on a single qubit. It equates to a rotation around the Z-axis of the Bloch sphere by ${\displaystyle \pi }$ radians. Thus, it is a special case of a phase shift gate (which are described in a next subsection) with ${\displaystyle \phi =\pi }$. It leaves the basis state ${\displaystyle |0\rangle }$ unchanged and maps ${\displaystyle |1\rangle }$ to ${\displaystyle -|1\rangle }$. Due to this nature, it is sometimes called phase-flip. It is represented by the Pauli Z matrix:

${\displaystyle Z={\begin{bmatrix}1&0\\0&-1\end{bmatrix}}}$.

The square of a Pauli matrix is the identity matrix.

${\displaystyle I^{2}=X^{2}=Y^{2}=Z^{2}=-iXYZ=I}$

Quantum circuit diagram of a square-root-of-NOT gate

The square root of NOT gate (or square root of Pauli-X, ${\displaystyle {\sqrt {X}}}$) acts on a single qubit. It maps the basis state ${\displaystyle |0\rangle }$ to ${\displaystyle {\frac {(1+i)|0\rangle +(1-i)|1\rangle }{2}}}$ and ${\displaystyle |1\rangle }$ to ${\displaystyle {\frac {(1-i)|0\rangle +(1+i)|1\rangle }{2}}}$.

${\displaystyle {\sqrt {X}}={\sqrt {NOT}}={\frac {1}{2}}{\begin{bmatrix}1+i&1-i\\1-i&1+i\end{bmatrix}}}$.

${\displaystyle X=({\sqrt {NOT}})^{2}={\begin{bmatrix}0&1\\1&0\end{bmatrix}}}$.

Squared root-gates can be constructed for all other gates by finding a unitary matrix that, multiplied by itself, yields the gate one wishes to construct the squared root gate of. All rational exponents of all gates can be found similarly.

This is a family of single-qubit gates that leave the basis state ${\displaystyle |0\rangle }$ unchanged and map ${\displaystyle |1\rangle }$ to ${\displaystyle e^{i\phi }|1\rangle }$. The probability of measuring a ${\displaystyle |0\rangle }$ or ${\displaystyle |1\rangle }$ is unchanged after applying this gate, however it modifies the phase of the quantum state. This is equivalent to tracing a horizontal circle (a line of latitude) on the Bloch sphere by ${\displaystyle \phi }$ radians.

${\displaystyle R_{\phi }={\begin{bmatrix}1&0\\0&e^{i\phi }\end{bmatrix}}}$

where ${\displaystyle \phi }$ is the phase shift. Some common examples are the T gate where ${\displaystyle \phi ={\frac {\pi }{4}}}$, the phase gate (written S, though S is sometimes used for SWAP gates) where ${\displaystyle \phi ={\frac {\pi }{2}}}$ and the Pauli-Z gate where ${\displaystyle \phi =\pi }$.

Circuit representation of SWAP gate

The swap gate swaps two qubits. With respect to the basis ${\displaystyle |00\rangle }$, ${\displaystyle |01\rangle }$, ${\displaystyle |10\rangle }$, ${\displaystyle |11\rangle }$, it is represented by the matrix:

${\displaystyle {\mbox{SWAP}}={\begin{bmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{bmatrix}}}$.

Circuit representation of ${\displaystyle {\sqrt {\mbox{SWAP}}}}$ gate

The ${\displaystyle {\sqrt {\mbox{SWAP}}}}$ gate performs half-way of a two-qubit swap. It is universal such that any many-qubit gate can be constructed from only ${\displaystyle {\sqrt {\mbox{SWAP}}}}$ and single qubit gates. The ${\displaystyle {\sqrt {\mbox{SWAP}}}}$ gate is not, however maximally entangling; more than one application of it is required to produce a Bell state from product states. With respect to the basis ${\displaystyle |00\rangle }$, ${\displaystyle |01\rangle }$, ${\displaystyle |10\rangle }$, ${\displaystyle |11\rangle }$, it is represented by the matrix:

${\displaystyle {\sqrt {\mbox{SWAP}}}={\begin{bmatrix}1&0&0&0\\0&{\frac {1}{2}}(1+i)&{\frac {1}{2}}(1-i)&0\\0&{\frac {1}{2}}(1-i)&{\frac {1}{2}}(1+i)&0\\0&0&0&1\\\end{bmatrix}}}$.

