# Tag Info

## New answers tagged field-theory

-1

In quantum (or classical) electrodynamics we are free to make gauge transformations, which change the form of terms in the Feynmann diagrams (or the potentials) without affecting any physical observable. This is sometimes viewed as a flaw in the theory. I've not heard anybody else say that. Can you give a reference? A similar freedom exists in ...

2

He uses that the action is dimensionless so that \begin{align} [ d^d x \left(\partial\varphi\right)^2] &= 0 \\ &=[d^d x]+2[\partial\varphi]\\ &= -d +2 + 2[\varphi] \end{align} using that $[dx]=-1$ and $[\partial\varphi] = [\partial] + [\varphi] = 1+[\varphi]$. This gives $[\varphi] = (d-2)/2$

2

Quantum fields cannot be turned on or off. The field itself exists for all time and space. It is possible to excite various modes of a quantum field at various spacetime points. These field excitations are interpreted as particles. When no excitations are present (i.e. no particles are present) the quantum field is in the vacuum state. Particles do not act ...

0

Nicest question since a long time. Your argumentation is excellent. You are the only one who give a comment to my last question - about the measured maximum distance of the electron influence. I suppose here, that electric fields are finite. That follows immediately if one agree that the electric field is quantized. A quanta has to have a finite energy and ...

1

Comments to the question (v8): Let us here for simplicity consider point mechanics. The generalization to field theory is straightforward and left to the reader. Given an (off-shell) action functional $$\tag{1} I[q]~=~\int_{t_i}^{t_f} \! dt~L,$$ it seems that OP in the first half of his post mainly confirms that the functional/variational derivative ...

2

Why $m^2$ in front of $\phi^2$ and why is $m$ the mass? Fist of all, from dimensional analysis the prefactor to the $\phi^2$ term in the Lagrangian must have mass-dimension$^1$ $2$ in $3+1$ dimensions since the Lagrangian has mass-dimension $4$ and $\phi$ has mass-dimension $1$. This just tells us that we can write the term as $m^2\phi^2$ where $m$ is ...

3

Take the commutator acting on a function $f$. Then \begin{split} [ P_i , P_j ] f &= [ - i \partial_i - q A_i , - i \partial_j - q A_j ]f \\ &= ( i \partial_i + q A_i )( i \partial_j + q A_j ) f -( i \partial_j + q A_j ) ( i \partial_i + q A_i ) f \\ &= - \partial_i \partial_j + i q A_i \partial_j \, f + i q \partial_i ( ...

0

If $$\tag{1} \delta\varphi~=~\varepsilon$$ is a global shift symmetry, we can gauge the symmetry, i.e. enhance it to a local symmetry by (i) introducing a gauge field $A_{\mu}$ with gauge symmetry $$\tag{2} \delta A_{\mu} ~=~\partial_{\mu}\varepsilon,$$ and (ii) replace partial derivatives $\partial_{\mu}\varphi$ with covariant derivatives $$\tag{3} ... 1 The fields of a supersymmetric theory form a representation of the super Poincare algebra. When this representation is restricted to a specific value of the mass operator P^{\mu}P_{\mu} = m^2, the representation is called an on shell representation multiplet. On shell representations are characterized by the equality of the number of bosonic and ... 1 In the following calculation, I ignore some coefficients. According to J(x)=\int d^4 k_1 e^{ik_1 x} , J(y)=\int d^4 k_2 e^{ik_2 y} and D(x-y)=\int d^4k \frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon} We have$$W(J) = \int d^4x d^4y d^4 k d^4 k_1 d^4 k_1 J(k_1)e^{ik_1 x} J(k_2)e^{ik_2 x} \frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon} W(J)=\int d^4x d^4y d^4 k ...

1

A n-particle reducible diagram is a diagram that can be cut into two pieces if one cuts n or less lines. Conversely, a n-particle irreducible (n-PI) diagram cannot be cut into two pieces if one cuts n lines. The sunset diagram is 3-particle reducible, since it has 3 internal lines, but it is both 1-PI and 2-PI, and contributes to the self-energy (which ...

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