Why does the QCD vacuum have zero momentum?

Question

Today in the hadron physics tutorial we were talking about the Fabri-Picasso theorem where one step in the proof involves $\hat P |0\rangle$ which is the four-momentum operator on the vacuum. And from the notation I think that it is the non-interacting vacuum instead of the interacting vacuum $|\Omega\rangle$.

So why is this the case?

The first thought is that there are no particles and therefore nothing can carry a momentum. This seems a bit too easy as QFT knows vacuums that do interact. So $\hat P |\Omega\rangle$ could be non-zero, right?

One argument that I came up with is the transformation behavior under Lorentz transformations. If the vacuum had a non-zero total momentum that would be a distinguished direction. Under a spatial rotation it would transform as a vector. But the vacuum has been a scalar under the Lorentz transformation, so it should be a scalar. And a scalar can only have zero three-momentum.

Another idea I had was writing $\mathbf 1 = \sum_p | p \rangle\langle p|$ and applying that to the vacuum. The overlap $\langle p| 0 \rangle$ should be zero again as a state with a particle with a definite momentum should be orthogonal to the vacuum.

We could not find a convincing argument which did not look like circular reasoning or a mere “vacuum is defined that way”. Defining the total momentum as the sum of the particle's momenta and then saying that the vacuum has particle number zero would do the trick.

Is there some fundamental reason for the vacuum to have vanishing total momentum or is it just a matter of definition?

Answer 1

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 $\lvert 0 \rangle$.

Answer 2

@ACuriousMind 's answer is pretty straightforward. As an alternate proof consider the following:

• Note that $\hat P^\mu$ is time-independent$^1$, which means that it commutes with $\hat H$. So it commutes with $\exp[-i\hat HT]\ \forall T\in \mathbb C$.
• We know$^2$ that $|\Omega\rangle \propto \lim_{T\to\infty} \exp[-i\hat HT]|0\rangle$. Using this, its easy to see that $$ \hat P^\mu |\Omega\rangle\propto \lim_{T\to\infty} \hat P^\mu\mathrm e^{-i\hat HT}|0\rangle=\lim_{T\to\infty} \mathrm e^{-i\hat HT}\hat P^\mu|0\rangle=0 $$ where I used $[\hat H,\hat P^\mu]=0$ and $\hat P^\mu|0\rangle=0$.

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$^1$ here, $\hat P^\mu$ and $\hat H$ are the total (interacting) momentum-hamiltonian operators. They are time-independent because they are the charges of the Noether current associated to space-time translations.

$^2$ See Quantum Field Theory I, lecture notes by Timo Weigand, page 57.