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Proof of relationship between Hamiltonian and Energy

  1. Oct 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Prove the relationship
    $$\left(\frac{\partial H}{\partial\lambda}\right)_{nn} = \frac{\partial E_{nn}}{\partial\lambda},$$
    where ##\lambda## is a parameter in the Hamiltonian. Using this relationship, show that the average force exerted by a particle in an infinitely deep potential well ##(0\le x\le a)##, on the right "wall" can be written as $$\langle F_\textrm{right}\rangle_n = \frac{\partial E_n}{\partial a},$$ where ##n## is the energy level. Calculate the force and compare it with the classical expression.

    2. Relevant equations

    For the first part: ##\hat{H}_\lambda|\psi_\lambda\rangle = E_\lambda|\psi_\lambda\rangle##

    ##\langle \psi_\lambda|\psi_\lambda\rangle = 1##
    ##\frac{d}{d\lambda}\langle \psi_\lambda|\psi_\lambda\rangle = 0##

    For the well: ##E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}##


    3. The attempt at a solution
    Okay so here is what I am thinking for the first part:

    $$\begin{eqnarray}
    \frac{d E_\lambda}{d\lambda} & = & \frac{d}{d\lambda}\langle\psi_\lambda|\hat{H}_\lambda|\psi_\lambda\rangle \\
    & = & \langle \frac{d\psi_\lambda}{d\lambda}|\hat{H}_\lambda|\psi_\lambda\rangle + \langle\psi_\lambda |\hat{H}_\lambda|\frac{d\psi_\lambda}{d\lambda}\rangle + \langle\psi_\lambda|\frac{d\hat{H}_\lambda}{d\lambda}|\psi_\lambda\rangle \\
    & = & E_\lambda \langle\frac{d\psi_\lambda}{d\lambda}|\psi_\lambda\rangle + E_\lambda\langle\psi_\lambda|\frac{d\psi_\lambda}{d\lambda}\rangle + \langle\psi_\lambda|\frac{d\hat{H}_\lambda}{d\lambda}|\psi_\lambda\rangle\\
    & = & \langle \psi_\lambda |\frac{d\hat{H}_\lambda}{d\lambda}|\psi_\lambda\rangle
    \end{eqnarray}$$

    I am not sure if this is the correct direction to go but I figured it was worth a try. Any help or comments are appreciated.
     
  2. jcsd
  3. Oct 13, 2014 #2

    Orodruin

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    Staff Emeritus
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    Homework Helper
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    Yes, your approach looks reasonable and will lead you correct. One small note on notation: I would write
    $$
    \frac{d\left| \psi\right>}{d\lambda}
    $$
    instead of
    $$
    \left|\frac{d\psi}{d\lambda}\right>.
    $$
     
  4. Oct 13, 2014 #3
    Okay thank you for your reply. Now for the second part I have
    $$\frac{\partial E_n}{\partial a} = -\frac{n^2\pi^2\hbar^2}{m a^3}$$. Now I need the average force on the right "wall", and my thought to get that is using ##-\vec{\nabla} U##, then
    $$\begin{eqnarray}
    \langle F_\textrm{right}\rangle &= & - \int \psi_\lambda^*\vec{\nabla}\psi_\lambda\,dq\\
    & = & -\int\psi_\lambda^* \frac{dU}{dx}\psi_\lambda\, dx
    \end{eqnarray}$$
    So my thought here is to jsut use the wavefunction for the infinitely deep potential well and more less plug-and-chug, but I could see some problems with that. However, I suppose in some sense this method is using the knowledge from part a.
     
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