Question

The electron in a hydrogen atom occupies the combined spin and position state$R_{21}\left(\sqrt{1/3}{Y}_1^0{\chi }_{+}+\sqrt{2/3}{Y}_1^1{\chi }_{-}\right)$

(a) If you measured the orbital angular momentum squared $\left(L^2\right)$, what values might you get, and what is the probability of each?

(b) Same for the component of orbital angular momentum $\left(L_z\right)$.

(c) Same for the spin angular momentum squared$\left(S^2\right)$ .

(d) Same for the component of spin angular momentum $\left(S_z\right)$.

Let $\mathbf{J}\equiv \mathbf{L}+\mathbf{S}$be the total angular momentum.

(e) If you measureddata-custom-editor="chemistry" $J^2$ , what values might you get, and what is the probability of each?

(f) Same for$J_z$ .

(g) If you measured the position of the particle, what is the probability density for finding it at$r$ , $\theta$,$\varphi$ ?

(h) If you measured both the component of the spin and the distance from the origin (note that these are compatible observables), what is the probability density for finding the particle with spin up and at radius ?

Short answer

1. The probability is 1 .
2. The eigen value and probability of$L_z$are$\hslash$and 1 .
3. The probability is 1 for both$s=\frac{1}{2}$.
4. The probability for$s_z=\frac{1}{2}\hslash$is$\frac{1}{3}$and the probability for$s_z=-\frac{1}{2}h$is$\frac{2}{3}$ .
5. The probability for$J^2=\frac{15}{4}{\hslash }^2$is$\frac{8}{9}$and the probability for$J^2=\frac{3}{4}{\hslash }^2$is$\frac{1}{9}$.
6. The probability is 1 .
7. the probability density is$\frac{1}{96\pi a^5}{r}^2{e}^{\frac{-r}{a}}$ .
8. the probability density for finding the particle with spin-up is $\frac{1}{72a^5}{r}^2{e}^{\frac{r}{a}}$.

Step-by-step solution

Definition of normalization factor.

Normalisation is the process of scaling wave functions to the point where all probabilities equal one. The probabilistic description of quantum mechanics only makes sense when all of the probabilities add up to one.

The probability.

(a)

The following is how the given position and spin state are expressed:

$R_{21}\left(\sqrt{\frac{1}{3}}{Y}_1^0{\chi }_{+}+\sqrt{\frac{2}{3}}{Y}_1^1{\chi }_{-}\right)$

In both terms$I=1$, compare the term of the form$Y_l^m$.

The square of the angular momentum is,

$\begin{array}{rcl}{L}^2& =& l(l+1){\hslash }^2\\ & =& 1(1+1){\hslash }^2\\ & =& 2{\hslash }^2\\ & & \end{array}$

For both the states $I=1$and hence the probability is 1 .

The eigen value and probability of  Lz.

(b)

The formula for Z component of orbital angular momentum$L_z$is expressed as follows:

$L_z=m_l\hslash$

The magnetic quantum number is$m_l$in this case.

The wave function has two spin states in the given issue, hence these states with twovalues are 0 and 1 .

Substitute 0 for$m_l$in the above equation.

$$\begin{gathered} L_z=\left(0\right)\hslash \\ =0 \end{gathered}$$

Substitute 1 for$m_l$in the equation$L_z=m_l\hslash$.

$$\begin{gathered} L_z=\left(1\right)\hslash \\ =\hslash \end{gathered}$$

The probability of$L_z=0$is,

$P_1=\frac{C_1^2}{\sqrt{C_1^2+C_2^2}}$

Here,$C_1$and$C_2$are the coefficients.

Substitute $\sqrt{\frac{1}{3}}$for $C_1$and $\sqrt{\frac{2}{3}}$for $C_2$in the above equation.

$$\begin{gathered} P_1=\frac{{\left(\sqrt{\frac{1}{3}}\right)}^2}{\sqrt{{\left(\sqrt{\frac{1}{3}}\right)}^2+{\left(\sqrt{\frac{2}{3}}\right)}^2}} \\ =\frac{1}{3} \end{gathered}$$

The probability of$L_z=\hslash$is,

$P_2=\frac{C_2^2}{\sqrt{C_1^2+C_2^2}}$

Substitute $\sqrt{\frac{1}{3}}$for $C_1$and$\sqrt{\frac{2}{3}}$ for $C_2$in the above equation.

$$\begin{gathered} P_2=\frac{\left(\sqrt{\frac{2}{3}}\right)}{\sqrt{{\left(\sqrt{\frac{1}{3}}\right)}^2+{\left(\sqrt{\frac{2}{3}}\right)}^2}} \\ =\frac{2}{3} \end{gathered}$$

The probability of the wave function forcomponent of orbital angular momentum is,

$P=P_1+P_2$

Substitute $\sqrt{\frac{1}{3}}$for $P_1$and $\sqrt{\frac{2}{3}}$for$P_2$ in the above equation.

$$\begin{gathered} P=\frac{1}{3}+\frac{2}{3} \\ =1 \end{gathered}$$

Therefore, the eigen value and probability of $L_z$are$\hslash$ and 1 .

