Hydrogen atom – wave function examples

Required math: calculus

Required physics: 3-d Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 4.11.

A few more examples of working out the hydrogen atom wave functions. Using the formulas in the last example, we can get {R_{20}}. The recursion formula for {n=2}, {l=0} is

\displaystyle  c_{j+1}=\frac{2(j+1)-4}{(j+1)(j+2)}c_{j} \ \ \ \ \ (1)

The series has 2 terms, and we get {c_{1}=-c_{0}}, so

\displaystyle   R_{20}\left(r\right) \displaystyle  = \displaystyle  \frac{1}{r}u_{20}\left(r\right)\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{r}\rho e^{-\rho}v_{20}(\rho)\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2a}e^{-r/2a}c_{0}\left(1-\frac{r}{2a}\right) \ \ \ \ \ (4)

To find {c_{0}} we normalize the radial function:

\displaystyle   \int_{0}^{\infty}r^{2}|R_{20}(r)|^{2}dr \displaystyle  = \displaystyle  \int_{0}^{\infty}c_{0}^{2}r^{2}\left[\frac{1}{2a}\left(1-\frac{1}{2a}\right)\right]^{2}e^{-r/a}dr\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \frac{a}{2}c_{0}^{2}\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  1 \ \ \ \ \ (7)

So {c_{0}=\sqrt{2/a}} and {R_{20}(r)} is

\displaystyle  R_{20}\left(r\right)=\frac{1}{\sqrt{2}a^{3/2}}e^{-r/2a}\left(1-\frac{r}{2a}\right) \ \ \ \ \ (8)

The complete wave function is then

\displaystyle   \psi_{200} \displaystyle  = \displaystyle  R_{20}(r)Y_{00}(\phi,\theta)\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\sqrt{8\pi}}a^{-3/2}\left(1-\frac{r}{2a}\right)e^{-r/2a} \ \ \ \ \ (10)

For {R_{21}\left(r\right)}, we have

\displaystyle  c_{j+1}=\frac{2(j+2)-4}{(j+1)(j+4)}c_{j} \ \ \ \ \ (11)

This time, there is only a single term in the series, so we have

\displaystyle   R_{21}\left(r\right) \displaystyle  = \displaystyle  \frac{1}{r}u_{21}\left(r\right)\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{r}\rho^{2}e^{-\rho}v_{21}(\rho)\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \frac{r}{\left(2a\right)^{2}}e^{-r/2a}c_{0} \ \ \ \ \ (14)

Doing the normalization integral for {R_{21}(r)} gives {c_{0}=\sqrt{2/3a}} which gives the final result

\displaystyle  R_{21}\left(r\right)=\frac{r}{2\sqrt{6}a^{5/2}}e^{-r/2a} \ \ \ \ \ (15)

There are 3 wave functions corresponding to {n=2,\; l=1}, for which we need the spherical harmonics

\displaystyle   Y_{1}^{1} \displaystyle  = \displaystyle  -\left(\frac{3}{8\pi}\right)^{1/2}\sin\theta e^{i\phi}\ \ \ \ \ (16)
\displaystyle  Y_{1}^{-1} \displaystyle  = \displaystyle  \left(\frac{3}{8\pi}\right)^{1/2}\sin\theta e^{-i\phi}\ \ \ \ \ (17)
\displaystyle  Y_{1}^{0} \displaystyle  = \displaystyle  \left(\frac{3}{4\pi}\right)^{1/2}\cos\theta \ \ \ \ \ (18)

The three wave functions are thus

\displaystyle   \psi_{211} \displaystyle  = \displaystyle  -\frac{1}{8\sqrt{\pi}}\frac{r}{a^{5/2}}e^{-r/2a}\sin\theta e^{i\phi}\ \ \ \ \ (19)
\displaystyle  \psi_{21-1} \displaystyle  = \displaystyle  \frac{1}{8\sqrt{\pi}}\frac{r}{a^{5/2}}e^{-r/2a}\sin\theta e^{-i\phi}\ \ \ \ \ (20)
\displaystyle  \psi_{210} \displaystyle  = \displaystyle  \frac{1}{4\sqrt{2\pi}}\frac{r}{a^{5/2}}e^{-r/2a}\cos\theta \ \ \ \ \ (21)

3 thoughts on “Hydrogen atom – wave function examples

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