Kramers’ relation
Kramers’ relation, named after the Dutch physicist Hans Kramers, is a relationship between expectation values of nearby powers of $r$ for the hydrogen-like atoms:
\[\boxed{ \frac{s+1}{n^2}\expct{r^s} -(2s+1)\left(\frac{a_0}{Z}\right)\expct{r^{s-1}} + \frac{s}{4}\left[ (2l+1)^2-s^2 \right]\left(\frac{a_0}{Z}\right)^2\expct{r^{s-2}} = 0 }\]
The relation is very important when computing, specifically, perturbative corrections to the hydrogen spectrum, as those computations require expectation values of the radial Hamiltonian and potentially perturbative corrections that have higher negative powers of $r$. Having this relation handy makes a lot of those calculations go more quickly.
Proof
Let’s remind how we solved the Schrödinger equation for hydrogen-like atoms; the radial differential equation is typically written as:
\[ -\frac{\hbar^2}{2\mu}\frac{\partial^2u}{\partial r^2} + \left[ \frac{\hbar^2l(l+1)}{2\mu r^2}-\frac{Ze^2}{4\pi\epzro r} \right]u = Eu \]
where $u=rR(r)$. Since
\[ a_0=\frac{4\pi\epzro\hbar^2}{\mu e^2} ,\; E=-\frac{\hbar^2}{2\mu}\frac{Z^2}{n^2a_0^2} \]
we can simplify the equation:
\[ u\rq\rq=\left[ \frac{l(l+1)}{r^2}-\frac{2Z}{a_0r}+\frac{Z^2}{n^2a_0^2} \right]u \label{1}\tag{1}\]
If we multiply both sides by $r^s$, then we’ll get exactly the powers of $r$ that we need for Kramer’s relation:
\[ r^s u\rq\rq=\left[ l(l+1)r^{s-2}-\frac{2Z}{a_0}r^{s-1}+\frac{Z^2}{n^2a_0^2}r^s \right]u \]
Now what we need to do is multiply both sides by $u^\ast=u$ and integrate. We’ll get the expected value on RHS since variables are separated, and each is normalized.
\[ \int_0^\infty ur^su\rq\rq dr = l(l+1)\expct{r^{s-2}}-\frac{2Z}{a_0}\expct{r^{s-1}}+\frac{Z^2}{n^2a_0^2}\expct{r^s} \label{2}\tag{2}\]
Let’s integrate $\ref{2}$ in parts.
\[ \begin{align*} \int_0^\infty ur^su\rq\rq dr &= \bigg[ ur^s u\rq \bigg]_0^\infty -\int_0^\infty \left( u\rq r^s+sur^{s-1} \right)u\rq dr \nl &= \underbrace{ -\int_0^\infty u\rq r^s u\rq dr } _{\dps I_1} + \underbrace{ -s\int_0^\infty ur^{s-1}u\rq dr } _{\dps I_2} \end{align*} \]
Let’s calculate $I_1$ integrating by parts.
\[ \begin{align*} I_1 &= -\left[ \frac{r^{s+1}}{s+1} u\rq^2 \right]_0^\infty + \frac{2}{s+1}\int_0^\infty r^{s+1}u\rq u\rq\rq dr \nl &= \frac{2}{s+1}\int_0^\infty r^{s+1}u\rq u\rq\rq dr \end{align*} \]
Here, we substitute $\ref{1}$ to $u\rq\rq$.
\[ I_1 = \frac{2}{s+1}\int_0^\infty u\rq\left[ l(l+1)r^{s-1}-\frac{2Z}{a_0}r^s-\frac{Z^2}{n^2a_0^2}r^{s+1} \right]u dr \]
Meanwhile,
\[ \begin{align*} \int_0^\infty r^s uu\rq dr &= \bigg[r^su^2\bigg]_0^\infty -\int_0^\infty\left( sr^{s-1}u^2 + r^s uu\rq \right)dr \nl &= -s\expct{r^{s-1}} -\int_0^\infty r^s uu\rq dr \end{align*} \]
\[ \therefore \int_0^\infty r^suu\rq dr = -\frac{s}{2}\expct{r^{s-1}} \label{3}\tag{3} \]
According to $\ref{3}$, we get $I_1$:
\[ I_1 = -\frac{l(l+1)(s-1)}{s+1}\expct{r^{s-2}} + \frac{2Zs}{(s+1)a_0}\expct{r^{s-1}} - \frac{Z^2}{n^2a_0^2}\expct{r^s} \]
We can also compute $I_2$ using $\ref{3}$:
\[ I_2 = \frac{s(s-1)}{2}\expct{r^{s-2}} \]
Substituting $I_1$ and $I_2$ to $\ref{2}$, we finally get the Kramers’ relation.
\[ \therefore \frac{s+1}{n^2}\expct{r^s} -(2s+1)\left(\frac{a_0}{Z}\right)\expct{r^{s-1}} + \frac{s}{4}\left[ (2l+1)^2-s^2 \right]\left(\frac{a_0}{Z}\right)^2\expct{r^{s-2}} = 0 \]
Expectation values
By using Hellmann–Feynman theorem and Kramers’ relation, we can induce various expectation values of hydrogen-like atoms.