The probability for the spin angular momentum squared S2 .

(c)

For a spin-half particle,$s=\frac{1}{2}$

$s^2$is the square of the spin angular momentum.

$s^2=s(s+1){\hslash }^2$

Substitute$\frac{1}{2}$for S.

$$\begin{gathered} s^2=\frac{1}{2}\left(\frac{1}{2}+1\right){\hslash }^2 \\ =\frac{3}{4}{\hslash }^2 \end{gathered}$$

The probability is 1 for both $s=\frac{1}{2}$.

The probability for the  component of spin angular momentum  Sz.

(d)

The Z -component of spin angular momentum is,

$S_z=\pm \frac{1}{2}\hslash$

The probability for$s_z=\frac{1}{2}\hslash$is,

$P_1=\frac{C_1^2}{\sqrt{C_1^2+C_2^2}}$

Here,$C_1$and$C_2$are the coefficients.

Substitute $\sqrt{\frac{1}{3}}$for $C_1$ and $\sqrt{\frac{2}{3}}$for $C_2$in the above equation.

$$\begin{gathered} P_1=\frac{{\left(\sqrt{\frac{1}{3}}\right)}^2}{\sqrt{{\left(\sqrt{\frac{1}{3}}\right)}^2+{\left(\sqrt{\frac{2}{3}}\right)}^2}} \\ =\frac{1}{3} \end{gathered}$$

The probability for$s_z=\frac{1}{2}\hslash$ is$\frac{1}{3}$.

The probability for$s_z=-\frac{1}{2}\hslash$is,

$P_2=\frac{C_2^2}{\sqrt{C_1^2+C_2^2}}$

Substitute$\sqrt{\frac{1}{3}}$for$C_1$and$\sqrt{\frac{2}{3}}$for$C_2$in the above equation.

$$\begin{gathered} P_2=\frac{\left(\sqrt{\frac{2}{3}}\right)}{\sqrt{{\left(\sqrt{\frac{1}{3}}\right)}^2+{\left(\sqrt{\frac{2}{3}}\right)}^2}} \\ =\frac{2}{3} \end{gathered}$$

The probability for$s_z=-\frac{1}{2}\hslash$ is$\frac{2}{3}$ .

The probability for J2 .

(e)

The possible values of$j$is,

$$\begin{gathered} j=L+S \\ =1+\frac{1}{2},1-\frac{1}{2} \\ =\frac{3}{2},\frac{1}{2} \end{gathered}$$

The square of the total angular momentum is,

$J^2=j(j+1){\hslash }^2$

Substitute$\frac{3}{2}$for J .

$$\begin{gathered} J^2=\frac{3}{2}\left(\frac{3}{2}+1\right){\hslash }^2 \\ =\frac{15{\hslash }^2}{4} \end{gathered}$$

Substitute $\frac{1}{2}$for J .

$$\begin{gathered} J^2=\frac{1}{2}\left(\frac{1}{2}+1\right){\hslash }^2 \\ =\frac{3}{4}{\hslash }^2 \end{gathered}$$

From the $1\times \frac{1}{2}$Clebsch-Gordan table, we have

$\begin{array}{rcl}\frac{1}{\sqrt{3}}\left|\frac{1}{2}\frac{1}{2}\right.>|10⟩+\sqrt{\frac{2}{3}}\left|\frac{1}{2}-\frac{1}{2}\right.>|11⟩& =& \frac{1}{\sqrt{3}}\left[\sqrt{\frac{2}{3}}\left|\frac{3}{2}\frac{1}{2}\right.>-\frac{1}{\sqrt{3}}\left|\frac{1}{2}\frac{1}{2}\right.>\right]+\sqrt{\frac{2}{3}}\left[\frac{1}{\sqrt{3}}\left|\frac{3}{2}\frac{1}{2}\right.>+\sqrt{\frac{2}{3}}\left|\frac{1}{2}\frac{1}{2}\right.>\right]\\ & =& \left[\frac{2\sqrt{2}}{3}\left|\frac{3}{2}\frac{1}{2}\right.>+\frac{1}{3}\left|\frac{1}{2}\frac{1}{2}\right.>\right]\\ & & \end{array}$

The probability for$J^2=\frac{15}{4}{\hslash }^2$is,

$$\begin{gathered} P=\frac{{\left(\frac{2\sqrt{2}}{3}\right)}^2}{\sqrt{{\left(\frac{2\sqrt{2}}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2}} \\ =\frac{8}{9} \end{gathered}$$

