# Sum of squares of 2 non-commutating operators

• I
• Swamp Thing
In summary, Prof Adams discusses the factorization of operators and reminds us of the identity for complex scalars, ##c^2+d^2=(c-id)(c+id)##. When applied to operators, this formula results in an additional commutator due to the non-commutativity of operators. However, the cross-terms in the factorization do not cancel, but instead result in a constant term which can be calculated.

#### Swamp Thing

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}## ?

Swamp Thing said:
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##?
Of course, and this is exactly the reason why you get an additional commutator when you try to generalize this formula to non-commuting operators.

Thanks, but I'm still confused.

I'm not sure how you can equate/replace ##\frac{\hat{X}^2}{X_0^2}+\frac{\hat{P}^2}{P_0^2}##

with

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

in the first place, when you know that the cross terms in the latter are not going to cancel?

Swamp Thing said:
Thanks, but I'm still confused.

I'm not sure how you can equate/replace ##\frac{\hat{X}^2}{X_0^2}+\frac{\hat{P}^2}{P_0^2}##

with

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

in the first place, when you know that the cross terms in the latter are not going to cancel?

They don't cancel, but the cross-terms are a constant. So:

$\frac{\hat{X}^2}{X_0^2}+\frac{\hat{P}^2}{P_0^2} = (\frac{\hat{X}}{X_0}-i\frac{\hat{P}}{P_0})(\frac{\hat{X}}{X_0}+i\frac{\hat{P}}{P_0}) + K$

You can work out what the constant $K$ is.

Thank you.