사용자:종잇조각/연습장: 두 판 사이의 차이
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==1.==
===12.5.3===
====(1)====
Since <math> J_x </math>, <math> J_y </math>, and <math> J_z</math> are Hermitian,
<math> J_x = J_x ^\dagger </math>, <math> J_y = J_y ^\dagger </math>, and <math> J_z = J_z ^\dagger </math>, all of their eigenvalues are real, and there is a set of eigenvectors that forms a complete basis of the Hilbert space.<br />
For any Hermitian operator, say <math>\hat A</math>, let its eigenbasis, <math> \left \{ | a_1 \rangle , | a_2 \rangle , \cdots \right \} </math>. Then for any <math> | \psi \rangle </math>,
<math> \langle \psi | \hat A | \psi \rangle = \sum_{i,j} \langle \psi | a_i \rangle \langle a_i | \hat A | a_j \rangle \langle a_j | \psi \rangle
= \sum_{i,j} a_j \langle a_i | \psi \rangle \langle a_i | a_j \rangle \langle a_j | \psi \rangle
= \sum_{i,j} a_j \delta_{ij} \langle a_i | \psi \rangle \langle a_j | \psi \rangle
= \sum_i a_i \left | \langle a_i | \psi \rangle \right | ^2 \in \mathbb{R}</math><br />
<math>
\begin{align}
\langle J_x \rangle & = \langle jm | J_x | jm \rangle = \langle jm | -{i \over \hbar} \left [ J_y , J_z \right ] | jm \rangle
= -{i \over \hbar} \left ( \langle jm | J_y J_z | jm \rangle - \langle jm | J_z J_y | jm \rangle \right ) \\
& = -{i \over \hbar} \left ( m \hbar \langle jm | J_y | jm \rangle - \langle jm | J_y^\dagger J_z^\dagger | jm \rangle ^\dagger \right )
= -{i \over \hbar} \left ( m \hbar \langle jm | J_y | jm \rangle - \langle jm | J_y J_z | jm \rangle ^\dagger \right ) \\
& = -{i \over \hbar} \left ( m \hbar \langle jm | J_y | jm \rangle - m \hbar \langle jm | J_y | jm \rangle ^\dagger \right )
= m \ \text{Im} \left \{ \langle jm | J_y | jm \rangle \right \} = 0 \left ( \because \langle jm | J_y | jm \rangle \in \mathbb{R} \right )
\end{align} </math><br />
<math>
\begin{align}
\langle J_y \rangle & = \langle jm | J_y | jm \rangle = \langle jm | -{i \over \hbar} \left [ J_z , J_x \right ] | jm \rangle
= - {i \over \hbar} \left ( \langle jm | J_z J_x | jm \rangle - \langle jm | J_x J_z | jm \rangle \right ) \\
& = {i \over \hbar} \left ( m \hbar \langle jm | J_x | jm \rangle - \langle jm | J_x^\dagger J_z^\dagger | jm \rangle ^\dagger \right )
= {i \over \hbar} \left ( m \hbar \langle jm | J_x | jm \rangle - \langle jm | J_x J_z | jm \rangle ^\dagger \right ) \\
& = {i \over \hbar} \left ( m \hbar \langle jm | J_x | jm \rangle - m \hbar \langle jm | J_x | jm \rangle ^\dagger \right )
= -m \ \text{Im} \left \{ \langle jm | J_x | jm \rangle \right \} = 0 \left ( \because \langle jm | J_x | jm \rangle \in \mathbb{R} \right )
\end{align} </math>
====(2)====
<math>\langle J_x^2 \rangle + \langle J_y^2 \rangle = \langle J_x^2 + J_y^2 \rangle = \langle J^2 - J_z^2 \rangle = \langle jm | J^2 - J_z^2 | jm \rangle = j \left ( j + 1 \right ) \hbar ^2 - m^2 \hbar ^2 </math>
By symmetry, <math> \langle J_x^2 \rangle = \langle J_y^2 \rangle </math>.<br />
:<math>\therefore \langle J_x^2 \rangle = \langle J_y^2 \rangle = {1 \over 2} \hbar^2 \left [ j \left ( j + 1 \right ) - m^2 \right ] </math>
====(3)====
<math>\left ( \Delta J_x \right ) ^2 = \langle J_x^2 \rangle - \langle J_x \rangle ^2 = {1 \over 2} \hbar^2 \left [ j \left ( j + 1 \right ) - m^2 \right ] </math>
and
<math>\left ( \Delta J_y \right ) ^2 = \langle J_y^2 \rangle - \langle J_y \rangle ^2 = {1 \over 2} \hbar^2 \left [ j \left ( j + 1 \right ) - m^2 \right ] </math>.<br />
Since <math>J_x = { {J_+ + J_-}\over 2 }</math> and <math>J_y = {{J_+ - J_-}\over {2i}}</math>,<br />
