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

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

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.