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How are fields quantized

Anonymous Wed 02 Dec 19:24:08 2020 No.12412198 View Reply Original Report

Quoted By: >>12412436 >>12415779 >>12416216

Quantization means assigning an operator to an observable. So is a field an (one) observable? Or is it just normal QM but with a system of infinite observables with infinite operators? Because describing a field takes infinite numbers i guess.

Anonymous

Anonymous Wed 02 Dec 2020 20:31:46 No.12412436 Report

Quoted By: >>12412589

>>12412198
Search klein-gordon equations

Anonymous

Anonymous Wed 02 Dec 2020 21:19:02 No.12412589 Report

Quoted By: >>12413625

>>12412436
I know what it is but that doesn't answer. You mean its supposed to be an easy toy model? Ok. Are we formally having infinite operators for infinite observables or is "the field" a single observable?

Anonymous

Anonymous Thu 03 Dec 2020 03:31:56 No.12413625 Report

Quoted By:

>>12412589
the observables (operators) in QFT correspond, more or less, to local quantities in whatever basis you choose. therefore you basically have one operator for every point in phase space. like for example “is there a particle at this point in space”. so in that example you have one operator for every point in space(time)

Anonymous

Anonymous Thu 03 Dec 2020 04:34:35 No.12413838 Report

Quoted By: >>12413845

Wtf is you guy's background? Cause I'm a chem PhD student and have no idea wtf you guys are talking about

Anonymous

Anonymous Thu 03 Dec 2020 04:36:10 No.12413845 Report

Quoted By:

>>12413838
this is physics

Anonymous

Anonymous Thu 03 Dec 2020 18:25:47 No.12415779 Report

Quoted By:

>>12412198
Yes, the field becomes an operator-valued field.

Anonymous

Anonymous Thu 03 Dec 2020 20:01:08 No.12416197 Report

Quoted By:

Your guess is correct. We essentially place a harmonic oscillator at every point in space, so the overall field has infinitely many dof. (Canonical) quantization just takes this classical field, says it is an operator, and then imposes some commutator relations between it and its conjugate momentum.

Anonymous

Anonymous Thu 03 Dec 2020 20:04:39 No.12416216 Report

Quoted By:

>>12412198
If you've ever worked with ladder operators (also called creation and annihilation operators) in SHO or angular momentum you'll get the idea. The quantization condition is basically that these operators allow us to jump between quantum states of a field discretely.

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