# Hydrogen atom – series solution and Bohr energy levels

Required math: calculus

Required physics: Schrödinger equation in 3-d

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Sec 4.2.1.

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercises 13.1.1 – 13.1.2.

[This page follows the derivation given in Griffiths. The discussion in Shankar’s chapter 13 is similar, but he uses Gaussian units, so the answer looks different. However, I can’t be bothered going through the whole derivation again with different units, since the steps are essentially the same.]

We saw in an earlier post that the radial part of the three-dimensional Schrödinger equation for the hydrogen atom can be reduced to the differential equation

$\displaystyle \rho\frac{d^{2}v}{d\rho^{2}}+2(l+1-\rho)\frac{dv}{d\rho}+(\rho_{0}-2l-2)v=0 \ \ \ \ \ (1)$

where

 $\displaystyle u(\rho)$ $\displaystyle =$ $\displaystyle \rho^{l+1}e^{-\rho}v(\rho)\ \ \ \ \ (2)$ $\displaystyle u(r)$ $\displaystyle \equiv$ $\displaystyle rR(r)\ \ \ \ \ (3)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \kappa r\ \ \ \ \ (4)$ $\displaystyle \rho_{0}$ $\displaystyle =$ $\displaystyle \frac{me^{2}}{2\pi\epsilon_{0}\hbar^{2}\kappa}\ \ \ \ \ (5)$ $\displaystyle \kappa$ $\displaystyle =$ $\displaystyle \frac{\sqrt{-2mE}}{\hbar} \ \ \ \ \ (6)$

and ${R(r)}$ is the radial part of the three-dimensional wave function.

Our task here is to solve 1 by using the same method as for the harmonic oscillator. We propose a solution of the form

$\displaystyle v(\rho)=\sum_{j=0}^{\infty}c_{j}\rho^{j} \ \ \ \ \ (7)$

and attempt to determine the coefficients ${c_{j}}$. The two derivatives needed in the equation are

 $\displaystyle \frac{dv}{d\rho}$ $\displaystyle =$ $\displaystyle \sum_{j=0}^{\infty}jc_{j}\rho^{j-1}\ \ \ \ \ (8)$ $\displaystyle \frac{d^{2}v}{d\rho^{2}}$ $\displaystyle =$ $\displaystyle \sum_{j=0}^{\infty}j(j-1)c_{j}\rho^{j-2} \ \ \ \ \ (9)$

We now plug these back into 1 and fiddle with the summation indexes so that every term in every sum is a multiple of ${\rho^{j}}$.

$\displaystyle \sum_{j=0}^{\infty}j(j-1)c_{j}\rho^{j-1}+2(l+1)\sum_{j=0}^{\infty}jc_{j}\rho^{j-1}-2\sum_{j=0}^{\infty}jc_{j}\rho^{j}+(\rho_{0}-2l-2)\sum_{j=0}^{\infty}c_{j}\rho^{j}=0 \ \ \ \ \ (10)$

The two terms containing ${\rho^{j-1}}$ can be converted to sums over ${\rho^{j}}$ by shifting the summation index from ${j}$ to ${j+1}$. This means that the sum becomes

$\displaystyle \sum_{j=-1}^{\infty}(j+1)jc_{j+1}\rho^{j}+2(l+1)\sum_{j=-1}^{\infty}(j+1)c_{j+1}\rho^{j}-2\sum_{j=0}^{\infty}jc_{j}\rho^{j}+(\rho_{0}-2l-2)\sum_{j=0}^{\infty}c_{j}\rho^{j}=0 \ \ \ \ \ (11)$

Note that the term with ${j=-1}$ in the first two sums is zero because of the ${(j+1)}$ factor, so we can start the sum at ${j=0}$. Since ${\rho^{j}}$ is now a common factor in all sums we can write the overall sum as

$\displaystyle \sum_{j=0}^{\infty}\left[(j+1)jc_{j+1}+2(l+1)(j+1)c_{j+1}-2jc_{j}+(\rho_{0}-2l-2)c_{j}\right]\rho^{j}=0 \ \ \ \ \ (12)$

Because each power series is unique (a mathematical theorem), the only way this sum can be valid for all values of ${\rho}$ is if all the coefficients are zero. That is

$\displaystyle (j+1)jc_{j+1}+2(l+1)(j+1)c_{j+1}-2jc_{j}+(\rho_{0}-2l-2)c_{j}=0 \ \ \ \ \ (13)$

This can be rewritten as a recursion relation:

$\displaystyle c_{j+1}=\frac{2(j+l+1)-\rho_{0}}{(j+1)(j+2(l+1))}c_{j} \ \ \ \ \ (14)$

[This equation is essentially the same as Shankar’s 13.1.11 if you replace ${j\rightarrow k}$ and use Gaussian units in ${\rho_{0}}$.]

