What do smeared field operators act on?

If we have a fock space and a vacuum state $$|0\rangle$$, it is obvious what a field operator does. It acts on this as though a position eigenstate and creates a particle, which is made more rigorous by expanding $$\phi(\vec{x}, t)$$ into a superposition of creation and annihilation operators.

Now I'm confused about how we can translate this into the smeared case. When we take $$\int dx f(x) \phi(x)$$, how does

$$\phi(f)|0\rangle$$

look? Does it allow the same expansion? I don't know much about operator valued distributions in QFT, only reading up to second quantization. Any help is appreciated.

It acts on this as though a position eigenstate and creates a particle [...]

You must be careful here - as you should recall from nonrelativistic quantum mechanics, position eigenstates don't actually exist as physical states. Their "wavefunctions" are delta functions, which are not square-integrable and therefore do not correspond to allowed states. The same is true for quantum field operators, where $$\hat\phi(x)$$ is a singular object for the same reason that $$|x\rangle$$ is.

Sweeping such subtleties under the rug, $$\hat\phi^\dagger(x)$$ should be thought of as a position-space creation operator which acts on the vacuum $$|0\rangle$$ to produce $$|x\rangle$$. In that sense, it would be somewhat more consistent to call this operator $$\hat \phi^\dagger_x$$, but conventions vary.

As such, the operator-valued distribution $$\Phi^\dagger$$ eats a (square-integrable) function $$f$$ and produces a single-particle state with $$f$$ as its wavefunction:

$$\Phi^\dagger(f)|0\rangle:=\int dx\ f(x) \hat\phi^\dagger(x) |0\rangle = \int dx \ f(x) |x\rangle\equiv |f\rangle$$

More generally, it acts on an $$n$$-particle state to produce an $$(n+1)$$-particle state by adding $$|f\rangle$$ in the appropriately symmetrized or antisymmetrized way. Explicitly, if we are working with a bosonic field, the action of $$\Phi^\dagger(f)$$ on the one-particle state $$|\psi\rangle = \int dx \ \psi(x)|x\rangle$$ would be

$$\Phi^\dagger(f)|\psi\rangle = \frac{1}{2}\left(|f\rangle\otimes |\psi\rangle + |\psi\rangle\otimes |f\rangle\right) = \int dx \int dy \ \frac{f(x)\psi(y)+\psi(x)f(y)}{2}|x\rangle\otimes |y\rangle$$

The adjoint operator $$\Phi(f)$$ would annihilate the vacuum; its action on one and two-particle states would be $$\Phi(f)|\psi\rangle = \langle f|\psi\rangle$$ $$\Phi(f)\left[\frac{|\psi\rangle\otimes|\rho\rangle +|\rho\rangle\otimes |\psi\rangle}{2}\right]= \frac{|\rho\rangle \cdot \langle f|\psi\rangle+ |\psi\rangle \cdot \langle f|\rho\rangle }{2}$$

One can show more generally that $$[\Phi(f),\Phi^\dagger(g)] = i\hbar \int dx\ f^*(x) g(x)$$; if we plug in delta functions, we find

$$[\hat \phi(x),\hat \phi^\dagger(y)] = [\Phi(\delta_x),\Phi^\dagger(\delta_y)]=i\hbar \int dz\ \delta(x-z) \delta (y-z) = i\hbar \delta(x-y)$$

in accordance with the field-operator commutation relations we already know.

• Oh nice, conceptually this makes much more sense. The only thing I'm still confused about is what the action of the operator actually looks like. How do you get from LHS to RHS of $$\phi(f)|0\rangle = \int dx f(x) | x \rangle?$$ Jul 15 '21 at 18:17
• Or maybe have I misinterpreted the math? Jul 15 '21 at 18:17
• @rage_man If you're comfortable with the idea that $\hat\phi(x)|0\rangle=|x\rangle$, then that expression follows as an immediate consequence. Jul 15 '21 at 18:31
• Sorry, I'm not seeing how to make this work. I see the relationship between integrating the Dirac function and getting a position ket from a localized operator, so I see how the above equation would arise from the unsmeared case when $f=\delta$. But it still isn't clear to me how the equation works mathematically. Jul 15 '21 at 22:32
• Probably I'm missing something obvious, but taking the previous superposition of states in terms of creation and annihilation operators seems much more complicated now that it is inside of an integral and multiplied by an arbitrary $L_2$ function. Jul 15 '21 at 22:35