### Quantum Mechanics - Orbital Angular Momentum

Which is the following is the orbital angular momentum eigenfuction $Y_l^m&space;(\theta,\phi&space;)$ in a state for which the operators $L^2$ and $L_z$ have eigenvalues $6\hbar^2$ and $-\hbar$ respectively?

$\\A.&space;\hspace{5&space;mm}&space;Y_2^1&space;(\theta,\phi)&space;\\\\B.&space;\hspace{5&space;mm}&space;Y_2^{-1}&space;(\theta,\phi)&space;\\\\C.&space;\hspace{5&space;mm}&space;\frac{1}{\sqrt{2}}[Y_2^1&space;(\theta,\phi)+Y_2^{-1}&space;(\theta,\phi)]&space;\\\\D.&space;\hspace{5&space;mm}&space;Y_2^3&space;(\theta,\phi)&space;\\\\E.&space;\hspace{5&space;mm}&space;Y_3^{-1}&space;(\theta,\phi)$
(GR0177 #81)
Solution:

Orbital angular momentum: $L=\sqrt{l(l+1)}&space;\hbar$

\begin{aligned}&space;L^2=l(l+1)&space;\hbar^2&=6\hbar^2&space;\\l(l+1)=l^2+l&=6&space;\\l^2+l-6&=0&space;\\(l+3)(l-2)&=0&space;\\l&=-3;2&space;\end{aligned}

Allowed values of  l:

$l=\left\{\begin{matrix}&space;0&space;1&space;2&space;3&space;4&space;...&space;\\&space;s&space;p&space;d&space;f&space;g&space;...&space;\end{matrix}\right.$

l cannot be negative. So in this case, $l=2$

The z-component of L:  $L_z=m_l&space;\hbar=-\hbar\rightarrow&space;m=-1$

Thus,  $Y_l^m&space;(\theta,\phi)=Y_2^{-1}&space;(\theta,\phi&space;)$

### Quantum Mechanics - Spin

Let  $|\alpha\rangle$ represent the state of an electron with spin up and  $|\beta\rangle$ the state of an electron with spin down. Valid spin eigenfunctions for a triplet state $(^3S)$ of a two-electron atom include which of the following?

$\\I.&space;&space;\hspace{5&space;mm}&space;|\alpha\rangle_1|\alpha\rangle_2&space;\\\\II.&space;&space;\hspace{4&space;mm}&space;\frac{1}{\sqrt{2}}&space;\left(&space;|\alpha\rangle_1|\beta\rangle_2-|\alpha\rangle_2|\beta\rangle_1\right&space;)&space;\\\\III.&space;&space;\hspace{3&space;mm}&space;\frac{1}{\sqrt{2}}&space;\left(&space;|\alpha\rangle_1|\beta\rangle_2+|\alpha\rangle_2|\beta\rangle_1\right&space;)$

A. I only
B. II only
C. III only
D. I and III
E. II and III
GR0177 #82)
Solution:

Singlet state is anti-symmetric: $\psi(1,2)=-\psi(2,1)$
Triplet state is symmetric: $\psi&space;(1,2)=\psi(2,1)$

$I.&space;&space;\hspace{5&space;mm}&space;|\alpha\rangle_1|\alpha\rangle_2$  is symmetric, because

$\\&space;\psi(1,2)=|\alpha\rangle_1|\alpha\rangle_2&space;\\&space;\psi(2,1)=|\alpha\rangle_2|\alpha\rangle_1=|\alpha\rangle_1|\alpha\rangle_2=\psi(1,2)$

$II.&space;&space;\hspace{4&space;mm}&space;\frac{1}{\sqrt{2}}&space;\left(&space;|\alpha\rangle_1|\beta\rangle_2-|\alpha\rangle_2|\beta\rangle_1\right&space;)$  is antisymmetric, because
\begin{aligned}&space;\\&space;\psi(1,2)&=\frac{1}{\sqrt{2}}&space;\left(&space;|\alpha\rangle_1|\beta\rangle_2-|\alpha\rangle_2|\beta\rangle_1\right&space;)&space;\\&space;\psi(2,1)&=\frac{1}{\sqrt{2}}&space;\left(&space;|\alpha\rangle_2|\beta\rangle_1-|\alpha\rangle_1|\beta\rangle_2\right&space;)\\&=-\frac{1}{\sqrt{2}}&space;\left(&space;|\alpha\rangle_1|\beta\rangle_2-|\alpha\rangle_2|\beta\rangle_1\right&space;)&space;=-\psi(1,2)&space;\end{aligned}

