# Tag Info

26

Yes, in scalar field theory, $\langle 0 | T\{\phi(y) \phi(x)\} | 0 \rangle$ is the amplitude for a particle to propagate from $x$ to $y$. There are caveats to this, because not all QFTs admit particle interpretations, but for massive scalar fields with at most moderately strong interactions, it's correct. Applying the operator $\phi({\bf x},t)$ to the ...

24

You wouldn't think it, from how easy it is to pose this question, but it is ridiculously nontrivial. As it happens, it is entirely impossible to find the position-basis matrix elements of this propagator. So far you've done good, and the identification $$U\left( t_{2},t_{1}\right) =e^{\frac {-i} {\hbar }H\left( t_{2}-t_{1}\right) } =\sum_{n=-\infty}^\... 19 Keep in mind that it is nearly impossible to explain how perturbative QFT calculations follow from Lagrangians such that the answer is both relatively short and detailed. So I am going to write an introductory answer. If you want more details on any of its part, you can look up textbooks, or you can let me know in the comments, in which case I will consider ... 18 Quantum mechanics and quantum field theory are different in how they treat their wave equations. The usage of the common term “propagator” could be traced back to the “relativistic wave equation” approach—i. e. people really used to think of the Schrödinger and the KG operators as belonging to the same class of “quantum operators”, but the modern ... 17 It has been many years since you asked this question. I assume that over time you have compiled meaning definitions and distinctions for the other terms in your list. However, there are terms not defined by @josh's response (A response which I have relied on multiple times, thank you for posting it @josh). Personally, my background is in Lattice QCD, which ... 16 I) OP is right, ideologically speaking. Ideologically, OP's first eq.$$ \tag{1} \left| \int_{\mathbb{R}}\! \mathrm{d}x_f~K(x_f,t_f;x_i,t_i) \right| ~\stackrel{?}{=}~1 \qquad(\leftarrow\text{Turns out to be ultimately wrong!}) $$is the statement that a particle that is initially localized at a spacetime event (x_i,t_i) must with probability 100% be ... 16 First, the term "propagator" is usually defined as the Green's function of the first type, not the second type, i.e. as a solution to the diffential equation \hat L G = \delta. At any rate, those definitions are ultimately equivalent – when the details are correctly written down – because the Green's function defined as the correlator in the second ... 15 For a given quantum system, the kernel of the path integral is, in fact, the kernel of an integral transform as you explicitly write down. It is the transform that governs time evolution of the system as is manifest in your first equation. For this reason, it is often referred to as the propagator of a given system. For example, for a single, non-... 15 Your question has been answered again and again, and again, albeit indirectly and elliptically--I'll just be more direct and specific. The point is you skipped variables: in this case, t, and so the expression you wrote ("according to books, \hat{L}K(x;x') = \delta(x-x')~"), is nonsense, as you already properly found out; unless you included t in the ... 14 Källén-Lehmann representation is just a way to expand in the momentum basis the two point correlation function of a local operator \hat{O}(x), it holds true for massive and massless theories alike except for non Abelian gauge theory in which the situation is a bit more complicated. Let's demonstrate the K-L formula with a more general proof: let's start ... 13 No, ⟨0|T{ϕ(y)ϕ(x)}|0⟩ is NOT the probability amplitude for a particle to propagate from x to y, even for a free scalar field. It seems to be a common false belief that it is. There is one obvious reason and one deep reason why it cannot be. The obvious reason is that the square of this value, which is supposed to be the probability density, does not ... 13 The first step is to recognize that equation is invariant under d-dimensional rotations around \mathbf{x} - \mathbf{x}' = \mathbf{0} and simultaneous identical translations of \mathbf{x} and \mathbf{x}', so we can make the following step:$$\begin{align}(-\nabla_d^2 + m^2)G(\mathbf{x}, \mathbf{x}') &= A \delta(\mathbf{x} - \mathbf{x}') \\ & \...

12

This can be seen by partial integration $$\frac{\partial}{\partial \rho}\sqrt{\rho^2-m^2}=\frac{\rho}{\sqrt{\rho^2-m^2}}$$ OP edit: More explicitly, we use this to write $(3)$ as \begin{align} D(x-y) &= \frac{1}{(2\pi)^2r}\int^\infty_m d\rho \frac{\partial}{\partial \rho}\sqrt{\rho^2-m^2} e^{-\rho r} \\ &= \frac{1}{(2\pi)^2r}\left[\sqrt{\rho^2-...

12

Here's the gist of it: If your field lives in a vector space $V$, then the propagator is a map $V\to V$, i.e., it lives in $V\otimes V^*$. In more down-to-earth terms, if your field has a certain index $i$, its propagator has a pair of such indices: $$\psi^i\quad\Longrightarrow\quad G^i{}_j$$ The reason is that, by definition, $G$ measures the difference ...

11

The factor of $1/2\pi$ is an artifact of the normalization convention being used for the momentum eigenstates. To begin to see how this is so, let us note that the choice of normalization of a Dirac-orthogonal continuous basis completely determines the form of the resolution of the identity. Writing an arbitrary state $|\psi\rangle$ in a given ...

11

Propagator in quantum mechanics is just a different name for Green's function for time-dependent Schroedinger equation. It is a unique function that enables us to write, for any time $t_0$, $$\psi(x,t) = \int G(x,t;x',t_0)\psi(x',t_0)\,dx'$$ for all $x$ and all $t$. This means the $\psi$ function of $x$ (at any time $t$) can be written as a result of ...

