6
$\begingroup$

Can we show that the integral over all Haar states $|\psi \rangle $ is $$ \int |\psi \rangle \, \mathrm{d}\psi = 0~. $$

This is an integral over Haar vectors

Reference to a post about what is Haar state

$\endgroup$

1 Answer 1

6
$\begingroup$

Yes, we can show this using the unitary invariance of the Haar measure on states. In more detail, we have $$ U \int |\psi\rangle\, \mathrm{d}\psi = \int U|\psi\rangle\, \mathrm{d}\psi = \int |\psi\rangle\, \mathrm{d}(U^\dagger\psi) = \int |\psi\rangle\, \mathrm{d}\psi. $$ Hence, the integral is a fixed point of any $U\in U(d)$. However, the only vector fixed by all unitaries is the zero vector (since $U(d)$ acts irreducibly), hence $$ \int |\psi\rangle\, \mathrm{d}\psi = 0. $$

PS: Geometrically, this is basically an integral over the complex sphere in $\mathbb C^d$, so it might be intuitively clear that it has to be zero by symmetry. This is exactly the unitary symmetry I have used above. I say "basically" because $\psi$ is actually a ray in $\mathbb C^d$, so it should live in complex projective space $P\mathbb C^{d-1}$. However, this detail does not matter here.

$\endgroup$
1
  • $\begingroup$ This is beautiful. $\endgroup$
    – R.W
    Mar 10, 2022 at 10:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.