The Feynman-Hellmann Theorem

A new physicist has entered the story: Richard Feynman.

The next topic in the usual sequence would be the fine structure of hydrogen, but first I need one very useful result: the Feynman–Hellmann theorem. I will reach it through three exercises—first the three-dimensional virial theorem, then the theorem itself, and finally two applications to hydrogen.

1. The virial theorem and the hydrogen expectation value

Problem 6.12

Use the virial theorem from Problem 4.40 to prove

where $a$ is the Bohr radius associated with the mass used in the Hamiltonian and $n$ is the principal quantum number. Throughout the hydrogen calculations, $m$ denotes the electron–proton reduced mass; in the infinite-proton-mass approximation it may be replaced by the electron mass. Accordingly,

$$ a=\frac{4\pi\epsilon_0\hbar^2}{me^2}; $$

with the electron-mass convention, this is the usual Bohr radius $a_0$. To get there, we first need the three-dimensional virial theorem.

Problem 4.40

For a stationary state, prove

The hint sends us back to Problem 3.31 and the general expectation-value equation.

Problem 3.31

Starting from

show that

For a stationary state, the left-hand side vanishes, so

This is the one-dimensional virial theorem. We now repeat the argument in three dimensions.

Let $Q=xp$ in one dimension. The general equation gives

In native notation, this step is

$$ \frac{d}{dt}\langle xp\rangle =\frac{i}{\hbar}\langle[H,xp]\rangle {}+\left\langle\frac{\partial(xp)}{\partial t}\right\rangle. $$

The only real work is the commutator $[H,xp]$:

Use the product rule for commutators,

so that

The first basic commutator is

and therefore

The second is

Putting these pieces together gives

and hence the stationary-state relation

The two displayed steps above are, in native form,

$$ \begin{aligned} \frac{d}{dt}\langle xp\rangle &=\frac{i}{\hbar} \left\langle-\frac{i\hbar}{m}p^2 {}+i\hbar x\frac{\partial V}{\partial x}\right\rangle {}+\left\langle\frac{\partial(xp)}{\partial t}\right\rangle\\ &=\frac1m\langle p^2\rangle -\left\langle x\frac{\partial V}{\partial x}\right\rangle {}+\left\langle\frac{\partial(xp)}{\partial t}\right\rangle, \end{aligned} $$

and, for the stationary-state step shown in the raster,

$$ \frac1m\langle p^2\rangle -\left\langle x\frac{\partial V}{\partial x}\right\rangle=0, \qquad 2\langle T\rangle =\left\langle x\frac{\partial V}{\partial x}\right\rangle. $$

The earlier uncertainty-principle post contains the same expectation-value identity in a different context:

In three dimensions we replace $xp$ by $\mathbf r\cdot\mathbf p$. For this stationary-state argument, assume the bound eigenstate belongs to the common operator or quadratic-form domain needed for the products below, the relevant expectations are finite, and the boundary terms vanish. Equivalently, one may use the symmetric dilation generator $D=(\mathbf r\cdot\mathbf p+\mathbf p\cdot\mathbf r)/2$; it differs from $\mathbf r\cdot\mathbf p$ only by a constant and therefore gives the same commutator with $H$.

The native equation is

$$ [H,\mathbf r\cdot\mathbf p] =[H,xp_x]+[H,yp_y]+[H,zp_z]. $$

The commutator splits into one copy for each Cartesian component,

and the result is

The commutator and the raster’s nested-minus expression give the native equality chain

$$ \begin{aligned} [H,\mathbf r\cdot\mathbf p] &=-\frac{i\hbar}{m}p^2+i\hbar\,\mathbf r\cdot\nabla V,\\ \frac{d}{dt}\langle\mathbf r\cdot\mathbf p\rangle &=\left\langle\frac{p^2}{m}\right\rangle {}-\langle\mathbf r\cdot\nabla V\rangle\\ &=2\langle T\rangle-\langle\mathbf r\cdot\nabla V\rangle. \end{aligned} $$

Thus, for a stationary state,

$$ 2\langle T\rangle=\left\langle \mathbf r\cdot\nabla V\right\rangle. $$

For the Coulomb potential,

we have $\mathbf r\cdot\nabla V=-V$. Therefore

Since $E=\langle T\rangle+\langle V\rangle$, it follows that $\langle V\rangle=2E$. Substituting the hydrogen energy levels then yields

or, accessibly,

$$ \boxed{\left\langle\frac1r\right\rangle=\frac{1}{a n^2}}. $$

2. The Feynman–Hellmann theorem

Problem 6.32

Let $H(\lambda)$ be a differentiable self-adjoint family on a suitable common operator or quadratic-form domain. Follow a differentiable eigenvalue branch $E_n(\lambda)$ with normalized eigenstates $|\psi_n(\lambda)\rangle$, written as

with eigenvalue equation

The Feynman–Hellmann theorem states

That is,

$$ \boxed{\frac{\partial E_n}{\partial\lambda} =\left\langle\psi_n(\lambda)\middle| \frac{\partial H}{\partial\lambda} \middle|\psi_n(\lambda)\right\rangle}. $$

At a degeneracy, let $P$ project onto the degenerate eigenspace. The first derivatives are obtained by diagonalizing the projected operator $P(\partial H/\partial\lambda)P$ and following the corresponding differentiable eigenbranches; repeated eigenvalues of that projected operator can leave the first-order basis nonunique.

