What is a path integral? [closed]

I was reading about path integrals because someone told me about it in this question. I read some articles about path integrals but couldn't understand it. Can you please explain path integral for me? What does it represent? What is the formulation?

closed as too broad by John Rennie, JamalS, ACuriousMind♦, Rob Jeffries, Kyle KanosDec 2 '14 at 14:24

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Mathematically, a path integral is a generalization of a multi-dimensional integral. In usual $N$-dimensional integrals, one integrates $$\int dx_1 dx_2 \dots dx_N$$ over a subspace of ${\mathbb R}^N$, an $N$-dimensional integral. A path integral is an infinite-dimensional integral $$\int {\mathcal D}f(y)\, Z[f(y)]$$ over all possible functions $f(y)$ of a variable $y$, which may be a real number or a vector. The values of the functions $f(0)$, $f(0.1)$, $f(0.2)$ etc. play the same role as the variables $x_1$, $x_2$ etc. in the usual multi-dimensional integral.

Because the index $i$ of $x_i$ was taking values in the finite set $1,2,\dots N$, and now it is replaced by the continuous variable $y$, the path integral is an infinite-dimensional integral.

Rigorous mathematicians see lots of problems preventing one from defining the infinite-dimensional path integral using the measure theory. But physicists know that similar integrals may be dealt with. There are some "ultraviolet divergences" etc. one experiences when trying to calculate them but they may be dealt with. In essence, one wants to use all the natural rules that apply to the finite-dimensional integrals. For example, the (path) integrals of a sum of two functions is the sum of two (path) integrals, and so on.

Two most important applications of path integrals in physics are in Feynman's approach to quantum mechanics, especially quantum field theory; and statistical mechanics.

In (classical) statistical mechanics, one wants to compute the partition sum $$Z = \sum_C \exp(-\beta E_c)$$ over all configurations $c$ of the physical system. But because the configurations are often labeled by whole functions $f(y)$ – infinitely many values at all allowed values of the argument $y$ – the sum isn't really a "sum". It isn't even a finite-dimensional integral. It is a path integral.

In quantum mechanics, the complex probability amplitudes etc. are calculated as $${\mathcal A}_{fi} = \int {\mathcal D}\phi(y)\,\exp(iS[\phi(y)]/ \hbar)$$ i.e. as the path integral over all configurations of the variables $\phi(y)$ etc. The integrand is a phase – a number whose absolute value is one – and the phase angle depends on the classical action evaluated from the possible history $\phi(y)$. The initial and final states $i,f$ are incorporated by integrating over those configurations in the "intermediate times" that obey the appropriate boundary conditions.

Almost all of quantum field theory may be expressed as a calculation of some path integrals. So in this sense, learning "everything" about a path integral is equivalent to learning almost all of quantum mechanics and quantum field theory, which may require between a semester and 10 years of intense study, depending on how deeply you want to get. It surely can't be covered in one allowed-size answer on this server.

The calculation of the path integrals with the Gaussian i.e. $\exp({\rm bilinear})$ integrand, perhaps with polynomial prefactors in the integration variables, is perhaps the most important or "simplest" example of a non-trivial path integral we actually need in physics.

In quantum mechanics, the path integral represents the explicit final formula for any probability amplitude. The amplitude for any transition from the state $|i\rangle$ to the state $|f\rangle$ may be directly expressed as a path integral, and the probability is the absolute value of the probability amplitude squared. Everything that quantum mechanics allows to calculate boils down to these probabilities - so the path integral represents "everything" in quantum mechanics. (This paragraph was originally posted as a comment of mine, and the user who proposed this edit had a good reason to do so.)

• +1, but I wouldn't say the values of the functions, $f(0), f(1)$, and so on play the role of $x_1,x_2$ etc. Since the functional maps entire functions to numbers, it's an entire function $f$ which replaces the role of a value of $x_1, x_2,$ etc. – JamalS Dec 2 '14 at 9:00
• I don't understand, @JamalS, which is a very diplomatic way of saying that I think that you don't understand. ;-) There is only one entire function $f$ but there are many variables $x_1,x_2$. The function carries even more (infinitely times more) information than several numbers $x_1,\dots , x_N$. In your last sentence, what is the conjunction in between $x_1,x_2$? If it's "or", then it's wrong because one has to specify all values of all $x_i$ to talk about the integrand. If it's "and", then OK, but then you are just trying to obscure the fact that the path in. is a multi-dimensional one. – Luboš Motl Dec 2 '14 at 9:28
• My objection is only to the analogy you state between the finite dimensional case, and the path integral. The way you've written it, you're saying the values of the function $f$ at different points "play the same role as the variables $x_1,x_2$ etc." Now, I agree, there's only one function $f$, and we are summing over all possible functions. So my point is, it's the different functions which are analogous to summing over different values of a scalar variable, $x$. I don't see how you've been able to extrapolate I think only smooth functions contribute from my single comment... – JamalS Dec 2 '14 at 9:38
• I only wrote that $\int D\phi(y)$ may be defined as the continuum limit of the multi-dimensional integral $\int \dots d\phi(-0.02)d\phi(-0.01) d\phi(0)d\phi(0.01)d\phi(0.02)\dots$ for $0.01$ sent to zero. I don't believe there can be anything controversial about this claim. It's really the essence of the answer. If you only say that "it is an integral over all values of a function everywhere", you are not moving by an epsilon to answering the question by the OP and explaining what an "integral over functions" actually is. An integral, in the pre-path-integral sense, is always finite-dim. – Luboš Motl Dec 2 '14 at 9:45
• Dear @TAbraham, it represents the explicit final formula for any probability amplitude. The amplitude for any transition from the state "i" to the state "f" may be directly expressed as a path integral, and the probability is the absolute value of the probability amplitude squared. Everything that quantum mechanics allows to calculate boils down to these probabilities - so the path integral represents "everything" in quantum mechanics. – Luboš Motl Dec 3 '14 at 7:07