Quantum Mechanics in Three Dimension: Angular Momentum and Probabilities

Hananish Joy Odarve and Edmar Pantohan

In this article, we want to show that given the state occupied by an electron in a hydrogen atom, one can measure the operators $L^2$ , $S^2$ , $L_z$ , $S_z$ , $J^2$ and $J_z$ and its corresponding probabilities.

Lets look at an example to get a clearer picture of these topic. The electron in a hydrogen atom occupies the combined spin and position state given by

$\Psi = R_{21} \;\; \left[\sqrt{\frac{1}{3}} \;\; {Y^0_1} \;\; \chi_+ \;\;\;\; + \;\;\;\; \sqrt{\frac{2}{3}} \;\; {Y^1_1} \;\; \chi_- \right]$

FOR $L^2$ :

First, we measure the orbital angular momentum squared given by

$L^2 = l(l+1) {\hbar}^2$

From the given state, we consider ${Y^0_1}$ where $l=1$ and $m=o$ , so

$L^2 =1(1 + 1)\hbar}^2 \\= 2{\hbar}^2$

and

$P = \frac{1}{3}$ .

Next, we consider ${Y^1_1}$ where $l=1$ and $m=1$ , so

$L^2 =1(1 + 1)\hbar}^2 \\= 2{\hbar}^2$

and

$P = \frac{2}{3}$ .

Therefore, for the state given, the value of $L^2 = 2{\hbar}^2$ and the total probability is:

$P =\frac{1}{3} + \frac{2}{3} = 1$ .

FOR $L_z$ :

Now, we measure the $z$ component of the orbital angular momentum which is given by

$L_z = m\hbar$ .

So, for ${Y^0_1}$ where $m=o$

$L_z = 0\hbar =0$ with $P = \frac{1}{3}$

and for ${Y^1_1}$ where $m=1$

$L_z = 1\hbar =\hbar$ with $P = \frac{2}{3}$ .

FOR $S^2$ :

Next, we measure the spin angular momentum squared given by

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

Since $s=\frac{1}{2}$ , there is only one possible value and this is

$S^2 = \frac{1}{2}\left(\frac{1}{2}+1\right) {\hbar}^2 \\ = \frac{1}{2}\left(\frac{3}{2}\right){\hbar}^2 \\ = \frac{3}{4}{\hbar}^2$

and the probabilty then is $P = 1$ .

FOR $S_z$ :

Here, we measure the $z$ component of the spin angular momentum which is given by

$S_z = m_s\hbar$ .

For $s=\frac{1}{2}$ , the value of $m_s=\pm \frac{1}{2}$ so for $m_s=+\frac{1}{2}$ :

$S_z = +\frac{1}{2}\hbar$ with $P = \frac{1}{3}$

and for $m_s=-\frac{1}{2}$ :

$S_z = -\frac{1}{2}\hbar$ with $P = \frac{2}{3}$ .

FOR $J^2$ :

Now, to calculate for the total angular momentum, we rewrite the given state of the system as:

$\sqrt{\frac{1}{3}}|\frac{1}{2}\;\;\frac{1}{2}\rangle \left|10\right \rangle + \sqrt{\frac{2}{3}}|\frac{1}{2}\;\;-\frac{1}{2}\rangle \left|11\right \rangle$

This is for particles of spin 1 and spin $\frac{1}{2}$ so we use Clebsch-Gordan table for $1\times\frac{1}{2}$ and so we have:

= $\sqrt{\frac{1}{3}}\;\;\left[\sqrt{\frac{2}{3}}|\frac{3}{2}\;\;\frac{1}{2}\rangle - \sqrt{\frac{1}{3}}|\frac{1}{2}\;\;\frac{1}{2}\rangle \right ]$ + $\sqrt{\frac{2}{3}}\;\;\left[\sqrt{\frac{1}{3}}|\frac{3}{2}\;\;\frac{1}{2}\rangle + \sqrt{\frac{2}{3}}|\frac{1}{2}\;\;\frac{1}{2}\rangle \right ]$

= $\left(2\frac{\sqrt{2}}{3}\right) |\frac{3}{2}\;\;\frac{1}{2}\rangle$ + $\left(\frac{1}{3}\right) |\frac{1}{2}\;\;\frac{1}{2}\rangle$

from here, we can say that $j=\frac{3}{2}$ or $\frac{1}{2}$ and so for $j=\frac{3}{2}$ :

$J^2 = \frac{3}{2}\left(\frac{3}{2}+1\right) {\hbar}^2 =\frac{3}{2}\left(\frac{5}{2}\right){\hbar}^2 =\frac{15}{4}{\hbar}^2$

with

$P = \frac{1}{3}$ .

And for $j=\frac{1}{2}$ :

$J^2 = \frac{1}{2}\left(\frac{1}{2}+1\right) {\hbar}^2 =\frac{1}{2}\left(\frac{3}{2}\right){\hbar}^2 =\frac{3}{4}{\hbar}^2$

with

$P = \frac{2}{3}$ .

FOR $J_z$ :

Now, we measure the $z$ component of the total angular momentum which is given by

$J_z = m_j\hbar$

and

$m_j= m+m_s$ .

for ${Y^0_1} \;\; \chi_+$ , $m=0$ and $m_s=\frac{1}{2}$ , $m_j=\frac{1}{2}$ and so

$J_z = \frac{1}{2}\hbar$ .

for ${Y^1_1} \;\; \chi_-$ , $m=1$ and $m_s=-\frac{1}{2}$ , $m_j=\frac{1}{2}$ and so it is still

$J_z = \frac{1}{2}\hbar$ .

Therefore, the probability of getting the value of $J_z = \frac{1}{2}\hbar$ is

$P =1$ .

FOR the Probabilty density of finding the particle at $\left(r, \theta, \phi\right )$ :

$\left|\Psi\right|^2 = \Psi*\Psi = {R_{21}}^* \left[\sqrt{\frac{1}{3}}{Y^0_1}^* \chi_+ + \sqrt{\frac{2}{3}}{Y^1_1}^* \chi_- \right]$ $R_{21} \left[\sqrt{\frac{1}{3}}{Y^0_1} \chi_+ + \sqrt{\frac{2}{3}}{Y^1_1} \chi_- \right]$

$= \left|R_{21}\right|^2\; \frac{1}{3}\left|Y^0_1\right|^2\left(\chi^\dagger_+\chi_+\right)$ + $\left|R_{21}\;\right|^2\frac{\sqrt 2}{3} \left[{Y^0_1}^*{Y^1_1}\left(\chi^\dagger_+\chi_-\right) + {Y^1_1}^*{Y^0_1}\left(\chi^\dagger_-\chi_+\right) \right]$ + $\left|R_{21}\;\right|^2 \frac{2}{3}\left|Y^1_1\right|^2\left(\chi^\dagger_-\chi_-\right)$

$= \frac{1}{3}\left|R_{21}\right|^2 \left( \left|Y^0_1\right|^2 + 2\left|Y^1_1\right|^2\right)$ .

\rangle – \sqrt{\frac{1}{3}}|\frac{1}{2}\;\;\frac{1}{2}\rangle \right]

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One Response to “Quantum Mechanics in Three Dimension: Angular Momentum and Probabilities”

Angie Says:
February 10th, 2010 at 8:51 pm
Great layout you have here! Clear and straightforward to read. But a line or two about what’s R_21 or the spherical harmonics for the less advanced reader would help…