Question

(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?

Short answer

(a) The wave function for hydrogen in the given states is

${\psi }_{433}=-\frac{1}{6144\sqrt{\pi a^9}}{r}^3{e}^{-r/4a}{\sin }^3\left(\theta \right)e^{3i\varphi }.$

(b) The expectation value in the state is 18a.

(c ) The value is$3{ħ}^2$ .

Step-by-step solution

Define the wave function

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,$\mathit{\psi }$.

Step 2: (a) Construct the wave function for hydrogen

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 .

(b) Determine the expectation value of r

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.

(c) Find the probability.

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$3{ħ}^2$