# I Boundary terms for field operators

#### kelly0303

Hello! In several of the derivations I read so far in my QFT books (M. Schawarz, Peskin and Schroeder) they use the fact that "we can safely assume that the fields die off at $x=\pm \infty$" in order to drop boundary terms. I am not sure I understand this statement in terms of QFT. A field in QFT is an operator which, when applied to vacuum, creates a particle at position $x$ i.e. $<0|\phi(x)|p>=e^{ipx}$ (ignoring the normalizations). But the vacuum is the same everywhere, so how can one assume that the field operator vanishes at infinity. The way I read that is that $\phi(\pm \infty)|0> = |0>$, or something like this. But from the point of the view of the vacuum all the points are the same so there is no infinity, the operator should behave the same at infinity as it does anywhere else. Can someone explain this to me? Thank you!

• maline
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#### mfb

Mentor
Nothing should happen at infinity, including no creation of particles.

#### MathematicalPhysicist

Gold Member
Nothing should happen at infinity, including no creation of particles.
How can we know this?
I thought to myself that this condition is brought from the mathematical tools that are used in QFT like Schwartz space and rigged Hilbert space.
But I don't think you can have a physical explanation, since you cannot really know what happens at infinity.

#### vanhees71

Science Advisor
Gold Member
Note that "plane waves" (momentum "eigenstates") are no true eigenstates as in quantum mechanics. They are only normalizable to a $\delta$-distribution. So "proper states" are those normalizable to 1 ("wave packets").

#### mfb

Mentor
How can we know this?
Go there and check .
As far as I know it doesn't matter what is there as long as it doesn't have anything too weird. Your reaction in infinite space without anything else going on is just the case that makes calculations easier.

#### Michael Price

Hello! In several of the derivations I read so far in my QFT books (M. Schawarz, Peskin and Schroeder) they use the fact that "we can safely assume that the fields die off at $x=\pm \infty$" in order to drop boundary terms. I am not sure I understand this statement in terms of QFT. A field in QFT is an operator which, when applied to vacuum, creates a particle at position $x$ i.e. $<0|\phi(x)|p>=e^{ipx}$ (ignoring the normalizations). But the vacuum is the same everywhere, so how can one assume that the field operator vanishes at infinity. The way I read that is that $\phi(\pm \infty)|0> = |0>$, or something like this. But from the point of the view of the vacuum all the points are the same so there is no infinity, the operator should behave the same at infinity as it does anywhere else. Can someone explain this to me? Thank you!
Can we ignore the surface terms because they grow as R squared, whereas volume terms grow as R cubed? So in the limit as R goes to infinity the surface terms are irrelevant, almost no matter how the fields behave?

#### MathematicalPhysicist

Gold Member
Go there and check .
As far as I know it doesn't matter what is there as long as it doesn't have anything too weird. Your reaction in infinite space without anything else going on is just the case that makes calculations easier.
I would if I had the time... #### kelly0303

Nothing should happen at infinity, including no creation of particles.
I am not sure I understand this. In a free theory (for which that statement should still work, according to the authors), the creation operator $a^\dagger_p$ is the same in all space time, so it has no position dependence. Also, the vacuum should be the same everywhere, too, as it also doesn't have a position dependence. So if the equation $|p>=a^\dagger_p|0>$ holds somewhere, it should hold everywhere, including at infinity, as there is no x dependence on it. And as any field can be written in terms of the creation and annihilation operators, any field should behave the same everywhere. I am just confused as to where do we take the limit to infinity and get zero.

• maline

#### kelly0303

Note that "plane waves" (momentum "eigenstates") are no true eigenstates as in quantum mechanics. They are only normalizable to a $\delta$-distribution. So "proper states" are those normalizable to 1 ("wave packets").
Thank you for this! I am still confused mathematically. In the free theory (for simplicity) the field (for a scalar, again for simplicity) is $$\phi(x)=\int{\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}[a_p e^{-ipx}+a_p^\dagger e^{ipx}]}$$ In a free theory the creation and annihilation operators have no space time dependence, so the only space dependence is in that exponential. Classically, if we have a field, let's say the intensity of the light emitted by a point source at origin, it makes perfect sense (to me) to write $$lim_{x\to\infty}I(x)=0$$ which simply means that infinitely far away you receive no energy from the source. But in the case of $\phi(x)$, writing $$lim_{x\to\infty}\phi(x)=0$$ makes no sense to me. As far as I can tell it doesn't even converge. So in which sense do we take this limit at infinity, such that we can say that the field vanishes there? Thank you!

#### vanhees71

Science Advisor
Gold Member
I think the answer is that boundary conditions are "weak conditions", i.e., they are valid for expectation values taken with proper (pure or mixed) states.

