# Eigenstates in QFT and amplitude of a field operator

I've seen in different posts (such as here) that given a field $$\hat{\phi}(x)$$, its eigenstates $$|\phi\rangle$$ are of the form:

$$|\phi\rangle\ = e^{\int dx\phi(x)\hat{\phi}(x)}|0\rangle\tag1$$

I don't understand where this came from.

Also, I have another question. If I had a one particle state $$|\vec{p}\rangle$$, and I want to know the amplitude of some shape of the field, let's call it $$f(x)$$, I guess I have to compute

$${\rm amplitude} =\ \langle f(x)|\vec{p}\rangle \ =\ \langle f(x)|a_{\vec{p}}^\dagger\ |0\rangle \tag2$$

Where $$|0\rangle$$ is the vacuum state. But, could I write equation (2) as:

$${\rm amplitude} = 4\sqrt{\pi}f(x)\exp\{-\frac{1}{2}\int\ d^3d^3yf(x)\Omega_{x, y}f(y)\}\tag3$$

In analogy with harmonic ($$4\sqrt{\pi}f(x) =$$ norm of state in Harmonic oscillator times the 1-degree Hermite's polynomial) and with $$\Omega_{x, y}$$ is the kernel that you can see in Eq. (3) of Interaction term in QFT

• physics.stackexchange.com/q/109343 , for instance Oct 31, 2018 at 15:27
• Yes, sorry. I forgot the vacuum state in Eq. (1). But I still have that questions Oct 31, 2018 at 15:32
• How exactly did you define $|f(x)\rangle$? Is it a function of x? Nov 1, 2018 at 0:52
• @CosmasZachos It's the set of eigenstates of the field $\hat{\phi}$, amin $\hat{\phi}(x)|f(x)\rangle = f(x)|f(x)\rangle$ where $f(x)$ gives you the shape of the field as a function of every point in space Nov 1, 2018 at 18:53
• In anyone of those posts is proven Eq. (1) so I still have that doubt Nov 1, 2018 at 18:56

Your best bet might be Itzykson & Zuber's overdetailed QFT text, insert in 3.1.2. They explain that your overly naive expression (1) yields eigenstates of $$\hat \phi (x)_-$$, not $$\hat \phi (x)$$. The answer you are citing distinctly warns you this is not a field operator eigenstate. (By the way, your $$|f(x)\rangle$$ should really be $$|f\rangle$$, as it covers all xs. The argument is the function, not its value at some x.)

Rather than perish in a nightmare of sequential Fourier transforms, rampant indexing and normalizations to enforce Lorentz invariance and Hermiticity, I'll give you an easy hint.

I'll eliminate the entire infinite array of oscillators of QFT, retaining only one, and remind you of the basic coherent state maneuvers, adumbrating the trail map for your proof involving an infinity of oscillators.

So, just keep a single oscillator, and go to here ; here; and here for technical details. $$\hat \phi$$ here corresponds to $$\hat x$$, $$\hat \phi_+$$ to $$a^\dagger$$, but $$\hat p$$ to the canonical conjugate field $$\pi$$; distinctly not your oscillator label, fluff, for your $$|\vec p\rangle$$, which has reduced to a single value here (along with your x fluff label).

Recall $$[a , a^\dagger ]=1$$. Then,taking x=f for your classical configuration, here, location, $$|f\rangle = N(f) ~ \exp\left ( -\frac{(a^\dagger -f\sqrt{2})^2} {2} \right )|0\rangle \Longrightarrow \hat{~x}~|f\rangle=f|f\rangle,\\ |p\rangle= N(p) ~ \exp \left( \frac{(a^\dagger +ip\sqrt{2})^2}{2} \right ) |0\rangle \Longrightarrow \hat{~p}~|p\rangle=p|p\rangle ~.$$ You may try to fix $$N(x)=e^{x^2/2}/\pi^{1/4}$$ from $$\langle 0|x\rangle$$, the Schrödinger ground state of the oscillator. (Indeed, N is the inverse of the Gaussian!) It is then shown that $$\langle p |x\rangle= e^{ixp}/ \sqrt{2\pi}.$$ The basic maneuver is that a acts like a derivative operator on the $$a^\dagger$$s in the exponentials.

The analog of the conjugate of your amplitude (2) here is $$\langle 0| a |f\rangle = \langle 0|f\sqrt{2}-a^\dagger|f\rangle= f\sqrt{2} ~ \langle 0 |f\rangle =f\sqrt{2}~e^{-f^2/2}/\pi^{1/4}.$$ I am unsure what you would learn from this for an isolated oscillator, and, a fortiori, in scalar field theory, but there it is.

Note that if you had only kept the cross terms in the exponent, i.e., if you had chucked away the quadratic term in the exponent, you would have gotten the very same answer for (2), as the operative shift function of the a is the very same for coherent states in QM!

Of course, the real McCoy here is but the state itself, $$\frac{ a +a^\dagger}{\sqrt{2}} |f\rangle = f |f\rangle .$$ This might help illuminate the role of the quadratic term in the exponent: dropping that quadratic term in the exponent would yield an eigenstate of merely a.

• I just learned the word "adumbrating". Nov 2, 2018 at 4:06
• @DanielSank Crepuscular hijinks... Nov 2, 2018 at 10:48
• Ok, I will study it and try to understand. Really thanks for the answer Nov 7, 2018 at 21:23