Sum of squares of 2 non-commutating operators

  • #1

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}## ?
 

Answers and Replies

  • #2
DrDu
Science Advisor
6,023
755
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.
 
  • #3
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
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,401
2,578
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:

[itex]\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[/itex]

You can work out what the constant [itex]K[/itex] is.
 
  • #5
Thank you.
 

Related Threads on Sum of squares of 2 non-commutating operators

Replies
2
Views
693
  • Last Post
Replies
1
Views
5K
Replies
12
Views
3K
Replies
1
Views
623
Replies
3
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
15
Views
1K
  • Last Post
Replies
8
Views
11K
Top