91影视

Probability is a way to gauge how likely something is to happen. Many things are difficult to forecast with absolute confidence. Using it, we can only make predictions about how probable an event is to happen, or its chance of happening.

The arbitrary State is Given by$\cos 蟿{蠄}_{2,0,0}+\sin 蟿{蠄}_{2,1,0}$. Let the Probability of finding it in +z =P

$$\begin{gathered} P={鈭�}_0^{\frac{蟺}{2}}\sin 胃.d胃{鈭�}_0^{2蟺}d蠒\left(\cos 蟿{蠄}_{2,0,0}+\sin 蟿{蠄}_{2,1,0}\right)\left(\cos 蟿{{蠄}^{*}}_{2,0,0}+\sin 蟿{{蠄}^{*}}_{2,1,0}\right) \\ {\cos }^2蟿{鈭�}_0^{\frac{蟺}{2}}\sin 胃.d胃{鈭�}_0^{2蟺}d蠒{\left|{蠄}_{2,0,0}\right|}^2+{\sin }^2蟿{鈭�}_0^{\frac{蟺}{2}}\sin 胃.d胃{鈭�}_0^{2蟺}d蠒{\left|{蠄}_{2,1,0}\right|}^2 \\ +\cos 蟿\sin 蟿{鈭�}_0^{\frac{蟺}{2}}\sin 胃.d胃{鈭�}_0^{2蟺}d蠒({蠄}_{2,0,0}{{蠄}^{*}}_{2,1,0}+{蠄}_{2,1,0}{{蠄}^{*}}_{2,0,0}){鈭�}_0^{鈭�}{r}^{2}dr{R}_{2,0}\left(r\right)R_{2,3}\left(r\right) \end{gathered}$$

We will only perform the integral for the radial direction for the cross terms since otherwise it is due to normalisation.

localid="1659782296366" $$\begin{gathered} {鈭�}_0^{\frac{蟺}{2}}\sin 胃.d胃{鈭�}_0^{2蟺}d蠒{\left|{蠄}_{2,0,0}\right|}^2={鈭�}_0^{\frac{蟺}{2}}\sin 胃.d胃{鈭�}_0^{2蟺}d蠒\frac{1}{4蟺}=\frac{1}{2} \\ {鈭�}_0^{\frac{蟺}{2}}\sin 胃.d胃{鈭�}_0^{2蟺}d蠒\left({蠄}_{2,0,0}{{蠄}^{*}}_{2,1,0}+{蠄}_{2,1,0}{{蠄}^{*}}_{2,0,0}\right) \\ ={鈭�}_0^{\frac{蟺}{2}}\sin 胃.d胃{鈭�}_0^{2蟺}d蠒2\left(\sqrt{\frac{1}{4蟺}}\right)\left(\sqrt{\frac{3}{4蟺}}\cos 胃\right) \\ =\sqrt{3}{鈭�}_0^{\frac{蟺}{2}}\sin 胃\cos 胃.d胃 \\ =\frac{\sqrt{3}}{4}{鈭�}_0^{\frac{蟺}{2}}\sin 2胃.d胃 \\ =\frac{\sqrt{3}}{2} \\ {鈭�}_0^{鈭�}{r}^{2}dr{R}_{2,0}\left(r\right)R_{2,3}\left(r\right)={鈭�}_0^{鈭�}{r}^{2}dr\frac{1}{{\left(2a_0\right)}^{{\displaystyle \frac{3}{2}}}}2(1-\frac{2}{2a_0})e^{\frac{-r}{2a_0}}\frac{1}{{\left(2a_0\right)}^{{\displaystyle \frac{3}{2}}}}x\frac{1}{{\left(2a_0\right)}^{{\displaystyle \frac{3}{2}}}}x\frac{r}{\sqrt{3a_0}}{e}^{\frac{-r}{2a_0}} \\ =\frac{1}{8\sqrt{3}}鈭�dr\left(2r^3-r^4\right)e^{-r} \\ =-\frac{\sqrt{3}}{2} \end{gathered}$$

Probability equals

$$\begin{gathered} P=\frac{1}{2}{\cos }^2蟿+\frac{1}{2}{\sin }^2蟿+\frac{\sqrt{3}}{2}(-\frac{\sqrt{3}}{2})\cos 蟿\sin 蟿 \\ =\frac{1}{2}-\frac{3}{4}\cos 蟿\sin 蟿 \\ =\frac{1}{2}-\frac{3}{4}\sin 2蟿 \end{gathered}$$

So, The probability for an electron in this state in the +z hemisphere is$\frac{1}{2}-\frac{3}{8}\sin 2蟿$

The maximum of the probability is taken when $蟿=-45掳$,the maximum is $\frac{1}{2}+\frac{3}{8}=\frac{7}{8}$.

In this Case $\cos 蟿=-\sin 蟿=\frac{\sqrt{2}}{2}$,thus the corresponding ratio of two wave function is 1:-1

The Probability density as a function of$r,胃$is equal to, the radius is in units of${伪}_0$鈥�

$$\begin{gathered} P\left(r,胃\right)=\left({蠄}_{2,0,0}-{蠄}_{2,1,0}\right)\left({蠄}_{2,0,0}-蠄x_{2,1,0}\right) \\ 鈭�(2{(1-\frac{r}{2})e^{\frac{-r}{2}}-r\cos 胃e^{\frac{-r}{2}})}^2={\left(2-r-\cos 胃\right)}^2{e}^{-r} \end{gathered}$$

The Plot is as follows,where$胃$changes from 0 to$蟺$from Top to bottom. The x-axis is theRadius in term of${伪}_0$.From the plot we can see that the probability density is largest when$胃$is between 0 and$\frac{蟺}{2}$,that is +z axis..