91影视

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.

(a)

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

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

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

The square of the angular momentum is,

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

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

(b)

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

$L_z=m_l\mathrm{鈩�}$

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.

$$\begin{gathered} L_z=\left(0\right)\mathrm{鈩�} \\ =0 \end{gathered}$$

Substitute 1 for$m_l$in the equation$L_z=m_l\mathrm{鈩�}$.

$$\begin{gathered} L_z=\left(1\right)\mathrm{鈩�} \\ =\mathrm{鈩�} \end{gathered}$$

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$L_z=\mathrm{鈩�}$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$\mathrm{鈩�}$ and 1 .

(c)

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

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

$s^2=s(s+1){\mathrm{鈩�}}^2$

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

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

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

(d)

The Z -component of spin angular momentum is,

$S_z=卤\frac{1}{2}\mathrm{鈩�}$

The probability for$s_z=\frac{1}{2}\mathrm{鈩�}$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$s_z=\frac{1}{2}\mathrm{鈩�}$ is$\frac{1}{3}$.

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

(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,

$J^2=j(j+1){\mathrm{鈩�}}^2$

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

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

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

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

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

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

The probability for$J^2=\frac{15}{4}{\mathrm{鈩�}}^2$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 $J^2=\frac{3}{4}{\mathrm{鈩�}}^2$is,

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

(f)

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

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

From the above expression,

$m_j=\frac{1}{2}\mathrm{鈩�}$

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 $m_j=\frac{1}{2}\mathrm{鈩�}$and hence the probability is 1 .

(g)

The probability density is given by

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

Using the orthonormality of the spin states,

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

Using the values,

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

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

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

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

(h)

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

$|蠄{|}^2=\frac{1}{3}{\left|R_{21}\right|}^2鈭�{\left|Y_1^0\right|}^2\sin 胃d胃d蠒$

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

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

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