Swap test

The swap test is a procedure in quantum computation that is used to check how much two quantum states differ, appearing first in the work of Barenco et al.¹ and later rediscovered by Harry Buhrman, Richard Cleve, John Watrous, and Ronald de Wolf.² It appears commonly in quantum machine learning, and is a circuit used for proofs-of-concept in implementations of quantum computers.³⁴

Formally, the swap test takes two input states ${\displaystyle |\phi \rangle }$ and ${\displaystyle |\psi \rangle }$ and outputs a Bernoulli random variable that is 1 with probability ${\displaystyle \textstyle {\frac {1}{2}}-{\frac {1}{2}}{|\langle \psi |\phi \rangle |}^{2}}$ (where the expressions here use bra–ket notation). This allows one to, for example, estimate the squared inner product between the two states, ${\displaystyle {|\langle \psi |\phi \rangle |}^{2}}$, to ${\displaystyle \varepsilon }$ additive error by taking the average over ${\displaystyle O(\textstyle {\frac {1}{\varepsilon ^{2}}})}$ runs of the swap test.⁵ This requires ${\displaystyle O(\textstyle {\frac {1}{\varepsilon ^{2}}})}$ copies of the input states. The squared inner product roughly measures "overlap" between the two states, and can be used in linear-algebraic applications, including clustering quantum states.⁶

Consider two states: ${\displaystyle |\phi \rangle }$ and ${\displaystyle |\psi \rangle }$. The state of the system at the beginning of the protocol is ${\displaystyle |0,\phi ,\psi \rangle }$. After the Hadamard gate, the state of the system is ${\displaystyle {\frac {1}{\sqrt {2}}}(|0,\phi ,\psi \rangle +|1,\phi ,\psi \rangle )}$. The controlled SWAP gate transforms the state into ${\displaystyle {\frac {1}{\sqrt {2}}}(|0,\phi ,\psi \rangle +|1,\psi ,\phi \rangle )}$. The second Hadamard gate results in $${\displaystyle {\frac {1}{2}}(|0,\phi ,\psi \rangle +|1,\phi ,\psi \rangle +|0,\psi ,\phi \rangle -|1,\psi ,\phi \rangle )={\frac {1}{2}}|0\rangle (|\phi ,\psi \rangle +|\psi ,\phi \rangle )+{\frac {1}{2}}|1\rangle (|\phi ,\psi \rangle -|\psi ,\phi \rangle )}$$

The measurement gate on the first qubit ensures that it's 0 with a probability of

$${\displaystyle P({\text{First qubit}}=0)={\frac {1}{2}}{\Big (}\langle \phi |\langle \psi |+\langle \psi |\langle \phi |{\Big )}{\frac {1}{2}}{\Big (}|\phi \rangle |\psi \rangle +|\psi \rangle |\phi \rangle {\Big )}={\frac {1}{2}}+{\frac {1}{2}}{|\langle \psi |\phi \rangle |}^{2}}$$

when measured. If ${\displaystyle \psi }$ and ${\displaystyle \phi }$ are orthogonal ${\displaystyle ({|\langle \psi |\phi \rangle |}^{2}=0)}$, then the probability that 0 is measured is ${\displaystyle {\frac {1}{2}}}$. If the states are equal ${\displaystyle ({|\langle \psi |\phi \rangle |}^{2}=1)}$, then the probability that 0 is measured is 1.²

In general, for ${\displaystyle P}$ trials of the swap test using ${\displaystyle P}$ copies of ${\displaystyle |\phi \rangle }$ and ${\displaystyle P}$ copies of ${\displaystyle |\psi \rangle }$, the fraction of measurements that are zero is ${\displaystyle 1-\textstyle {\frac {1}{P}}\textstyle \sum _{i=1}^{P}M_{i}}$, so by taking ${\displaystyle P\rightarrow \infty }$, one can get arbitrary precision of this value.

Below is the pseudocode for estimating the value of ${\displaystyle |\langle \psi |\phi \rangle |^{2}}$ using P copies of ${\displaystyle |\psi \rangle }$ and ${\displaystyle |\phi \rangle }$:

Inputs P copies each of the n qubits quantum states ${\displaystyle |\psi \rangle }$ and ${\displaystyle |\phi \rangle }$
Output An estimate of ${\displaystyle |\langle \psi |\phi \rangle |^{2}}$

for j ranging from 1 to P:
    initialize an ancilla qubit A in state ${\displaystyle |0\rangle }$
    apply a Hadamard gate to the ancilla qubit A
    for i ranging from 1 to n: 
        apply CSWAP to ${\displaystyle \psi _{i}}$ and ${\displaystyle \phi _{i}}$ (the ith qubit of the jth copy of ${\displaystyle |\psi \rangle }$ and ${\displaystyle |\phi \rangle }$), with A as the control qubit
    apply a Hadamard gate to the ancilla qubit A
    measure A in the ${\displaystyle Z}$ basis and record the measurement Mj as either a 0 or 1
compute ${\displaystyle s=1-\textstyle {\frac {2}{P}}\textstyle \sum _{i=1}^{P}M_{i}}$.
return ${\displaystyle s}$ as our estimate of ${\displaystyle |\langle \psi |\phi \rangle |^{2}}$