Identity gate

The identity gate is the identity matrix, usually written as I, and is defined for a single qubit as

${\displaystyle I={\begin{bmatrix}1&0\\0&1\end{bmatrix}},}$

where I is basis independent and does not modify the quantum state. The identity gate is most useful when describing mathematically the result of various gate operations or when discussing multi-qubit circuits.

Pauli gates (X,Y,Z)

Quantum gates (from top to bottom): Identity gate, NOT gate, Pauli Y, Pauli Z

The Pauli gates ${\displaystyle (X,Y,Z)}$ are the three Pauli matrices ${\displaystyle (\sigma _{x},\sigma _{y},\sigma _{z})}$ and act on a single qubit. The Pauli X, Y and Z equate, respectively, to a rotation around the x, y and z axes of the Bloch sphere by ${\displaystyle \pi }$ radians.^{[lower-alpha 2]}

The Pauli-X gate 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 axis on the Bloch sphere. It is sometimes called a bit-flip as it maps ${\displaystyle |0\rangle }$ to ${\displaystyle |1\rangle }$ and ${\displaystyle |1\rangle }$ to ${\displaystyle |0\rangle }$. Similarly, the Pauli-Y maps ${\displaystyle |0\rangle }$ to ${\displaystyle i|1\rangle }$ and ${\displaystyle |1\rangle }$ to ${\displaystyle -i|0\rangle }$. Pauli Z leaves the basis state ${\displaystyle |0\rangle }$ unchanged and maps ${\displaystyle |1\rangle }$ to ${\displaystyle -|1\rangle }$. Due to this nature, Pauli Z is sometimes called phase-flip.

These matrices are usually represented as

${\displaystyle X=\sigma _{x}=\operatorname {NOT} ={\begin{bmatrix}0&1\\1&0\end{bmatrix}},}$

${\displaystyle Y=\sigma _{y}={\begin{bmatrix}0&-i\\i&0\end{bmatrix}},}$

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

The Pauli matrices are involutory, meaning that the square of a Pauli matrix is the identity matrix.

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

The Pauli matrices also anti-commute, for example ${\displaystyle ZX=iY=-XZ.}$

The matrix exponential of a Pauli matrix ${\displaystyle \sigma _{j}}$ is a rotation operator, often written as ${\displaystyle e^{-i\sigma _{j}\theta /2}.}$

Controlled gates

Circuit representation of controlled-U gate

Controlled gates act on 2 or more qubits, where one or more qubits act as a control for some operation.^[2] 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 Hermitian unitary matrix:

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

The CNOT (or controlled Pauli-X) gate can be described as the gate that maps the basis states ${\displaystyle |a,b\rangle \mapsto |a,a\oplus b\rangle }$, where ${\displaystyle \oplus }$ is XOR.

The CNOT can be expressed in the Pauli basis as:

${\displaystyle {\mbox{CNOT}}=e^{i{\frac {\pi }{4}}(I-Z_{1})(I-X_{2})}=e^{-i{\frac {\pi }{4}}(I-Z_{1})(I-X_{2})}.}$

Being a Hermitian unitary operator, CNOT has the property that ${\displaystyle e^{i\theta U}=(\cos \theta )I+(i\sin \theta )U}$ and ${\displaystyle U=e^{i{\frac {\pi }{2}}(I-U)}=e^{-i{\frac {\pi }{2}}(I-U)}}$, and is involutory.

More generally if U is a gate that operates on a single qubit 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 diagrams of controlled Pauli gates (from left to right): CNOT (or controlled-X), controlled-Y and controlled-Z.

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

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

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

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

The matrix representing the controlled U is

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

When U is one of the Pauli operators, X,Y, Z, the respective terms "controlled-X", "controlled-Y", or "controlled-Z" are sometimes used.^{[4]: 177–185} Sometimes this is shortened to just CX, CY and CZ.

In general, any single qubit unitary gate can be expressed as ${\displaystyle U=e^{iH}}$, where H is a Hermitian matrix, and then the controlled U is ${\displaystyle CU=e^{i{\frac {1}{2}}(I-Z_{1})H_{2}}.}$

Control can be extended to gates with arbitrary number of qubits^[2] and functions in programming languages.^[10] Functions can be conditioned on superposition states.^[11][12]

Classical control

Further information: Deferred Measurement Principle

Example: The qubit ${\displaystyle \phi }$ is measured, and the result of this measurement is a Boolean value, which is consumed by the classical computer. If ${\displaystyle \phi }$ measures to 1, then the classical computer tells the quantum computer to apply the U gate on ${\displaystyle \psi }$.
In circuit diagrams, single lines are qubits, and doubled lines are bits.

