Timeline for What is a path integral?
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
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| Dec 15, 2014 at 8:24 | comment | added | Luboš Motl | I've already answered how you use path integrals. | |
| Dec 15, 2014 at 8:24 | comment | added | Luboš Motl | No, it is a function - a mathematically similar object - but it is meant to be a classical field, and a classical field is something else than a quantum wave function. The value of a classical field - in the whole spacetime - describes a history of a physical system (the field). The integral is a "sum over histories". Except that it is not a sum of a small/finite number of terms. It is an integral, a path integral which is even more "integral-like" sum than the ordinary integral. | |
| Dec 15, 2014 at 0:21 | comment | added | TanMath | Also, is ϕ(y) the wavefunction? | |
| Dec 14, 2014 at 21:50 | comment | added | TanMath | how do you use path integrals? | |
| Dec 14, 2014 at 10:17 | comment | added | Luboš Motl | Dear @TAbraham, it is a functional which needs to be explained more operationally to be useful in the calculation of the path integral, and in my explanation (or lattice regularization of the path integral), $S[\phi(y)]$ is a function $S$ of finitely many variables called $\phi(0),\phi(0.01)$, and $\phi(y)$ for $y$ in the finite set that approximates the real interval. | |
| Dec 14, 2014 at 2:06 | comment | added | TanMath | @LubošMotl what does $S[ϕ(y)]$ mean but? | |
| Dec 3, 2014 at 13:29 | comment | added | MBN | @JamalS: I think you are both saying the same thing. What he says is that f() plays the role of x, and y the role of the subscript. | |
| Dec 3, 2014 at 13:18 | history | edited | Luboš Motl | CC BY-SA 3.0 |
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| Dec 3, 2014 at 7:10 | review | Suggested edits | |||
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| Dec 3, 2014 at 7:07 | comment | added | Luboš Motl | 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. | |
| Dec 2, 2014 at 19:58 | comment | added | TanMath | Thanks @Luboš-Motl , but what does it represent in quantum mechanics? | |
| Dec 2, 2014 at 18:19 | vote | accept | TanMath | ||
| Dec 2, 2014 at 9:45 | comment | added | Luboš Motl | 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. | |
| Dec 2, 2014 at 9:44 | comment | added | JamalS | I know perfectly well what an entire function is, if you read my last comment, you'll see I meant it in the non-technical sense. | |
| Dec 2, 2014 at 9:43 | comment | added | Luboš Motl | No, you're wrong. It's the values of $f(y)$ for individual values of $y$ which are the variabes that are being integrated over just like the variables $x_1$ or $x_3$ in the finite-dimensional integral. The reason why you wrote that only smooth functions contribute is that you wrote "entire function". Check what an "entire function" actually means: en.wikipedia.org/wiki/Entire_function It surely has to be smooth. | |
| Dec 2, 2014 at 9:42 | comment | added | JamalS | However, I apologise for my use of the word entire; I didn't mean it in the sense of an integral function, it was in the every day English sense :) | |
| Dec 2, 2014 at 9:38 | comment | added | JamalS | 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... | |
| Dec 2, 2014 at 9:37 | comment | added | Luboš Motl | Imagining that only smooth functions are enough to contribute to the path integral is totally incorrect. In the physics interpretation, this opinion directly conflicts with the uncertainty principle. The fact that the integral originates in unsmooth functions is absolutely needed to explain why $xp-px\neq 0$ using path integrals, see e.g. motls.blogspot.com/2012/06/… | |
| Dec 2, 2014 at 9:33 | comment | added | Luboš Motl | BTW if the essence of your comment is the word "entire" in front of the function, and you mean this adjective in the calculus sense, then you're completely wrong. The function $f$ isn't holomorphic (it's a function of a real $y$ in general) or analytic in any sense. It's not even smooth. In fact, the functions that are differentiable contribute by zero to the path integral. The whole value of the path integral comes from Brownian-motion-like chaotic functions that are not differentiable almost anywhere. | |
| Dec 2, 2014 at 9:28 | comment | added | Luboš Motl | 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. | |
| Dec 2, 2014 at 9:00 | comment | added | JamalS | +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. | |
| Dec 2, 2014 at 8:39 | history | answered | Luboš Motl | CC BY-SA 3.0 |