$III.&space;&space;\hspace{3&space;mm}&space;\frac{1}{\sqrt{2}}&space;\left(&space;|\alpha\rangle_1|\beta\rangle_2+|\alpha\rangle_2|\beta\rangle_1\right&space;)$  is symmetric, because
\begin{aligned}&space;\\&space;\psi(1,2)&=\frac{1}{\sqrt{2}}&space;\left(&space;|\alpha\rangle_1|\beta\rangle_2+|\alpha\rangle_2|\beta\rangle_1\right&space;)&space;\\&space;\psi(2,1)&=\frac{1}{\sqrt{2}}&space;\left(&space;|\alpha\rangle_2|\beta\rangle_1+|\alpha\rangle_1|\beta\rangle_2\right&space;)\\&=\frac{1}{\sqrt{2}}&space;\left(&space;|\alpha\rangle_1|\beta\rangle_2+|\alpha\rangle_2|\beta\rangle_1\right&space;)&space;=\psi(1,2)&space;\end{aligned}

Thus, II is singlet state, I and III are triplet state.

### Nuclear & Particle Physics - Selection Rules

An energy-level diagram of the n = 1 and n = 2 levels of atomic hydrogen (including the effect of spin-orbit coupling and relativity) is shown in the figure. Three transitions are labeled A, B, and C. Which of the transitions will be possible electric-dipole transition?

A. B only
B. C only
C. A & C only
D. B & C only
E. A, B, & C
(GR0177 #84)
Solution:

Selection Rules:

\begin{aligned}&space;\\\Delta&space;l&=\pm&space;1&space;\\\Delta&space;s&=0&space;\\\Delta&space;m_l&=0,\pm&space;1&space;\\\Delta&space;j&=0,\pm&space;1&space;\hspace{5&space;mm}&space;\text&space;{but&space;}&space;\hspace{5&space;mm}&space;j=0&space;\not\rightarrow&space;j=0&space;\end{aligned}

Transition A:
$l=0&space;\hspace{5&space;mm}\text&space;{to}\hspace{5&space;mm}&space;l=0&space;\hspace{5&space;mm}\rightarrow&space;\hspace{5&space;mm}\Delta&space;l=0$.  NOT ALLOWED

Transition B:
$l=1&space;\hspace{5&space;mm}\text&space;{to}\hspace{5&space;mm}&space;l=0&space;\hspace{5&space;mm}\rightarrow&space;\hspace{5&space;mm}\Delta&space;l=-1$.  ALLOWED $j=3/2&space;\hspace{5&space;mm}\text&space;{to}\hspace{5&space;mm}&space;j=1/2&space;\hspace{5&space;mm}\rightarrow&space;\hspace{5&space;mm}\Delta&space;j=-1$. ALLOWED

Transition C:
$j=1/2&space;\hspace{5&space;mm}\text&space;{to}\hspace{5&space;mm}&space;j=1/2&space;\hspace{5&space;mm}\rightarrow&space;\hspace{5&space;mm}\Delta&space;j=0$.  ALLOWED

### Electromagnetism - Kirchoff's Law

One end of Nichrome wire of length 2L and cross-sectional area A is attached to an end of another Nichrome wire of length L and cross-sectional area 2A. If the free end of the longer wire is at an electric potential of 8.0 volts, and the free end of the shorter wire is at an electric potential of 1.0 volt, the potential at the junction of the two wires is most nearly equal to

A. 2.4 V
B. 3.3 V
C. 4.5 V
D. 5.7 V
E. 6.6 V
(GR0177 #85)
Solution:

Ohm’s Law:   $V=IR$

Kirchoff's Law #1:   $I_{\text{in}}=I_{\text{out}}=I$

$\\I_{\text{in}}=(V_1-V)/R_1&space;\\I_{\text{out}}=(V-V_2)/R_2$

Resistance:  $R=\frac{\rho&space;L}{A}$

Both are Nichrome wires, $\rho_1=\rho_2=\rho$

$\\R_1=\rho&space;(2L/A)=2R&space;\\R_2=\rho&space;(L/2A)=0.5R$

$I_{\text{in}}&=I_{\text{out}}&space;\\\\\frac{8-V}{2}&=\frac{V-1}{0.5}&space;\rightarrow&space;V=\frac{6}{2.5}=2.4&space;&space;V$