Hydrogen atom – a sample wave function

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercise 13.1.3.

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The wave function for the hydrogen atom can be obtained by a series solution of the differential equation, leading to the result (which I’ve rewritten in Shankar’s notation, although my original post used Griffiths’s notation):

\displaystyle  \psi_{nlm}\left(r,\theta,\phi\right)=\frac{U_{El}\left(r\right)}{r}Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (1)

Here, we have

\displaystyle   U_{El} \displaystyle  = \displaystyle  e^{-\rho}v_{El}\ \ \ \ \ (2)
\displaystyle  v_{El} \displaystyle  = \displaystyle  \rho^{l+1}\sum_{k=0}^{\infty}C_{k}\rho^{k}\ \ \ \ \ (3)
\displaystyle  \rho \displaystyle  = \displaystyle  \sqrt{\frac{-2mE}{\hbar^{2}}}r \ \ \ \ \ (4)

The energy levels of the hydrogen atom are

\displaystyle  E=-\frac{me^{4}}{2\hbar^{2}n^{2}} \ \ \ \ \ (5)

where {n=1,2,3,\ldots}. The coefficients {C_{k}} in 3 are given by a recursion relation

\displaystyle   C_{k+1} \displaystyle  = \displaystyle  \frac{-e^{2}\lambda+2\left(k+l+1\right)}{\left(k+l+2\right)\left(k+l+1\right)-l\left(l+1\right)}C_{k}\ \ \ \ \ (6)
\displaystyle  \lambda \displaystyle  = \displaystyle  \sqrt{-\frac{2m}{\hbar^{2}E}} \ \ \ \ \ (7)

Combining {\lambda} and {E}, the formula becomes, for a given {n}

\displaystyle  C_{k+1}=\frac{2\left(k+l+1\right)-2n}{\left(k+l+2\right)\left(k+l+1\right)-l\left(l+1\right)}C_{k}

The coefficient {C_{0}} which starts everything off is determined by normalization.

As an example, we can find the wave function {\psi_{210}}. In this case {n=2} and {l=1} so the first term in the recursion, with {k=0} gives {k+l+1=2} and {C_{1}=0}. The full wave function is then

\displaystyle   \psi_{210} \displaystyle  = \displaystyle  \frac{1}{r}\rho^{2}e^{-\rho}C_{0}Y_{1}^{0} \ \ \ \ \ (8)

To evaluate {\rho} we use the energy for {n=2}:

\displaystyle  E_{2}=-\frac{me^{4}}{8\hbar^{2}} \ \ \ \ \ (9)

This gives

\displaystyle  \rho=\sqrt{\frac{2m^{2}e^{4}}{8\hbar^{4}}}r=\frac{me^{2}}{2\hbar^{2}}r=\frac{r}{2a_{0}} \ \ \ \ \ (10)

where {a_{0}} is the Bohr radius

\displaystyle  a_{0}\equiv\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (11)

Plugging everything into 8, using {Y_{1}^{0}=\sqrt{\frac{3}{4\pi}}\cos\theta}, we have

\displaystyle  \psi_{210}=\sqrt{\frac{3}{4\pi}}\frac{C_{0}}{4a_{0}^{2}}re^{-r/2a_{0}}\cos\theta \ \ \ \ \ (12)

Normalizing gives the condition

\displaystyle  \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\psi_{210}^{2}r^{2}\sin\theta dr\;d\theta\;d\phi=1 \ \ \ \ \ (13)

Working out the integral (using software or tables) gives

\displaystyle   \frac{3}{2}a_{0}C_{0}^{2} \displaystyle  = \displaystyle  1\ \ \ \ \ (14)
\displaystyle  C_{0} \displaystyle  = \displaystyle  \sqrt{\frac{2}{3a_{0}}} \ \ \ \ \ (15)

So the final wave function is

\displaystyle  \psi_{210}=\frac{1}{\sqrt{32\pi a_{0}^{3}}}\frac{r}{a_{0}}e^{-r/2a_{0}}\cos\theta \ \ \ \ \ (16)

which agrees with Shankar’s equation 13.1.27.

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