Hellman-Feynman Theorem and its Applications – Quantum Mechanics

Apr 23, 2017
Manas Sharma

Hellman-Feynman Theorem investigates how the energy of a system varies as the Hamiltonian varies.
If a system is characterised by a Hamiltonian H that depends on a parameter \lambda , then the Hellmann-Feynman theorem states that,

\left<\frac{\partial H}{\partial \lambda}\right>=\frac{d E_n}{d \lambda}

where, E_n   is the energy of the system.

Proof:

H\left|\psi_n\right>=E_n\left|\psi_n\right>

\implies \left<\psi_n\right|H\left|\psi_n\right> = \left<\psi_n\right|E_n\left|\psi_n\right>

\implies \left<\psi_n\right|H\left|\psi_n\right> = E_n\left<\psi_n|\psi_n\right>

Since the wavefunctions must be normalised,

\implies \left<\psi_n\right|H\left|\psi_n\right> = E_n

Differentiating both sides w.r.t some parameter \lambda

\implies \frac{d }{d \lambda}\left<\psi_n\right|H\left|\psi_n\right> = \frac{d E_n}{d \lambda}

\implies \left<\frac{\partial \psi_n}{\partial \lambda}|H|\psi_n\right>+\left<\psi_n|\frac{\partial H}{\partial \lambda}|\psi_n\right>+ \left<\psi_n|H|\frac{\partial \psi_n}{\partial \lambda}\right> = \frac{d E_n}{d \lambda}

Since, H\left|\psi_n\right>=E_n\left|\psi\right> ,

\implies E_n\left<\frac{\partial \psi_n}{\partial \lambda}|\psi_n\right>+\left<\psi_n|\frac{\partial H}{\partial \lambda}|\psi_n\right>+ E_n\left<\psi_n|\frac{\partial \psi_n}{\partial \lambda}\right> = \frac{d E_n}{d\lambda}

\implies \left<\psi_n|\frac{\partial H}{\partial \lambda}|\psi_n\right>+E_n\left<\frac{\partial \psi_n}{\partial \lambda}|\psi_n\right>+E_n\left<\psi_n|\frac{\partial \psi_n}{\partial \lambda}\right> = \frac{d E_n}{d \lambda}

The 2nd and 3rd terms on the L.H.S. can be replaced by E_n\frac{d}{d \lambda}\left<\psi_n|\psi_n\right>

\implies \left<\psi_n|\frac{\partial H}{\partial \lambda}|\psi_n\right> + E_n\frac{d}{d \lambda}\left<\psi_n|\psi_n\right>  =  \frac{d E_n}{d \lambda}

Since \left<\psi_n|\psi_n\right>=1 (Normalisation condition)

\implies \left<\psi_n|\frac{\partial H}{\partial \lambda}|\psi_n\right>=  \frac{ d E_n}{d \lambda}

\implies \left<\frac{\partial H}{\partial \lambda}\right>=\frac{d E_n}{d \lambda}

Applications:

  • Expectation value of 1/r : \left<\frac{1}{r}\right> for Hydrogen, using Hellmann-Feynam Theorem:

Hamiltonian:
H=\frac{p^2}{2m} - \frac{e^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}

and  Energy:
E_n=-\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\frac{1}{n^2}

Taking, \lambda=e , in the Hellmann-Feynman theorem,

 \left<\frac{\partial H}{\partial e}\right>=\frac{d E_n}{d e}

\implies -\left<\frac{2e}{4\pi\epsilon_0 r}\right> =  -\frac{m}{2\hbar^2}\frac{4e^3}{(4\pi\epsilon_0)^2}\frac{1}{n^2}

\implies \frac{2e}{4\pi\epsilon_0}\left<\frac{1}{r}\right> = \frac{m}{2\hbar^2}\frac{4e^3}{(4\pi\epsilon_0)^2}\frac{1}{n^2}

\implies \left<\frac{1}{r}\right> = \frac{m}{\hbar^2}\frac{e^2}{4\pi\epsilon_0}\frac{1}{n^2}

Since, Bohr radius, a_0= \frac{4\pi\epsilon_0\hbar^2}{m e^2}

Therefore,
\implies \left<\frac{1}{r}\right> = \frac{1}{a_0 n^2}

  • Expectation value of 1/r^2 : \left<\frac{1}{r^2}\right> for Hydrogen, using Hellmann-Feynam Theorem:

Hamiltonian:
H=\frac{p^2}{2m} - \frac{e^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}

and  Energy:
E_n=-\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\frac{1}{(j_{max}+l+1)^2}

Taking, \lambda=l , in the Hellmann-Feynman theorem,

 \left<\frac{\partial H}{\partial l}\right>=\frac{d E_n}{d l}

  • Harmonic Oscillator:

Relation between \left<V\right> and E_n :
Take \lambda=\omega

Relation between \left<T\right> and E_n :
Take \lambda=\hbar

Relation between \left<T\right> and \left<V\right> :
Take \lambda=m

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Manas Sharma
I'm a physicist specializing in theoretical, computational and experimental condensed matter physics. I like to develop Physics related apps and softwares from time to time. Can code in most of the popular languages. Like to share my knowledge in Physics and applications using this Blog and a YouTube channel.



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