Circuit representation of controlled NOT gate

Controlled gates act on 2 or more qubits, where one or more qubits act as a control for some operation. For example, the controlled NOT gate (or CNOT or cX) acts on 2 qubits, and performs the NOT operation on the second qubit only when the first qubit is ${\displaystyle |1\rangle }$, and otherwise leaves it unchanged. With respect to the basis ${\displaystyle |00\rangle }$, ${\displaystyle |01\rangle }$, ${\displaystyle |10\rangle }$, ${\displaystyle |11\rangle }$, it is represented by the matrix:

${\displaystyle {\mbox{CNOT}}=cX={\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}}}$.

The CNOT (or controlled-X) gate can be described as the gate that maps ${\displaystyle |a,b\rangle }$ to ${\displaystyle |a,a\oplus b\rangle }$.

More generally if U is a gate that operates on single qubits with matrix representation

${\displaystyle U={\begin{bmatrix}u_{00}&u_{01}\\u_{10}&u_{11}\end{bmatrix}}}$,

then the controlled-U gate is a gate that operates on two qubits in such a way that the first qubit serves as a control. It maps the basis states as follows.

Circuit representation of controlled-U gate

${\displaystyle |00\rangle \mapsto |00\rangle }$

${\displaystyle |01\rangle \mapsto |01\rangle }$

${\displaystyle |10\rangle \mapsto |1\rangle \otimes U|0\rangle =|1\rangle \otimes \left(u_{00}|0\rangle +u_{10}|1\rangle \right)}$

${\displaystyle |11\rangle \mapsto |1\rangle \otimes U|1\rangle =|1\rangle \otimes \left(u_{01}|0\rangle +u_{11}|1\rangle \right)}$

The matrix representing the controlled U is

${\displaystyle {\mbox{C}}(U)={\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&u_{00}&u_{01}\\0&0&u_{10}&u_{11}\end{bmatrix}}}$.

controlled X-, Y- and Z- gates

controlled-X gate

controlled-Y gate

controlled-Z gate

When U is one of the Pauli matrices, σ_x, σ_y, or σ_z, the respective terms "controlled-X", "controlled-Y", or "controlled-Z" are sometimes used.[1] Sometimes this is shortened to just cX, cY and cZ.

Circuit representation of Toffoli gate

The Toffoli gate, named after Tommaso Toffoli; also called CCNOT gate or Deutsch ${\displaystyle D(\pi /2)}$ gate; is a 3-bit gate, which is universal for classical computation but not for quantum computation. The quantum Toffoli gate is the same gate, defined for 3 qubits. If we limit ourselves to only accepting input qubits that are ${\displaystyle |0\rangle }$ and ${\displaystyle |1\rangle }$, then if the first two bits are in the state ${\displaystyle |1\rangle }$ it applies a Pauli-X (or NOT) on the third bit, else it does nothing. It is an example of a controlled gate. Since it is the quantum analog of a classical gate, it is completely specified by its truth table. The Toffoli gate is universal when combined with the single qubit Hadamard gate.[2]

It can be also described as the gate which maps ${\displaystyle |a,b,c\rangle }$ to ${\displaystyle |a,b,c\oplus ab\rangle }$.

Circuit representation of Fredkin gate

The Fredkin gate (also CSWAP or cS gate), named after Edward Fredkin, is a 3-bit gate that performs a controlled swap. It is universal for classical computation. It has the useful property that the numbers of 0s and 1s are conserved throughout, which in the billiard ball model means the same number of balls are output as input.

The Ising gate (or XX gate) is a 2-qubit gate that is implemented natively in some trapped-ion quantum computers.[3][4] It is defined as

${\displaystyle XX_{\phi }=\cos(\phi )I\otimes I-i\sin(\phi )\sigma _{x}\otimes \sigma _{x}={\begin{bmatrix}\cos(\phi )&0&0&-i\sin(\phi )\\0&\cos(\phi )&-i\sin(\phi )&0\\0&-i\sin(\phi )&\cos(\phi )&0\\-i\sin(\phi )&0&0&\cos(\phi )\\\end{bmatrix}}}$,

${\displaystyle YY_{\phi }={\begin{bmatrix}\cos(\phi )&0&0&i\sin(\phi )\\0&\cos(\phi )&-i\sin(\phi )&0\\0&-i\sin(\phi )&\cos(\phi )&0\\i\sin(\phi )&0&0&\cos(\phi )\\\end{bmatrix}}}$,