• Demystifier and Auto-Didact

#### samalkhaiat

Science Advisor
A field in QFT is an operator which, when applied to vacuum,
No, the field is not an operator (valued function on spacetime) because the norm $\lVert \varphi (t,x) |0 \rangle \rVert^{2}$ is divergent. A densely defined operator on Hilbert space $\mathcal{H}$ can be constructed by smearing the field $\varphi (t,x)$ (which is an operator-valued distribution) with a nice test function $f (x) \in \mathcal{S} (\mathbb{R}^{3})$: $$\varphi (0 , f ) = \int d^{3}x \ \varphi (0 , x) f(x),$$$$\varphi (0 , f ): \mathcal{H} \to \mathcal{H} .$$ For example, choosing $f(x) = e^{-x^{2}}$, the state $$| \Psi \rangle = \varphi ( 0 , e^{-x^{2}} ) | 0 \rangle = \int d^{3}x \ e^{-x^{2}} |x \rangle ,$$ is a normalizable state $\lVert |\Psi \rangle \rVert^{2} = \sqrt{ \pi / 2}$.

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#### kelly0303

I think the answer is that boundary conditions are "weak conditions", i.e., they are valid for expectation values taken with proper (pure or mixed) states.
I am sorry, but I am not sure I understand what you mean

#### kelly0303

No, the field is not an operator (valued function on spacetime) because the norm $\lVert \varphi (t,x) |0 \rangle \rVert^{2}$ is divergent. A densely defined operator on Hilbert space $\mathcal{H}$ can be constructed by smearing the field $\varphi (t,x)$ (which is an operator-valued distribution) with a nice test function $f (x) \in \mathcal{S} (\mathbb{R}^{3})$: $$\varphi (0 , f ) = \int d^{3}x \ \varphi (0 , x) f(x),$$$$\varphi (0 , f ): \mathcal{H} \to \mathcal{H} .$$ For example, choosing $f(x) = e^{-x^{2}}$, the state $$| \Psi \rangle = \varphi ( 0 , e^{-x^{2}} ) | 0 \rangle = \int d^{3}x \ e^{-x^{2}} |x \rangle ,$$ is a normalizable state $\lVert |\Psi \rangle \rVert^{2} = \sqrt{ \pi / 2}$.
I am a bit confused. In both QFT books I read they mention almost every 2 pages that the fields in QFT are operators, and I don't remember reading anywhere (not even a footnote), that they are not, rigorously speaking, operators. Is this just because from a physics point of view it doesn't make much of a difference or I should read a more advanced QFT book (Weinberg maybe?) to get a better understanding of your explanation? And relating to your example, what is that $f(x)$ in the case of QFT? Thank you!

#### romsofia

QFT assumes special relativity, and thanks to special relativity, there is a universal speed limit. Things occurring at infinity cannot influence us due to this, so we say that all fields vanish at infinity.

#### vanhees71

Science Advisor
Gold Member
I am a bit confused. In both QFT books I read they mention almost every 2 pages that the fields in QFT are operators, and I don't remember reading anywhere (not even a footnote), that they are not, rigorously speaking, operators. Is this just because from a physics point of view it doesn't make much of a difference or I should read a more advanced QFT book (Weinberg maybe?) to get a better understanding of your explanation? And relating to your example, what is that $f(x)$ in the case of QFT? Thank you!
Samalkait is arguing from the point of view of axiomatic QFT. This you don't find in Weinberg's book. It's an interesting approach with the aim to make QFT mathematically rigorous. The success after decades is a bit meager if you ask me since it's not yet even possible to formulate QED nor any other "realistic" QFT in (1+3) dimensions rigorously.

The point can be understood already in usual non-relativistic QT. In the usual physicists' sloppy math we deal with "eigenvectors" of self-adjoint operators with "eigenvalues" in the continuous spectrum and "normalized" them to "$\delta$ distributions". Of course these are not true Hilbert-space vectors since their norm is not well defined (but only normalized to "$\delta$ distributions). Take e.g., the momentum and position operators in non-relativistic QM, defined via the Heisenberg algebra commutation relations,
$$[\hat{x},\hat{p}]=\mathrm{i} \hat{1}.$$
This implies that both $\hat{x}$ and $\hat{p}$ have the entire real axis as "eigenvalues" and the "eigenfunctions" are generalized eigenfunctions (living in the dual of a dense subspace where the position and momentum operators are defined). The normalization is usually chosen as
$$\langle \vec{x}|\vec{y} \rangle=\delta^{(3)}(\vec{x}-\vec{y}), \quad \langle \vec{p}|\vec{k} \rangle =\delta^{(3)}(\vec{p}-\vec{q}).$$
In nature you cannot prepare any particle to have a definite position or momentum but you can of course use square-integrable wave functions describing a particle with a pretty well defined position (then momentum is rather unprecisely determined) or momentum (then position is rather unprecisely determined). So you have to construct wave packets to describe a particle with either a quite well defined position or a well defined momentum.

