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16

The fundamental equation which serves as the basis for the path-integral formulation of finance and many physical problems is the Chapman-Kolmogorov equation. $$p(X_f|X_i)=\int p(X_f|X_k)p(X_k|X_i) dX_k$$ This is analogous to the following equation for amplitudes in quantum mechanics $$\langle X_f|X_i \rangle=\int \langle X_f|X_k\rangle\langle ... 16 In order to use Lagrangians in QM, one has to use the path integral formalism. This is usually not covered in a undergrad QM course and therefore only Hamiltonians are used. In current research, Lagrangians are used a lot in non-relativistic QM. In relativistic QM, one uses both Hamiltonians and Lagrangians. The reason Lagrangians are more popular is that ... 16 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 ... 11 There are rigorous constructions of QFTs in infinite volume. Glimm & Jaffe's book does this for interacting 2d scalars (with the assumption that the interactions are not too strong). I'm sure you can find other examples in the literature (or perhaps someone else will point you to them). However, restricting yourself to a lattice doesn't really buy ... 10 Path integral is indeed very problematic on its own. But there are ways to almost capturing it rigorously. Wiener process One way is to start with Abstract Wiener space that can be built out of the Hamiltonian and carries a canonical Wiener measure. This is the usual measure describing properties of the random walk. Now to arrive at path integral one has ... 10 The book Quantum dissipative systems by Weiss dedicates a subsection to the Feynman Vernon method, see also the original reference. See also this article and chapter 18.8 of the book by Kleinert. It's applied to the Caldeira-Leggett model, which is a toy model for a particle in contact with a heat bath. There are a number of mesoscopic systems out there in ... 8 There are already several good answers. Here I will only answer the very last question, i.e., if the Boltzmann factor in the path integral is f(S(t_f,t_i)), with action S(t_f,t_i)=\int_{t_i}^{t_f} dt \ L(t), why is the function f:\mathbb{R}\to\mathbb{C} an exponential function, and not something else? Well, since the Feynman "sum over histories" ... 8 One of the avenues to search for an answer is the so-called Keldysh formalism which is used extensively in condensed matter, in particular in mescopic physics, to define and study steady-state and time-dependent quantum phenomena in systems with infinitely many degrees of freedom. A recent comprehensive review is given by Kamenev and Levchenko, ... 7 Nope, Feynman's path integral formulation of quantum mechanics is a method to directly calculate the complex probability amplitudes and all objects that appear in its formalism - not counting proofs of equivalence with other approaches to quantum mechanics - are c-numbers representing classical observables. In particular, the exponent in the path integral ... 7 In 2-dimensional space-time, Feynman path integrals are perfectly well-defined, though understanding how this is done rigorously is somewhat heavy-going. But everything is spelled out in the book ''Quantum Physics: A Functional Integral Point of View'' by Glimm and Jaffe. http://www.amazon.com/Quantum-Physics-Functional-Integral-Point/dp/0387964770 In 4 ... 7 Yes, for this particular Matsubara sum you mentioned, taking different limits will lead to two results differed by 1. This is because the summation in consideration does not converge, which can be seen from the following integral (the continuous limit of the sum) considering the ultraviolet divergence (the large \omega behavior) ... 7 Let's do what Heidar says and write it with indices, and identify the Lagrangian.$$ L=\frac{1}{2}(\vec{\nabla}\times \vec{A})^2 = \frac{1}{2}\epsilon_{ijk}\partial_j A_k \epsilon_{ilm}\partial_l A_m $$where, if you haven't heard of it yet, you pretend there is a summation symbol for each repeated index. Then since there are no bare A_i sitting by ... 7 The easiest way to see imaginary time used is in elementary quantum mechanics in one dimension. (This is the explanation cribbed from wikipedia). Suppose we're looking at a tunneling-through-a-barrier problem. We start with the Schrodinger equation:$$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x) = E\psi(x) $$Make the ansatz$$\psi(x) = ...

6

Our plan of an answer is as follows. Firstly, we will introduce Planck's constant $\hbar$ so that the particular value $\hbar=1$ corresponds to the original problem. Secondly, we mention a connection to (what physicists often calls) the group property of Feynman path integrals. Thirdly, we will show that the sought-for formula happens to be the classical ...

