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I Kinetic energy operator

  1. Dec 13, 2016 #1
    Hello everybody,

    A special problem constrain me to make a variable change in my Hamiltonian operator, so with the kinetic energy operator, I have a doubt.
    The variable change is: ## \theta \longrightarrow (\theta + k) ## (where ##k## is a constant).
    And the kinetic energy operator change as :
    $$ \hat{T}_{old}=\frac{-\hbar^2}{2m} \frac{\partial}{\partial \theta^2} \,\, \longrightarrow \,\, \hat{T}_{new}=\frac{-\hbar^2}{2m} \frac{\partial}{\partial (\theta+k)^2} $$

    When ##\hat{T}_{old}## and ##\hat{T}_{new}## operate respectively onto a function, say ##\psi(\theta)##, will I have two different results?


    Thank you very much.
     
  2. jcsd
  3. Dec 13, 2016 #2

    hilbert2

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    If ##\theta## is an angular variable, such as the ##\theta## in cylindrical coordinates, then that is not the correct expression of kinetic energy. If ##\theta## is a normal cartesian position coordinate, then those operators produce the same result when operating on a function, as you can deduce from the equation ##\frac{d}{d\theta} = \frac{d(\theta+k)}{d\theta}\frac{d}{d(\theta + k)}=\frac{d}{d(\theta + k)}##
     
  4. Dec 13, 2016 #3
    This hamiltonian is for a plane rotator (polar coordinate whith fixed-##r##). I should have mentioned inertia ##I## instead of mass ##m##.
    So it is about an angular variable ##\theta##. Is something change with respect to your answer in the Cartesian coordinate case?

    Thanks.
     
  5. Dec 13, 2016 #4

    hilbert2

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    Yes, if you use ##I## instead of ##m##, the expression is correct. The derivative ##\frac{d}{d\theta}## is equal to ##\frac{d}{d(\theta + k)}## for any variable ##\theta## and constant ##k##.
     
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