I've been reading the wikipedia page on the SWAP test, and am particularly confused on the last step of the explanation of the circuit.

Swap test explanation

I understand every step, except for the last one:

$P(0) = \frac{1}{4} \left ( \left < \phi \right | \left < \psi \right | + \left < \psi \right | \left < \phi \right | \right ) \left ( \left | \phi \right > \left | \psi \right > + \left | \psi \right > \left | \phi \right > \right )$.

How exactly are we going from a quantum state to an inner product of two quantum states? Can someone provide a more mathematically full explanation of this step?



1 Answer 1


Please look at picture of the circuit.

enter image description here

The state in the second last step, is the state after the second $H$ gate. Now, we only have to measure, and we are interested in the probability of the two possible outcomes $0$ and $1$.

This is done via the formalism of the partial measurement. Let $(n+1)$-qubits be in state $$ |0\rangle|\Psi_0\rangle_n + |1\rangle|\Psi_1\rangle_n, $$ where $\langle \Psi_0 |\Psi_0\rangle + \langle \Psi_1|\Psi_1\rangle = 1$.

If we measure the first qubit, then the probability of outcome $0$ is given by $$ P(0) = \langle\Psi_0|\Psi_0\rangle $$ and, similarly for outcome $1$.

This is exactly what is happening here, with $$ |\Psi_0\rangle = \frac{1}{2}(|\psi\rangle|\phi\rangle + |\phi\rangle|\psi\rangle). $$

With this, the computation is simple.

\begin{align*} P(0) &= \langle\Psi_0|\Psi_0\rangle, \\ &= \frac{1}{2}(\langle\psi|\langle\phi| + \langle\phi|\langle\psi|) \frac{1}{2}(|\psi\rangle|\phi\rangle + |\phi\rangle|\psi\rangle). \end{align*} Here, the product of the first and the third term is 1, the product of the second and fourth term is also 1, and the two cross terms have product $\langle \psi|\phi\rangle \langle \psi|\phi\rangle = |\langle \psi|\phi\rangle|^2$.

Putting this all together, we get $$ P(0) = \frac{1}{4}(1 + 1 + 2 |\langle \psi|\phi\rangle|^2), \\ = \frac{1}{2} + \frac{1}{2} |\langle \psi|\phi\rangle|^2. $$

  • $\begingroup$ Oh, I've never heard of the partial measurement formalism. Thanks. $\endgroup$
    – wavosa
    Commented Mar 8, 2023 at 1:50
  • $\begingroup$ I have added additional details. $\endgroup$ Commented Mar 8, 2023 at 2:55
  • 1
    $\begingroup$ "...where $|\Psi_0\rangle_n, |\Psi_0\rangle_n$ are normalized." First, I think you have a typo where one of the $0$ symbols should be a $1$ symbol. Second, these kets are not individually normalized. If that were literally the case you would have both $P(0) = 1$ and $P(1) = 1$. I think what you mean to say is that $\langle\Psi_0|\Psi_0\rangle+\langle\Psi_1|\Psi_1\rangle=1$ $\endgroup$
    – hft
    Commented Mar 8, 2023 at 5:26
  • 1
    $\begingroup$ @hft Thank you for the correction. Fixed. $\endgroup$ Commented Mar 8, 2023 at 6:41

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