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

I Confused about operator action

  1. Jul 14, 2016 #1

    dyn

    User Avatar

    Hi. I have come across the following equation < x | p | ψ > = - iħ d/dx < x | ψ > where p is the momentum operator. I'm confused as to how p which acts to the right on | ψ >can then be taken outside the inner product so it now also acts on the bra < x | ?
    Thanks
     
  2. jcsd
  3. Jul 14, 2016 #2

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    The inner product ##\left<x|\psi\right>## is the wave function ##\psi(x)## that corresponds to the abstract state vector ##\left|\psi\right>##. That equation just says that the momentum operator in the position representation is the differential operator ##-i\hbar\frac{d}{dx}##.
     
  4. Jul 14, 2016 #3

    dyn

    User Avatar

    Still confused. What is p | ψ > then ? and why does the differential operator not act on the < x | bra ?
     
  5. Jul 14, 2016 #4

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    ##p\left|\psi\right>## is the abstract state vector that has a corresponding wave function ##\left<x|p|\psi\right>=-i\hbar\frac{d\psi(x)}{dx}##.
     
  6. Jul 14, 2016 #5

    dyn

    User Avatar

    Does p | ψ > = -iħ d/dx | ψ > ?
     
  7. Jul 14, 2016 #6

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    No it isn't, because ##\left|\psi\right>## is not a function of ##x##, it is a vector. The inner product ##\left<x|\psi\right>## is a function of ##x## and you can act on it with differential operators. There is a difference between the abstract operator ##p## that acts on state vectors, and the corresponding differential operator ##-i\hbar\frac{d}{dx}## that acts on wavefunctions.
     
  8. Jul 14, 2016 #7

    dyn

    User Avatar

    Thanks. So does p | ψ > have any meaning on its own ? Can it be evaluated as it is ?
     
  9. Jul 15, 2016 #8

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    You can't really make anything of the expression ##\left|\psi\right>## unless you represent it in some basis. The position representation, where you convert it into a function ##\psi(x)## is one possible basis. Another way would be to give the complex numbers ##\left<0|\psi\right>,\left<1|\psi\right>,\left<2|\psi\right>,\dots##, where the ##\left|0\right>,\left|1\right>,\left|2\right>,\dots## are a complete set of energy eigenstates.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted