1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum Mechanics: Raising and Lowering Operators

  1. Apr 25, 2015 #1
    1. The problem statement, all variables and given/known data

    Consider a particle in an energy eigenstate ##|n\rangle.##

    Calculate ##\langle x\rangle## and ##\langle p_x\rangle## for this state.

    2. Relevant equations

    ##x = \sqrt{\frac{\hbar}{2m\omega}}(a+a^{\dagger})##

    3. The attempt at a solution

    ##\langle x\rangle = \sqrt{\frac{\hbar}{2m\omega}}\langle n(a + a^{\dagger})|n\rangle = \sqrt{\frac{\hbar}{2m\omega}}(\sqrt{n}\langle n|n-1\rangle+\sqrt{n+1}\langle n|n+1\rangle).##

    But why does it equal zero?
     
  2. jcsd
  3. Apr 25, 2015 #2

    ShayanJ

    User Avatar
    Gold Member

    The energy eigenstates form an orthonormal set and here is no different. So ## \langle n|m\rangle=\delta_{nm} ##.
     
  4. Apr 25, 2015 #3
    So, ##\langle n|n+1\rangle = 0## and ##\langle n|n-1\rangle## equal zero from the property of the dirac function?
     
  5. Apr 25, 2015 #4

    ShayanJ

    User Avatar
    Gold Member

  6. Apr 25, 2015 #5

    ShayanJ

    User Avatar
    Gold Member

    And that is Kronecker delta, not Dirac delta!
     
  7. Apr 26, 2015 #6
    Opps! Thanks for clarifying.
     
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: Quantum Mechanics: Raising and Lowering Operators
Loading...