Expectation values in the hydrogen ground state

The function is independent of the angles, and the angular integral gives $4\pi$:

$$\int|\psi_{1s}|^2\,d^3\mathbf r=\frac{4\pi}{\pi a^3}\int_0^\infty r^2e^{-2r/a}\,dr=\frac{4}{a^3}\times2!\Bigl(\frac a2\Bigr)^3=\frac{4}{a^3}\cdot\frac{a^3}{4}=1 .$$

The probability of finding the electron in the shell between the spheres of radii $r$ and $r+dr$ is $|\psi_{1s}|^2$ multiplied by the volume $4\pi r^2dr$ of this shell:

$$p(r)=\frac{4r^2}{a^3}\,e^{-2r/a}.$$

To find its maximum, we set the derivative to zero: $p'(r)=\frac{4}{a^3}\bigl(2r-\frac{2r^2}{a}\bigr)e^{-2r/a}=0$, which gives $r=a$. The most probable distance is the Bohr radius.

Using the radial density and the integral given above, with $b=a/2$:

$$\langle r\rangle=\frac{4}{a^3}\int_0^\infty r^3e^{-2r/a}dr=\frac{4}{a^3}\times3!\Bigl(\frac a2\Bigr)^4=\frac{4\times6}{16}\,a=\frac32\,a, \langle r^2\rangle=\frac{4}{a^3}\int_0^\infty r^4e^{-2r/a}dr=\frac{4}{a^3}\times4!\Bigl(\frac a2\Bigr)^5=\frac{4\times24}{32}\,a^2=3a^2 .$$

Thus $\sigma_r^2=3a^2-\frac94a^2=\frac34a^2$ and $\sigma_r=\frac{\sqrt3}{2}a\simeq0{,}87\,a$. The mean distance is greater than the most probable distance because the distribution has a long tail towards large values of $r$. The standard deviation is of the same order as the mean distance: the electron—nucleus distance is very poorly defined, and the picture of a circular orbit of radius $a$ is meaningless.

We have

$$\Bigl\langle\frac1r\Bigr\rangle=\frac{4}{a^3}\int_0^\infty r\,e^{-2r/a}dr=\frac{4}{a^3}\times1!\Bigl(\frac a2\Bigr)^2=\frac1a .$$

Note that $\langle1/r\rangle\neq1/\langle r\rangle$. The mean potential energy is therefore $\langle V\rangle=-\kappa/a$, and since $E_1=-\kappa/(2a)$, we have $\langle V\rangle=2E_1\simeq-27{,}2$ eV. The mean kinetic energy is $\langle T\rangle=E_1-2E_1=-E_1=\kappa/(2a)\simeq13{,}6$ eV. We indeed verify that $2\langle T\rangle=\kappa/a=-\langle V\rangle$.

We can also calculate $\langle T\rangle$ directly. Integration by parts gives $\langle T\rangle=\frac{\hbar^2}{2m_e}\int|\nabla\psi_{1s}|^2d^3\mathbf r$, and since $\psi_{1s}$ depends only on $r$, $|\nabla\psi_{1s}|=|\partial_r\psi_{1s}|=\psi_{1s}/a$. Thus $\langle T\rangle=\frac{\hbar^2}{2m_ea^2}\int|\psi_{1s}|^2=\frac{\hbar^2}{2m_ea^2}$. With $a=\hbar^2/(m_e\kappa)$, we obtain $\frac{\hbar^2}{2m_ea^2}=\frac{\kappa}{2a}$, in agreement with the previous result.

Classically, the electron can be found only where its kinetic energy $E_1-V(r)$ is positive, that is, where $-\kappa/r\leq-\kappa/(2a)$, or $r\leq2a$. The region $r>2a$ is classically forbidden. Let us calculate the probability of finding the electron beyond a distance $R$. Setting $u=2r/a$, we have $\frac{4r^2}{a^3}dr=\frac{u^2}{2}du$, and

$$P(r>R)=\int_{2R/a}^\infty\frac{u^2}{2}e^{-u}du=e^{-u_0}\Bigl(1+u_0+\frac{u_0^2}{2}\Bigr),\qquad u_0=\frac{2R}{a},$$

where we have integrated by parts twice. For $R=2a$, $u_0=4$, and

$$P(r>2a)=e^{-4}(1+4+8)=13\,e^{-4}\simeq0{,}24 .$$

Nearly a quarter of the probability lies in the classically forbidden region: this is the three-dimensional analogue of wave-function penetration outside a finite well.

From the previous question, $P(r<R)=1-e^{-u_0}(1+u_0+u_0^2/2)$. For $R=r_p$, $u_0=2r_p/a\simeq3\times10^{-5}$ is very small, and we can expand the exponential: $e^{-u_0}(1+u_0+u_0^2/2)\simeq1-u_0^3/6$. We can also reason directly: the volume probability density is practically constant, equal to its value at the centre $1/(\pi a^3)$, inside the proton, hence

$$P(r<r_p)\simeq\frac{1}{\pi a^3}\cdot\frac43\pi r_p^3=\frac43\Bigl(\frac{r_p}{a}\Bigr)^3\simeq\frac43\Bigl(\frac{0{,}84\times10^{-15}}{0{,}529\times10^{-10}}\Bigr)^3\simeq5\times10^{-15}.$$

This probability is extremely small, which justifies treating the proton as a point charge. It is nevertheless non-zero, and is greater the more the orbital is concentrated near the nucleus: this small overlap is responsible for measurable corrections to the energy levels, such as the hyperfine structure, which affect almost exclusively the $s$ orbitals, as these do not vanish at the nucleus.