<math>
\begin{align}
J_x | jm \rangle & = {1 \over 2} \left ( J_+ | jm \rangle + J_- | jm \rangle \right ) = {1 \over 2} \left ( J_+ | jm \rangle + J_- | jm \rangle \right ) \\
& = {1 \over 2} \left [ \hbar \left \{ \left ( j - m \right ) \left ( j + m + 1 \right ) \right \} ^{1/2} | j, m+1 \rangle +
\hbar \left \{ \left ( j + m \right ) \left ( j - m + 1 \right ) \right \} ^{1/2} | j, m-1 \rangle \right ] \\
J_y | jm \rangle & = {1 \over 2i} \left ( J_+ | jm \rangle - J_- | jm \rangle \right ) = {1 \over 2i} \left ( J_+ | jm \rangle - J_- | jm \rangle \right ) \\
& = {1 \over 2i} \left [ \hbar \left \{ \left ( j - m \right ) \left ( j + m + 1 \right ) \right \} ^{1/2} | j, m+1 \rangle -
\hbar \left \{ \left ( j + m \right ) \left ( j - m + 1 \right ) \right \} ^{1/2} | j, m-1 \rangle \right ]
\end{align}</math>
<math>\langle j, m+1 | j, m+1 \rangle = \langle j, m - 1 | j, m - 1 \rangle = 1, \langle j, m-1 | j, m+1 \rangle = \langle j, m + 1 | j, m - 1 \rangle = 0 \Rightarrow </math>
<math>
\begin{align}
\langle jm | J_x J_y | jm \rangle & = \left ( J_x ^\dagger | jm \rangle \right ) ^\dagger J_y | jm \rangle = \left ( J_x | jm \rangle \right ) ^\dagger J_y | jm \rangle \\
& = {1 \over 2} \left [ \hbar \left \{ \left ( j - m \right ) \left ( j + m + 1 \right ) \right \} ^{1/2} \langle j, m+1 | +
\hbar \left \{ \left ( j + m \right ) \left ( j - m + 1 \right ) \right \} ^{1/2} \langle j, m-1 | \right ] \times \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {1 \over 2i} \left [ \hbar \left \{ \left ( j - m \right ) \left ( j + m + 1 \right ) \right \} ^{1/2} | j, m+1 \rangle -
\hbar \left \{ \left ( j + m \right ) \left ( j - m + 1 \right ) \right \} ^{1/2} | j, m-1 \rangle \right ] \\
& = {{\hbar^2} \over {4i}} \left \{ \left ( j - m \right ) \left ( j + m + 1 \right ) - \left ( j + m \right ) \left ( j - m + 1 \right ) \right \} \\
& = - {{m \hbar^2} \over {2i}}
\end{align}
</math>
<math>
\begin{align}
\langle J_x \rangle ^2 \langle J_y \rangle ^2 - \left | \langle jm | J_x J_y | jm \rangle \right | ^2 & = {\hbar^4 \over 4}\left [ j \left ( j + 1 \right ) - m^2 \right ]^2 - {{m^2 \hbar^4} \over {4}}
= {\hbar^4 \over 4} \left \{ j \left ( j + 1 \right ) - m^2 + m \right \} \left \{ j \left ( j + 1 \right ) - m^2 - m \right \} \\
& = {\hbar^4 \over 4} \left ( j + m \right ) \left ( j - m + 1 \right ) \left ( j + m + 1 \right ) \left ( j - m \right )
= {\hbar^4 \over 4} \left ( j^2 - m^2 \right ) \left \{ \left ( j + 1 \right ) ^2 - m^2 \right \}
\end{align}
</math>
<math> j \ge 0 </math> and <math> j \ge m \ge -j </math>. So <math> \left ( j + 1 \right ) ^2 \ge j^2 \ge m^2 </math>
:<math> \therefore \langle J_x \rangle ^2 \langle J_y \rangle ^2 \ge \left | \langle jm | J_x J_y | jm \rangle \right | ^2 </math>
====(4)====
When <math> m = \pm j </math>, <math>j^2 - m^2 = 0 </math>. Then <math>\langle J_x \rangle ^2 \langle J_y \rangle ^2 - \left | \langle jm | J_x J_y | jm \rangle \right | ^2 = 0 </math>
:<math> \therefore \langle J_x \rangle ^2 \langle J_y \rangle ^2 = \left | \langle j, \pm j | J_x J_y | j, \pm j \rangle \right | ^2 </math>
===12.5.10===
<math>- \hbar^2 \left ( {1 \over {\sin \theta}}{\partial \over {\partial \theta}}\sin \theta {\partial \over {\partial \theta}} + {1\over {\sin^2 \theta }}{\partial^2 \over {\partial \phi^2}} \right ) \psi_{\alpha\beta}(\theta, \phi) = \alpha \psi_{\alpha\beta}(\theta, \phi) </math>