Feynman obtained the result independently in his 1939 MIT undergraduate thesis, after Hellmann’s earlier work.

A quick perturbative proof

Choose a reference parameter

and call the corresponding Hamiltonian

Now shift the parameter by a small amount:

The new Hamiltonian differs slightly from the original,

so the perturbation is

To first order,

and ordinary first-order perturbation theory gives

Taking $d\lambda\to0$ proves the theorem.

There is also a direct proof: differentiate $H|n\rangle=E_n|n\rangle$ and multiply on the left by $\langle n|$. Self-adjointness supplies the eigenbra relation that cancels the state-derivative terms, while normalization supplies $\langle n|n\rangle=1$, leaving exactly the boxed equation above.

3. Application: the one-dimensional harmonic oscillator

For the harmonic oscillator,

$$ E_n=\left(n+\frac12\right)\hbar\omega, \qquad H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac12m\omega^2x^2. $$

(a) Choose $\lambda=\omega$

Differentiate both $E_n$ and $H$ with respect to $\omega$:

$$ \frac{\partial E_n}{\partial\omega} =\left(n+\frac12\right)\hbar, \qquad \frac{\partial H}{\partial\omega}=m\omega x^2. $$

Then the Feynman–Hellmann theorem gives

$$ \left(n+\frac12\right)\hbar =\langle m\omega x^2\rangle, \qquad \frac12\left(n+\frac12\right)\hbar\omega =\left\langle\frac12m\omega^2x^2\right\rangle =\langle V\rangle. $$

because $\partial H/\partial\omega=m\omega x^2=2V/\omega$. Therefore

$$ \boxed{\langle V\rangle=\frac{E_n}{2}}. $$

(b) Choose $\lambda=\hbar$

Now regard $\hbar$ as the parameter:

The derivatives are

$$ \begin{aligned} \frac{\partial E_n}{\partial\hbar} &=\left(n+\frac12\right)\omega,\\ \frac{\partial H}{\partial\hbar} &=-\frac{\hbar}{m}\frac{d^2}{dx^2},\\ \left(n+\frac12\right)\omega &=\left\langle-\frac{\hbar}{m}\frac{d^2}{dx^2}\right\rangle,\\ \frac12\left(n+\frac12\right)\hbar\omega &=\left\langle-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\right\rangle =\langle T\rangle. \end{aligned} $$

Since $\partial H/\partial\hbar=2T/\hbar$, we obtain

$$ \boxed{\langle T\rangle=\frac{E_n}{2}}. $$

So the two choices reproduce the familiar oscillator result $\langle T\rangle=\langle V\rangle=E_n/2$.

4. Application: hydrogen expectation values

Problem 6.33

For the reduced radial function $u(r)=rR(r)$, the effective Hamiltonian is

$$ H=-\frac{\hbar^2}{2m}\frac{d^2}{dr^2} {}+\frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2} -\frac{e^2}{4\pi\epsilon_0}\frac1r. $$

and the bound-state energy can be written as

$$ E_n=-\frac{me^4} {32\pi^2\epsilon_0^2\hbar^2(j_{\max}+\ell+1)^2}. $$

The radial Schrödinger equation begins as

$$ -\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} {}+\left(V+\frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}\right)u =Eu. $$

where

$$ V_{\mathrm{eff}}=V+\frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}. $$

is the effective potential. In other words, the entire operator in braces is the radial Hamiltonian:

$$ \left(-\frac{\hbar^2}{2m}\frac{d^2}{dr^2} {}+V+\frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}\right)u=Eu. $$

For hydrogen, substituting $V(r)=-e^2/(4\pi\epsilon_0r)$ gives

$$ \left(-\frac{\hbar^2}{2m}\frac{d^2}{dr^2} -\frac{e^2}{4\pi\epsilon_0}\frac1r {}+\frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}\right)u=Eu. $$

The earlier hydrogen-atom post used this restored illustration:

To recall how the spectrum arises, introduce the dimensionless quantities

Using the source raster’s notation,

$$ K=\frac{\sqrt{-2mE}}{\hbar}, \qquad \rho=Kr, \qquad \rho_0=\frac{me^2}{2\pi\epsilon_0\hbar^2K}. $$

and solve the series recurrence

$$ C_{j+1}= \frac{2(j+\ell+1)-\rho_0}{(j+1)(j+2\ell+2)}C_j. $$

Normalizability forces the series to terminate:

$$ n\equiv j_{\max}+\ell+1, \qquad \rho_0=2n. $$

which gives

The preserved raster is source evidence, but it contains two errors: the first equality is missing the square on $K$, and its intermediate coefficient $1/32$ is inconsistent with the stated definition of $\rho_0$. The authoritative native correction is

