# Sum of squares of 2 non-commutating operators

## 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 $XP\ne{PX}$ ?

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DrDu
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?

stevendaryl
Staff Emeritus
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.