<math>\Rightarrow - \hbar^2 \left ( {1 \over {\sin \theta}}{\partial \over {\partial \theta}}\sin \theta {\partial \over {\partial \theta}} + {1\over {\sin^2 \theta }}{\partial^2 \over {\partial \phi^2}} \right ) \left ( P_\alpha^\beta (\theta ) e^{i\beta\phi} \right ) = \alpha P_\alpha^\beta (\theta ) e^{i\beta\phi} </math>
<math>\Rightarrow -{{\hbar^2} \over {\sin \theta}}{\partial \over {\partial \theta}}\sin \theta {\partial \over {\partial \theta}} P_\alpha^\beta (\theta ) e^{i\beta\phi}
+ {{\beta^2 \hbar^2}\over {\sin^2 \theta }} P_\alpha^\beta (\theta ) e^{i\beta\phi} = \alpha P_\alpha^\beta (\theta ) e^{i\beta\phi} </math>
<math>\Rightarrow {1 \over {\sin \theta}}{\partial \over {\partial \theta}}\sin \theta {\partial \over {\partial \theta}} P_\alpha^\beta (\theta )
- {{\beta^2}\over {\sin^2 \theta }} P_\alpha^\beta (\theta ) = - {\alpha \over {\hbar^2}} P_\alpha^\beta (\theta )</math>
<math>\Rightarrow \left ( {1 \over {\sin \theta}}{\partial \over {\partial \theta}}\sin \theta {\partial \over {\partial \theta}} + {\alpha \over {\hbar^2}}
- {{\beta^2}\over {\sin^2 \theta }} \right ) P_\alpha^\beta (\theta ) = 0 </math>
:<math>\therefore \left ( {1 \over {\sin \theta}}{\partial \over {\partial \theta}}\sin \theta {\partial \over {\partial \theta}} + {\alpha \over {\hbar^2}}
- {{m^2}\over {\sin^2 \theta }} \right ) P_\alpha^m (\theta ) = 0 </math>
<math>{d \over {d\theta}} P_\alpha^m = {{du} \over {d\theta}} {d \over {du}} P_\alpha^m = -\sin \theta {d \over {du}} P_\alpha^m</math>
<math>
\begin{align}
\Rightarrow {1 \over {\sin \theta}}{d \over {d \theta}}\sin \theta {d \over {d \theta}} P_\alpha^m & =
-{1 \over {\sin \theta}}{d \over {d \theta}}\sin^2 \theta {d \over {du}} P_\alpha^m
=-{1 \over {\sin \theta}}{d \over {d \theta}} \left \{ \left ( 1 - u^2 \right ) {d \over {du}} P_\alpha^m \right \} \\
& = - {1 \over {\sin \theta}} {{du} \over {d\theta}} {d \over {du}}\left \{ \left ( 1 - u^2 \right ) {d \over {du}} P_\alpha^m \right \}
= \left ( 1 - u^2 \right ) {d^2 \over {du^2}} P_\alpha^m - 2u {d \over {du}} P_\alpha^m
\end{align}</math>
For <math> m = 0</math>,
<math>\left ( {1 \over {\sin \theta}}{\partial \over {\partial \theta}}\sin \theta {\partial \over {\partial \theta}} + {\alpha \over {\hbar^2}} \right ) P_\alpha^0 (\theta ) = \left ( 1 - u^2 \right ) {d^2 \over {du^2}} P_\alpha^0 - 2u {d \over {du}} P_\alpha^0 + {\alpha \over {\hbar^2}} P_\alpha^0 = 0 </math>
<math>\Rightarrow \left ( 1 - u^2 \right ) \sum_{n=2}^\infty C_n n \left ( n - 1 \right ) u^{n-2} - 2u \sum_{n=1}^\infty C_n n u^{n-1} + {\alpha \over {\hbar^2}} \sum_{n=0}^\infty C_n u^n = 0</math>
<math>\Rightarrow \sum_{n=0}^\infty C_{n+2} \left ( n + 2 \right ) \left ( n + 1 \right ) u^n - \sum_{n=2}^\infty C_n n \left ( n - 1 \right ) u^n - 2 \sum_{n=1}^\infty C_n n u^n + {\alpha \over {\hbar^2}} \sum_{n=0}^\infty C_n u^n = 0</math>
<math>\Rightarrow \sum_{n=0}^\infty \left \{ C_{n+2} \left ( n + 2 \right ) \left ( n + 1 \right ) - C_n n \left ( n - 1 \right ) - 2 C_n n + {\alpha \over {\hbar^2}} C_n \right \} u^n = 0 \left ( \because n = 1 \Rightarrow n \left ( n - 1 \right ) = 0 , n = 0 \Rightarrow n \left ( n - 1 \right ) = n = 0 \right ) </math>
<math>\Rightarrow C_{n+2} \left ( n + 2 \right ) \left ( n + 1 \right ) = \left ( n^2 + n - {\alpha \over {\hbar^2}} \right ) C_n </math>
:<math>\therefore \lim_{n \to \infty} {C_{n+2} \over C_n} = \lim_{n \to \infty} { n^2 + n - \alpha / \hbar^2 \over \left ( n + 2 \right ) \left ( n + 1 \right )} = 1
</math>
Let <math> a_0 = 1 </math>, and <math> a_{n+1} = {4n^2 + 2n - \alpha / \hbar^2 \over { \left ( 2n + 2 \right ) \left ( 2n + 1 \right ) }}a_n</math>.
