# I Sum of squares of 2 non-commutating operators

1. Jun 21, 2016

### 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 $XP\ne{PX}$ ?

2. Jun 21, 2016

### DrDu

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.

3. Jun 21, 2016

### Swamp Thing

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?

4. Jun 21, 2016

### stevendaryl

Staff Emeritus
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.

5. Jun 21, 2016

Thank you.