- $ \Expct{\dfrac{1}{r}} = \dfrac{Z}{n^2a_0} \label{1.}$
Let’s use the fundamental charge $e$ as a parameter for the Hellman-Feynman theorem. \[ \hat{\hamiltonian}(e) = -\frac{\hbar^2}{2\mu}\nabla^2 - \frac{Ze^2}{4\pi\epzro r} \] Then, \[ \frac{\partial\hat{\hamiltonian}}{\partial e} = -\frac{Ze}{2\pi\epzro r} \] The energy of the electron dependent on $e$ is: \[ E_n(e) = -\frac{\mu}{2\hbar^2}\left( \frac{Ze^2}{4\pi\epzro} \right)^2 \] Then, \[ \frac{\partial E_n}{\partial e} = -\frac{1}{2\pi\epzro}\frac{Z^2}{n^2a_0} \] By the Hellmann-Feynman theorem, we get: \[ \begin{align*} \frac{\partial E_n}{\partial e} &= \Expct{\frac{\partial\hat{\hamiltonian}}{\partial e}} \nl &= -\frac{Ze}{2\pi\epzro}\Expct{\frac{1}{r}} \end{align*} \] Therefore, \[ \therefore \Expct{\frac{1}{r}} = \frac{Z}{n^2a_0} \] It can be also derived from substituting $s=0$ to the Kramers’ relation.
- $ \Expct{\dfrac{1}{r^2}} = \dfrac{2Z^2}{(2l+1)n^3a_0^2} $
Let’s use the angular momentum quantum number $l$ as a parameter for the Hellman-Feynman theorem. \[ \hat{\hamiltonian}(l) = \frac{\hbar^2}{2\mu r}\frac{\partial^2}{\partial r^2}r + \frac{\hbar^2l(l+1)}{2\mu r^2} - \frac{Ze^2}{4\pi\epzro r} \] Then, \[ \frac{\partial\hat{\hamiltonian}}{\partial l} = \frac{\hbar^2(2l+1)}{2\mu r^2} \] We showed that the principle quantum number $n$ and the angular momentum quantum number $l$ has the following relationship finding the eigenfunction of hydrogen-like atoms: \[ n=l+N+1 \;(N\in\N_0) \] The energy of the electron dependent on $l$ is: \[ E_n(l) = -\frac{\hbar^2}{2\mu} \frac{Z^2}{(l+N+1)^2a_0^2} \] Then, \[ \frac{\partial E_n}{\partial l} = \frac{\hbar^2}{\mu} \frac{Z^2}{(l+N+1)^3a_0^2} \] By the Hellmann-Feynman theorem, we get: \[ \begin{align*} \frac{\partial E_n}{\partial l} &= \Expct{\frac{\partial\hat{\hamiltonian}}{\partial l}} \nl &= \frac{\hbar^2(2l+1)}{2\mu} \Expct{\frac{1}{r^2}} \end{align*} \] Therefore, \[ \therefore \Expct{\frac{1}{r^2}} = \frac{2Z^2}{(2l+1)n^3a_0^2} \]
- $ \Expct{\dfrac{1}{r^3}} = \dfrac{1}{l(l+\frac{1}{2})(l+1)} \dfrac{Z^3}{n^3a_0^3} $
Let’s put $s=-1$ for the Kramers’ relation. \[ \frac{a_0}{Z}\Expct{\frac{1}{r^2}} - \frac{1}{4}\left[ (2l+1)^2-1 \right]\left(\frac{a_0}{Z}\right)^2\Expct{\frac{1}{r^3}} = 0 \] \[ \therefore \Expct{\frac{1}{r^3}} = \frac{1}{l(l+\frac{1}{2})(l+1)} \frac{Z^3}{n^3a_0^3} \]
- $ \expct{r} = \dfrac{1}{2}\Big[ 3n^2-l(l+1) \Big]\dfrac{a_0}{Z} $
It can be similarly shown by the Kramers’ relation.
- $ \expct{r^2} = \dfrac{1}{2}\Big[ 5n^2+1-3l(l+1) \Big]\dfrac{n^2a_0^2}{Z^2} $
Either.
- $ \Expct{\hamiltonian} = -\dfrac{Z^2e^2}{8\pi\epzro n^2a_0} $
We showed it here.
- $ \Expct{V} = -\dfrac{Z^2e^2}{4\pi\epzro n^2a_0} $
It is trivial from 1.
- $ \Expct{T} = \dfrac{Z^2e^2}{8\pi\epzro n^2a_0} $
It is trivial from 6 and 7.
We can also check that classical virial theorem, which is here $ \expct{T}=-\dfrac{1}{2}\expct{V} $, is satisfied.