${\displaystyle ZZ_{\phi }={\begin{bmatrix}e^{i\phi /2}&0&0&0\\0&e^{-i\phi /2}&0&0\\0&0&e^{-i\phi /2}&0\\0&0&0&e^{i\phi /2}\\\end{bmatrix}}}$[5],

The Deutsch (or ${\displaystyle D_{\theta }}$) gate, named after physicist David Deutsch is a three-qubit gate. It is defined as

${\displaystyle |a,b,c\rangle \mapsto {\begin{cases}i\cos(\theta )|a,b,c\rangle +\sin(\theta )|a,b,1-c\rangle &{\mbox{for }}a=b=1\\|a,b,c\rangle &{\mbox{otherwise.}}\end{cases}}}$

Unfortunately, a working Deutsch gate has remained out of reach, due to lack of a protocol. However, a method was proposed to realize such a Deutsch gate with dipole-dipole interaction in neutral atoms.

Two gates Y and X in series. The order in which they appear on the wire is reversed when multiplying them together.

Assume that we have two gates A and B, that both act on ${\displaystyle n}$ qubits. When B is put after A (a series circuit), then the effect of the two gates can be described as a single gate C.

${\displaystyle C=B\cdot A}$

Where ${\displaystyle \cdot }$ is the usual matrix multiplication. The resulting gate C will have the same dimensions as A and B. The order in which the gates would appear in a circuit diagram is reversed when multiplying them together.

For example, putting the Pauli X gate after the Pauli Y gate, both of which act on a single qubit, can be described as a single combined gate C:

${\displaystyle C=X\cdot Y={\begin{bmatrix}0&1\\1&0\end{bmatrix}}\cdot {\begin{bmatrix}0&-i\\i&0\end{bmatrix}}={\begin{bmatrix}i&0\\0&-i\end{bmatrix}}}$

The product symbol (${\displaystyle \cdot }$) is often omitted.

Two gates ${\displaystyle Y}$ and ${\displaystyle X}$ in parallel is equivalent to the gate ${\displaystyle Y\otimes X}$

The tensor product (or kronecker product) of two quantum gates is the gate that is equal to the two gates in parallel.

The gate ${\displaystyle H_{2}=H\otimes H}$ is the Hadamard gate (${\displaystyle H}$) applied in parallel on 2 qubits. It can be written as ${\displaystyle H_{2}=H\otimes H={\frac {1}{\sqrt {2}}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}\otimes {\frac {1}{\sqrt {2}}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}={\frac {1}{2}}{\begin{bmatrix}1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1\end{bmatrix}}}$

This "two-qubit parallel Hadamard gate" will when applied to, for example, the two-qubit zero-vector (${\displaystyle |00\rangle }$), create a quantum state that have equal probability of being observed in any of its four possible outcomes; 00, 01, 10 and 11. We can write this operation as:

${\displaystyle H_{2}|00\rangle ={\frac {1}{2}}{\begin{bmatrix}1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1\end{bmatrix}}{\begin{bmatrix}1\\0\\0\\0\end{bmatrix}}={\frac {1}{2}}{\begin{bmatrix}1\\1\\1\\1\end{bmatrix}}={\frac {1}{2}}|00\rangle +{\frac {1}{2}}|01\rangle +{\frac {1}{2}}|10\rangle +{\frac {1}{2}}|11\rangle ={\frac {|00\rangle +|01\rangle +|10\rangle +|11\rangle }{2}}}$

Here the amplitude for each measurable state is ${\displaystyle {\frac {1}{2}}}$. The probability to observe any state is the square of the absolute value of the measurable states amplitude, which in the above example means that there is one in four that we observe any one of the individual four cases. See measurement for details.

${\displaystyle H_{2}}$ performs the Hadamard transform on two qubits. Similarly the gate ${\displaystyle \underbrace {H\otimes H\otimes \dots \otimes H} _{n{\text{ times}}}=\bigotimes _{1}^{n}H=H^{\otimes n}=H_{n}}$ performs a Hadamard transform on a register of ${\displaystyle n}$ qubits.

When applied to a register of ${\displaystyle n}$ qubits all initialized to ${\displaystyle |0\rangle }$, the Hadamard transform puts the quantum register into a superposition with equal probability of being measured in any of its ${\displaystyle 2^{n}}$ possible states:

${\displaystyle \bigotimes _{0}^{n-1}H\bigotimes _{0}^{n-1}|0\rangle ={\frac {1}{\sqrt {2^{n}}}}{\begin{bmatrix}1\\1\\\vdots \\1\end{bmatrix}}={\frac {1}{\sqrt {2^{n}}}}{\Big (}|0\rangle +|1\rangle +\dots +|2^{n}-1\rangle {\Big )}={\frac {1}{\sqrt {2^{n}}}}\sum _{i=0}^{2^{n}-1}|i\rangle }$

This state is a uniform superposition and it is generated as the first step in some search algorithms, for example in amplitude amplification and phase estimation.