<math>\sum a_n</math> diverges if for any <math>N \in \mathbb{N}</math>, there is <math> n > N </math> such that <math>a_n \ne 0</math>.
<math>
\begin{align}
C_0C_1 >0 & \Rightarrow \left | P_\alpha^0 (0) \right | \ge \sum \left | C_0 \right | a_n \\
C_0C_1 < 0 & \Rightarrow \left | P_\alpha^0 (\pi ) \right | \ge \sum \left | C_1 \right | a_n \\
C_0 = 0 & \Rightarrow \left | P_\alpha^0 (\pi ) \right | \ge \sum \left | C_1 \right | a_n \\
C_1 = 0 & \Rightarrow \left | P_\alpha^0 (0 ) \right | \ge \sum \left | C_0 \right | a_n \\
\end{align}
</math>
Since <math>C_0</math> and <math>C_1</math> cannot both be zero at the same time, one of <math>P_\alpha^0 (0)</math> or <math>P_\alpha^0 (\pi )</math> diverges unless the series terminates.
<math>
\begin{align}
{\alpha \over {\hbar^2}} = l \left ( l + 1 \right ) & \Rightarrow C_{n+2} \left ( n + 2 \right ) \left ( n + 1 \right ) = \left ( n^2 + n - l^2 - l \right ) C_n \\
& \Rightarrow C_{n+2} \left ( n + 2 \right ) \left ( n + 1 \right ) = \left ( n + l + 1 \right ) \left ( n - l \right ) C_n \\
& \Rightarrow \text{for all k} \in \mathbb{N}, C_{l+2k} = 0
\end{align}
</math>
If <math> l </math> is even and <math> C_1 = 0 </math> , <math>P_\alpha^0 </math> is even. If <math> l </math> is odd and <math> C_0 = 0 </math> , <math>P_\alpha^0 </math> is odd. For other values of <math> C_0</math> or <math> C_1</math> the series diverges and lacks any physical meaning.
For <math> l = 0</math>, <math> C_{2k - 1} = C_{2k} = 0</math> for all <math> k\in \mathbb{N} \Rightarrow P_0 = 1</math>.
For <math> l = 1</math>, <math> C_{2k-2} = C_{2k + 1} = 0</math> for all <math> k\in \mathbb{N} \Rightarrow P_1 = u</math>.
For <math> l = 2</math>, <math> C_{2k+2} = C_{2k - 1} = 0</math> for all <math> k\in \mathbb{N}</math> and <math> 2C_2 = -6C_0\Rightarrow P_2 = 3u^2 - 1</math>.
They are equal to <math>Y_0^0 = \left ( 4\pi \right ) ^{\text{-}1/2}, Y_1^0 = \sqrt{3 \over {4\pi}} u, Y_2^0 = \sqrt{5 \over {16\pi}} \left ( 3u^2 -1 \right ) </math> if we ignore the scaling factors.
===12.6.9===
The eigenvalue equation on <math>U</math> is
<math>\left [ -{ \hbar ^2\over {2\mu}} {d^2\over{dr^2}} + V(r) + {l\left ( l + 1 \right ) \hbar ^2 \over {2\mu r^2}} \right ] U = EU</math>
Since <math> l = 0</math>,
<math>
\begin{cases}
-{\hbar ^2\over {2\mu}} {d^2\over{dr^2}}U = EU & r < r_0 \\
-{\hbar ^2\over {2\mu}} {d^2\over{dr^2}}U = \left ( E -V_0 \right ) U & r>r_0
\end{cases}
</math>
Let <math> k' = {\sqrt{2\mu E}\over \hbar}</math>, <math> \kappa = {\sqrt{2\mu \left ( V_0 -E \right ) }\over \hbar} </math>, and
<math>U _E (r) = \begin{cases} U_1 & r < r_0 \\ U_2 & r > r_0 \end{cases} </math>.
The general solutions of each differential equations are <math> U_1 = A e^{ik'r} + B e^{-ik'r} </math> and <math> U_2 = C e^{\kappa r} + D e^{-\kappa r}</math>.
