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.

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