21

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 ...


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 ...


14

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 ...


13

Saying that $\delta(0) = 0$ is completely non-sensical since the Dirac delta function is not a function to begin with. When we physicists write $$ \int \delta(x)f(x) \mathrm{d}x = f(0) \tag{1}$$ when that's all the "definition" of the delta "function" you actually need. Formally, the $\delta$ function is a tempered distribution, something that assigns ...


12

Let me expand a little more on what Craig Thone just said : Consider the energy/frequency-dependent Green function : $$ \tilde{G}(\omega)=\frac{1}{\omega-(a-\mathrm{i}b)} $$ with one single pole in $\omega=a-\mathrm{i}b$ (with $b>0$), which is Fourier transform of the time-dependent $G(t)$ Green function such as : $$ G(t)=\int\frac{\mathrm{d}\omega}{2\pi}...


12

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-...


12

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}') \\ & \...


11

I) Notational issues: Greens function vs. kernel. First of all, be aware that Ref. 1 between eq. (4-27) and eq. (4-28) effectively introduces the retarded Greens function/propagator $$\begin{align} G(x_2,t_2;x_1,t_1)~=~&\theta(\Delta t)~K(x_2,t_2;x_1,t_1), \cr \Delta t~:=~&t_2-t_1,\end{align}\tag{A}$$ rather than the kernel/path integral $$\begin{...


11

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 ...


11

@ArnoldNeumaier and @dushya have both pointed out correct solutions, but I want to elaborate a bit. The easy approach is the one dushya suggested. (You can also do what Arnold Neumaier suggests: First define the time-ordered product of operator-valued distributions, and then take expectation values.) Begin by recalling how Wightman functions are defined. ...


11

In axiomatic approaches to quantum field theory, the basic field operators are usually realized as operator-valued distributions. That's what Wightman fields are: operator-valued distributions satisfying the Wightman Axioms. Wightman functions are the correlation functions of Wightman fields, nothing more. There's a nice theorem that says if you have a ...


10

This is a somewhat broad question, because there are a number of different Green's functions in quantum physics. Perhaps the simplest one is the resolvent Green's function for a single-particle system. Its definition is $$G(\omega^{\pm})=\lim_{\delta \rightarrow 0^+}\left[ \omega \pm i\delta - H \right]^{-1}\equiv \frac{1}{\omega\pm i\delta - H},$$ where $H$ ...


10

The in- and out-states are free states, and the S-matrix definition of Mandl and Shaw is perfectly valid (with an appropriate notion of Texp). It is the one used in rigorous mathematical physics; see the treatise by Reed and Simon. It is also the one from which the LSZ formula is derived. It is the only way to define the S-matrix rigorously. The $+i\...


10

Weinberg, QFT 2, in Section 16.1 in a footnote 2 refers to Coleman, Aspects of Symmetry, p. 135-6, which features the $\hbar$/loop expansion. See also Refs. 3 & 4 for a similar idea. In this answer we provide a non-inductive argument along these lines. A nice feature of this argument is that we do not have to deal explicitly with pesky combinatorics and ...


10

(1) Your definition of strongly correlated system is correct "single-particle fails." We can still use ARPES to study strong correlated systems, we just do not see features that would be present in a weakly correlated system. The most prominent feature in a weakly correlated system is a sharp peak at certain energy and momentum. If you track this peak in ...


10

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 ...


10

What you want to find is the parameters $\theta=(C, \omega_0, \gamma)$ that minimizes the difference between $\nu(\omega|\theta)$ (the curve given the parameters) and the measured $\nu_i$ values. The most popular method is least mean square fitting, which minimizes the sum of the squares of the differences. One can also do it by formulating the normal ...


10

The correlation length of the 2d Ising model has been computed explicitly. You can find the expression in the famous book by McCoy and Wu. Here's a plot of the inverse correlation length (i.e., $1/\xi$) at various temperatures, taken from this recent review paper: This is only to show the directional dependence, as the radial scale is not the same for all ...


9

They are different. 'Schwinger terms' arise as central or abelian (or even non-abelian) extensions of current algebras of operators, very often as a consequence of the regularization procedure. They are source of anomalies. The original paper is short and readable: Field theory and commutators by J. Schwinger. You can read more in Moshe's answer to my ...


9

T-matrix elements are not the same as fully connected diagrams. Remember T-matrix is defined as $$T=S-I,$$ where $S$ is the S-matrix and $I$ is the identity matrix. S-matrix part includes all possible diagrams except vacuum bubbles and unamputated diagrams. Now we just need to ask, does subtracting an identity matrix remove all the disconnected diagram? ...


9

In addition to the correct mathematical interpretation appearing in the other answer by ACuriousMind, perhaps a good physically minded viewpoint is to observe that objects like $\delta(t)$ have always to be interpreted in the sense of the average value, using some smearing or averaging function (in QFT we use the same interpretation regarding field ...


8

The correlation function measures, as you would expect, how correlated two random variables are. That is, how often two random variables have similar values. We can construct such a function very simply. Say you are flipping coins, and you want to know if their results are correlated. To quantify things, call "heads" $+1$ and "tails" $-1$. To make ...


8

Don't try using any general-purpose curve fitting algorithm for this. The form of your function looks like a frequency response function, with the two unknown parameters $\omega_0$ and $\gamma$ - i.e. the resonant frequency, and the damping parameter. The function you specified omits an important feature if this is measured data, namely the relative phase ...


8

Let me write the Hamiltonian $$ H = -J \sum_i S_i^z S_{i+1}^z. $$ This choice will avoid some annoying (and irrelevant) signs. One way to formulate the statement in the OP precisely is as follows. Consider the variables $\delta_i=S_i^zS_{i+1}^z$. Since $\delta_i=1$ when the spins at $i$ and $i+1$ agree and $\delta_i=-1$ when the spins at $i$ and $i+1$ ...


7

I think they do contribute to the S-matrix.The amplitude of disconnected diagrams is the product of the amplitudes of all disconnected pieces. For example, putting two connected 2-particle scattering diagrams will give you a 4-particle scattering process, but it's not that physically interesting because this process is not a "genuine" 4-particle process in ...


7

The primary utility in introducing the generating functional is in using it to compute correlation functions of the given quantum field theory. Let's restrict the discussion to that of a theory of a single, real scalar field on Minkowski space, and let $x_1, \dots, x_n$ denote spacetime points. Of central importance are time-ordered vacuum expectation ...


7

Although there is already an accepted answer, I'll provide a different point of view here. It is important to keep in mind that the imaginary time is periodic, which means fields satisfy periodic(or anti-periodic) boundary conditions (the reason is because we use imaginary time path integral to represent the partition function, which is $\mathrm{Tr}\, e^{-\...


7

This is called the Fabri-Picasso theorem. Their argument requires both the vacuum and the charge $Q$ to be translationally invariant: $P |0\rangle = 0$, and $[P, Q] = 0$. The argument goes as follows: Since the charge originates from a symmetry, then according to Noether's theorem: $$Q = \int d^3x j_0(x)$$ Consider the correlation function of the charge ...


7

First note that, as you say, the 2-point function $\langle \sigma_i \sigma_j\rangle$ does not tend to zero as $\|j-i\|\to\infty$ when $T<T_{\rm c}$; namely, $$ \lim_{\|j-i\|\to\infty} \langle \sigma_i \sigma_j\rangle^+ = (m^*(T))^2, $$ where $m^*(T)=\langle\sigma_0\rangle^+$ is the spontaneous magnetization (the $+$ superscript indicates that the ...


Only top voted, non community-wiki answers of a minimum length are eligible