Quantum Mechanics in Three Dimension: Angular Momentum and Probabilities | Quantum Science Philippines

## 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 [eq]L^2[/eq] , [eq]S^2[/eq] , [eq]L_z[/eq], [eq]S_z[/eq], [eq]J^2[/eq] and [eq]J_z[/eq] 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

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

FOR [eq]L^2[/eq]:

First, we measure the orbital angular momentum squared given by

[eq]L^2 = l(l+1) {\hbar}^2 [/eq]

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

[eq]L^2 =1(1 + 1)\hbar}^2 \\= 2{\hbar}^2[/eq]

and

[eq]P = \frac{1}{3}[/eq].

Next, we consider [eq]{Y^1_1}[/eq] where [eq]l=1[/eq] and [eq]m=1[/eq], so

[eq]L^2 =1(1 + 1)\hbar}^2 \\= 2{\hbar}^2[/eq]

and

[eq]P = \frac{2}{3}[/eq].

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

[eq]P =\frac{1}{3} + \frac{2}{3} = 1[/eq].

FOR [eq]L_z[/eq]:

Now, we measure the [eq]z[/eq] component of the orbital angular momentum which is given by

[eq]L_z = m\hbar[/eq].

So, for [eq]{Y^0_1}[/eq] where [eq]m=o[/eq]

[eq]L_z = 0\hbar =0[/eq] with [eq]P = \frac{1}{3}[/eq]

and for [eq]{Y^1_1}[/eq] where [eq]m=1[/eq]

[eq]L_z = 1\hbar =\hbar[/eq] with [eq]P = \frac{2}{3}[/eq].

FOR [eq]S^2[/eq]:

Next, we measure the spin angular momentum squared given by

[eq]S^2 = s(s+1) {\hbar}^2 [/eq]

Since [eq]s=\frac{1}{2}[/eq], there is only one possible value and this is

[eq]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[/eq]

and the probabilty then is [eq]P = 1[/eq].

FOR [eq]S_z[/eq]:

Here, we measure the [eq]z[/eq] component of the spin angular momentum which is given by

[eq]S_z = m_s\hbar [/eq].

For [eq]s=\frac{1}{2}[/eq], the value of [eq]m_s=\pm \frac{1}{2}[/eq] so for [eq]m_s=+\frac{1}{2}[/eq]:

[eq]S_z = +\frac{1}{2}\hbar [/eq]  with  [eq]P = \frac{1}{3}[/eq]

and for [eq]m_s=-\frac{1}{2}[/eq]:

[eq]S_z = -\frac{1}{2}\hbar [/eq]  with  [eq]P = \frac{2}{3}[/eq].

FOR [eq]J^2[/eq]:

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

[eq]\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[/eq]

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

= [eq]\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 ][/eq] + [eq]\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 ][/eq]

= [eq]\left(2\frac{\sqrt{2}}{3}\right) |\frac{3}{2}\;\;\frac{1}{2}\rangle [/eq] + [eq]\left(\frac{1}{3}\right) |\frac{1}{2}\;\;\frac{1}{2}\rangle [/eq]

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

[eq]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[/eq]

with

[eq]P = \frac{1}{3}[/eq].

And for [eq]j=\frac{1}{2}[/eq]:

[eq]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[/eq]

with

[eq]P = \frac{2}{3}[/eq].

FOR [eq]J_z[/eq]:

Now, we measure the [eq]z[/eq] component of the total angular momentum which is given by

[eq]J_z = m_j\hbar [/eq]

and

[eq]m_j= m+m_s[/eq].

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

[eq]J_z = \frac{1}{2}\hbar [/eq].

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

[eq]J_z = \frac{1}{2}\hbar [/eq].

Therefore, the probability of getting the value of [eq]J_z = \frac{1}{2}\hbar [/eq] is

[eq]P =1[/eq].

FOR the Probabilty density of finding the particle at [eq]\left(r, \theta, \phi\right )[/eq]:

[eq]\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] [/eq][eq]R_{21} \left[\sqrt{\frac{1}{3}}{Y^0_1} \chi_+ + \sqrt{\frac{2}{3}}{Y^1_1} \chi_- \right] [/eq]

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

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

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

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

1. Angie Says:

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…