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Partial derivative of Psi function

  1. Sep 3, 2007 #1
    1. The problem statement, all variables and given/known data
    Calculate [tex]d\left\langle p\right\rangle/dt.[/tex] Answer:

    2. Relevant equations
    [tex]d\left\langle p\right\rangle/dt = \left\langle -\partial V / \partial x\right\rangle[/tex]

    3. The attempt at a solution
    I've been through the rigor down to getting
    [tex]\left\langle -\partial V / \partial x\right\rangle + \int \hbar ^{2}/2m (\Psi ^{*} \partial ^{3}/\partial x^{3} \Psi - \partial ^{2} \Psi ^{*}/\partial x^{2} \partial \Psi /\partial x) dx[/tex]

    So I have what I'm looking for (ie. [tex]d\left\langle p\right\rangle/dt = \left\langle -\partial V / \partial x\right\rangle[/tex]); but I need to prove the rest of it (what's inside the integrand) as being equal to zero. So I'm wondering if there is anything which proves that [tex]\partial ^{2}\Psi ^{*}/\partial x^{2} = \partial ^{2}\Psi /\partial x^{2}[/tex]?

    If not, someone else showed me another method which uses the old [tex]d/dt (A*B) = (d/dt A *B) + (A * d/dt B)[/tex].

    Any thoughts?
  2. jcsd
  3. Sep 4, 2007 #2
    Is my question not very clear maybe? I'm just wondering how I can prove that what's inside the integral is equal to zero. I was trying to do it by proving that [tex]\partial ^{2} \Psi ^{*}/\partial x^{2} = \partial ^{2} \Psi /\partial x^{2}[/tex]; but maybe there is another way to approach it? I just need a direction to head in and I can do the work.
  4. Sep 4, 2007 #3
    Try integration by parts.
  5. Sep 4, 2007 #4
    Psi* d3(Psi) = d(Psi*d2(Psi))-d(Psi*)d2(Psi)

    The first term will vanish since Psi at infinity vanishes. Play around with the second term. Sorry, I'm i a bit of a rush. I'll look at the problem later this evening.
  6. Sep 4, 2007 #5
    ChaoticOrder - Ok, so [tex]\Psi ^{*}\partial ^{3}\Psi /\partial x^{3} = \partial /\partial x(\Psi ^{*}\partial ^{2}/\partial x^{2}(\Psi ))-\partial /\partial x(\Psi ^{*})\partial^{2}/\partial x^{2}(\Psi )[/tex]?
  7. Sep 4, 2007 #6


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    Look at it this way. It's much simpler. Use the canonical quantization scheme in the second postulate. Start with the classical problem. For a classical hamiltonian [itex] H(x,p)=\frac{p^{2}}{2m}+V(x) [/itex], the Hamilton motion equation is

    [tex]\frac{dp}{dt}=\left[p,H(x,p)\right]_{PB} = -\frac{dV(x)}{dx} [/tex]

    Quantize using the second postulate. By this you land in the Heisenberg picture, Schrödinger representation and, at an operator level

    [tex] \frac{d\hat{p}}{dt}= - \frac{dV(x)}{dx} \hat{1}_{L^{2}(\mathbb{R},dx)} [/tex]

    Take matrix element for an arbitrary state and you're there.
  8. Sep 4, 2007 #7


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    Right. The first term is a total derivative so the integral will give zero (assuming the usual condition that the wavefunction is normalized so it must go to zero faster than bla bla bla). Now, do an integration by parts again of the second term. Again, a surface term will go to zero and in the end you will find that,

    [tex] \int \Psi^* \partial^3 \Psi = \int (\partial^2 ~\Psi^*) \partial \Psi [/tex]
    so this will cancel the other term you had in your first post.
  9. Sep 5, 2007 #8
    dextercioby - Wow, so that approach is starting from my initial probelm right? I'm not sure what the "second postulate" is (my class has just started); but I can check that out. It looks pretty straight forward and easy to show. Thanks.

    nrqed - Ah, thank you for pointing that out. I now understand what my friend was saying for the other method he used to solve it. Thanks for the help.
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