Timeline for How precisely the Klein-Gordon equation is derived?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 18, 2016 at 21:01 | comment | added | Phoenix87 | In second quantisation $\psi(x)$ is a system of infinitely, but countably many quantum harmonic oscillators and therefore it is interpreted as a field rather than a wave-function | |
| Oct 18, 2016 at 0:27 | comment | added | user129968 | @Phoenix87 I will be grateful if you just kindly answer to my last question (just by a few words). Just I'd like be sure. Thanks a lot. | |
| Oct 17, 2016 at 22:40 | comment | added | user129968 | @Phoenix87 Yes, the question was about the second quantization. I see.. So by the above argument, in fact, we precisely obtain a single-particle massive field equation (which quantum mechanically can describe a single particle)? | |
| Oct 17, 2016 at 22:22 | comment | added | Phoenix87 | I'm not sure what you are alluding to. Perhaps second quantisation? In that case one starts from the single-particle Hilbert space, which is precisely the $H$ above, and then performs the standard Fock representation construction. The result is an operator-valued distribution whose kernel (usually referred to as The Quantum Field by physicists) can be shown to formally satisfy the Klein-Gordon equation, but the story is a bit long. Details are in Streater-Wightman. | |
| Oct 17, 2016 at 22:00 | comment | added | user129968 | @Phoenix87 Fine; in addition, just one more note. I read that the Klein-Gordon (and other similar equations such as Dirac equations and so on) is subject to an additional quantization as well, because it is not a complete quantum mechanical equation (?). Would you please explain this? | |
| Oct 17, 2016 at 21:48 | comment | added | Phoenix87 | Quasi-invariant means that, after a transformation, you get a measure which is only equivalent to the starting one, but not exactly the same measure. The above argument can be seen as the mathematically precise way of supporting the argument that energy and momentum can be replaced by the corresponding differential operators in the relation $p^2-m^2=0$. | |
| Oct 17, 2016 at 21:44 | comment | added | user129968 | @Phoenix87 Thank you. Is this procedure which you wrote, practically, equivalent to this that in relativistic energy-momentum relation one formally replaces p, E and so on by their corresponding quantum operators, and once she writes the "wave function" (which is really a field) behind these operators, she will get the massive free Klein-Gordon equation? | |
| Oct 17, 2016 at 21:38 | comment | added | user129968 | @CountTo10 Thanks. Is this procedure which you wrote, practically, equivalent to this that in relativistic energy-momentum relation one formally replace p, E and so on by their corresponding quantum operators, and once she writes the "wave function" (which is really a field) behind these operators, she will get the massive free Klein-Gordon equation? | |
| Oct 17, 2016 at 21:10 | comment | added | user108787 | I just have to ask, and congrats on a different approach, what is the difference between an invariant and a quasi invariant, is it duration.? | |
| Oct 17, 2016 at 20:53 | history | answered | Phoenix87 | CC BY-SA 3.0 |