# Why momentum operator anticommutes with parity?

In studying continuous geometrical symmetries, we found that the conservation of momentum is a consequence of translational symmetry. In quantum mechanics it means that the momentum operator is a generator of translation, whose infinitesimal transformation is $$U = 1 + (i/\hbar) \ \epsilon p$$

• Are you cool with $\Pi p \Pi^{-1} =-p$? Commented Apr 9, 2020 at 18:29

Parity is the transformation that makes $$x\to-x$$, $$y\to-y$$ and $$z\to-z$$, so the gradient changes as $$\nabla=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\to-\nabla=\left(\frac{\partial}{\partial (-x)},\frac{\partial}{\partial (-y)},\frac{\partial}{\partial (-z)}\right).$$
Then, since the momentum operator in coordinate representation is $$\hat{P}\psi(x,y,z)=-i\hbar\nabla\psi$$
the parity transformation on the momentum changes it to $$\hat{P}\to-\hat{P}\sim-i\hbar(-\nabla)$$.
Edit: perhaps an easier way of seeing this would be looking at the action of parity $$\Pi$$ on momentum eigenstates: Since $$\Pi|x\rangle=|-x\rangle$$
$$\Pi|p\rangle=\Pi\int|x\rangle\langle x|p\rangle\mathrm{d}x=\int|-x\rangle\langle x|p\rangle\mathrm{d}x=\int|x\rangle\langle -x|p\rangle\mathrm{d}x,$$ and since $$\langle x|p\rangle\sim\exp(-ip\cdot x),$$ we have $$\langle -x|p\rangle=\langle x|-p\rangle$$ and then $$\Pi|p\rangle=|-p\rangle.$$
• It does, as $(\partial_x,\partial_y)\to(-\partial_x,-\partial_y)$. I edited my answer to add another approach. Commented Apr 10, 2020 at 21:00