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Computing the expectation value of momentum, kinetic energy, and compute

  1. Sep 16, 2012 #1
    1. consider this wave function ψ(x)=(√(30/L^5))(L-x) if 0≤x≤L
    and 0 else



    2. Compute the expectation value of the momentum. Compute the expectation value of the kinetic energy.
    Compute Δ p⋅Δ x for*this*wavefunction


    3. I have already proven that the wave function is normalized so it's equal to one. I just have no idea how to do the following. Thank you so much in advance!!!!
     
  2. jcsd
  3. Sep 16, 2012 #2
    Hi laser123,

    Recall that the definition of the expectation value [itex] \langle A \rangle [/itex] for any operator A is defined as [itex] \langle A \rangle = \langle \psi | A | \psi \rangle [/itex]. Do you know the operator forms of momentum, position, and kinetic energy in the position basis?

    Additionally, recall that [itex] \Delta A = \sqrt{ \langle A^2 \rangle - \langle A \rangle^2 }[/itex]. Once you calculate the square of the expectation value and the expectation value of the square of the position and momentum, you can do some simple calculations to solve for [itex] \Delta x \Delta p[/itex].

    Hope this helps!
     
  4. Sep 16, 2012 #3
    I am familiar with the momentum operator. Not so much with the kinetic energy or position operator

    Thank You!!
     
  5. Sep 16, 2012 #4
    Remember that, for the transition from classical dynamical variables to quantum dynamical variables, we replace the respective variable with its operator (where applicable).

    In the X basis, when we act on a ket [itex] | \psi \rangle [/itex] with the position operator, we obtain the position. As an eigenvalue equation, [itex] X | \psi \rangle = x | \psi \rangle [/itex].

    The classical representation of kinetic energy is [tex] KE = \frac{1}{2} m v^2 = \frac{1}{2} \frac{p^2}{2m} [/tex] The transition to quantum mechanics, then, requires inserting the momentum operator [itex] \hat{p} = -i \hbar \frac{\partial}{\partial x}[/itex]. (This is for the 1-D momentum operator in the position basis.) The kinetic energy then becomes [tex] KE = \frac{1}{2} \frac{\hat{p}^2}{2m} = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} [/tex]

    Now simply use the expectation value equations in order to solve for the expectation values of momentum and kinetic energy.
     
  6. Sep 16, 2012 #5
    I calculated the expectation value of the momentum to be <p> to be 0. The kinetic energy i calculated this to be (5/L^3)(ħ^2/2m). I have no idea how to do the last part! A dummy walk through on this uncertainty principle part would be awesome :D. Thank you!

    Thanks jmcleve
     
  7. Sep 17, 2012 #6
    Before we do the dummy walkthrough, can you show me how you computed the expectation values for the momentum and kinetic energy? Since this wavefunction is a little weird, I'm wondering if I'm calculating it correctly myself.
     
  8. Sep 17, 2012 #7
    For momentum i set up <P>=∫((√30/L^5)) P((√30/L^5)x(L-x))/1[/STRIKE] because the function is already normalized. So i subbed -iℏd/dx for p and distributed and applied the derivative function. Then I took the actual integral from 0 to L of what was left over. it all cancelled very nicely, but took like a whole page of work.

    For kinetic I did the same thing <T> = ∫((√30/L^5)) T((√30/L^5)x(L-x))/1. instead i subbed what you gave me inside of it (−ℏ^2/2m)(d^2/dx^2) and literally did the same as above but it cancelled to (5/L^3)(ℏ^2/2m)
     
  9. Sep 17, 2012 #8
    Sorry -- is the wavefunction supposed to be [itex] \psi(x) = \sqrt{\frac{30}{L^5}} x (L-x) [/itex]? From the calculation you just did, it looks like that. But your original post is [itex] \psi(x) = \sqrt{\frac{30}{L^5}} (L-x) [/itex]. Frankly, I like the former more than the latter.
     
  10. Sep 17, 2012 #9

    vela

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    The forum rules require you show some effort at attempting the problem yourself. Jmceive's already given you enough information in this thread for you to make some attempt at solving the last part of the problem.
     
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