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A question from Srednicki's QFT textbook

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I have a question in Srednicki's QFT textbook. In order to compute the vacuum to vacuum transition amplitude given by : $$\left \langle 0|0 \right \rangle_{J}~=~\int \left [ d\varphi \right ]e^{i\int dt ~L\left ( \varphi ,\partial _{\mu }\varphi \right )},$$

one has to Fourier transform the field to the momentum space

$$\varphi \left ( x \right )~=~ \int d^{4}k~e^{ikx}\phi \left ( k \right ).$$

But what is the motivation to do so? And what other methods that could be used to compute $\left \langle 0|0 \right \rangle_{J}$ in the path integral formalism? I feel that although I understand the derivation, many steps are done in many places in this book but I don't understand what are the motivations for taking them.

This post imported from StackExchange Physics at 2015-03-30 13:53 (UTC), posted by SE-user nabill
asked Jul 7, 2012 in Theoretical Physics by nabill (20 points) [ no revision ]
BTW, you don't have $J$ on the right

This post imported from StackExchange Physics at 2015-03-30 13:53 (UTC), posted by SE-user Kostya

3 Answers

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Fourier transform is usually used to get rid of derivative terms. In the exponential the Lagrangian contains derivatives of field $\varphi$ w.r.t. time as well as space. After Fourier transformation you will get an algebraic expression in terms of $\varphi(k)'s$. This will simplify matters for now the expression that you have to integrate is just (exponential of) some algebraic expression in the variables over which you integrate. After all these formal calculations you will get to some diagrammatic rules for computing vacuum to vacuum transition amplitude. One can also use Hamiltonian point of view to get to the same diagrammatic rules as for example is done in Peskin-Schroeder and some other books. However the point that I would like to stress is that most modern QFT textbooks aim to teach QFT only at a practical level, rather than trying to go very deep into clarification of the arguments they use. In fact (interacting) QFT is mathematically not a well defined subject and so most (though not all:-) of arguments presented in textbooks are only at heuristic and formal level. So don't worry if you do not find these arguments very convincing or do not see much motivation behind them. It only means that you are beginning to understand QFT :-)

This post imported from StackExchange Physics at 2015-03-30 13:53 (UTC), posted by SE-user user10001
answered Jul 7, 2012 by user10001 (635 points) [ no revision ]
Thanks for this very good answer, But if one is doing a research , Could these heuristic methods lead to completely wrong answers ? Even if they seem plausible to most physicists?

This post imported from StackExchange Physics at 2015-03-30 13:53 (UTC), posted by SE-user nabill
As far as I understand it, most of interacting QFT is based on arguments which mathematically do not make much sense; but still (somewhat magically) physicists have had tremendous success in applying their QFT ideas to physics and mathematics. However, though I am not so knowledgeable as to give you any concrete examples, but yes logically there is always some possibility of getting to wrong conclusions if one uses seemingly correct but anyway wrong arguments.

This post imported from StackExchange Physics at 2015-03-30 13:53 (UTC), posted by SE-user user10001
The arguments in interacting QFT are frequently phrased in a way which makes them sound like mathematical gibberish. However, if you look at what the physicists are actually doing (making calculations in an effective field theory) instead of listening to what they're saying (blather about Lebesgue measure on infinite dimensional spaces), it's fairly sensible. This is not to say that it's rigorous, but it has not looked like magic since Wilson's work in the 70s.

This post imported from StackExchange Physics at 2015-03-30 13:53 (UTC), posted by SE-user user1504
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The essential motivation to transform to momentum space is that the translation invariance of the dynamics and of the vacuum state requires the 2-point Vacuum Expectation Value to be diagonal in the momentum variable. Thus, for the free field we have $$\left<0\right|\hat\phi(x)\hat\phi(y)\left|0\right>=G(x-y),$$ for some function $G(x-y)$ of the separation $x-y$, where the Fourier transform of this is $$\tilde G(k)=2\pi\delta(k^2-m^2)\theta(k_0).$$ $G$ has to be a function of $x-y$, not of $x$ and $y$ separately, for it to be translation invariant. In contrast, in real space, $$G(x-y)=\int 2\pi\delta(k^2-m^2)\theta(k_0)\mathrm{e}^{-\mathrm{i}k\cdot(x-y)}\frac{\mathrm{d}^4k}{(2\pi)^4},$$ which is manifestly not diagonal in the $x$ and $y$ "indices". If one considers two general "vectors" in the space of operators, $\hat\phi_f=\int\hat\phi(x)f(x)\mathrm{d}^4x$ and a similarly defined $\hat\phi_g=\int\hat\phi(y)g(y)\mathrm{d}^4y$, in the $f$ and $g$ "directions", one obtains $$\left<0\right|\hat\phi_f^\dagger\hat\phi_g\left|0\right>=\int f^*(x) G(x-y)g(y)\mathrm{d}^4x\mathrm{d}^4y=\int \tilde f^*(k)\tilde G(k)\tilde g(k)\frac{\mathrm{d}^4k}{(2\pi)^4},$$ which hopefully makes clear the Hermitian structure of the inner product this gives us. Another fact that is vitally important to the Hilbert space structure is manifest in the Fourier space presentation, that the function $G(x-y)$ is positive semi-definite in the Fourier co-ordinates (and hence it is positive semi-definite, but not diagonal, in all coordinates).

Working in the appropriate co-ordinates as always makes computations more compact, and keeps the computations relatively closer to the underlying conceptual structure, but one can use whatever coordinates one likes, or, better, that one can justify using.

This post imported from StackExchange Physics at 2015-03-30 13:53 (UTC), posted by SE-user Peter Morgan
answered Jul 7, 2012 by Peter Morgan (1,125 points) [ no revision ]
+ 1 like - 0 dislike

Sredniecki probably has his eye on perturbation theory, where you start with a collection of free fields, and then add interactions through formal power series.

The Fourier basis is useful for treating free fields because a) it diagonalizes the free field Hamiltonian, and because b) the Fourier modes have zero coupling to one another. In path integrals, this means that the path integral measure factorizes into a product of one-dimensional integrals, which makes it possible to evaluate it explicitly.

This post imported from StackExchange Physics at 2015-03-30 13:53 (UTC), posted by SE-user user1504
answered Jul 7, 2012 by user1504 (1,100 points) [ no revision ]

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