# Is the norm of a inner product symmetric?

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.

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

## 1 Answer

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)

• 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. Apr 25, 2022 at 4:37
• 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 Apr 25, 2022 at 7:51