Measuring this state results in a random number between ${\displaystyle |0\rangle }$ and ${\displaystyle |2^{n}-1\rangle }$. How random the number is depends on the fidelity of the logic gates. If not measured, it is a quantum state with equal probability amplitude ${\displaystyle {\frac {1}{\sqrt {2^{n}}}}}$ for each of its possible states.

The Hadamard transform acts on a register ${\displaystyle |\psi \rangle }$ with ${\displaystyle n}$ qubits such that ${\displaystyle |\psi \rangle =\bigotimes _{i=0}^{n-1}|\psi _{i}\rangle }$ as follows: ${\displaystyle \bigotimes _{0}^{n-1}H|\psi \rangle =\bigotimes _{i=0}^{n-1}{\frac {|0\rangle +(-1)^{\psi _{i}}|1\rangle }{\sqrt {2}}}={\frac {1}{\sqrt {2^{n}}}}\bigotimes _{i=0}^{n-1}{\Big (}|0\rangle +(-1)^{\psi _{i}}|1\rangle {\Big )}=H|\psi _{0}\rangle \otimes H|\psi _{1}\rangle \otimes \cdots \otimes H|\psi _{n-1}\rangle }$

If two or more qubits are viewed as a single quantum state, this combined state is equal to the tensor product of the constituent qubits. Any state that can be written as a tensor product from the constituent subsystems are called separable states. On the other hand, an entangled state is any state that cannot be tensor-factorized, or in other words: An entangled state can not be written as a tensor product of its constituent qubits states. Special care must be taken when applying gates to constituent qubits that make up entangled states.

If we have a set of N qubits that are entangled and wish to apply a quantum gate on M < N qubits in the set, we will have to extend the gate to take N qubits. This can be done by combining the gate with an identity matrix such that their tensor product becomes a gate that act on N qubits. The identity matrix (${\displaystyle I}$) is a representation of the gate that maps every state to itself (i.e., does nothing at all). In a circuit diagram the identity gate or matrix will appear as just a wire.

The example given in the text. The Hadamard gate ${\displaystyle H}$ only act on 1 qubit, but ${\displaystyle |\psi \rangle }$ is an entangled quantum state that spans 2 qubits. In our example, ${\displaystyle |\psi \rangle ={\frac {|00\rangle +|11\rangle }{\sqrt {2}}}}$

For example, the Hadamard gate (${\displaystyle H}$) acts on a single qubit, but if we for example feed it the first of the two qubits that constitute the entangled Bell state ${\displaystyle {\frac {|00\rangle +|11\rangle }{\sqrt {2}}}}$, we cannot write that operation easily. We need to extend the Hadamard gate ${\displaystyle H}$ with the identity gate ${\displaystyle I}$ so that we can act on quantum states that span two qubits:

${\displaystyle K=H\otimes I={\frac {1}{\sqrt {2}}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}\otimes {\begin{bmatrix}1&0\\0&1\end{bmatrix}}={\frac {1}{\sqrt {2}}}{\begin{bmatrix}1&0&1&0\\0&1&0&1\\1&0&-1&0\\0&1&0&-1\end{bmatrix}}}$

The gate ${\displaystyle K}$ can now be applied to any two-qubit state, entangled or otherwise. The gate ${\displaystyle K}$ will leave the second qubit untouched and apply the Hadamard transform to the first qubit. If applied to the Bell state in our example, we may write that as:

${\displaystyle K{\frac {|00\rangle +|11\rangle }{\sqrt {2}}}={\frac {1}{\sqrt {2}}}{\begin{bmatrix}1&0&1&0\\0&1&0&1\\1&0&-1&0\\0&1&0&-1\end{bmatrix}}{\frac {1}{\sqrt {2}}}{\begin{bmatrix}1\\0\\0\\1\end{bmatrix}}={\frac {1}{2}}{\begin{bmatrix}1\\1\\1\\-1\end{bmatrix}}={\frac {|00\rangle +|01\rangle +|10\rangle -|11\rangle }{2}}}$

The time complexity for multiplying two ${\displaystyle n\times n}$-matrices is at least ${\displaystyle \Omega (n^{2}log(n))}$[8]. Because the size of a gate that operates on ${\displaystyle q}$ qubits is ${\displaystyle 2^{q}\times 2^{q}}$ it means that the time complexity for simulating a step in a quantum circuit (by means of multiplying the gates) that operates on generic entangled states is ${\displaystyle \Omega ({2^{q}}^{2}log({2^{q}}))}$. For this reason it is believed to be intractable to simulate large entangled quantum systems using classical computers. The Clifford gates is an example of a set of gates that however can be efficiently simulated on classical computers.

