# Why $Q|0\rangle=0$ where $Q$ generates a symmetry?

Quote: "Concepts of Elementary Particle Physics" by Michael E. Peskin

In quantum mechanics with a finite number of coordinates, it can be shown that, if $$Q$$ generates a symmetry of the theory, then the ground states of the theory $$|0\rangle$$ must obey $$Q|0\rangle =0$$.

I suppose that $$[Q,H]=0$$, but what did the book mean by $$Q$$ (an operator?) generated symmetry? And how to prove that $$Q|0\rangle=0$$?

• The key is "if $Q$ generates a symmetry". If we take $Q$ that generates a symmetry, what disqualifies $Q+qI$ from generating a symmetry, with $I$ the identity and $q$ a scalar with the right units? – Sean E. Lake Dec 5 '19 at 7:46

Most likely here $$Q$$ is the generator of infinitesimal transformations that is a symmetry of your Hamiltonian. A finite transformation is then given by
$$g(q) = \exp(i q Q) = I + q Q + \mathcal O(q^2)$$ for some real number $$q$$. For example if $$Q$$ is the momentum operator, $$g(q)$$ is the translation operator (translation of distance $$q$$). If $$Q$$ is angular momentum along the $$z$$-axis, $$g(q)$$ is a rotation along $$z$$-axis of angle $$q$$, etc.
If the ground-state is invariant under this symmetry then we must have $$g(q)|0\rangle = |0\rangle,$$ for any $$q$$. This is true if $$Q|0\rangle = 0.$$
In more mathematical language, $$Q$$ belongs to the Lie Algebra of your symmetry group while $$g(q)$$ belongs to the Lie Group. The fact that the ground-state is invariant, means it transforms as the trivial representation of the Lie Group.