In the Hadamard test (e.g., page 40 of these lecture notes) we have:

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But if you look at standard textbook reference, like Nielsen and Chuang, there's an example for how to compute the measurement probability of a single qubit in a multi-qubit system:

For the two qubit state: $$\alpha_{00} |00\rangle + \alpha_{01} |01\rangle + \alpha_{10} |10\rangle + \alpha_{11} |11\rangle$$ the measurement probability for the first qubit to be zero is $p(0) = |\alpha_{00}|^2 + |\alpha_{01}|^2.$

Essentially, you take the amplitudes in front of the terms with $|0\rangle$ in the first qubit and sum their measurement probabilities.

So, back to the Hadamard test, we have the state: $$ \frac{1}{2} (|0\rangle \otimes (|\psi\rangle + U|\psi\rangle)) + \frac{1}{2} (|1\rangle \otimes (|\psi\rangle - U|\psi\rangle)) \\ = \frac{1}{2}|0\rangle|\psi\rangle + \frac{1}{2} |0\rangle U |\psi\rangle + \frac{1}{2}|1\rangle|\psi\rangle - \frac{1}{2}|1\rangle U |\psi\rangle $$

Generalizing from the textbook example, why is it not the case that the measurement probability for the first qubit to be zero is: $$ p(0) = \bigg|\frac{1}{2}\bigg|^2 + \bigg|\frac{1}{2}\bigg|^2 = \frac{1}{2}, $$ but instead is $p(0) = \frac{1}{2} (1 + \text{Re} \langle\psi| U |\psi\rangle)$?


1 Answer 1


In both cases you compute the probability by computing the square norm of the state after projection on the first qubit outcome you're considering.

In the first example, you can rewrite your state as $$|0\rangle(\alpha_{00}|0\rangle+\alpha_{01}|1\rangle)+ \cdots$$ Thus correspondingly to the first qubit being $|0\rangle$ the second qubit is in the state $(\alpha_{00}|0\rangle+\alpha_{01}|1\rangle)$. Being $|0\rangle$ and $|1\rangle$ orthonormal, the norm of this vector is just the sum of the squared moduli of the coefficients: $$ \| (\alpha_{00}|0\rangle+\alpha_{01}|1\rangle) \|^2 = |\alpha_{00}|^2+|\alpha_{01}|^2. $$ Same identical reasoning goes for the Hadamard test. The state of the second register conditional to the first one being $|0\rangle$ is, $$\frac12(|\psi\rangle + U|\psi\rangle).$$ The squared norm of this object gives the probability you're looking for. Though in this case the expression is not written as a sum over orthonormal basis states, hence the norm isn't just the sum of the square moduli of the coefficients. In this case you can write the norm observing that $$\||\psi\rangle+U|\psi\rangle\|^2 = \| (U+I)|\psi\rangle\|^2 \equiv \langle\psi|(U+I)^\dagger(U+I)|\psi\rangle \\ = 2 + \langle \psi|U^\dagger+U|\psi\rangle,$$ from which you get the standard expression.

In summary, the rule you mention only works when the coefficients are attached to orthonormal states, and $|\psi\rangle$ and $U|\psi\rangle$ aren't generally orthogonal.

  • $\begingroup$ Thanks for your answer! I can now see how the math works out. Conceptually though, I still feel a little confused. If I measure the state $\frac{1}{2}|0\rangle|\psi\rangle + \frac{1}{2} |0\rangle U |\psi\rangle + \frac{1}{2}|1\rangle|\psi\rangle - \frac{1}{2}|1\rangle U |\psi\rangle$ and I get the first qubit to be |0>, then the second qubit is either in state $| \psi \rangle$ or $U | \psi \rangle$, which are both normalized. Why is it that the possibility for the 2nd qubit to be in one of two non-orthonormal states affects the measurement probability of the 1st qubit? $\endgroup$
    – Count Ably
    Commented Mar 30 at 17:18
  • 1
    $\begingroup$ @CountAbly it's not in either of those states. It's in their superposition, that is, $|\psi\rangle+U|\psi\rangle$ (properly renormalised) $\endgroup$
    – glS
    Commented Mar 30 at 17:23

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