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## Main Question or Discussion Point

Prof Adams does something rather strange, starting from 14:35 minutes in this lecture -- http://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013/lecture-videos/lecture-9/

He reminds us that for complex scalars, ##c^2+d^2=(c-id)(c+id)## and then proceeds to do the same with operators,

factorizing ##\frac{\hat{X}^2}{X_0^2}+\frac{\hat{P}^2}{P_0^2}##

in this way :

##=(\frac{\hat{X}}{X_0}-i\frac{\hat{P}}{P_0})(\frac{\hat{X}}{X_0}+i\frac{\hat{P}}{P_0})##

which he re-expands into a sum of squares plus a NON-ZERO commutator.

Is it not true that the identity he started with, i.e. ##c^2+d^2=(c-id)(c+id)## for complex scalars - is valid precisely when (and because) ##icd=idc##? So how does this apply to the operators where ##XP\ne{PX}## ?

He reminds us that for complex scalars, ##c^2+d^2=(c-id)(c+id)## and then proceeds to do the same with operators,

factorizing ##\frac{\hat{X}^2}{X_0^2}+\frac{\hat{P}^2}{P_0^2}##

in this way :

##=(\frac{\hat{X}}{X_0}-i\frac{\hat{P}}{P_0})(\frac{\hat{X}}{X_0}+i\frac{\hat{P}}{P_0})##

which he re-expands into a sum of squares plus a NON-ZERO commutator.

Is it not true that the identity he started with, i.e. ##c^2+d^2=(c-id)(c+id)## for complex scalars - is valid precisely when (and because) ##icd=idc##? So how does this apply to the operators where ##XP\ne{PX}## ?