Example: The unitary inverse of the Hadamard-CNOT product. The three gates ${\displaystyle H}$, ${\displaystyle I}$ and ${\displaystyle CNOT}$ are their own unitary inverses.

Because all quantum logical gates are reversible, any composition of multiple gates is also reversible. All products and tensor products of unitary matrices are also unitary matrices. This means that it is possible to construct an inverse of all algorithms and functions, as long as they contain only gates. (Initialization, measurement, I/O and spontaneous decoherence are side effects in quantum computers. Gates however are purely functional and bijective.)

If a function ${\displaystyle F}$ is a product of ${\displaystyle m}$ gates (${\displaystyle F=A_{1}\cdot A_{2}\cdot ...\cdot A_{m}}$), the unitary inverse of the function, ${\displaystyle F^{\dagger }}$, can be constructed:

Because ${\displaystyle (UV)^{\dagger }=V^{\dagger }U^{\dagger }}$ we have, after recursive application on itself

${\displaystyle F^{\dagger }=\left(\prod _{0<i\leq m}A_{i}\right)^{\dagger }=\prod _{0\leq i<m}A_{m-i}^{\dagger }=A_{m}^{\dagger }\cdot ...\cdot A_{2}^{\dagger }\cdot A_{1}^{\dagger }}$

Similarly if ${\displaystyle G=A\otimes B}$ (the function ${\displaystyle G}$ consists of two gates ${\displaystyle A}$ and ${\displaystyle B}$ in parallel) then ${\displaystyle G^{\dagger }=(A\otimes B)^{\dagger }=A^{\dagger }\otimes B^{\dagger }}$

The dagger operator (${\displaystyle \dagger }$) is the complex conjugate of the transpose. It is also called the Hermitian adjoint.

Gates that are their own unitary inverses are called Hermitian or self-adjoint operators. Some elementary gates such as the Hadamard and the Pauli gates are Hermitian operators, while others like the phase shift (e.g. S, T) and the Ising (XX) gates are not.

Since ${\displaystyle F}$ is a unitary matrix, ${\displaystyle F^{\dagger }F=FF^{\dagger }=I}$ and ${\displaystyle F^{\dagger }=F^{-1}}$

For example, an algorithm for addition can in some situations be used for subtraction, if it is being "run in reverse", as its unitary inverse. The inverse quantum fourier transform is the unitary inverse. Unitary inverses can also be used for uncomputation. Programming languages for quantum computers, such as Microsofts Q#[9] and Bernhard Ömer's QCL, contain function inversion as programming concepts.

The Hadamard-CNOT gate, which when given the input ${\displaystyle |00\rangle }$ produces a Bell state.

If two quantum registers are entangled (their combined state cannot be expressed as a tensor product), measurement of one register affects or reveals the state of the other register by partially or entirely collapsing its state too. This effect can be used for computation, and is used in many algorithms.

The Hadamard-CNOT combination acts on the zero-state as follows: ${\displaystyle CNOT(H\otimes I)|00\rangle ={\Bigg (}{\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}}{\Big (}{\frac {1}{\sqrt {2}}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}\otimes {\begin{bmatrix}1&0\\0&1\end{bmatrix}}{\Big )}{\Bigg )}{\begin{bmatrix}1\\0\\0\\0\end{bmatrix}}={\frac {1}{\sqrt {2}}}{\begin{bmatrix}1\\0\\0\\1\end{bmatrix}}={\frac {|00\rangle +|11\rangle }{\sqrt {2}}}}$

This resulting state is the Bell state ${\displaystyle {\frac {|00\rangle +|11\rangle }{\sqrt {2}}}={\frac {1}{\sqrt {2}}}{\begin{bmatrix}1\\0\\0\\1\end{bmatrix}}}$. It cannot be described as a tensor product of two qubits. There is no solution for

${\displaystyle {\begin{bmatrix}a\\b\end{bmatrix}}\otimes {\begin{bmatrix}c\\d\end{bmatrix}}={\frac {1}{\sqrt {2}}}{\begin{bmatrix}ac\\ad\\bc\\bd\end{bmatrix}}={\frac {1}{\sqrt {2}}}{\begin{bmatrix}1\\0\\0\\1\end{bmatrix}}}$

because for example ${\displaystyle c}$ needs to be both non-zero and zero in the case of ${\displaystyle ac}$ and ${\displaystyle bc}$.

