# Why do we require quantum fields to vanish at infinity?

Classical fields, like the electrical field must vanish at infinity, because otherwise their energy would be infinite. This can be used in computations to exclude certain solutions.

In quantum mechanics, the wave function must be normalized, because of the probalistic interpretation. (The probability for finding the particle, can't be bigger than $100$%). A wave function spreading out all over space can't be normalized.

In quantum field theory, it's a commenly used trick to integrate by parts and neglect the boundary term "because fields vanish at infinity"? At first sight, this sounds reasonable and logical, but I was never able to nail it really down. Whats the exact reason, we can or must assume that quantum fields vanish at infinity?

Take for example the electron field, which is responsible for the creation and destruction of electrons. Why shouldn't there be electrons everywhere? I know that we have a probabilistic interpretation in qft, too. Nevertheless, I can't put it together why this means here that our quantum field must vanish.

• A mathematical reason is that the quantum fields are "operator valued distributions", i.e. objects that make sense as operators only when applied to a "single particle wavefunction" (a vector in some Hilbert space). The usual single-particle space is an $L^2$-space, and functions in $L^2$ vanish at infinity. Hence it is possible to neglect boundary terms in an integration by parts of the type $\int [(\nabla a)^*(x)f(x)+(\nabla a)(x)\bar{f}(x)]dx$ where $a^\#(x)$ are the anihilation/creation operator(-valued distribution)s and $f\in L^2$. – yuggib Jan 23 '15 at 10:21
• However I think physicists don't care too much and manipulate quantum fields freely, simply assuming that they are always allowed to neglect boundary terms. (To be mathematically precise a term like the one above defines the distribution, and the integration by parts is just formal, like if you "integrate by parts" in $\int \delta'(x)f(x)dx$, where $\delta'(x)$ is the derivative of Dirac's delta). – yuggib Jan 23 '15 at 10:24
• @yuggib I think this is a very nice answer, thank you! Nevertheless, I hope that maybe someone comes up with something more "intuitive", like for classical fields etc. – Tim Jan 23 '15 at 10:42
• Physically I think that this is consequence of the thermodynamic limit: Space is so large that what is happening in front of you can not be affected by what is happening on its boundary. Then, if it does not matter how you specify your fields at infinity, you might as well choose something that is easy to work with, i.e. zero. – Steven Mathey Jan 23 '15 at 11:53

Quantum fields are tempered (operator-valued) distributions on space-time, because the machinery of QFT requires distributions that possess a Fourier transform - else you get no creators and annihilators.

Their test functions are the Schwartz functions, which are the functions rapidly vanishing at infinity.

Since integration by parts for distributions is defined through the integration by parts when applied to any test function, and the boundary terms vanish there since the Schwartz functions are rapidly decreasing, all boundary terms for these distributions vanish.

Sometimes, you will encounter a quantum field configuration that seems to not vanish at infinity. This is usually due to a "classical background", i.e. the fields is $\phi = \phi_\text{cl} + \phi_\text{q}$, and only $\phi_\text{q}$ is the quantum field being (path)-integrated over, while $\phi_\text{cl}$ is a "vacuum" (not the true vacuum) around which the theory is considered.

One can also try to get the integrations by parts to vanish without requiring that fields vanish at infinity: Just take the theory inside a finite volume of spacetime and impose periodic boundary condition. For such periodic situations, integration by parts has no boundary terms. Take the limit of infinite extent of the volume to recover the usual theory.

A more physical, handwavy reasoning goes like this: We do not expect field configurations infinitely far away to influence what happens at finite spacetime coordinates, so we might as well say the field is zero there. This doesn't really work because functions don't have a "value at infinity", but rather a limiting behaviour, and one needs nice statements like the rapid decrease of Schwartz functions to really conclude that the boundary terms vanish.

• Why quantum fields should make sense as operators only when acting on functions of rapid decrease? There are (commonly used) quantizations, such as the one by Segal, that associate a quantum field to any vector of the one-particle Hilbert space, and the Fock representation (hence the introduction of creation-annihilation operators) is perfectly legitimate. – yuggib Jan 23 '15 at 15:29
• @yuggib: I think Segal quantization also produces a quantum field that is a tempered distribution. At least, the construction in "Mathematical Aspects of Quantum Field Theory" by de Faria&de Melo, section 6.3.4 produces at the end a quantum field that is defined with Schwartz functions as test functions, and it is precisely the Fourier transform existence that is the reason for that. – ACuriousMind Jan 23 '15 at 16:16
• I think that this requirement is just a technical and temporary restriction to do some calculations. Given a complex Hilbert space $H$ with inner product $(\cdot,\cdot)_c$, the Single Particle (Symplectic) Structure $\Sigma(H)$ is the pair $\{K,B(\cdot,\cdot)\}$; where $K$ is $H$ considered as a real Hilbert space with product $(\cdot,\cdot)=\mathrm{Re}(\cdot,\cdot)_c$ and $B(\cdot,\cdot)=\mathrm{Im}(\cdot,\cdot)_c$ is the symplectic form. The (Segal) quantization is a linear map $R(\cdot)$ from $K$ to the self-adjoint operators on some complex Hilbert space – yuggib Jan 23 '15 at 17:32
• such that: 1) $e^{iR(\cdot)}$ is weakly continuous when restricted to any finite dimensional subspace of $K$; 2) $e^{iR(z_1)}e^{iR(z_2)}=e^{-iB(z_1,z_2)/2}e^{iR(z_1+z_2)}$ for any $z_1,z_2\in K$ (Weyl relations). The $R(\cdot)$ is the quantum field, and it is indeed an operator valued distribution, in the sense that it makes sense as an operator only when it acts on a vector of the single particle structure $\Sigma(H)$. If $\mathrm{dim}=\infty$, there are uncountably many irreducible unitarily inequivalent quantizations of $\Sigma(H)$ – yuggib Jan 23 '15 at 17:37
• one of them is the so-called Fock representation where the creation and annihilation operators are defined, and related to the field $R(z)$ by $R(z)=\frac{1}{\sqrt{2}}(a^*(z)+a(z))$. – yuggib Jan 23 '15 at 17:43