Boundary conditions:
(1) <math>U \to 0 \text{ as } r \to 0</math>
: <math> A + B = 0 </math>
(2) <math>U \to 0 \text{ as } r \to \infty</math>
: <math> C = 0 </math>
(3) continuity
: <math> A e^{ik'r_0} - A e^{-ik'r_0} = D e^{-\kappa r_0}</math>
(4) continuity of the derivative
: <math> ik'A e^{ik'r_0} + ik'A e^{-ik'r_0} = -\kappa D e^{-\kappa r_0}</math>
<math>
\begin{align}
ik'A e^{ik'r_0} + ik'A e^{-ik'r_0} = -\kappa \left ( A e^{ik'r_0} - A e^{-ik'r_0} \right )
& \Rightarrow ik' e^{ik'r_0} + ik' e^{-ik'r_0} = -\kappa \left ( e^{ik'r_0} - e^{-ik'r_0} \right ) \\
& \Rightarrow 2ik' \cos \left ( k'r_0 \right ) = -2i\kappa \sin \left ( k'r_0 \right )
\end{align} </math>
:<math>\therefore k'/\kappa = -\tan \left ( k'r_0 \right ) </math>
Since <math> k'^2 + \kappa ^2 = {{2\mu V_0}\over \hbar ^2}</math>, if we take <math> \alpha = k'r_0 </math> and <math> \beta = \kappa r_0 </math>, then <math> \alpha ^2 + \beta ^2 = {{2\mu V_0 {r_0}^2}\over \hbar ^2}</math> and <math> \beta = -\alpha \cot \alpha </math>.
The minimum value of <math> \alpha ^2 + \beta ^2 </math> on the first quadrant of the graph <math> \beta = -\alpha \cot \alpha </math> is <math> \pi^2 \over 4</math>.
<math>\alpha ^2 + \beta^2 = {{2\mu V_0 {r_0}^2}\over \hbar ^2} = {\pi^2 \over 4} \Rightarrow V_0 = {{\pi^2 \hbar ^2}\over {8\mu {r_0}^2}}</math>
:<math>\therefore</math> If <math>V_0 < {{\pi^2 \hbar ^2}\over {8\mu {r_0}^2}}</math> there are no bound states.
===12.6.11===
====(1)====
<math>v' = \left ( l + 1 \right ) y^l \sum_{n=0}^\infty C_n y^n + y^{l + 1} \sum_{n=1}^\infty n C_n y^{n -1}
= y^l \sum_{n = 0}^\infty \left ( n + l + 1 \right ) C_n y^n </math>
<math> v'' = l y^{l - 1} \sum_{n = 0}^\infty \left ( n + l + 1 \right ) C_n y^n + y^l \sum_{n = 1}^\infty n \left ( n + l + 1 \right ) C_n y^{n-1}
= y^{l-1} \sum_{n = 0}^\infty \left ( n + l \right ) \left ( n + l + 1 \right ) C_n y^n</math>
<math>
\begin{align}
v'' & - 2yv' + \left [ 2 \lambda - 1 - {{l \left ( l + 1 \right )}\over y^2} \right ] v = 0 \\
& \Rightarrow y^{l-1} \sum_{n = 0}^\infty \left ( n + l \right ) \left ( n + l + 1 \right ) C_n y^n - 2y^{l+1} \sum_{n = 0}^\infty \left ( n + l + 1 \right ) C_n y^n +
\left (2 \lambda - 1\right ) y^{l+1}\sum_{n=0}^\infty C_n y^n - l\left ( l + 1 \right ) y^{l-1}\sum_{n=0}^\infty C_n y^n = 0 \\
& \Rightarrow \sum_{n = 0}^\infty \left ( n + l \right ) \left ( n + l + 1 \right ) C_n y^n - 2 \sum_{n = 0}^\infty \left ( n + l + 1 \right ) C_n y^{n+2} +
\left (2 \lambda - 1\right ) \sum_{n=0}^\infty C_n y^{n+2} - l\left ( l + 1 \right ) \sum_{n=0}^\infty C_n y^n = 0 \\
& \Rightarrow \left ( l + 1 \right ) \left ( l + 2 \right ) C_1 y - l \left ( l + 1 \right ) C_1 y + \sum_{n = 2}^\infty \left ( n + l \right ) \left ( n + l + 1 \right ) C_n - 2 \left ( n + l + 1 \right ) C_{n-2} y^n + \left (2 \lambda - 1\right ) C_{n-2} y^n - l\left ( l + 1 \right ) C_n y^n = 0 \\
& \Rightarrow 2 \left ( l + 1 \right ) C_1 y + \sum_{n = 2}^\infty \left \{ n \left ( n + 2l + 1 \right ) C_n - \left ( 2n + 2l -2\lambda +3 \right ) C_{n-2} \right \} y^n = 0 \\
& \Rightarrow C_1 = 0, \ n \left ( n + 2l + 1 \right ) C_n = \left ( 2n + 2l -2\lambda +3 \right ) C_{n-2} \\
& \Rightarrow \text{for any } k \in \mathbb{N}, C_{2k - 1} = 0
\end{align}
</math>
If <math> C_0 = 0</math>, then for every <math> n</math>, <math> C_n = 0</math>, which leads to a nonphysical zero wave function.
