I was reading about the Inversion Test and during the derivation (in Machine Learning with Quantum Computers, Schuld and Petruccione) I find the follwing:

Assume we have $|a\rangle = A|0\rangle$ and $|b\rangle = B|0\rangle$. For the inversion test, run $B^\dagger A|0\rangle$ and measure. The probability of measuring a $|0\rangle$ is given by $|\langle 0 | (B^\dagger A|0\rangle)|^2$. Hence, the expectation value of the projective measurement $M = |0\rangle \langle 0|$ would be (this is a bit confusing, but not the point of this question): $$ \langle 0 |A^\dagger B (|0\rangle \langle 0) | B^\dagger A | 0 \rangle \\ = \langle 0 |A^\dagger B|0\rangle \; \langle 0 | B^\dagger A | 0 \rangle \\ = |\langle 0 | B^\dagger A |0\rangle|^2 \\ = |\langle a| b \rangle|^2 $$

It is the last step I find puzzling, because I believe it should be (since $A|0\rangle = a$ and $\langle 0| B^\dagger = \langle b|$): $$ ... = |\langle b| a \rangle|^2 $$

Is the norm of the inner product symmetric?! I wouldn't know how to prove that.

  • 1
    $\begingroup$ The right word is symmetric, and yes, the absolute value of a (Hermitian) inner product is symmetric. $\endgroup$ Apr 25, 2022 at 7:25
  • $\begingroup$ Thanks! I guess I meant to say 'commutative' but I will use symmetric from here on. $\endgroup$
    – rhundt
    Apr 25, 2022 at 15:20

1 Answer 1


The inner product has the property $\langle a | b\rangle = \langle b | a \rangle^*$, but this is taken care of when you take the norm (of a complex number)

  • $\begingroup$ Just to double check, you are saying that: $|\langle a | b\rangle|^2$ is $(\langle a | b\rangle)^\dagger \langle a | b\rangle$ is $\langle b | a \rangle \langle a | b \rangle$. The inner products are just complex numbers and their order can be changed. Hence, yes, the norm of that inner product is indeed associative. $\endgroup$
    – rhundt
    Apr 25, 2022 at 4:37
  • $\begingroup$ Yes, that's another way to see it. Ultimately its the relationship that $(\langle a | b\rangle)^* = \langle b | a\rangle$ that gives you this property $\endgroup$
    – xzkxyz
    Apr 25, 2022 at 7:51

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