How to show that the integral over all Haar states vanishes: $\int|\psi\rangle\,{\rm d}\psi = 0 $?

Question

How can one show that the integral over all Haar states $|\psi \rangle $ is $$ \int |\psi \rangle \, \mathrm{d}\psi = 0\ ? $$

Answer 1

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.