# What does it mean to apply a creation or annihilation operator to a free field, e.g. $\langle 0|a(p)\varphi(x)| 0 \rangle$?

I am self studying Quantum Field Theory, and I am starting to get a little lost. So far, I have studied free fields and some basic computations involving them, such as creation and annihilation operators. My understanding is that the free field, $$\varphi$$, models a collection of particles that do not interact. Hence if $$a$$, $$a^\dagger$$ respectively, represent the annihilation and creation operators, then $$a\varphi$$ lowers the number of particles in $$\varphi$$ by 1 and similarly for the creation operator.

In Lemma 13.7.2 of Talagrand's book What Is a Quantum Field Theory?, he gives the formulas $$\langle 0|a(p)a^\dagger(p')| 0 \rangle = (2\pi)^3\delta^{(3)}(p-p') \\ \langle 0|a(p)\varphi(x)| 0 \rangle = \frac{1}{\sqrt{2 \omega_p}}\exp(i(x,p)) \\ \langle 0|\varphi(x)a^\dagger(p)| 0 \rangle = \frac{1}{\sqrt{2 \omega_p}}\exp(-i(x,p)).$$

He provides a proof of these statements, but I am trying to interpret their physical meaning (if they have one at all). For example, in the first case it seems we are calculating the probability that we annihilate a particle of momentum $$p$$ that has been created with momentum $$p'$$? For the last two I have no guess on what they physically mean.

Part of my confusion could be due to the fact that I am a mathematics student, and hence bra-ket notation is very unfamiliar to me.

• Note that an annihilation operator acting to the left, creates a particle. Apply the same logic to the other expressions. Dec 19, 2022 at 1:12
• They are equivalent because they are the complex conjugate of each other since $\phi(x)$ is Hermitian.
– Gold
Dec 19, 2022 at 1:37
• Indeed $\langle p|\phi(x)|0\rangle$ is the probability amplitude of creating a particle with momentum $p$ from the vacuum by applying $\phi(x)$ to it. I'm just not sure if there's something really physical about it. This quantity is useful in some contexts though (it gives you the position space wavefunction of a state, which in QFT is different from the one you met in non-relativistic QM. This will be important latter on in the context of LSZ reduction).
– Gold
Dec 19, 2022 at 1:58
• Well, for starters, as more detailed in the linked answer by @CosmasZachos, $\phi(x)|0\rangle$ is not a state containing a particle localized at $x$ as sometimes people seem to think. In fact, we do not even have a position operator in QFT in order to be able to talk about localized particles. Secondly, the most accurate view on quantum fields is that they are building blocks for observables. In particular they are most clearly introduced as a mathematical construct which allows one to easily write down relativistic interactions. This is the point of view in Weinberg's textbook for example.
– Gold
Dec 19, 2022 at 2:29
• I admit I don't understand at all what you mean by "maps a state into its state into the field". Also, I didn't mean that to measure an observable you somehow need to apply the field to the state. I meant observables, like the Hamiltonian, are constructed from $\phi$. They are functions of $\phi$ if you will.
– Gold
Dec 19, 2022 at 3:27

When you hit $$|0\rangle$$ with $$\hat{a}^{\dagger}(p)$$, it injects a particle into the system moving with momentum $$p$$. Likewise, when you hit something with $$\hat{a}(p)$$, it removes a particle that already is moving with that momentum.

Note that you can then represent a state containing a single particle with general momental wave function $$\psi_p$$ via

$$|\psi\rangle := \int_{p=-\infty}^{\infty} \psi_p(p) [\hat{a}^{\dagger}(p) |0\rangle ]\ dp$$

Think about it like using a paintbrush: you "paint" the wave function onto the vacuum by sweeping it over the whole $$p$$-space while weighting with the weight of the function you want to make (how hard your "paintbrush" pushes in, if you could push a brush complexly hard, with which this brush you can, ordinary paintbrushes you can't).

If you took that state, and tried to annihilate any single momentum value via the corresponding annihilator $$\hat{a}(p)$$, it would only make it impossible for it to have that exact value of momentum. The field would still contain 1 particle. To erase it all, you'd have to do this:

$$|\text{0-equivalent}\rangle := \int_{p=-\infty}^{\infty} \hat{a}(p) |\psi\rangle\ dp$$

think like you're "cleaning up" the particle's wave function like using an eraser or mop. No probability is left anywhere for it to be after this, so the particle has been completely removed. (The "equivalent" label is because I think there will be some constant factors out front; so the actual state in strict terms, which is a ray or even better an [extremal] density operator, is the same, even if the ket vector is not)

As for inner products, $$\langle \phi | \psi\rangle$$, the meaning is the same as for regular QM: it means whether if you tried to measure the field state $$|\phi\rangle$$ for whether it was the field state $$|\psi\rangle$$, what would be the probability to obtain "yes". When we put a pair of operators in there like

$$\langle 0|\hat{a}(p_2) \hat{a}^{\dagger}(p_1)|0\rangle$$

you have to be a bit careful: you say "is it the probability to annihilate a particle with one momentum after creating it with another". No, because you have to remember that in

$$\langle \phi | \psi\rangle$$

the thing on the left is actually a bra - it's a dual vector, living in the Hilbert dual space $$H^{*}$$. That is, $$\langle \phi|$$, is acting, in its linear-functional way, on the ket vector $$|\psi\rangle$$. And that little $$\dagger$$ notation there on the creation and annihilation operator is not just for show: that actually literally means that the creation operator is the Hermitian conjugate of the annihilation operator, thus it follows that when it comes to the dual space, $$\hat{a}(p)$$ actually acts as creation operator, and $$\hat{a}^{\dagger}(p)$$ acts as annihilation operator. That is to say, the roles are exactly reversed!

