The figure of a circuit and the state are as follows.

The final state before the measurement is $|O_{out}\rangle=\frac{1}{2}|0\rangle(|\phi\rangle|\psi\rangle+|\psi\rangle|\phi\rangle)+\frac{1}{2}|1\rangle(|\phi\rangle|\psi\rangle-|\psi\rangle|\phi\rangle)$.

Measuring the first qubit of this state produces outcome 1, how can I get the probability $\frac12(1-|\langle\phi|\psi\rangle|^2)$?

figure of a circuit

the description of the state


2 Answers 2


It's a fantastic question because the typical measurement intuition we apply no longer is sufficient - it's really necessary to formalize measurement.

Specifically, we create a set of nonlinear operators $M_\psi = |\psi \rangle \langle \psi |$, where the probability of measuring $\psi$ on an arbitrary state $|\phi\rangle $ is $\langle \phi | M^\dagger M | \phi \rangle$.

In our case, we have a measurement operator $M_1$ we are interested in. However, we can actually apply $M_0$ for simplicity, and then subtract this probability from 1. Thus, where $| \varphi \rangle $ is the state provided above:

\begin{align} \langle \varphi | M^\dagger M | \varphi \rangle &= \langle \varphi |0\rangle \langle 0| 0\rangle \langle 0|\varphi\rangle \\ &= \langle \varphi|0\rangle\langle 0|\varphi\rangle \\ &= \frac{1}{4} (\langle \phi|\langle \psi| + \langle \psi|\langle \phi|)( |\phi \rangle |\psi \rangle + |\psi \rangle |\phi \rangle) \\ &= \frac{1}{4}(2 \langle\phi| \langle\psi|\phi\rangle |\psi\rangle + 2) \\ &= \frac{1}{2}\Big(|\langle\psi|\phi\rangle|^2 + 1 \Big) \end{align}

Thus, because this is the zero probability, we have:

$$ 1 - \frac{1}{2}\Big(|\langle\psi|\phi\rangle|^2 + 1 \Big) = \frac{1 - |\langle \psi|\phi \rangle|^2}{2} = \frac{1 - |\langle \phi|\psi \rangle|^2}{2} $$

As desired.

  • $\begingroup$ Thank you, can you specify how can get the result $|\langle\phi|\psi\rangle|^2$ from $2 \langle\phi| \langle\psi|\phi\rangle |\psi\rangle$? That's what I'm confused. $\endgroup$
    – karry
    Oct 19, 2020 at 9:19
  • $\begingroup$ $ \langle \phi | \langle \psi | \phi \rangle |\psi \rangle = \langle \psi |\phi \rangle \langle \phi | \psi \rangle = \langle \phi | \psi \rangle (\langle \phi | \psi \rangle)^*= |\langle \phi | \psi \rangle|^2 $ $\endgroup$
    – C. Kang
    Oct 19, 2020 at 15:36
  • $\begingroup$ That's the point. $ \langle \phi | \langle \psi | \phi \rangle |\psi \rangle = \langle \psi |\phi \rangle \langle \phi | \psi \rangle $ just because $ \langle \phi | \langle \psi | $ are in two separate direct product space? $\endgroup$
    – karry
    Oct 20, 2020 at 2:28
  • 1
    $\begingroup$ $\langle \phi | \langle \psi | \phi \rangle |\psi \rangle = \langle \psi |\phi \rangle \langle \phi | \psi \rangle$ because $\langle \psi | \phi \rangle$ is just a number and so it can be moved in front of $\langle \phi |$ $\endgroup$ Oct 20, 2020 at 14:51
  • $\begingroup$ After my checking calculation, I found that the results of $\langle \phi | \langle \psi | \phi \rangle |\psi \rangle = \langle \psi |\phi \rangle \langle \phi | \psi \rangle$ and $\langle \phi | \langle \psi | \phi \rangle |\psi \rangle = \langle\phi|\phi\rangle\langle\psi|\psi\rangle$ are different, while the calculation result $\langle \phi | \langle \psi | \phi \rangle |\psi \rangle =\langle\phi|\otimes\langle\psi|\cdot|\phi\rangle\otimes|\psi\rangle= \langle\phi|\phi\rangle\langle\psi|\psi\rangle=1$ seems to be right. $\endgroup$
    – karry
    Oct 23, 2020 at 2:52

I assume you're happy with the idea that the state before measurement is $$|O_{out}\rangle=\frac12|0\rangle(|\phi\rangle|\psi\rangle+|\psi\rangle|\phi\rangle)+\frac{1}{2}|1\rangle(|\phi\rangle|\psi\rangle-|\psi\rangle|\phi\rangle).$$ Now you want to measure qubit 1 in the 0/1 basis. There's a couple of different ways you might approach this.

  1. Define the two measurement projectors to be $P_0=|0\rangle\langle 0|\otimes I\otimes I$ (i.e. measure first qubit in 0, and do nothing to the other two), and $P_1=|1\rangle\langle 1|\otimes I\otimes I$. The probability of getting the 0 answer is $\langle O_{out}|P_0|O_{out}\rangle$.

  2. Alternative, rewrite your state as $$ |O_{out}\rangle=\gamma_0|0\rangle|\sigma\rangle+\gamma_1|1\rangle|\tau\rangle, $$ where $|\sigma\rangle$ and $|\tau\rangle$ are properly normalised vectors, and the coefficients $\gamma_{0/1}$ compensate for that normalisation. The probability of getting the answer 0 when you measure the first qubit is $|\gamma_0|^2$. So, we have $$ \gamma_0|\sigma\rangle=\frac12(|\phi\rangle|\psi\rangle+|\psi\rangle|\phi\rangle). $$ Take the inner product of that equation with itself and you get \begin{align*} |\gamma_0|^2&=\frac14(\langle\phi |\langle\psi |+\langle\psi |\langle\phi|)(|\phi\rangle|\psi\rangle+|\psi\rangle|\phi\rangle) \\ &=\frac14(2+2|\langle\phi|\psi\rangle|^2). \end{align*} The actual question requires the probability of getting answer 1. This is left as an exercise for the reader.

  • $\begingroup$ Nice answer, but the question requires to obtain the expression for $|\gamma_1|^2$, not for $|\gamma_0|^2$. $\endgroup$ Oct 19, 2020 at 8:00
  • 3
    $\begingroup$ @DavitKhachatryan Since this could easily be a homework question, the actual calculation is left for the reader (or let's pretend that's what I inteneded ;) ) $\endgroup$
    – DaftWullie
    Oct 19, 2020 at 8:46

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