11

The free scalar and fermion propagator is $$G_\psi(x,y) = \int \frac{d^dp}{(2\pi)^d} \frac{-i(\gamma^\mu p_\mu + m)}{ p^2 + m^2 - i \epsilon} e^{- i p \cdot ( x - y ) }$$ The scalar propagator is $$G_\phi(x,y) = \int \frac{d^dp}{(2\pi)^d} \frac{-i}{ p^2 + m^2 - i \epsilon} e^{- i p \cdot ( x - y ) }$$ Clearly, $$G_\psi(x,y) = ( i \gamma^\mu \partial_\... 10 Suppose you want to compute a correlation say in Euclidean signature$$ \frac{1}{Z}\int D\phi\ \prod_i \phi(x_i)\ \exp\left(-\frac{1}{\hbar}S(\phi)\right) $$with$$ S(\phi)=\frac{1}{2}(\phi,A\phi)+g\int dx\ \phi(x)^4 $$where (\phi,A\phi)=\int\ dx\ dy\ \phi(x)A(x,y)\phi(y) for some "matrix" or rather kernel A. One usually does that by expanding the ... 10 I will discuss the closed string propagator because this case is pictorially closer to the scalar propagator in quantum field theory case. The closed string analog of the (two-leg amputated) line of propagation of a scalar field in a Feynman diagram is a cylinder of finite height s and twist angle \theta. At this point you must notice the analogue with ... 9 I) Conceptually, OP's original eq. (1)$$\int_{\mathbb{R}}\! \mathrm{d}x_f~ \left| K(x_f,t_f;x_i,t_i) \right|^2 ~\stackrel{?}{=}~1 \qquad(\leftarrow\text{Wrong!})\tag{1} $$clashes (as OP independently realized) with the fundamental principle of the Feynman path integral that the amplitude$$K( x_f ,t_f ; x_i ,t_i )~=~\sum_{\rm hist.}\ldots$$is a sum of ... 9 There is an integration formula (see "Table of integrals, series and products" 7ed, p337 section3.324 1st integral)$$\int_0^\infty d\beta \exp\left[-\frac{A}{4\beta}-B\beta\right]=\sqrt{\frac{A}{B}}K_1\left(\sqrt{AB}\right)\qquad [\mathrm{Re}A\ge0, \mathrm{Re}B>0].$$If \mathrm{Re}A\ge0, \mathrm{Re}B>0 is violated, the integral will be divergence. ... 9 My own attempt at this: the first result is wrong, and the second one is right but incomplete. Feynman: in the diagram, all the lines are external, so there is no propagator in the diagram. Therefore, Feynman rules give \mathcal A=1. Check up of this: In canonical quantisation, the amplitude is given by \langle p|q\rangle=\langle 0|a_p a^\dagger_q|0\... 9 Method One: \begin{eqnarray*} & & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}+i\epsilon}e^{-ik\cdot x}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\int dk_{0}e^{-ik_{0}x_{0}}\frac{1}{[k_{0}+(|\mathbf{k}|-i\epsilon)][k_{0}-(|\mathbf{k}|-i\epsilon)]}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\... 9 The propagator of an arbitrary vector field is [ref.1] $$\langle A_\mu A_\nu\rangle=\frac{-\eta_{\mu\nu}+p_\mu p_\nu/m_1^2}{p^2-m_1^2}-\frac{p_\mu p_\nu/m_1^2}{p^2-m_0^2}\tag1$$ for a pair of masses m_0,m_1. This propagator is usually called the Stückelberg propagator, and A_\mu a Stückelberg field. The Stückelberg field ... 8 The probability density$$\lim_{x_1\to x_0} P(x_1, t_1; x_0, 0)is basically the probability density for a particle at position x_0 to be back at that position after a time t_1 has passed. 8 Virtual particles are not real. A virtual particle is essentially defined by being associated to a propagator. It is, formally, nothing more than such a propagator. The idea of "virtual particle" doesn't even exist before you notice that you can draw pretty Feynman diagrams as a succinct representation of the way QFT amplitudes are calculated. This is ... 8 To get a better intuition consider a field with a general spin, this can be written as \begin{align*} \psi_\ell &\propto \sum_\sigma \int d^3 p \left( u_\ell (\vec{p},\sigma) e^{i p\cdot x}a(\vec{p},\sigma) + v_\ell (\vec{p},\sigma) e^{-i p\cdot x}a^\dagger (\vec{p},\sigma) \right) \end{align*} where \ell is the spin index, and \sigma summed over all ... 8 Consider \phi^4 theory: \mathcal L=\frac12 Z_1(\partial\phi)^2-\frac12 Z_m m^2\phi^2-\frac{1}{4!}\lambda_0\phi^4 $$There are two approaches to perturbation theory: First The propagator is given by$$ \Delta=\frac{1}{Z_1p^2-Z_m m^2} $$and there is one type of vertex, with value$$ -i\lambda_0 $$Second The propagator is given by$$ \Delta=\frac{1}{...

8

This is a common (and good) question, which is the result of taking introductory texts more seriously than one should. Reading such texts, one gets the impression that the philosophy is We proceed as naively as possible and, when we find a divergence, we regulate it. On the other hand, the correct (and much more useful) attitude is We take things ...

8

When we integrate the propagator with respect to $k^0$ (i.e. the energy), we encounter two poles: one at $\omega_{\mathbf{k}}=\sqrt{\mathbf{k}^2+m^2}$ and one at $-\omega_{\mathbf{k}}=-\sqrt{\mathbf{k}^2+m^2}$, where $\mathbf{k}$ is the 3-momentum. In order to regularize this integral, we move these poles slightly off of the real line by adding or ...

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