The argument at this point is again similar to that for the harmonic oscillator: we examine the behaviour for large ${j}$. In that case, we can ignore the ${l+1}$ and ${\rho_{0}}$ terms and write

 $\displaystyle c_{j+1}$ $\displaystyle \sim$ $\displaystyle \frac{2j}{j(j+1)}c_{j}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{j+1}c_{j} \ \ \ \ \ (16)$

(We could also ignore the 1 in the denominator, but keeping it makes the argument easier, as we will see.) If we took this as an exact recursion relation, then starting with some initial constant ${c_{0}}$, we get

 $\displaystyle c_{1}$ $\displaystyle =$ $\displaystyle \frac{2}{1}c_{0}\ \ \ \ \ (17)$ $\displaystyle c_{2}$ $\displaystyle =$ $\displaystyle \frac{2^{2}}{2\times1}c_{0}\ \ \ \ \ (18)$ $\displaystyle c_{3}$ $\displaystyle =$ $\displaystyle \frac{2^{3}}{3\times2\times1}c_{0}\ \ \ \ \ (19)$ $\displaystyle c_{j}$ $\displaystyle =$ $\displaystyle \frac{2^{j}}{j!}c_{0}\ \ \ \ \ (20)$ $\displaystyle v(\rho)$ $\displaystyle =$ $\displaystyle c_{0}\sum_{j=0}^{\infty}\frac{2^{j}}{j!}\rho^{j}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}e^{2\rho} \ \ \ \ \ (22)$

In the last line we used the series expansion for the exponential function.

Returning for a moment to the original definition of ${v(\rho)}$, we get

 $\displaystyle u(\rho)$ $\displaystyle =$ $\displaystyle \rho^{l+1}e^{-\rho}v(\rho)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\rho^{l+1}e^{\rho} \ \ \ \ \ (24)$

Thus the infinite series solution gives a value for ${u}$ that increases exponentially for large ${\rho}$, which isn’t normalizable, so isn’t a valid solution. The only way to resolve this problem is again the same as in the harmonic oscillator case, which is to require the series to terminate after a finite number of terms. That is, we must have, for some value of ${j}$,

$\displaystyle 2(j+l+1)=\rho_{0} \ \ \ \ \ (25)$

That is, ${\rho_{0}}$ must be an even integer, which we can define as ${2n}$. Recalling the definition of ${\rho_{0}}$ from above, we therefore have the condition which quantizes the energy levels in the hydrogen atom:

 $\displaystyle \rho_{0}$ $\displaystyle =$ $\displaystyle \frac{me^{2}}{2\pi\epsilon_{0}\hbar^{2}\kappa}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2n \ \ \ \ \ (27)$

so

$\displaystyle \kappa=\frac{me^{2}}{4\pi\epsilon_{0}\hbar^{2}n} \ \ \ \ \ (28)$

But ${\kappa=\frac{\sqrt{-2mE}}{\hbar}}$, so for the energy levels, we get

$\displaystyle E=-\frac{1}{n^{2}}\frac{me^{4}}{2\hbar^{2}(4\pi\epsilon_{0})^{2}} \ \ \ \ \ (29)$

This is the Bohr formula (although Bohr got the formula without using the Schrödinger equation) for the energy levels of hydrogen. [Again, this is equivalent to Shankar’s 13.1.16 if you use Gaussian units, so that the ${(4\pi\epsilon_{0})^{2}}$ factor becomes 1.]

The degeneracy of each energy level is found by noting that for a given value of ${n}$, any value of ${l}$ is possible such that ${j+l+1=n}$. Since ${j}$ is just the index on the series coefficient ${c_{j}}$, this means that ${l}$ can be any value from 0 up to ${n-1}$. For each ${l}$, the ${z}$ component of angular momentum can have any value from ${m=-l}$ up to ${m=+l}$, which gives ${2l+1}$ possibilities for each ${l}$. Thus the degeneracy for energy state ${E_{n}}$ is

 $\displaystyle d\left(n\right)$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{n-1}\left(2l+1\right)\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\frac{1}{2}\left(n-1\right)n+n\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n^{2} \ \ \ \ \ (32)$

where we’ve used the formula

$\displaystyle \sum_{l=1}^{N}l=\frac{1}{2}N\left(N+1\right) \ \ \ \ \ (33)$

Before leaving the series solution, we need to point out that the polynomials produced by 14, with the constraint that ${\rho_{0}=2n}$, are known mathematically as the associated Laguerre polynomials. They can be written as derivatives. First we define the ordinary Laguerre polynomials ${L_{q}}$:

$\displaystyle L_{q}(x)=e^{x}\frac{d^{q}}{dx^{q}}(e^{-x}x^{q}) \ \ \ \ \ (34)$

Now the associated Laguerre polynomials ${L_{q-p}^{p}}$ which depend on two parameters can be defined in terms of the ordinary Laguerre polynomials:

$\displaystyle L_{q-p}^{p}(x)=(-1)^{p}\frac{d^{p}}{dx^{p}}(L_{q}(x)) \ \ \ \ \ (35)$

A more useful formula for the associated Laguerre polynomials is

$\displaystyle L_{n}^{k}(x)=\sum_{j=0}^{n}\frac{(-1)^{j}(n+k)!}{(n-j)!(k+j)!j!}x^{j} \ \ \ \ \ (36)$

In terms of associated Laguerre polynomials, the solution of 1 is (apart from normalization)

$\displaystyle v(\rho)=L_{n-l-1}^{2l+1}(2\rho) \ \ \ \ \ (37)$

We can verify that this is the solution of 1 by direct substitution. First, we plug in the correct indexes into 36:

$\displaystyle L_{n-l-1}^{2l+1}(2\rho)=\sum_{j=0}^{n-l-1}\frac{(-1)^{j}2^{j}(n+l)!}{(n-l-j-1)!(2l+j+1)!j!}\rho^{j} \ \ \ \ \ (38)$

Now we define the coefficients in the polynomial and show that the recurrence relation 14 is valid:

 $\displaystyle c_{j}$ $\displaystyle =$ $\displaystyle \frac{(-1)^{j}2^{j}(n+l)!}{(n-l-j-1)!(2l+j+1)!j!}\ \ \ \ \ (39)$ $\displaystyle \frac{c_{j+1}}{c_{j}}$ $\displaystyle =$ $\displaystyle \frac{-2(n-l-1-j)}{(j+1)(2l+j+2)} \ \ \ \ \ (40)$

This is the same recurrence relation provided ${\rho_{0}=2n}$. However, this isn’t enough to verify the solution since other definitions of ${c_{j}}$ would give the same relation (for example, we could leave out the ${(n+l)!}$ factor in the numerator and still get the same recurrence relation). To verify that the polynomials are in fact solutions, we can work out their derivatives and plug them into 1 directly.

We get

 $\displaystyle \sum_{j=0}^{n-l-1}\left[c_{j}(j-1)j\rho^{j-1}+2(l+1-\rho)c_{j}j\rho^{j-1}+2(n-l-1)c_{j}\rho^{j}\right]$ $\displaystyle =$ $\displaystyle \sum_{j=0}^{n-l-1}\left[c_{j}(j-1)j\rho^{j-1}+2(l+1)c_{j}j\rho^{j-1}+-2jc_{j}\rho^{j}+2(n-l-1)c_{j}\rho^{j}\right] \ \ \ \ \ (41)$

We can now shift the summation index for the first two terms so that we sum over ${j+1}$ instead of ${j.}$ This results in

$\displaystyle \sum_{j=-1}^{n-l-2}\left[c_{j+1}j(j+1)+2(l+1)(j+1)c_{j+1}\right]\rho^{j}+\sum_{j=0}^{n-l-1}\left[-2jc_{j}+2(n-l-1)c_{j}\right]\rho^{j} \ \ \ \ \ (42)$

In the first sum, the ${j=-1}$ term is zero due to the ${(j+1)}$ factor, so we can start both sums from ${j=0}$. Thus for all values of ${j}$ from 0 to ${n-l-2}$, we can examine the coefficient of ${\rho^{j}}$:

$\displaystyle c_{j+1}(j+1)(j+2l+2)+c_{j}(-2j+2n-2l-2) \ \ \ \ \ (43)$

Using the relation between ${c_{j}}$ and ${c_{j+1}}$ above, we get

 $\displaystyle \frac{c_{j+1}}{c_{j}}(j+1)(j+2l+2)+(-2j+2n-2l-2)$ $\displaystyle =$ $\displaystyle 2(j+l+1-n)+2(-j+n-l-1)\ \ \ \ \ (44)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (45)$

For the one remaining term in the second sum where ${j=n-l-1}$ we note that this term is zero on its own, since ${(-j+n-l-1)=0}$ in this case. Thus the overall sum satisfies the original differential equation 1.