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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.

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  • $\begingroup$ 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 $|\langle\psi|\phi\rangle|=1$. $\endgroup$ – Mark S Oct 17 at 19:56
  • $\begingroup$ Hey @MarkS thanks. I will give improve my post $\endgroup$ – User5 Oct 17 at 20:17
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The key to understanding many quantum protocols and circuits is in the following circuit: enter image description here

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.

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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.

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