The quantum state spans the two qubits. This is called entanglement. The GHZ state is a similar entangled quantum state that spans three or more qubits. Measuring one of the two qubits that make up this Bell state will result in that the other qubit logically must have the same value, both must be the same: Either it will be found in the state ${\displaystyle |00\rangle }$, or in the state ${\displaystyle |11\rangle }$. If we measure one of the qubits to be for example ${\displaystyle |1\rangle }$, then the other qubit must also be ${\displaystyle |1\rangle }$, because their combined state became ${\displaystyle |11\rangle }$. Measurement of one of the qubits collapses the entire quantum state, that span the two qubits.

This type of value-assignment occurs instantaneously over any distance and this has as of 2018 been experimentally verified by QUESS for distances of up to 1200 kilometers.[11][12][13] That the phenomena appears to happen instantaneously as opposed to the time it would take to traverse the distance separating the qubits at the speed of light is called the EPR paradox, and it is an open question in physics how to resolve this. Originally it was solved by giving up the assumption of local realism, but other interpretations have also emerged. For more information see the Bell test experiments. The no-communication theorem proves that this phenomena cannot be used for faster-than-light communication of classical information.

The effect of a unitary transform F on a register A that is in a superposition of ${\displaystyle 2^{n}}$ states and pairwise entangled with the register B. Here, ${\displaystyle n}$ is 3 (each register has 3 qubits).

A register is a sequence of qubits, similar in analogy to a classic CPU register.

Take a register A with ${\displaystyle n}$ qubits all initialized to ${\displaystyle |0\rangle }$, and feed it through a parallel Hadamard gate ${\displaystyle \bigotimes _{1}^{n}H}$. Register A will then enter the state ${\displaystyle {\frac {1}{\sqrt {2^{n}}}}\sum _{i=0}^{2^{n}-1}|i\rangle }$ that have equal probability of when measured to be in any of its ${\displaystyle 2^{n}}$ possible states; ${\displaystyle |0\rangle }$ to ${\displaystyle |2^{n}-1\rangle }$. Take a second register B, also with ${\displaystyle n}$ qubits initialized to ${\displaystyle |0\rangle }$ and pairwise CNOT its qubits with the qubits in register A, such that for each ${\displaystyle i}$ the qubits ${\displaystyle A_{i}}$ and ${\displaystyle B_{i}}$ forms the state ${\displaystyle |A_{i}B_{i}\rangle ={\frac {|00\rangle +|11\rangle }{\sqrt {2}}}}$

If we now measure the qubits in register A, then register B will be found to contain the same value as A. If we however instead apply a quantum logic gate ${\displaystyle F}$ on A and then measure, then ${\displaystyle |A\rangle =F|B\rangle \iff F^{\dagger }|A\rangle =|B\rangle }$, where ${\displaystyle F^{\dagger }}$ is the unitary inverse of ${\displaystyle F}$.

Because of how unitary inverses of gates act, ${\displaystyle F^{\dagger }|A\rangle =F^{-1}(|A\rangle )=|B\rangle }$. For example, say ${\displaystyle F(x)=x+3{\pmod {2^{n}}}}$, then ${\displaystyle |B\rangle =|A-3{\pmod {2^{n}}}\rangle }$.

Even though the equalities holds, the probabilities of measuring the possible values may change as a result of applying ${\displaystyle F}$, as may be the intent in a quantum search algorithm.

The equality will hold no matter in which order measurement is performed (on the registers A or B). Measurement can even be randomly and concurrently interleaved qubit by qubit, since the measurements assignment of one qubit will limit the possible value-space from the other entangled qubits.

This effect of value-sharing via entanglement is used in Shor's algorithm and in phase estimation. Using the Fourier transform to amplify the probability amplitudes of the solution states for some problem is a generic method known as "Fourier fishing". See also amplitude amplification.

• Nielsen, Michael A.; Chuang, Isaac (2000). Quantum Computation and Quantum Information. Cambridge: Cambridge University Press. ISBN 0521632358. OCLC 43641333.