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Jun 19, 2016 at 10:57 vote accept sagittarius_a
Mar 7, 2015 at 21:50 comment added dolun See the edit $\uparrow$ for discussion purpose.
Mar 7, 2015 at 21:49 history edited dolun CC BY-SA 3.0
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Mar 7, 2015 at 20:50 comment added sagittarius_a If it's not in Heisenberg picture and you transform to it the frequency doubles
Mar 7, 2015 at 20:46 comment added sagittarius_a Okay. But as a matter of fact I get a time dependence in the EM field. So. Is it now an operator in Heisenberg picture or not?
Mar 7, 2015 at 20:44 comment added dolun Once again, you don't land in any peculiar picture while performing quantization. These are two different stories.
Mar 7, 2015 at 20:37 comment added sagittarius_a The problem is: in which picture do you land when you perform the quantization
Mar 7, 2015 at 20:36 comment added dolun Yes and what exactly is the problem with that? All different QM pictures are unitary equivalent since probabilities are conserved in the process. The way the field is quantized is independant of the picture one can use to treat time-dependance.
Mar 7, 2015 at 20:28 comment added sagittarius_a Exactly. If you quantize the EM field, all ladder operations appear already as operators in the Heisenberg picture would appear .
Mar 7, 2015 at 20:26 comment added dolun I'm confused. What are you exactly calling "Heisenberg operators"? Are you referring to the Heisenberg picture of the operators $a$ and $\sigma_\pm$?
Mar 5, 2015 at 10:32 comment added sagittarius_a Thank you for your elaborate answer. However I am not completely happy since I am still confused why the quantization procedure yields the Heisenberg operators already..
Mar 4, 2015 at 14:01 history edited dolun CC BY-SA 3.0
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Mar 4, 2015 at 13:42 history answered dolun CC BY-SA 3.0