(a) Construct the wave function for hydrogen in the state $\mathbf{n}\mathbf{=}\mathbf{4}\mathbf{,}\mathbf{I}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{m}\mathbf{=}\mathbf{3}$. Express your answer as a function of the spherical coordinates $\mathit{r}\mathbf{,}\mathit{\theta }\mathbf{}\mathbf{and}\mathbf{}\mathit{\varphi }$.

(b) Find the expectation value of role="math" localid="1658391074946" $\mathit{r}$in this state. (As always, look up any nontrivial integrals.)

(c) If you could somehow measure the observable ${\mathit{L}}_{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathit{L}}_{\mathbf{y}}^{\mathbf{2}}$on an atom in this state, what value (or values) could you get, and what is the probability of each?

The location of an electron at a specific place in space (defined by its x, y, and z coordinates) and the amplitude of its wave, which corresponds to its energy, are related by a mathematical function known as a wave function,$\mathit{\psi }$.

The equation for the spatial wave function of a hydrogen atom ,

${\psi }_{n/m}=\sqrt{{\left(\frac{2}{na}\right)}^3\frac{\left(n-I-1\right)!}{2n{\left[\left(n+I\right)!\right]}^3}}{e}^{-r/na}{\left(\frac{2r}{na}\right)}^I{L}_{n-I-1}^{2I+1}\left(\frac{2r}{na}\right)Y_I^m\left(\theta ,\varphi \right)$

HereL is an associated Laguerre polynomial andY is a spherical harmonic, and they are given as follow:

$$\begin{gathered} L_p^q\left(x\right)=c_0\sum _{j=0}^p\frac{{\left(-1\right)}^j\left(p+q\right)!}{\left(p-j\right)!\left(q+j\right)!j!}{x}^j{Y}_I^m\left(\theta ,\varphi \right) \\ ={\left[\frac{2I+1\left(I-m\right)!}{4\mathrm{\pi }\left(\mathrm{I}+\mathrm{m}\right)!}\right]}^{-1/2}{e}^{imo}{P}_I^m\left(\cos \left(\theta \right)\right) \end{gathered}$$

Here$P_I^m$is the associated Legendre function:

$P_l^m\left(x\right)=(1-x^2{)}^{\left|m\right|/2}{\left(\frac{d}{dx}\right)}^{\left|m\right|}{P}_l(x)$

And,

$P_l\left(x\right)=\frac{1}{2^{l}l!}{\left(\frac{d}{dx}\right)}^I(x^2-1{)}^l$

For n=4,I= 3 and m =3, determine $Y_3^3$. To find it construct $Y_I^I$

$Y_l^l={(-1)}^I\sqrt{\frac{(2\mathrm{l}+1)}{4\mathrm{\pi }}\frac{1}{\left(2I\right)!}}{e}^{i}l\varphi P_l^l\left(cos\right(\theta \left)\right)$

From (1), write$P_I^I$as:

$P_l^l\left(x\right)=(1-x^2{)}^{1/2}{\left(\frac{d}{dx}\right)}^I{P}_l(x)$

Replace from (5) with $P_I$

$P_l^l\left(x\right)=\frac{1}{2^{l}l!}(1-x^2{)}^{I/2}{\left(\frac{d}{dx}\right)}^{2I}(x^2-1{)}^l$

but $(x^2-1{)}^l=x^{2I}+\cdots$, and the remaining term has a power less than $2I$.So, when differentiate $(x^2-1{)}^l,2l$times all the terms vanishes except the first term with the power of , thus:

$P_l^l\left(x\right)=\frac{1}{2^{l}l!}(1-x^2{)}^{I/2}{\left(\frac{d}{dx}\right)}^2{x}^{2I}$

Now,

${\left(\frac{d}{dx}\right)}^n{x}^n=n!$

Hence:

$P_l^l=\frac{\left(2l\right)!}{2^{l}l!}(1-x^2{)}^{I/2}$

Next for $x=cos\left(\theta \right),1-x^2={\sin }^2\left(\theta \right)$:

$P_l^l=\frac{\left(2l\right)!}{2^{l}l!}si{n}^l\left(\theta \right)$

So:

$$\begin{gathered} Y_I^I={\left(-1\right)}^I\sqrt{\frac{\left(2I+1\right)}{4\mathrm{\pi }\left(2\mathrm{I}\right)!}}{e}^{iI\varphi }\frac{\left(2I\right)!}{2^{I}I!}{\sin }^i\left(\theta \right) \\ ={\left(-1\right)}^I\sqrt{\frac{\left(2I\right)!\left(2I+1\right)}{4\mathrm{\pi }}}{e}^{iI\varphi }\frac{1}{2^{I}I!}{\sin }^I\left(\theta \right) \\ =\frac{1}{I!}\sqrt{\frac{\left(2I+1\right)!}{4\mathrm{\pi }}}{\left(-\frac{1}{2}{e}^{i\varphi }\sin \theta \right)}^I \end{gathered}$$

Again, forI=3,

$y_3^3=-{\left(\frac{35}{64\mathrm{\pi }}\right)}^{\frac{1}{2}}{\sin }^3\left(\theta \right)e^{3i\varphi }$

Also, for $n=4,l=3$and $m=3$, use $L_0^7\left(x\right)=7!=5040.$Substitute $L_0^7\left(x\right)$and$Y_3^3$into the overall formula (1),

localid="1658397501310" $$\begin{gathered} {\psi }_{433}=\sqrt{{\left(\frac{1}{2a}\right)}^3\frac{1}{6\times {5040}^3}}{e}^{-r/4a}{\left(\frac{r}{2a}\right)}^3\left(5040\right)\left(-{\left(\frac{35}{64\mathrm{\pi }}\right)}^{\frac{1}{2}}{\sin }^3\left(\theta \right)e^{3i\varphi }\right) \\ =-\frac{1}{6144\sqrt{{\mathrm{\pi a}}^9}}{r}^3{e}^{-r/4a}{\sin }^3\theta e^{3i\varphi }{\psi }_{433} \end{gathered}$$

Therefore, the wave function for hydrogen in the given states as a function of the spherical coordinates$r,\theta$and$=-\frac{1}{6144\sqrt{{\mathrm{\pi a}}^9}}{r}^3{e}^{-r/4a}{\sin }^3\theta e^{3i\varphi }{\psi }_{433}$is .

Evaluate the expectation value of$r$, that is $\left(r\right)$, as:

localid="1658403974750" $$\begin{gathered} \left(r\right)=\int r{\left|\psi \right|}^2{d}^{3}r \\ =\frac{1}{{\left(6144\right)}^2{\mathrm{\pi a}}^9}\int r\left(r^6{e}^{-r/2a}{\sin }^6\left(\theta \right)\right)r^2\sin \left(\theta \right)drd\theta d\varphi \\ =\frac{1}{{\left(6144\right)}^2{\mathrm{\pi a}}^9}{\int }_0^{\infty }{r}^9{e}^{-r/2a}{\int }_0^{\mathrm{\pi }}\left(\theta \right)d\theta {\int }_0^{2\mathrm{\pi }}d\varphi \\ =\frac{1}{{\left(6144\right)}^2{\mathrm{\pi a}}^9}\left[9!{\left(2a\right)}^{10}\right]\left(2\frac{1.4.6}{3.5.7}\right)\left(2\mathrm{\pi }\right) \end{gathered}$$

Further evaluate and get,

$$\begin{gathered} =18a\left(r\right) \\ =18a \end{gathered}$$

Thus, the expectation value of r in the state is 18a.

Assume thatbe an eigen function of the operator with eigen value of $l(l+1){\hslash }^2,\mathrm{for}I=3$

Thus:

$$\begin{gathered} L^2=I\left(I+1\right){ħ}^2 \\ =12{ħ}^2 \end{gathered}$$

Supposerole="math" localid="1658398355830" ${\psi }_{433}$be an eigen function of the operator $L_z$with eigen value of $m{\hslash }_{tfor}m=3,$:

$L_z=3\hslash$

Hence:

role="math" localid="1658398913271" $$\begin{gathered} L_x^2+L_y^2=L^2-L_z^2 \\ =12{ħ}^2-9{ħ}^2 \\ =3{ħ}^2 \\ =3{ħ}^2 \end{gathered}$$

Thus, the required value is$3{ħ}^2$