Timeline for What is a quantum field?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jun 5, 2017 at 22:02 | comment | added | md2perpe | physics.stackexchange.com/q/336369 | |
| Apr 13, 2017 at 12:40 | history | edited | CommunityBot |
replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/
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| S Dec 30, 2014 at 0:51 | history | suggested | glS | CC BY-SA 3.0 |
edited links. As they currently stand they don't show up in the "linked" section
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| Dec 29, 2014 at 23:50 | review | Suggested edits | |||
| S Dec 30, 2014 at 0:51 | |||||
| Dec 29, 2014 at 17:35 | comment | added | Calmarius | Why is this marked as "too broad"? If I get rid of the last paragraph will it be ok? | |
| Dec 29, 2014 at 17:24 | comment | added | CuriousOne | @Phoenix87: I certainly hope that we can find a way to get rid of these ugly descriptions. Like a whole generation of physicists I tend to hang on to what stuck to my brain early, but I would like to see a clean, hopefully algebraic and geometric description of QFT that does not suffer the conceptual clutter of the early approaches. | |
| Dec 29, 2014 at 17:14 | review | Close votes | |||
| Dec 29, 2014 at 17:26 | |||||
| Dec 29, 2014 at 17:03 | comment | added | Phoenix87 | @CuriousOne In AQFT there is hardly the need to refer to distributions when defining fields. As a further example, the Dirac delta function was envisaged by Dirac and then mathematically formulated later on thanks to the theory of operator algebras. | |
| Dec 29, 2014 at 16:57 | comment | added | CuriousOne | @Phoenix: To the mathematically uninitiated that sounds a lot like the procedure of defining distributions as linear operators over spaces of test functions in functional analysis? I would very much expect that kind of procedure to be necessary to formalize the problems physicists like to gloss over both in qft as well as in classic field theory. | |
| Dec 29, 2014 at 16:46 | comment | added | Phoenix87 | @CuriousOne What I have in mind is something more similar to a Segal field, which acts on states of a C*-algebra. To make contact with the operator-valued distributions and test functions one can take some Fourier transforms that send the state to the test function and the Segal field to said distribution. The action of the Segal field on the state (or perhaps the action of the state on the Segal field), which is a well defined evaluation of a linear functional over an element of a C*-algebra, becomes the usual field smeared-out with a test function. | |
| Dec 29, 2014 at 16:46 | answer | added | CuriousOne | timeline score: 9 | |
| Dec 29, 2014 at 16:36 | comment | added | CuriousOne | @Phoenix87: You just defined a quantum field properly: it is a distribution of operator values over a pre-defined geometry. Mathematicians may not like this approach (it has very serious problems with regards to the usual concepts of calculus that physicists like to apply to this object), but that is what we do and we get the correct results (i.e. the predictions match the experiments). | |
| Dec 29, 2014 at 16:32 | comment | added | Phoenix87 | A quantum field is a ill-defined mathematical concept. It can be made rigorous, but it is not the way a physicist would like. So a rather good answer could be: a field is an operator-valued distribution that somehow describe the physical content of a relativistic quantum mechanical system. | |
| Dec 29, 2014 at 16:24 | history | asked | Calmarius | CC BY-SA 3.0 |