In QFT the field operators also create "generalized states", not true Hilbert-space states. To do so you have to "smear" the operators somewhat such that you get operators that create some analogue of wave packets, i.e., true normalizable Hilbert-space states.

The usual sloppy use of the field operators leads, in perturbation theory, to the problem of divergences of N-point functions, because it's not not allowed, from a rigorous mathematical point of view, to multiply the "distribution-like field operators", but that's what you do all the time. From a pragmatic point of view, this is "cured" by renormalization theory. Since you have to renormalize anyway somehow in perturbation theory, from a physicist's point of view not too much is lost. For a mathematician it's a pretty dirty business though.

There are some attempts to formulate QFT in a more careful way to define the successful sloppy perturbative treatment with renormalization more strictly. A good book is, e.g.,

https://www.amazon.com/dp/3642633455/?tag=pfamazon01-20

based on the Epstein-Glaser approach to (perturbative) QFT.

#### samalkhaiat

Science Advisor
In both QFT books I read they mention almost every 2 pages that the fields in QFT are operators,
They always mean operator-valued distribution. Operator valued functions satisfying the QFT axioms do not exist. So, we really need distributions. The physical reason is the uncertainty principle: Measuring the field $\varphi$ at a point $x$ causes very large fluctuations of energy and momentum, which prevent $\varphi (x)$ from being a well-defined operator.

And relating to your example, what is that $f(x)$ in the case of QFT? Thank you!
In QFT, the wrong statement “$\varphi (x)$ is an operator” corresponds exactly to another wrong statement in QM “the position eigen-state $|x \rangle$ is an element of the Hilbert space $L^{2}(\mathbb{R}^{3})$”. In QM, instead of $|x \rangle$ which is not square-integrable, we form the state $$| \Psi \rangle = \int d^{3}x \ \Psi (x) \ |x \rangle ,$$ which is a true element of $L^{2}(\mathbb{R}^{3})$. Similarly in QFT, instead of the naively-defined operator $\varphi (x)$, we form a true operator on $\mathcal{H}$ by writing $$\varphi (f) = \int d^{4}x \ f(x) \varphi (x) .$$ This can be viewed as the mathematical representation of the fact that only space-time averages of $\varphi (x)$ are “observable”. The class of (test) functions $f$ (for which $\varphi (f)$ is densely defined) is taken as the class $\mathcal{D}(\mathbb{R}^{4})$ of all infinitely differentiable functions of compact support in space-time to reflect the possibility of making field measurements in finite spatio-temporal regions and to control the behaviour at the boundary.

• dextercioby and weirdoguy

#### maline

When we "assume stuff goes to zero at infinity", it is usually in the context of integration by parts for an integral that gives the action, or various derivatives of the action. Now hopefully someone will correct me if I'm wrong, but I don't see any real reason to think of the Lagrangian and the action, or the fields that make them up, as operators!
What is the meaning of the action? In the context of the (bosonic) path integral, the action gives a phase for each "path" i.e. each field configuration over the full spacetime history. But these are classical, c-numbered field configurations, and the action is a classical functional on them. I do think that the path integral should only include configurations with good behavior at infinity.
In the canonical formalism, the purpose of the Lagrangian is to express a theory (i.e. the field equations that relate the operators at different points) in a concise form that allows us to directly examine the symmetries of the theory (in particular, Lorentz-covariance of the S-matrix). The passage from Lagrangian to field equations by way of the Euler-lagrange formalism is basically just a formalistic coding/decoding device, and the "fields going to zero at infinity" is just par of the rules of that game.
I think that similar comments apply to the fermionic path integral: I cannot make heads or tails of a "Grassman-number-valued field configuration", and anyway the "Berezin integral" has almost nothing in common with the standard use of the word "integral" (it is much more like a derivative- note how it behaves under a change of integration variable). So in this context as well, I see the Lagrangian as just a neat way of packaging the Feynman rules, and so I'm not worried whether dropping the boundary terms is "justified"- it is if we want it to be.
Finally, the action gives us conservation laws, through Noether's theorem. Conservation laws refer to measurable quantities, so they should be taken as referring to the expectation values of the currents involved, in some (normalizable) physical state. In this context, it should generally be possible to interpret the integrals as calculating expectation values, and the dropped boundary terms will be justified if the relevant expectation values go to zero in all physical states.

#### DarMM

Science Advisor
Gold Member
it's not yet even possible to formulate QED nor any other "realistic" QFT in (1+3) dimensions rigorously.
To mention the state of the art, continuum Yang Mills is known to exist rigorously. We need to show it is unique, has an infinite volume limit and its perturbation theory matches a physicist's typical formal calculations.

• weirdoguy

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