# How and why does swap test works?

I am having some trouble understanding why a SWAP test would work. I meant I read that and understood the concepts as follows:

• If the two input states are equal, the output register always results in a state of $$|1\rangle$$, so a $$1$$ outcome will be obtained when applying a READ to this register.

• However, as the two inputs become increasingly more different, the probability of READing a $$1$$ outcome in the output register decreases

But my question is why?

I meant how can a third qubit being $$|0\rangle$$ or $$|1\rangle$$ assure us about the probabilities of them being the same or different? It could only tell us if they are swapped or not.

• Hi User5! Welcome to QCSE. I think this is a fine question, and I've edited it lightly for readability; however, your comment that "you should find that if the two input states are equal, the output register always results in a state of $|1\rangle$" is a little nonstandard. For example, have you reviewed the Wiki article on the SWAP test? That emphasizes the third register (the control register) would be in $|0\rangle$ if $|\langle\psi|\phi\rangle|=1$. Oct 17, 2019 at 19:56
• Hey @MarkS thanks. I will give improve my post Oct 17, 2019 at 20:17

## What that cSWAP test does (and doesn't) do

The important thing about the controlled-SWAP test is that what it does isn't just to SWAP, or to not SWAP, the two inputs. The controlled-SWAP test involves a control qubit which is in a superposition of $$\def\ket#1{\lvert#1\rangle}\def\bra#1{\langle#1\rvert}\ket{0}$$ and $$\ket{1}$$: that is, we measure the first qubit of the state $$\mathrm{cSWAP} \,\Bigl(\ket{+} \otimes \ket{\alpha} \otimes \ket{\beta}\Bigr)$$ in the basis $$\ket{+}, \ket{-}$$; or to put it in a different way, we measure the first qubit of the state $$\bigl(H \otimes \mathbf 1 \otimes \mathbf 1\bigr)\,\mathrm{cSWAP}\,\bigl(H \otimes \mathbf 1 \otimes \mathbf 1\bigr) \,\Bigl(\ket{0} \otimes \ket{\alpha} \otimes \ket{\beta}\Bigr)$$ in the standard basis, to obtain information about how similar $$\ket{\alpha}$$ and $$\ket{\beta}$$ are.

Why do we do this? Because what this does is to measure how symmetric the state $$\ket{\alpha} \otimes \ket{\beta}$$ is.

## The cSWAP test as a simulated two-qubit measurement

The SWAP operator is Hermitian: that is, it has eigenvalues ±1. Obviously, any state of the form $$\ket{\alpha} \otimes \ket{\alpha}$$ is a +1 eigenstate: these span a space of dimension 3, which is spanned by $$\ket{0}\ket{0}$$, $$\ket{1}\ket{1}$$, and $$\ket{+}\ket{+}$$ — or, equivalently, $$\ket{00}$$, $$\ket{11}$$, and $$\tfrac{1}{\sqrt 2}\bigl(\ket{01} + \ket{10}\bigr)$$. The remaining state orthogonal to this space is $$\ket{\Psi^-} = \tfrac{1}{\sqrt 2}\bigl(\ket{01} - \ket{10}\bigr)$$ which, up to scalar factors, we can equivalently write as $$\ket{\Psi^-} \propto \tfrac{1}{\sqrt 2}\bigl(\ket{\alpha}\ket{\alpha^\perp} - \ket{\alpha^\perp}\ket{\alpha}\bigr) \,.$$ What the controlled-SWAP test does is to simulate a measurement of $$\bigl\{ \Pi_{\textrm{sym}}, \ket{\Psi^-}\!\bra{\Psi^-} \bigr\}$$ on the two qubits, where $$\Pi_{\textrm{sym}}$$ is the projector onto the span of all symmetric states $$\ket{\alpha}\ket{\alpha}$$ described above.

More specifically, the cSWAP test simulates measurement of the SWAP operator as an observable (as opposed to a unitary transformation). The SWAP operator has eigenvalues ±1, after all — in principle, we can measure the two qubits to see whether they are (or they collapse to) a +1 eigenstate.

The way that the cSWAP test does this is by using cSWAP in place of SWAP, to realise a phase kick, from the two qubits acted on up to the control. For the eigenstates of SWAP, this records the eigenvalue in the phase of the control; performing a Hadamard then converts this information into a $$\ket{0}$$ (for symmetric states) or a $$\ket{1}$$ (for the antisymmetric state $$\ket{\Psi^-}$$). More generally, it collapses the target states either to the symmetric subspace, or the state $$\ket{\Psi^-}$$. The procedure is literally performing a measurement of how symmetric (or antisymmetric) the state is, and producing the result $$\ket{0}$$ or $$\ket{1}$$ with some probability accordingly.

## The maximum probability of failing the cSWAP test

Considering the procedure in this way, it's easy to see that the maximum probability of obtaining the outcome $$\ket{1}$$ is 0.5: this is the probability of the state $$\ket{\alpha}\ket{\alpha^\perp}$$ collapsing to the antisymmetric state $$\ket{\Psi^-} \propto \tfrac{1}{\sqrt 2}\bigl(\ket{\alpha}\ket{\alpha^\perp} - \ket{\alpha^\perp}\ket{\alpha}\bigr)$$, upon measurement.

• Does $|{\Psi^-}\rangle\langle{\Psi^-}|$ span the set of all antisymmetric states? Nov 9, 2021 at 8:28

The key to understanding many quantum protocols and circuits is in the following circuit: This is especially true in the case where $$U^2=I$$, such that $$U$$ has eigenvalues $$\pm1$$. You can readily calculate that if the input, $$|\psi\rangle$$, of the second qubit has an amplitude $$\alpha_+$$ for being supported on the $$+1$$ eigenspace, then at the end of the algorithm, the probability of the first qubit being in the $$|0\rangle$$ state is $$|\alpha_+|^2$$.

So, consider the case of $$U$$ being swap. Swapping twice is certainly the identity, so swap has $$\pm1$$ eigenvalues. You'll also agree that two copies of the same state, when swapped, are the same as the input, so they correspond to a $$+1$$ eigenvalue. $$\text{SWAP}\cdot|\psi\rangle|\psi\rangle=|\psi\rangle|\psi\rangle$$ The more different the two input states, the greater the difference between the input and the output of the swap, i.e. the less support there is on the $$+1$$ eigenspace there is, and the more likely you are to get the 1 outcome on measurement: $$|\psi\rangle|\phi\rangle=\frac{1}{2}(|\psi\rangle|\phi\rangle+|\phi\rangle|\psi\rangle)+\frac{1}{2}(|\psi\rangle|\phi\rangle-|\phi\rangle|\psi\rangle)$$ In the equation above, I've split the input state into two terms: those from the symmetric (+1 eigenvalue) subspace and those from the antisymmetric (-1 eigenvalue) subspace. The lengths of the two vectors convey the probability amplitudes for being found in either subspace.