6

Jane, $\partial_\tau$ is clearly a derivative with respect to a bosonic time $\tau$, so it commutes with everything else (except for functions of $\tau$ itself, with which it has a nonzero commutator), rather than anticommutes. Only if both objects have a fermionic character (if both of them are Grassmann-odd), they anticommute with one another (or they have ...

6

When physicists say that a quantum field $\phi(x)$ is real-valued, they are usually referring to Feynman's path integral formulation of quantum field theory, which is equivalent to Schwinger's operator formulation. The values of a field $\phi(x)$ in the path integral formulations are numbers. E.g.: If the numbers are real, we say that the field $\phi(x)$ ...

6

The theory of deformation quantization provides a framework in which the quantum to classical transition can be carried out and understood. According to this theory, for (practically any) quantum system, one can find (may be nonuniquely) a Poisson manifold $\mathcal{M}$ (phase space) equipped with an associative product called the "star product" such that ...

5

the original strategy of Feynman and wheeler was really about the desire to get rid of all self-interactions. In the modern language, it would eliminate most loop diagrams. In particular, consider an electron propagator, in the modern language. One may attach a photon propagator on it. That modifies the electron's self-energy, and this is the kind of a term ...

5

One major difficulty with defining path-integrals (which is entirely mathematicians' fault)is that the mathematicians insist for no good reason (and many bad ones) that there are non-measurable subsets of R. This is a psychological artifact of early days of set theory, where ZFC ws not seen as a way of generating countable models of a set-theoretic universe, ...

5

Yes, the Schaden and Spruch interpretation is correct. The interpretation is not used much because it's not as well connected into how experiments are run. In the usual text books, the Fourier transform is taken over position and time $(\vec{x},t)$ to get energy and momentum $(E,\vec{p})$. This is done by four integrations, one each getting rid of one of ...

5

Sources for the path integral You can read any standard source, so long as you supplement it with the text below. Here are a few which are good: Feynman and Hibbs Kleinert (although this is a bit long winded) An appendix to Polchinski's string theory vol I Mandelstam and Yourgrau There are major flaws with other presentations, these are pretty much the ...

5

How can we understand the presence of on-shell symmetry after quantization from a path integral point of view? One can derive a Schwinger-Dyson equation associated with the current conservation, also known as a Ward identity; see e.g. Peskin and Schroeder, An Introduction to Quantum Field Theory, Section 9.6; or Srednicki, Quantum Field Theory, Chapter ...

5

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

In the present case I think that it is more convenient to perform the propagator computation covariantly (and not in components). The inverse propagator (in the momentum space) can be read from the Abelian Chern Simons action including the gauge fixing term as: $G^{-1}_{\mu\nu}(k) = \alpha q_{\mu} q_{\nu} + i \frac{\theta}{4} ... 5 There is a well established method to evaluate the Jacobian for path integrals. This most often appears in the literature as a very nice way of understanding anomalies. In a theory with an anomaly, the integrand (action) must be invariant under a certain transformation (classical symmetry), but the symmetry disappears at the quantum level due to the path ... 4 The path integral over a "thick layer" of spacetime always produces the transition amplitudes $$\langle {\rm final}| U | {\rm initial}\rangle$$ where$U$is the appropriate unitary evolution operator. This is already true in non-relativistic quantum mechanics where the equivalence between Feynman's path integral approach and the operator formalism is being ... 4 The path integral in quantum mechanics computes the evolution kernel, which is the matrix element of the evolution operator:$\mathrm{exp}(iH t) $, ($H$is the Hamiltonian), between two position eigenstates. The path integral expresses the evolution kernel as a sum over paths:$U(x,t, x_0) = \int_{x(0)=x_0}^{x(t)=x} \mathrm{exp}(\frac{iS}{\hbar}) ...

4

I am not going to disagree with Lubos, because his answer is mostly correct, but the quantities in the path integral can be also be interpreted as operators on the Hilbert space of states, if you like. They are classical quantities on each individual trajectory of the path integral (for bosonic fields), but they become operators after you integrate, when ...

4

You start by writing down the probability to find a particle at $y$ at time $t$ when it was at $x$ at time $0$, denoted as $K(y,t;x,0)$. You get this by solving the Schrödinger equation with the initial condition $\psi(y,0) = \delta(y-x)$. Then, $K(y,t;x,0) = \psi(y,t)$. Thus, to solve this, we need to know the time development of $\psi(y,t0$. Let us start ...

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