
#1
Feb2613, 11:27 AM

P: 160

Basically is it possible to take a time derivative of a creation/annhilation operator?




#2
Feb2613, 12:53 PM

P: 226

I think that's not possible. If you are talking about creation and annhilation operators usually found in QFT, they live in the momentum space, so they don't depend on x, but on p. However, I am not sure if in some other circumstances this could be possible.




#3
Feb2613, 12:56 PM

P: 160

yeah so for eample [itex] \frac{d \hat{a}}{dt} [/itex] would =0 if a was the standard QFT annihilation op




#4
Feb2613, 01:46 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

time derivative of creation/annhilation operators
Saying you cannot take the "time derivative" does NOT mean that derivative is 0. It means that the derivative does not exist at all.



Register to reply 
Related Discussions  
Creation and Annihilation Operators  Quantum Physics  2  
Creation/Ann operators acting on <xp>  Quantum Physics  3  
Time it takes for a particle creation/annhilation (according to QFT)?  High Energy, Nuclear, Particle Physics  1  
Creation/annihilation operators  Quantum Physics  16  
Ask for Help: Is there this creation operators?  Atomic, Solid State, Comp. Physics  1 