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.

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