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vacuum needn't be unique, actually
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ACuriousMind
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It's by definition. A vacuum state is defined to be Poincaré invariant, since there should be only one vacuum state and it should not depend on the frame (in special relativistic QFT; you get frame-dependent vacua in QFT in curved spacetime).

If it had non-zero momentum, it would not be invariant under rotations and boosts, for instance.

For the non-interacting vacuum, you can also easily see this: The vacuum is by definition the state that gives zero when any annihilation operator is applied to it - it is the "empty state". The mode expansion of the momentum operator of a non-interacting theory is $$ P^\mu = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}p^\mu a^\dagger(\vec p) a(\vec p)$$ (for a scalar field, neglecting a vacuum energy term for $P^0$), and so applying this to the non-interacting vacuum gives zero since the $a(\vec p)$ just give zero when acting on $\lvert 0 \rangle$.

A vacuum state is defined to be Poincaré invariant, since there should be only one vacuum state and it should not depend on the frame (in special relativistic QFT; you get frame-dependent vacua in QFT in curved spacetime).

If it had non-zero momentum, it would not be invariant under rotations and boosts, for instance.

It's by definition. A vacuum state is defined to be Poincaré invariant, since it should not depend on the frame (in special relativistic QFT; you get frame-dependent vacua in QFT in curved spacetime).

If it had non-zero momentum, it would not be invariant under rotations and boosts, for instance.

For the non-interacting vacuum, you can also easily see this: The vacuum is by definition the state that gives zero when any annihilation operator is applied to it - it is the "empty state". The mode expansion of the momentum operator of a non-interacting theory is $$ P^\mu = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}p^\mu a^\dagger(\vec p) a(\vec p)$$ (for a scalar field, neglecting a vacuum energy term for $P^0$), and so applying this to the non-interacting vacuum gives zero since the $a(\vec p)$ just give zero when acting on $\lvert 0 \rangle$.

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ACuriousMind
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A vacuum state is defined to be Poincaré invariant, since there should be only one vacuum state and it should not depend on the frame (in special relativistic QFT; you get frame-dependent vacua in QFT in curved spacetime).

If it had non-zero momentum, it would not be invariant under rotations and boosts, for instance.