Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Sum of squares of 2 non-commutating operators

  1. Jun 21, 2016 #1
    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. jcsd
  3. Jun 21, 2016 #2

    DrDu

    User Avatar
    Science Advisor

    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.
     
  4. Jun 21, 2016 #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?
     
  5. Jun 21, 2016 #4

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  6. Jun 21, 2016 #5
    Thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Sum of squares of 2 non-commutating operators
Loading...