:<math>\therefore C_0 \ne 0</math>
For <math> y = 1</math>, <math>v = \sum_{n=0}^\infty C_n = \sum_{k = 0}^\infty C_{2k}</math>.
Unless the series terminates,
<math>\text{as } k\to \infty , {C_{2k+2}\over C_{2k}} \to {1 \over {k+1}} \Rightarrow v \sim y^{l+1}e^{y^2/2}</math>, which goes to infinity as r increases.
This cannot be a physical solution, so the series must terminate at certain <math> k \Rightarrow 4k + 2l - 2\lambda + 3 = 0 </math>
:<math>\therefore E = \left ( 2k + l +3/2 \right ) \hbar \omega</math>
====(2)====
For each <math> n </math>, the allowed <math> l </math> values are <math> n, n-2, \cdots , 1 \text{ or } 0 </math>.
For each <math> l</math>, there are <math> 2l+1</math>-fold degeneracy, so the total degeneracy is <math>
\left ( 2n + 1 \right ) + \left ( 2n - 3 \right ) + \cdots + 3 \text{ or } 1 </math>.
If <math> n = 2k</math>, <math>
\left ( 4k + 1 \right ) + \left ( 4k - 3 \right ) + \cdots + 1 = {{\left ( 4k + 2 \right ) \left ( k + 1 \right ) } \over 2} = {{\left ( n + 1 \right ) \left ( n + 2 \right ) } \over 2} </math>.
If <math> n = 2k - 1</math>, <math>
\left ( 4k - 1 \right ) + \left ( 4k - 5 \right ) + \cdots + 3 = {{\left ( 4k + 2 \right ) k} \over 2} = {{\left ( n + 1 \right ) \left ( n + 2 \right ) } \over 2} </math>.
:<math>\therefore \psi_n</math> has <math> {{\left ( n + 1 \right ) \left ( n + 2 \right ) } \over 2}</math>-fold degeneracy
Also <math>Y_l^m</math> are odd if <math> l</math> is odd and even if <math> l </math> is even.
:<math>\therefore</math> parity of <math>\psi_n</math> is <math>\left ( \text{-}1 \right ) ^n</math>
The results are equal to the results obtained from the Cartesian solutions.
====(3)====
<math> n = 2k + l = 0 \Rightarrow l = k = m = 0 </math>
Then <math>v = C_0 y</math> because <math>C_k = 0</math> for all <math>k \in \mathbb{N}</math>.
To normalize <math>U_{00}</math>, <math>\int_0^\infty \left | U_{00} \right | ^2 dr = {C_0}^2 \int_0^\infty y^2 e^{-y^2} \left ( {\hbar \over {\mu\omega}} \right ) ^{1/2} dy = {C_0}^2 \left ( {\hbar \over {\mu\omega}} \right ) ^{1/2} {\sqrt{\pi}\over 4} = 1 \Rightarrow U_{00} = 2 \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{1/4} y e^{-y^2/2} </math>.
:<math>\because \int_0^\infty x^2 e^{-x^2} = \left [ -{x \over 2} e^{-x^2} \right ] _0^\infty + \int_0^\infty {1 \over 2} e^{-x^2}
= {\sqrt{\pi}\over 4}
</math>
<math>\psi_{000} = U_{00} / r Y_0^0 = 2 \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{1/4} \left ( {\mu \omega \over {4\pi \hbar}} \right ) ^{1/2} \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right ) = \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{3/4} \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right )</math>
<math> n = 2k + l = 1 \Rightarrow l = 1, k = 0, m = \text{-}1, 0, 1 </math>
Then <math>v = C_0 y^2</math> because <math>C_k = 0</math> for all <math>k \in \mathbb{N}</math>.
To normalize <math>U_{01}</math>, <math>\int_0^\infty \left | U_{01} \right | ^2 dr = {C_0}^2 \int_0^\infty y^4 e^{-y^2} \left ( {\hbar \over {\mu\omega}} \right ) ^{1/2} dy = {C_0}^2 \left ( {\hbar \over {\mu\omega}} \right ) ^{1/2} {{3\sqrt{\pi}}\over 8} = 1 \Rightarrow U_{01} = \sqrt{8\over 3} \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{1/4} y^2 e^{-y^2/2} </math>.