Gates can also be controlled by classical logic. A quantum computer is controlled by a classical computer, and behaves like a coprocessor that receives instructions from the classical computer about what gates to execute on which qubits.^{[13]: 42–43 [14]} Classical control is simply the inclusion, or omission, of gates in the instruction sequence for the quantum computer.^{[4]: 26–28 [1]: 87–88}

Phase shift gates

The phase shift is a family of single-qubit gates that map the basis states ${\displaystyle |0\rangle \mapsto |0\rangle }$ and ${\displaystyle |1\rangle \mapsto e^{i\varphi }|1\rangle }$. 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 constant latitude), or a rotation about the z-axis on the Bloch sphere by ${\displaystyle \varphi }$ radians. The phase shift gate is represented by the matrix:

${\displaystyle P(\varphi )={\begin{bmatrix}1&0\\0&e^{i\varphi }\end{bmatrix}}}$

where ${\displaystyle \varphi }$ is the phase shift with the period 2π. Some common examples are the T gate where ${\textstyle \varphi ={\frac {\pi }{4}}}$ (historically known as the ${\displaystyle \pi /8}$ gate), the phase gate (also known as the S gate, written as S, though S is sometimes used for SWAP gates) where ${\textstyle \varphi ={\frac {\pi }{2}}}$ and the Pauli-Z gate where ${\displaystyle \varphi =\pi }$.

The phase shift gates are related to each other as follows:

${\displaystyle Z={\begin{bmatrix}1&0\\0&e^{i\pi }\end{bmatrix}}={\begin{bmatrix}1&0\\0&-1\end{bmatrix}}=P\left(\pi \right)}$

${\displaystyle S={\begin{bmatrix}1&0\\0&e^{i{\frac {\pi }{2}}}\end{bmatrix}}={\begin{bmatrix}1&0\\0&i\end{bmatrix}}=P\left({\frac {\pi }{2}}\right)={\sqrt {Z}}}$

${\displaystyle T={\begin{bmatrix}1&0\\0&e^{i{\frac {\pi }{4}}}\end{bmatrix}}=P\left({\frac {\pi }{4}}\right)={\sqrt {S}}={\sqrt[{4}]{Z}}}$

Note that the phase gate ${\displaystyle P(\varphi )}$ is not Hermitian (except for all ${\displaystyle \varphi =n\pi ,n\in \mathbb {Z} }$). These gates are different from their Hermitian conjugates: ${\displaystyle P^{\dagger }(\varphi )=P(-\varphi )}$. The two adjoint (or conjugate transpose) gates ${\displaystyle S^{\dagger }}$ and ${\displaystyle T^{\dagger }}$ are sometimes included in instruction sets.^[15][16]

Hadamard gate

The Hadamard or Walsh-Hadamard gate, named after Jacques Hadamard (French: [adamaʁ]) and Joseph L. Walsh, acts on a single qubit. It maps the basis states ${\textstyle |0\rangle \mapsto {\frac {|0\rangle +|1\rangle }{\sqrt {2}}}}$ and ${\textstyle |1\rangle \mapsto {\frac {|0\rangle -|1\rangle }{\sqrt {2}}}}$ (it creates an equal superposition state if given a computational basis state). The two states ${\displaystyle (|0\rangle +|1\rangle )/{\sqrt {2}}}$ and ${\displaystyle (|0\rangle -|1\rangle )/{\sqrt {2}}}$ are sometimes written ${\displaystyle |+\rangle }$ and ${\displaystyle |-\rangle }$ respectively. The Hadamard gate performs a rotation of ${\displaystyle \pi }$ about the axis ${\displaystyle ({\hat {x}}+{\hat {z}})/{\sqrt {2}}}$ at the Bloch sphere, and is therefore involutory. It is represented by the Hadamard matrix:

Circuit representation of Hadamard gate

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

If the Hermitian (so ${\displaystyle H^{\dagger }=H^{-1}=H}$) Hadamard gate is used to perform a change of basis, it flips ${\displaystyle {\hat {x}}}$ and ${\displaystyle {\hat {z}}}$. For example, ${\displaystyle HZH=X}$ and ${\displaystyle H{\sqrt {X}}\;H={\sqrt {Z}}=S.}$

Swap gate

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

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

The swap gate can be decomposed into summation form:

${\displaystyle {\mbox{SWAP}}={\frac {I\otimes I+X\otimes X+Y\otimes Y+Z\otimes Z}{2}}}$

Toffoli (CCNOT) gate

Circuit representation of Toffoli gate

The Toffoli gate, named after Tommaso Toffoli and also called the CCNOT gate or Deutsch gate ${\displaystyle D(\pi /2)}$, is a 3-bit gate that 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 CC-U (controlled-controlled Unitary) 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.^[17]

The Toffoli gate is related to the classical AND (${\displaystyle \land }$) and XOR (${\displaystyle \oplus }$) operations as it performs the mapping ${\displaystyle |a,b,c\rangle \mapsto |a,b,c\oplus (a\land b)\rangle }$ on states in the computational basis.

