Hellman-Feynman Theorem and its Applications – Quantum Mechanics

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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One thought on “Hellman-Feynman Theorem and its Applications – Quantum Mechanics

  1. Thanks for your explanation, I found it really clear and helpful!

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