Finding the Expectation value for the ground state of a Hydrogen atom
John Paul Aseniero and Gibson T. Maglasang
For the particle in the state [eq]\Psi[/eq], the expectation value of x is expressed as
[eq]\langle x\rangle = \int_{-\infty}^{+\infty}x|\Psi(x,t)|^2dx[/eq]
where the expectation value is the average of repeated measurements on an ensemble of identically prepared systems.
In this article, we would like to find [eq]\langle r\rangle[/eq], [eq]\langle r^2\rangle[/eq], [eq]\langle x\rangle[/eq] and [eq]\langle x^2\rangle[/eq] for an electron in the ground state of a hydrogen atom and express it in Bohr radius.
a.) Calculating for the [eq]\langle r\rangle[/eq] and [eq]\langle r^2\rangle[/eq],
(i) Finding [eq]\langle r\rangle[/eq]
The ground state wavefunction for the Hydrogen atom is given by
[eq]\Psi_{100}(r\theta,\phi)=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}[/eq]
Now getting the expectation value of r, we have
[eq]\langle r\rangle = \int{r|\frac{1}{\sqrt{\pi a^3}}e^{-r/a}|^2}(r^2\sin\theta d\theta d\phi dr)[/eq]
[eq]=\frac{1}{\pi a^3}\int_{0}^{\infty}\int_0^{2\pi}\int_0^{\pi}r^3 e^{-2r/a} \sin\theta d\theta d\phi dr[/eq]
[eq]=\int_0^{\infty}r^3 e^{-2r/a}dr[/eq]
The above integration can now easily be facilitated by using the table of integral,
[eq]\int_0^{\infty}r^3 e^{-2r/a}dr = 3!(\frac{a}{2})^4[/eq]. (1)
Therefore,
[eq] \langle r\rangle = \Big(\frac{4}{a^3}\Big)3!\Big(\frac{a}{2}\Big)^4[/eq].
(ii) For [eq]\langle r^2\rangle[/eq]
Next is we find the value of [eq]\langle r^2\rangle[/eq] by using the same process employed in the previous exercise. We have,
[eq]\langle r^2\rangle = \int r^2 |\frac{1}{\sqrt{\pi a^3}}e^{-r/a}|^2(r^2\sin\theta d\theta d\phi dr)[/eq]
[eq]= \int_0^{\infty}\int_0^{2\pi}\int_0^{\pi}r^4\frac{1}{\pi a^3}e^{-2r/a}\sin\theta d\theta d\phi dr[/eq]
[eq]=\frac{4}{a^3}\int_0^{\infty}r^4e^{-2r/a}dr[/eq]
Using again the table of integral used in (i) given in equation 1 to facilitate the integration, we get
[eq]=\frac{4}{a^3}4!\Big(\frac{a}{5}\Big)^5[/eq].
Thus,
[eq]\langle r^2\rangle=3a^2[/eq].
b) In the case of [eq]\langle x\rangle[/eq] and [eq]\langle x^2\rangle[/eq], for electron in ground state of hydrogen atom, this requires no new integration since [eq]r^2=x^2 + y^2 + z^2 [/eq].
(i) For the calculation of [eq]\langle x\rangle[/eq]
Now we have,
[eq]\langle x\rangle = \int{x|\frac{1}{\sqrt{\pi a^3}}e^{-r/a}|^2}(r^2\sin\theta d\theta d\phi dr)[/eq]
but [eq] x= r\sin\theta \cos\phi [/eq], it implies that
[eq]\langle x\rangle=\frac{1}{\pi a^3}\int_{0}^{\infty}\int_0^{2\pi}\int_0^{\pi}(r\sin\theta\cos\phi)r^2 e^{-2r/a} \sin\theta d\theta d\phi dr[/eq]
Try to have a closer look at the integral of the [eq]\phi[/eq] part and evaluate it from 0 to [eq]2\pi[/eq]. Obviously we have,
[eq]\int_0^{2\pi}\cos\phi d\phi=[sin\phi]|_0^{2\pi}=0[/eq].
Therefore,
[eq]\langle x\rangle = 0[/eq].
(ii) However, for the [eq]\langle x^2\rangle[/eq]
To find for [eq]\langle x^2\rangle[/eq], we have the following calculation,
[eq]\langle x^2\rangle = \int{x^2|\frac{1}{\sqrt{\pi a^3}}e^{-r/a}|^2}(r^2\sin\theta d\theta d\phi dr)[/eq]
[eq]=\frac{1}{\pi a^3}\int_{0}^{\infty}\int_0^{2\pi}\int_0^{\pi}(r\sin\theta\cos\phi)^2r^2 e^{-2r/a} \sin\theta d\theta d\phi dr[/eq]
[eq]= \frac{1}{\pi a^3}\int r^4e^{-2r/a}\sin^3\theta\cos^2\phi d\theta d\phi dr[/eq]
[eq]= \frac{1}{\pi a^3}\int_0^{\infty}r^4e^{-2r/a}\int_0^{2\pi}\cos^2\phi d\phi\int_0^{2\pi}\sin^3\theta d\theta[/eq]
Note that the above integration is facilitated by the following integral formulas:
[eq]\int_0^{2\pi} \sin^3\theta = \frac{4}{3}[/eq],
[eq]\int_0^{2\pi} \cos^2\phi d\phi = \pi[/eq].
Therefore,
[eq]\langle x^2\rangle = \frac{1}{\pi a^3}\frac{4}{3}\pi \int_{0}^{\infty} r^4 e^{-2r/a}[/eq]
[eq]= \frac{1}{3}\frac{4}{a^3}\int_0^{\infty} r^4 e^{-2r/a}dr[/eq]
[eq]=\frac{1}{3}\langle r^2\rangle[/eq].
So,
[eq]\langle x^2\rangle=a^2[/eq].
The expectation value for a ground state hydrogen atom are explicitly shown in this paper. The readers are also enjoined to calculate for the expectation value for momentum and see how they compare and contrast.
May 5th, 2012 at 6:49 am
Thanks……..very interesting article.
It really helped me!
May 5th, 2012 at 6:51 am
How can I solve the expectation value for ; especially…for 1s, 2s and 2p?
July 25th, 2014 at 8:45 am
Please correct:
1.) For : please change (a/5)^5, correct form is (a/2)^5.
2.) Limits in : Change 2pi to pi for sin^3.