The Toffoli gate can be expressed using Pauli matrices as

${\displaystyle {\mbox{Toff}}=e^{i{\frac {\pi }{8}}(I-Z_{1})(I-Z_{2})(I-X_{3})}=e^{-i{\frac {\pi }{8}}(I-Z_{1})(I-Z_{2})(I-X_{3})}.}$

Deutsch gate

A single-gate set of universal quantum gates can also be formulated using the parametrized three-qubit Deutsch gate ${\displaystyle D(\theta )}$,^[21] named after physicist David Deutsch. It is a general case of CC-U, or controlled-controlled-unitary gate, and is defined as

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

Unfortunately, a working Deutsch gate has remained out of reach, due to lack of a protocol. There are some proposals to realize a Deutsch gate with dipole–dipole interaction in neutral atoms.^[22]

A universal logic gate for reversible classical computing, the Toffoli gate, is reducible to the Deutsch gate ${\displaystyle D(\pi /2)}$, thus showing that all reversible classical logic operations can be performed on a universal quantum computer.

There also exist single two-qubit gates sufficient for universality. In 1996, Adriano Barenco showed that the Deutsch gate can be decomposed using only a single two-qubit gate (Barenco gate), but it is hard to realize experimentally.^[1]: 93 This feature is exclusive to quantum circuits, as there is no classical two-bit gate that is both reversible and universal.^[1]: 93 Universal two-qubit gates could be implemented to improve classical reversible circuits in fast low-power microprocessors.^[1]: 93

Serially wired gates

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 in 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 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.^{[4]: 17–18,22–23,62–64 [5]: 147–169}

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:

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

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

Parallel gates

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.^{[4]: 71–75 [5]: 148}

If we, as in the picture, combine the Pauli-Y gate with the Pauli-X gate in parallel, then this can be written as:

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

Both the Pauli-X and the Pauli-Y gate act on a single qubit. The resulting gate ${\displaystyle C}$ act on two qubits.

Sometimes the tensor product symbol is omitted, and indexes are used for the operators instead.^[25]

Hadamard transform

Further information: Hadamard transform

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; ${\displaystyle |00\rangle }$, ${\displaystyle |01\rangle }$, ${\displaystyle |10\rangle }$, and ${\displaystyle |11\rangle }$. We can write this operation as:

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

Example: The Hadamard transform on a 3-qubit register ${\displaystyle |\psi \rangle }$.

Here the amplitude for each measurable state is 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:

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

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 ${\displaystyle |2^{n}-1\rangle }$.^{[lower-alpha 5]} 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 ${\textstyle |\psi \rangle =\bigotimes _{i=0}^{n-1}|\psi _{i}\rangle }$ as follows:

${\displaystyle \bigotimes _{0}^{n-1}H|\psi \rangle =\bigotimes _{i=0}^{n-1}{\frac {|0\rangle +(-1)^{\psi _{i}}|1\rangle }{\sqrt {2}}}={\frac {1}{\sqrt {2^{n}}}}\bigotimes _{i=0}^{n-1}{\Big (}|0\rangle +(-1)^{\psi _{i}}|1\rangle {\Big )}=H|\psi _{0}\rangle \otimes H|\psi _{1}\rangle \otimes \cdots \otimes H|\psi _{n-1}\rangle }$

Application on entangled states

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 application 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 often appear as just a bare 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 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:

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

Unitary inversion of gates

Example: The unitary inverse of the Hadamard-CNOT product. The three gates ${\displaystyle H}$, ${\displaystyle I}$ and ${\displaystyle \mathrm {CNOT} }$ 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 (i.e. series and parallel combinations) 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 ${\displaystyle U}$ is a unitary matrix, then ${\displaystyle U^{\dagger }U=UU^{\dagger }=I}$ and ${\displaystyle U^{\dagger }=U^{-1}}$. The dagger (${\displaystyle \dagger }$) denotes the conjugate transpose. It is also called the Hermitian adjoint.