:<math>\because \int_0^\infty x^4 e^{-x^2} = \left [ -{x^3 \over 2} e^{-x^2} \right ] _0^\infty + \int_0^\infty {3x^2 \over 2} e^{-x^2}
= \left [ -{3x \over 4} e^{-x^2} \right ] _0^\infty + \int_0^\infty {3 \over 4} e^{-x^2} = {{3\sqrt{\pi}}\over 8}
</math>
<math> \psi_{11\pm 1} = U_{01}/rY_1^{\pm1} = \mp \sqrt{8\over 3} \sqrt{3\over {8\pi}} \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{1/4} \left ( {\mu \omega \over {\hbar}} \right ) r \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right ) \sin \theta e^{\pm \phi},
\psi_{110} = U_{01}/rY_1^0 = \sqrt{8\over 3} \sqrt{3 \over {4\pi}} \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{1/4} \left ( {\mu \omega \over { \hbar}} \right ) r \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right ) \cos \theta, </math>
<math> \Rightarrow \psi_{11\pm 1} = \mp \sqrt{\pi} \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{5/4} r \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right ) \sin \theta e^{\pm i\phi}, \psi_{110} = \sqrt{2\pi} \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{5/4} r \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right ) \cos \theta </math>
Cartesian solutions for <math> n = 0 , 1</math> are:
<math> \psi '_{000} = \psi_0 (x) \psi_0 (y) \psi_0 (z) = {A_0}^3 \exp \left ( - {\mu \omega \over {2\hbar}} r^2 \right )
= \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{3/4} \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right )</math>
<math> \psi '_{100} = \psi_1 (x) \psi_0 (y) \psi_0 (z) = {A_0}^2 A_1 \left ( {\mu \omega \over {\hbar}} \right ) ^{1/2} x \exp \left ( - {\mu \omega \over {2\hbar}} r^2 \right )
= \sqrt{2 \pi} \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{5/4} x \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right )</math>
By the same logic,
<math> \psi '_{010} = \sqrt{2\pi} \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{5/4} y \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right )</math>
<math> \psi '_{001} = \sqrt{2\pi} \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{5/4} z \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right )</math>
So <math>\psi_{000} = \psi '_{000}</math> and
<math> \psi_{11\pm 1} = \mp \sqrt{\pi} \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{5/4} \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right ) r \sin \theta \left ( \cos \phi \pm i \sin \phi \right ) = \mp \sqrt{\pi} \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{5/4} \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right ) \left ( x \pm i y \right ) = \mp {1\over\sqrt{2}} \psi '_{100} - {i\over\sqrt{2}} \psi '_{010}</math>
<math> \psi_{110} = \sqrt{2\pi} \left ( {{\mu \omega} \over {\pi \hbar}} \right ) ^{5/4} z \exp \left ( - {\mu \omega \over {2 \hbar}} r^2 \right ) = \psi '_{001}</math>
:<math>\therefore \psi_{000} = \psi '_{000}, \psi_{111} = -{1\over\sqrt{2}} \psi '_{100} - {i\over\sqrt{2}} \psi '_{010}, \psi_{110} = \psi '_{001}, \psi_{11\text{-}1} = {1\over\sqrt{2}} \psi '_{100} - {i\over\sqrt{2}} \psi '_{010}
</math>
===13.1.5===
<math>\langle \dot{\Omega} \rangle = - { i\over \hbar} \langle \left [ \Omega, H \right ] \rangle </math>
<math> \left [ \Omega, H \right ] = \left [ XP_X + YP_Y + ZP_Z, H \right ] </math>
<math> \left [ XP_X, H \right ] = {1 \over {2m}} \left [ XP_X, P_X^2 + P_Y^2 + P_Z^2 \right ] - e^2 \left [ XP_X, {1\over R} \right ] </math>
<math> \begin{align} \left [ XP_X, P_X^2 + P_Y^2 + P_Z^2 \right ] & = \left [ XP_X, P_X^2 \right ] = XP_X^3 - P_X^2 X P_X \\
& = XP_X^3 - P_X X P_X^2 + P_X X P_X^2 - P_X^2 X P_X = \left [ X, P_X \right ] P_X^2 + P_X \left [ X, P_X \right ] P_X = 2i\hbar P_X^2 \end{align}</math>
<math> {i\over \hbar} \langle x', y', z' | \left [ XP_X, {1\over R} \right ] | x, y, z \rangle = x' \delta '(x-x') \delta (y-y') \delta (z-z'){1 \over \sqrt{x^2 + y^2 + z^2}} - {1 \over \sqrt{x'^2 + y'^2 + z'^2}} x' \delta '(x-x') \delta (y-y') \delta (z-z')</math>
<math>
\begin{align}
{i\over \hbar} \int \langle n, l, m & | x', y', z' \rangle \langle x', y', z' | \left [ XP_X, {1\over R} \right ] | x, y, z \rangle \langle x, y, z | n, l, m \rangle dxdydzdx'dy'dz' \\
& = \int \psi_{nlm} ^* x {\partial \over {\partial x}}{1 \over \sqrt{x^2 + y^2 + z^2}} \psi_{nlm} - \psi_{nlm}^*{x \over \sqrt{x^2 + y^2 + z^2}} {\partial \over {\partial x}} \psi_{nlm} dxdydz \\
& = \int \psi_{nlm} ^* x {\partial \over {\partial x}} \left ( {1\over r} \right ) \psi_{nlm} dxdydz \\
& = \int - \psi_{nlm} ^* {x^2 \over r^3} \psi_{nlm} dxdydz = - \langle {x^2\over r^3} \rangle
\end{align}</math>
<math> \langle \dot{\Omega} \rangle = - { i\over \hbar} \langle \left [ \Omega, H \right ] \rangle = {1\over m} \langle P^2 \rangle - e^2 \langle {1 \over r}\rangle = 2 \langle T \rangle + \langle V \rangle = 0 \left ( \because \langle {x^2\over r^3} \rangle + \langle {y^2\over r^3} \rangle + \langle {z^2\over r^3} \rangle = \langle {1 \over r} \rangle \right )</math>
:<math>\therefore \langle T \rangle = -{1\over 2} \langle V \rangle </math>
===13.2.1===
====(1)====
<math>\mathbf{n} = {{\mathbf{p} \times \mathbf{l}} \over m} - {e^2 \over r} \mathbf{r} = {{\mathbf{p} \times \left ( \mathbf{r} \times \mathbf{p} \right ) } \over m} - {e^2 \over r} \mathbf{r} = {{\mathbf{r} p^2 - \mathbf{p} \left ( \mathbf{p} \cdot \mathbf{r} \right )} \over m} - {e^2 \over r} \mathbf{r}
= \left ( {p^2 \over m} - {e^2 \over r} \right ) \mathbf{r} - {\mathbf{p} \cdot \mathbf{r} \over m} \mathbf{p} </math>
====(2)====
By energy conservation, <math> {p^2 \over {2m}} - {e^2 \over r} = E </math>.