Hence the inner product you give actually means "what is the probability [better: quantum amplitude] I will observe that a particle which has been created with momentum definitively equal to $$p_1$$, to actually instead be one created with a momentum equal to $$p_2$$?", because $$\hat{a}(p_2)$$, even though it lacks a dagger, creates a particle when the dual vector $$\langle 0|$$ hits it coming in from the left. And the expression on the right then should make perfect sense: it is probability zero so long as $$p_1 \ne p_2$$, because that just can't happen!

Exercise: tell me what $$\langle 0| \hat{a}^{\dagger}(p_1) \hat{a}^{\dagger}(p_2) |0\rangle$$ means. Leave a comment. (It may not be what you first think!)

FWIW, insofar as the field operator $$\hat{\phi}(x)$$ ... that's different. That's a Hermitian operator; and it can and does actually belong to the observable algebra. $$\hat{\phi}(x)$$ is the value the quantum field takes at $$x$$, understood as a quantum observable just like any other. Hence hitting $$|0\rangle$$ with it alone makes about as much sense as trying to interpret what $$\hat{p}|\psi\rangle$$ "physically means" in non-relativistic quantum mechanics. We use the operators for their algebraic properties, not their actual "action".

Thus, we can say that when we are thinking of $$\hat{a}^{\dagger}(p)$$ and $$\hat{a}(p)$$, we are thinking of particles, in momentum space representation. When we think of $$\hat{\phi}(x)$$, we are thinking of fields that fill position space. The really cool bit is how that the two go together!

Finally, regarding particles in position space ... position is funny, and you'd get a number of views on it, because there isn't just one mathematical way to relate to position space. You'll see Newton-Wigner operator, and then you'll see it has caveats (in particular, perfectly "localized" Newton-Wigner positions are not orthogonal, i.e. there is probability to measure one "localized" particle "at" position $$x_1$$ as being "at" position $$x_2 \ne x_1$$!), versus many who say "just forget about position at all". That's probably the best summary insofar as "consensus" goes.

(Nonetheless, I don't really like that :D The way I personally like to think about it may be a bit unusual, and I am not even entirely sure it truly works, so take this with salt. I won't call it "original", just "unusual", because especially H. Nikolic pretty much gave the gist [we have to talk of a "probability density function in space-time"], just not the precise details I've laid out, and moreover he was working in the context of Bohmian mechanics. Others [Stueckelberg? iirc] seem to "point" at it with brief footnotes about "promoting time to an operator", but then back off from it. And because I have nobody to bounce the ideas off, I have no way to know if or how valid my specific approach is, so I rather not just post dilettante junk here, hehe.)

• @The_Sympathizer I think part of the reason I am having trouble is that I am actually a student of mathematics, so the physics notation (in particular bra-ket notation) is very foreign to me. For example, I am unclear if in an expression such as $\langle 0 | a^\dagger a | 0 \rangle$ one should view the zero functional acting on the state $a^\dagger a | 0 \rangle$ or if we should view it as the functional $\langle 0 | a^\dagger$ acting on the state $a | 0 \rangle$ or the functional $\langle 0 | a^\dagger a$ acting on the state $| 0 \rangle$. It seems very ambiguous to me. Dec 20, 2022 at 0:23
• @CBBAM: Aha. What's going on, then, is that as you may know, the Hilbert space is a type of "good" inner product space. When we write $\langle \phi | \psi \rangle$, what is being done is that we are exploiting the fact then when an inner product, more "maths-y" written as $\left\langle |\phi\rangle, |\psi\rangle \right\rangle$, is present, it establishes a natural map between the vector space and its dual space. In particular, an inner product establishes the ability for us to "transpose" any vector. You know, the conversion from a "column vector" to a "row vector". (...) Dec 20, 2022 at 2:57
• Naively, transpose in this way requires a basis, and while an inner product does not inherently provide one, it does provide us with a preferred class of bases, the orthonormal bases, which turn out to be sufficient to generate the transpose operation uniquely. With transpose, we can then write the inner product simply as the evaluation of the linear functional correspondign to one vector upon the other. Hence the Dirac notation, $\langle \phi | \psi\rangle$, for the inner product of the vector (not dual vector) $|\phi\rangle$ with $|\psi\rangle$. Dec 20, 2022 at 2:57
• So I'm just saying in a really long way, "to find the meaning of the dual part of that Dirac sandwich, just transpose (or better, Hermitian transpose) it first :)" Dec 20, 2022 at 3:08
• @The_Sympathizer I think I got it now, thank you for all your help! Dec 20, 2022 at 17:55