In class 23

23.1 Determine the most probable r in the ground state H-atom.

Probability density and unknown R, solve for R Take the derivative, set to 0

$$P(r)=r^2\left[2{\left(\frac{1}{a_0}\right)}^{3/2}{e}^{-r/a_0}\right]\left[2{\left(\frac{1}{a_0}\right)}^{3/2}{e}^{-r/a_0}\right]$$ $$\frac{d}{dr}P(r)=\frac{8r{\mathrm{e}}^{-\frac{2r}{a_0}}}{a_0^3}-\frac{8r^2{\mathrm{e}}^{-\frac{2r}{a_0}}}{a_0^4}$$

with zeros at $r=0,a_0$. Therefore the most probable R is $a_0$

23.2 What is the expectation value < r > for the ground state of H-atom?

$$R_{1,0}(r)=2(\frac{1}{a_0}{)}^{3/2}{e}^{-r/a_0}$$ $$<r>=4{\left(\frac{1}{a_0}\right)}^2\int r^3{e}^{(-2/a_0)r}dr={\left[-\frac{a_0}{2}{r}^3{e}^{-2r/a_0}+\frac{3a_0}{2}{e}^{-2r/a_0}\left(-\frac{a_0{r}^2}{2}-\frac{a_0^{2}r}{2}-\frac{a_0^3}{4}\right)\right]}_0^{\infty }$$ $$4{\left(\frac{1}{a_0}\right)}^3\left[\frac{3a_0^4}{8}\right]$$ $$\frac{3}{2}{a}_0$$

23.3 At t=0 a hydrogen atom is in a mixed state given

$$\Psi (r,\theta ,\varphi ,t=0)=\frac{1}{\sqrt{2}}{\Psi }_{1,0,0}+\frac{1}{\sqrt{2}}{\Psi }_{(2,1,1)}$$

(a) Determine the wave function at t later.

$$\Psi (r,\theta ,\varphi ,t)=\frac{1}{\sqrt{2}}{\Psi }_{1,0,0}{e}^{-i{E}_{n}t/\hslash }+\frac{1}{\sqrt{2}}{\Psi }_{(2,1,1)}{e}^{-i{E}_{n}t/\hslash }$$

(b) Determine expectation value of energy < 𝐸 >.

$$<E>=\int (\frac{1}{\sqrt{2}}{\Psi }_{1,0,0}{e}^{-i{E}_{n}t/\hslash }+\frac{1}{\sqrt{2}}{\Psi }_{(2,1,1)}{e}^{-i{E}_{n}t/\hslash })\hat{H}(\frac{1}{\sqrt{2}}{\Psi }_{1,0,0}{e}^{-i{E}_{n}t/\hslash }+\frac{1}{\sqrt{2}}{\Psi }_{(2,1,1)}{e}^{-i{E}_{n}t/\hslash })$$ $$\frac{E_1+E_2}{2}$$

23.4 At t=0 a hydrogen atom is in a mixed state given:

$$\Psi (r,\theta ,\varphi ,t)=\frac{1}{\sqrt{2}}{\Psi }_{1,0,0}{e}^{-i{E}_{n}t/\hslash }+\frac{1}{\sqrt{2}}{\Psi }_{(2,1,1)}{e}^{-i{E}_{n}t/\hslash }$$ $$L^2=L_x^2+L_y^2+L_z^2,L_z=x{P}_y-y{P}_x$$ $$\Psi (r,\theta ,\phi ,t)=\frac{1}{\sqrt{2}}{e}^{-i{E}_{1}t/\hslash }{\psi }_{(1,0,0)}+\frac{1}{\sqrt{2}}{e}^{-i{E}_{1}t/\hslash }{\psi }_{(2,1,1)}$$ $$<L^2>=\frac{1}{2}\cdot 0{\hslash }^2+\frac{1}{2}(2{\hslash }^2)={\hslash }^2$$ $$<L_z>=\frac{1}{2}(0\hslash )+\frac{1}{2}(\hslash )=\frac{\hslash }{2}$$