The probability for $J^2=\frac{3}{4}{\hslash }^2$is,

$$\begin{gathered} P=\frac{{\left(\frac{1}{3}\right)}^2}{\sqrt{{\left(\frac{2\sqrt{2}}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2}} \\ =\frac{1}{9} \end{gathered}$$

The probability for  Jz

(f)

From the$1\times \frac{1}{2}$ Clebsch-Gordan table, we have

$\begin{array}{rcl}\frac{1}{\sqrt{3}}\left|\frac{1}{2}\frac{1}{2}\right.>|10⟩+\sqrt{\frac{2}{3}}\left|\frac{1}{2}-\frac{1}{2}\right.>|11⟩& =& \frac{1}{\sqrt{3}}\left[\sqrt{\frac{2}{3}}\left|\frac{3}{2}\frac{1}{2}\right.>-\frac{1}{\sqrt{3}}\left|\frac{1}{2}\frac{1}{2}\right.>\right]+\sqrt{\frac{2}{3}}\left[\frac{1}{\sqrt{3}}\left|\frac{3}{2}\frac{1}{2}\right.>+\sqrt{\frac{2}{3}}\left|\frac{1}{2}\frac{1}{2}\right.>\right]\\ & =& \left[\frac{2\sqrt{2}}{3}\left|\frac{3}{2}\frac{1}{2}\right.>+\frac{1}{3}\left|\frac{1}{2}\frac{1}{2}\right.>\right]\\ & & \end{array}$

From the above expression,

$m_j=\frac{1}{2}\hslash$

For both the states in the expression $\left[\frac{2\sqrt{2}}{3}\left|\frac{3}{2}\frac{1}{2}\right.>+\frac{1}{3}\left|\frac{1}{2}\frac{1}{2}\right.>\right]$ the value of $m_j=\frac{1}{2}\hslash$and hence the probability is 1 .

The probability density.

(g)

The probability density is given by

$|\psi {|}^2={\left|R_{21}\right|}^2\{\frac{1}{3}{\left|Y_1^0\right|}^2\left({\chi }_{+}^{†}{\chi }_{+}\right)+\frac{\sqrt{2}}{3}\left[Y_1^{0*}{Y}_1^1\left({\chi }_{+}^{†}{\chi }_{-}\right)+Y_1^{1^{*}}{Y}_0^1\left({\chi }_{-}^{†}{\chi }_{+}\right)\right]+\frac{2}{3}{\left|Y_1^1\right|}^2\left({\chi }_{-}^{†}{\chi }_{-}\right)\}$

Using the orthonormality of the spin states,

$|\psi {|}^2=\frac{1}{3}{\left|R_{21}\right|}^2\left[{\left|Y_1^0\right|}^2+2{\left|Y_1^1\right|}^2\right]$

Using the values,

$$\begin{gathered} Y_1^0=\sqrt{\frac{3}{4\pi }}\cos \theta \\ {Y}_1^{\text{'}}=-\sqrt{\frac{3}{8\pi }}\sin \theta e^{i\varphi } \end{gathered}$$

And $R_{21}=\frac{1}{\sqrt{24}}{a}^{-3/2}\frac{r}{a}\exp (-r/2a)$

$\begin{array}{rcl}|\psi {|}^2& =& \frac{1}{3}\left[\frac{1}{24}·\frac{1}{a^3}\frac{r^2}{a^2}{e}^{\frac{-r}{a}}\left(\frac{3}{4\pi }{\cos }^2\theta +2\left(\frac{3}{8\pi }\right){\sin }^2\theta \right)\right]\\ & =& \frac{1}{3.24·a^5}{r}^2{e}^{\frac{-r}{a}}\left(\frac{3}{4\pi }\left({\cos }^2\theta +{\sin }^2\theta \right)\right)\\ & =& \frac{1}{96\pi a^5}{r}^2{e}^{\frac{-r}{a}}\\ & & \end{array}$

Therefore, the probability density is $\frac{1}{96\pi a^5}{r}^2{e}^{\frac{-r}{a}}$.

The probability density for finding the particle with spin-up.

(h)

The probability density for finding a particle with a radius ofand spin-up is

$|\psi {|}^2=\frac{1}{3}{\left|R_{21}\right|}^2\int {\left|Y_1^0\right|}^2\sin \theta d\theta d\varphi$

The integral is 1 using the orthogonal condition of spherical harmonics.

$\begin{array}{rcl}|\psi {|}^2& =& \frac{1}{3}{\left|R_{21}\right|}^2\\ & =& \frac{1}{3}·\frac{1}{24a^3}{\left(\frac{r}{a}\right)}^2{e}^{\frac{-r}{a}}\\ & =& \frac{1}{72a^5}{r}^2{e}^{\frac{r}{a}}\\ & & \end{array}$

Therefore, the probability density for finding the particle with spin-up is $\frac{1}{72a^5}{r}^2{e}^{\frac{r}{a}}$.