# Proof of relationship between Hamiltonian and Energy

## Homework Statement

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.

## Homework 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}##

## 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.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Yes, your approach looks reasonable and will lead you correct. One small note on notation: I would write
$$\frac{d\left| \psi\right>}{d\lambda}$$
$$\left|\frac{d\psi}{d\lambda}\right>.$$
$$\frac{\partial E_n}{\partial a} = -\frac{n^2\pi^2\hbar^2}{m a^3}$$. Now I need the average force on the right "wall", and my thought to get that is using ##-\vec{\nabla} U##, then
$$\begin{eqnarray} \langle F_\textrm{right}\rangle &= & - \int \psi_\lambda^*\vec{\nabla}\psi_\lambda\,dq\\ & = & -\int\psi_\lambda^* \frac{dU}{dx}\psi_\lambda\, dx \end{eqnarray}$$