$$ E=-\frac{\hbar^2K^2}{2m} =-\frac{me^4}{8\pi^2\epsilon_0^2\hbar^2\rho_0^2}. $$

Because $\rho_0=2n$, the factor $1/(8\rho_0^2)$ becomes $1/(32n^2)$, giving the denominator $32$ in $E_n$. The termination condition also gives

$$ \rho_0^2=(2n)^2=4(j_{\max}+\ell+1)^2. $$

Substitution produces

$$ E_n=-\frac{me^4} {32\pi^2\epsilon_0^2\hbar^2(j_{\max}+\ell+1)^2}. $$

Equivalently, with $n=j_{\max}+\ell+1$,

$$ E_n=-\frac{m e^4}{2\hbar^2(4\pi\epsilon_0)^2n^2} =-\frac{\hbar^2}{2ma^2n^2}, \qquad a=\frac{4\pi\epsilon_0\hbar^2}{me^2}. $$

(a) Choose $\lambda=e$ to find $\langle 1/r\rangle$

Differentiating the energy and Hamiltonian with respect to $e$ gives

$$ \begin{aligned} E_n&=-\frac{me^4} {32\pi^2\epsilon_0^2\hbar^2(j_{\max}+\ell+1)^2},\\ \frac{\partial E_n}{\partial e} &=-\frac{me^3} {8\pi^2\epsilon_0^2\hbar^2(j_{\max}+\ell+1)^2},\\ \frac{\partial H}{\partial e} &=-\frac{e}{2\pi\epsilon_0}\frac1r. \end{aligned} $$

The Feynman–Hellmann equation then becomes

$$ -\frac{me^3} {8\pi^2\epsilon_0^2\hbar^2(j_{\max}+\ell+1)^2} =\left\langle-\frac{e}{2\pi\epsilon_0}\frac1r\right\rangle, \qquad \frac4eE_n=-\frac{e}{2\pi\epsilon_0} \left\langle\frac1r\right\rangle. $$

Rewriting $E_n$ with the Bohr radius,

immediately gives

$$ \boxed{\left\langle\frac1r\right\rangle=\frac{1}{a n^2}}. $$

This agrees with the virial-theorem result from the first exercise.

(b) Choose $\lambda=\ell$ to find $\langle 1/r^2\rangle$

Here we temporarily treat the angular-momentum quantum number $\ell$ as a continuous parameter along the regular radial branch with $j_{\max}$ held fixed, preserving the regular boundary condition at $r=0$ and the common quadratic-form domain. Thus $n(\ell)=j_{\max}+\ell+1$ varies during the differentiation; the physical principal quantum number is not held fixed. The useful energy form is

with $n(\ell)=j_{\max}+\ell+1$ on this fixed-$j_{\max}$ branch. Differentiating the energy gives

$$ \begin{aligned} E_n&=-\frac{me^4}{32\pi^2\epsilon_0^2\hbar^2} (j_{\max}+\ell+1)^{-2},\\ \frac{\partial E_n}{\partial\ell} &=\left(-\frac{me^4}{32\pi^2\epsilon_0^2\hbar^2}\right) \left[-2(j_{\max}+\ell+1)^{-3}\right]\\ &=\frac{2me^4} {32\pi^2\epsilon_0^2\hbar^2(j_{\max}+\ell+1)^3}. \end{aligned} $$

and differentiating the radial Hamiltonian gives

$$ \begin{aligned} H&=-\frac{\hbar^2}{2m}\frac{d^2}{dr^2} {}+\frac{\hbar^2}{2m}\frac{\ell^2+\ell}{r^2} -\frac{e^2}{4\pi\epsilon_0}\frac1r,\\ \frac{\partial H}{\partial\ell} &=\frac{\hbar^2}{2m}\frac{2\ell+1}{r^2}. \end{aligned} $$

Equating them through the Feynman–Hellmann theorem,

$$ \frac{2me^4} {32\pi^2\epsilon_0^2\hbar^2(j_{\max}+\ell+1)^3} =\left\langle \frac{\hbar^2}{2m}\frac{2\ell+1}{r^2} \right\rangle. $$

then substituting the Bohr-radius form of the energy,

The source raster rewrites the same equality as

$$ -2E_n(j_{\max}+\ell+1)^{-1} =\frac{\hbar^2}{2m}(2\ell+1) \left\langle\frac1{r^2}\right\rangle, $$

and, using $j_{\max}+\ell+1=n$ and $a=4\pi\epsilon_0\hbar^2/(me^2)$,

$$ \begin{aligned} \left\langle\frac1{r^2}\right\rangle &=\left(\frac1a\frac{e^2}{4\pi\epsilon_0}\frac1{n^3}\right) \left(\frac{2m}{\hbar^2}\frac1{2\ell+1}\right)\\ &=\frac{1}{a^2n^3(\ell+\tfrac12)}. \end{aligned} $$

produces

$$ \boxed{\left\langle\frac1{r^2}\right\rangle =\frac{1}{a^2n^3\left(\ell+\tfrac12\right)}}. $$

The next post will use the Kramers relation to connect a wider family of radial expectation values,

including $\langle r^{-3}\rangle$.