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

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

${\displaystyle F^{\dagger }=\left(\prod _{i=1}^{m}A_{i}\right)^{\dagger }=\prod _{i=m}^{1}A_{i}^{\dagger }=A_{m}^{\dagger }\cdot \dots \cdot A_{2}^{\dagger }\cdot A_{1}^{\dagger }}$

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

Gates that are their own unitary inverses are called Hermitian or self-adjoint operators. Some elementary gates such as the Hadamard (H) and the Pauli gates (I, X, Y, Z) are Hermitian operators, while others like the phase shift (S, T, P, CPhase) gates generally are not.

For example, an algorithm for addition can 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 Microsoft's Q#,^[10] Bernhard Ömer's QCL,^[13]: 61 and IBM's Qiskit,^[27] contain function inversion as programming concepts.

The effect of measurement on entangled states

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

If two quantum states (i.e. qubits, or registers) are entangled (meaning that 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:

${\displaystyle \operatorname {CNOT} (H\otimes I)|00\rangle =\left({\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}}\left({\frac {1}{\sqrt {2}}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}\otimes {\begin{bmatrix}1&0\\0&1\end{bmatrix}}\right)\right){\begin{bmatrix}1\\0\\0\\0\end{bmatrix}}={\frac {1}{\sqrt {2}}}{\begin{bmatrix}1\\0\\0\\1\end{bmatrix}}={\frac {|00\rangle +|11\rangle }{\sqrt {2}}}}$

The Bell state in the text is ${\displaystyle |\Psi \rangle =a|00\rangle +b|01\rangle +c|10\rangle +d|11\rangle }$ where ${\displaystyle a=d={\frac {1}{\sqrt {2}}}}$ and ${\displaystyle b=c=0}$. Therefore, it can be described by the plane spanned by the basis vectors ${\displaystyle |00\rangle }$ and ${\displaystyle |11\rangle }$, as in the picture. The unit sphere (in ${\displaystyle \mathbb {C} ^{4}}$) that represent the possible value-space of the 2-qubit system intersects the plane and ${\displaystyle |\Psi \rangle }$ lies on the unit spheres surface. Because ${\displaystyle |a|^{2}=|d|^{2}=1/2}$, there is equal probability of measuring this state to ${\displaystyle |00\rangle }$ or ${\displaystyle |11\rangle }$, and because ${\displaystyle b=c=0}$ there is zero probability of measuring it to ${\displaystyle |01\rangle }$ or ${\displaystyle |10\rangle }$.

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

${\displaystyle {\begin{bmatrix}x\\y\end{bmatrix}}\otimes {\begin{bmatrix}w\\z\end{bmatrix}}={\begin{bmatrix}xw\\xz\\yw\\yz\end{bmatrix}}={\frac {1}{\sqrt {2}}}{\begin{bmatrix}1\\0\\0\\1\end{bmatrix}},}$

because for example w needs to be both non-zero and zero in the case of xw and yw.

The quantum state spans the two qubits. This is called entanglement. 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.

The GHZ state is a similar entangled quantum state that spans three or more 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.^[34][35][36] 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 phenomenon cannot be used for faster-than-light communication of classical information.

Measurement on registers with pairwise entangled qubits

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, n is 3 (each register has 3 qubits).

Take a register A with n qubits all initialized to ${\displaystyle |0\rangle }$, and feed it through a parallel Hadamard gate ${\textstyle H^{\otimes n}}$. Register A will then enter the state ${\textstyle {\frac {1}{\sqrt {2^{n}}}}\sum _{k=0}^{2^{n}-1}|k\rangle }$ 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 n qubits initialized to ${\displaystyle |0\rangle }$ and pairwise CNOT its qubits with the qubits in register A, such that for each p the qubits ${\displaystyle A_{p}}$ and ${\displaystyle B_{p}}$ forms the state ${\displaystyle |A_{p}B_{p}\rangle ={\frac {|00\rangle +|11\rangle }{\sqrt {2}}}}$.

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 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 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 ${\displaystyle |B\rangle =|A-3{\pmod {2^{n}}}\rangle }$.

The equality will hold no matter in which order measurement is performed (on the registers A or B), assuming that F has run to completion. 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.

Even though the equalities holds, the probabilities for measuring the possible outcomes may change as a result of applying F, as may be the intent in a quantum search algorithm.

This effect of value-sharing via entanglement is used in Shor's algorithm, phase estimation and in quantum counting. Using the Fourier transform to amplify the probability amplitudes of the solution states for some problem is a generic method known as "Fourier fishing".^[37]