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Quantum Field Theory (QFT) is the theoretical framework describing the quantisation of classical fields which allows a Lorentz-invariant formulation of quantum mechanics. QFT is used both in high energy physics as well as condensed matter physics and closely related to statistical field theory. Use this tag for many-body quantum-mechanical problems and the theory of particle physics. Don’t combine with the [quantum-mechanics] tag.

15 votes
1 answer
1k views

What does it mean for a field to be defined by a measure?

In Quantum Physics by Glimm and Jaffe they mention on p. 90 that The Euclidean fields are defined by a probability measure $d\mu(\phi) = d\mu$ on the space of real distributions. Here $d\mu$ plays th …
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12 votes
2 answers
2k views

How to interpret quantum fields?

As an analogy of what I am looking for, suppose $f(x,t)$ represents a classical field. Then we may interpret this as saying at position $x$ and time $t$ the field takes on a value $f(x,t)$. In quantum …
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10 votes
3 answers
3k views

How do we know that all quantum fields are Fourier transforms of creation and annihilation o...

In Folland's book Quantum Field Theory, he says ...we start out with classical field equations and a relativistically invariant Lagrangian from which they are derived, then replace the classical fiel …
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9 votes
2 answers
890 views

Why are test functions always taken to be Schwartz in quantum field theory?

A quantum field theory is defined as a measure over tempered distributions. Why do we take the space of test functions to always be Schwartz space? Why don't we, for example, smear fields using $C^\in …
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7 votes
3 answers
926 views

How does the absence of quadratic terms in the Lagrangian imply massless quanta?

When studying gauge theory, I often see the statement that gauge invariance does not allow the Lagrangian of the theory to contain terms that are quadratic in the gauge field. For example, to quote Fo …
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7 votes
3 answers
682 views

How does one go from wavefunctions to fields?

The original motivation for the Klein-Gordon equation $$(\square + m^2)\varphi = 0$$ and the Dirac equation $$(i\hbar\gamma^\mu\partial_\mu - mc)\psi = 0$$ were such that $\varphi$ and $\psi$ were to …
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7 votes
2 answers
361 views

Hilbert space of free theory vs interacting theory

In view of Haag's Theorem, it seems the Hilbert spaces of a free theory and an interacting theory are not the same. Though it seems very believable, I could not find a result that states that this is …
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5 votes
1 answer
342 views

What does it mean to apply a creation or annihilation operator to a free field, e.g. $\langl...

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 operat …
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5 votes
1 answer
298 views

Trouble understanding Wilsonian renormalization

I have been attempting to go through Chapter 12 of Peskin & Schroeder, but I have been having a very tough time. In particular I have been having trouble following this chapter much beyond page 397 be …
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5 votes
1 answer
133 views

Why do we rescale momenta after integrating out high momenta in Wilsonian renormalization?

In Section 12.1 of Peskin & Schroeder they motivate Wilson's approach to renormalization by asking how a quantum field theory changes after changing the momentum scale. To answer this they start with …
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5 votes
2 answers
744 views

Why does causality lead to different contours when calculating propagators?

On the Wikipedia page for propagators they mention three types of Green's functions for the Klein-Gordon equation: The retarded propagator (taken when $x^0 > y^0$). The advanced propagator (taken whe …
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5 votes
1 answer
460 views

Why do different contours give different answers in the limit $\epsilon \rightarrow 0$ when ...

Let $\phi$ denote the Klein-Gordon field. Then its propagator $\langle 0 \mid [\phi(x), \phi(y)] \mid 0 \rangle$ can be calculated as $$\int \frac{\mathrm d^4p}{(2\pi)^3} \frac{-e^{-ip(x-y)}}{p^2 -m ^ …
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5 votes
1 answer
207 views

Why must Grassmann algebras for Fermionic theories be infinite dimensional?

A Fermionic field $\psi$ is defined as a field over spacetime taking values in a Grassmann algebra $\mathcal{G}$. Why is the Grassmann algebra $\mathcal{G}$ usually taken to be infinite dimensional? I …
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4 votes
1 answer
316 views

How is the source term chosen when using path integrals?

Suppose I would like to compute (time ordered) vacuum expectation values for a quantum field theory by using the path integral approach. Using the Lagrangian for the theory, we define a generating fun …
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4 votes
2 answers
462 views

Interpreting generating functional as sum of all diagrams

The generating functional is defined as: $$Z[J] = \int \mathcal{D}[\phi] \exp\Big[\frac{i}{\hbar}\int d^4x [\mathcal{L} + J(x)\phi(x)]\Big].$$ I know this object is used as a tool to generate correlat …
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