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24

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 ...


20

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 ...


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 ...


13

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 ...


12

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

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" ...


10

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 ...


10

If the functional derivative $$\tag{1} \frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)} $$ exists (wrt. to a certain choice of boundary conditions), it obeys infinitesimally $$\tag{2}\delta F ~:=~ F[\phi+\delta\phi]- F[\phi] ~=~\int_M \!dx\sum_{\alpha\in J} \frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)}\delta\phi^{\alpha}(x). $$ OP's functional integral ...


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 ...


9

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) ...


9

Comments to the question (v2): 1) The correspondence between Lagrangian (L) and Hamiltonian (H) theories is mired with subtleties. Some general tools for singular Legendre transformations are available, such as Dirac-Bergmann analysis, Faddeev-Jackiw method, etc. But rather than claiming complete understanding and existence of the L-H correspondence, it is ...


9

In the context of axiomatic quantum field theory, there is a theorem (see theorem 3-7 in PCT, Spin and Statistics, and All That by Streater and Wightman, who I will refer to as "SW"), which SW call the "reconstruction theorem," essentially stating that correlation functions serve to completely determine a corresponding field theory in the Hilbert Space ...


9

I think it will depend the kind of statistical mechanics. For classical statistical mechanics, there is no time, so it is really hard to imagine a nice physical picture of the propagation of something. But nevertheless we still talk of loops as propagating "particles" (we give the "momenta", for instance, which is conserved, etc.). Interestingly, ...


9

Good question; I remember spending hours trying to understand this when I first learned QFT. Let's address your two main points in turn. First, you say I don't understand how rhyme these two different pictures. Let's outline how to connect the two pictures in steps. It's a good exercise to try and work through all of the gory details yourself, so I ...


8

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 ...


8

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) ...


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

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

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

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

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 ...


6

Good question. It has made me think. Strictly speaking, it is not possible to compute $\theta(t-t')\langle \dot\phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\dot\phi (t)\rangle$ ( shorthand notation: $\langle\,\equiv \langle 0 |\,$. Also note that I'm omitting the spatial arguments of the fields ) using the Lagrangian version of the path integral, ...


6

As Weinberg points in his QFT book, in the Hamiltonian formalism it is easier to check the unitarity of the theory because unitarity is directly related to evolution, while in the Lagrangian formalism the symmetries that mix space with time are more explicit. Therefore the Hamiltonian formalism is usually more convenient in non-relativistic and galilean ...


6

First, the equivalence theorem refers to S-matrix elements rather than off-shell n-point functions, or their generator $Z[j]$, which are generally different. What you have to study is the LSZ formula that gives the relation between S-matrix elements and expectation values of time-ordered product of fields (off-shell n-point functions, what one gets after ...


6

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 ...


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

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 ...


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 ...



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