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 $U = 1 + (i/\hbar) \ \epsilon p$

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

1 Answer 1


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 $$ \Pi|p\rangle=|-p\rangle. $$

  • $\begingroup$ So parity only anticommutes with momentum in 3d but not in 2d? $\endgroup$
    – Zonova
    Commented Apr 10, 2020 at 2:31
  • $\begingroup$ It does, as $(\partial_x,\partial_y)\to(-\partial_x,-\partial_y)$. I edited my answer to add another approach. $\endgroup$ Commented Apr 10, 2020 at 21:00

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