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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 $Q|0\rangle=0$?

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    $\begingroup$ 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? $\endgroup$ – Sean E. Lake Dec 5 '19 at 7:46
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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.

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