# Tag Info

18

The main distinction you want to make is between the Green function and the kernel. (I prefer the terminology "Green function" without the 's. Imagine a different name, say, Feynman. People would definitely say the Feynman function, not the Feynman's function. But I digress...) Start with a differential operator, call it $L$. E.g., in the case of ...

17

This is just a property of Gaussian averaging analogous to the finite dimensional case: $\langle e^{ix} \rangle=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty} e^{ix}e^{-\frac{x^2}{2\sigma^2}}=e^{-\frac{\sigma^2}{2}}= e^{-\frac{\langle x^2 \rangle}{2}}$ The field can be decomposed into its independent Gaussian modes and integrated for each mode ...

16

The correlation function you wrote is a completely general correlation of two quantities, $$\langle f(X) g(Y)\rangle$$ You just use the symbol $x'$ for $Y$ and the symbol $x+x'$ for $X$. If the environment - the vacuum or the material - is translationally invariant, it means that its properties don't depend on overall translations. So if you change $X$ and ...

12

Cutology The math books aren't good for this, you need seat-of-the-pants intuition. The quick and dirty physicist answer is that a branch cut is best thought of as a continuum of poles densely spread over a line. You reproduce branch cuts by integrating poles spread over an interval, for example with a constant "residue density" (this is not the standard ...

11

a very intuitive example for correlation functions can be seen in laser speckle metrology. If you shine light on a surface which is rough compared to the wavelength, the resulting reflected signal will be somehow random. This can also be stated as that you cannot say from one point of a signal how a neighbouring one looks like - they are uncorrelated. Such ...

9

Excellent question, Kostya. Lubos already gave a detailed answer using general arguments in the language of QFT. In astrophysics and cosmology, however, there is another, and very simple, reason why we use the correlation functions all the time. It turns out that the mean value of the function $f(\vec{x})$, denoted $\langle f(\vec{x})\rangle$, can often not ...

9

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 ...

8

@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. ...

6

David Bar Moshe's derivation is of course right. Let me offer you a Taylor-expansion-based alternative proof: $$\left\langle e^{ix} \right \rangle = \left\langle \sum_{n=1}^\infty \frac{(ix)^n}{n!} \right \rangle = \left\langle \sum_{k=1}^\infty \frac{(ix)^{2k}}{(2k)!} \right \rangle$$ Here, I just used that by some odd-ness, the odd powers have a ...

5

For simplicity, let's restrict the discussion to that of a single particle moving in one dimension. Path integrals can be performed in much broader contexts like quantum field theory, but I think that would conceptually obscure the issue at this point. Let $H$ denote the (time-independent) quantum hamiltonian. Then the time evolution on the system is ...

5

The goal is to find the single particle propagator in the presence of interactions. This propogator will be the sum of all diagrams which have two external vertices. This sum of diagrams would be difficult to compute, but it turns out it easy to write this big sum of diagrams in terms of a sum of a smaller set of diagrams: the set of "one particle ...

4


3

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 ...

3

Consider the case of a free scalar field, governed by the usual Lagrangian, $$\mathcal{L} = \frac{1}{2}\partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2 \phi^2$$ The propagator, or equivalently Green's function for the theory is a function which can be though of as a response when we use a delta function as an input in the equations of motion, i.e. ...

3

A system's density matrix can be written in the form you cite if and only if it is a classical mixture of factorisable pure states. A factorisable pure state is, of course, one that can be written as a tensor product $\psi_A\otimes\psi_B$, where $\psi_A$ and $\psi_B$ are pure states in subsystems $A$ and $B$, respectively. Correlations between measurements ...

3

The latter route works as long as perturbative expansion works, i.e. states spectrum does not change. This can be checked by hand using said LSZ formula: you will end up calculating correlators $<\Phi(x_1)...>$ of free theory, which can be readily written in terms of creation-annihilation operators, nicely reproducing canonical perturbation series. ...

3

Lets start with the second question: Also, should I worry about taking $T\to\infty(1-i\epsilon)$ instead of $T\to \infty$ (which would the natural thing to do in (1))? And there lets start with asking ourselves "why" do we do the limit in the first place? The answer is -- we don't want the expectation for the $|\phi_a\rangle$ and $|\phi_b\rangle$ ...

3

I think some of your confusion stems from the fact that there are two different kinds of vacuua in QFT. First there is the vacuum of the free theory, usually denoted $|0\rangle$, second there is the full (interacting) vacuum, usually denoted $|\Omega \rangle$. What we want to calculate are the different quantities in the full theory like: ...

3

If you can calculate vacuum-to-vacuum transition amplitudes, you can calculate S-matrix elements, because the two are related by the LSZ reduction formula. The LSZ will in any case chop off the propagators for external lines that the generating functional inserts, so you will end up only needing to compute amputated diagrams.

3

Hint $\int_0^t \int_0^{t_1} dt_1 dt_2 \, a(t_1) a(t_2) = \frac{1}{2!} \int_0^t\int_0^t dt_1 dt_2\, \mathcal{T}\{ a(t_1) a(t_2) \}$ and so forth. You can see this by noting that the (square) integration region in the second integral can be split up into two triangular integration regions like in the first integral. This is one way to define the ...

3

Hints to the question (v1): Recall that the operator time ordering is symmetric $$\tag{1}T[A_1(t_1)\ldots A_n(t_n)]~=~T[A_{\pi(1)}(t_{\pi(1)})\ldots A_{\pi(n)}(t_{\pi(n)})],$$ where $\pi\in S_n$ is a permutation. (Here we assume for simplicity that all operators are Grassmann-even. Else there will be additional sign factors.) Recall that if $t_1> ... 3 Consider non-relativistic quantum mechanics of a point particle in 1 dimension with the classical Lagrangian $$\tag{A} L~:=~\frac{m}{2}\dot{x}^2-V(x).$$ Let$\Delta t:=t_f-t_i$and$\Delta x:=x_f-x_i$. OP's question concerns the following properties (B) and (C) of the kernel$K(x_f,t_f;x_i,t_i)$:$\$\tag{B} K(x_f,t_f;x_i,t_i) ...

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