Let the angle between <math>\mathbf{p}</math> and <math> \mathbf{r} </math> be <math> \theta</math>. i.e. <math>\mathbf{p} \cdot \mathbf{r} = pr\cos\theta</math>
<math>\mathbf{n} = \left ( 2E + {e^2\over r} \right ) \mathbf{r} - {\mathbf{p} \cdot \mathbf{r} \over m} \mathbf{p} = \left ( 2E + {e^2 \over r} \right ) \mathbf{r} - \left ( \dot{\mathbf{r}} \cdot \mathbf{r} \right ) \mathbf{p} = \left ( 2E + {e^2 \over r} \right ) \mathbf{r} - {1\over 2} {d\over {dt}} \left ( \mathbf{r} \cdot \mathbf{r} \right ) \mathbf{p} = \left ( 2E + {e^2 \over r} \right ) \mathbf{r} - {1\over 2} {d\over {dt}} \left ( r^2 \right ) \mathbf{p}</math>
When the particle arrives at the maximum or minimum distance from the origin, <math>{d\over {dt}} \left ( r^2 \right ) = 0</math>.
So <math> \mathbf{n} = \mathbf{r}_{\text{max}} \left ( 2E + {e^2 \over r_{\text{max}}} \right ) = \mathbf{r}_{\text{min}} \left ( 2E + {e^2 \over r_{\text{min}}} \right ) </math> and <math>\mathbf{n} \parallel \mathbf{r}_{\text{max}} \parallel \mathbf{r}_{\text{min}}</math>.
If <math>\mathbf{n}</math>, <math>\mathbf{r}_{\text{max}} </math>, and <math>\mathbf{r}_{\text{min}}</math> are all pointing to the same direction, <math>\mathbf{r}_{\text{max}} = r_{\text{max}} \hat{r}, \mathbf{r}_{\text{min}} = r_{\text{min}} \hat{r} </math>.
<math>\mathbf{n} = r_{\text{max}} \left ( 2E + {e^2 \over r_{\text{max}}} \right ) \hat{r} = \left ( 2E r_{\text{max}} + e^2 \right ) \hat{r} = r_{\text{min}} \left ( 2E + {e^2 \over r_{\text{min}}} \right ) = \left ( 2E r_{\text{min}} + e^2 \right ) \hat{r} \Rightarrow 2E r_{\text{max}} + e^2 = 2E r_{\text{min}} + e^2 \Rightarrow r_{\text{max}} = r_{\text{min}} </math>
This does not hold for general cases, so <math>\mathbf{r}_{\text{max}} = -r_{\text{max}} \hat{r}</math>.
<math>\mathbf{n} = - r_{\text{max}} \left ( 2E + {e^2 \over r_{\text{max}}} \right ) \hat{r} = - \left ( 2E r_{\text{max}} + e^2 \right ) \hat{r} = r_{\text{min}} \left ( 2E + {e^2 \over r_{\text{min}}} \right ) = \left ( 2E r_{\text{min}} + e^2 \right ) \hat{r} \Rightarrow -2E r_{\text{max}} - e^2 = 2E r_{\text{min}} + e^2 \Rightarrow r_{\text{max}} + r_{\text{min}} = -{e^2\over E} </math>
<math> 2E + {e^2 \over r_{\text{min}}} \ge 0 </math>, so <math>\mathbf{r}_{\text{max}}</math> and <math>\mathbf{r}_{\text{min}} </math> are antiparallel to each other.
For circular orbits, <math>r_{\text{max}} + r_{\text{min}} = 2r = -{e^2\over E} \Rightarrow 2E + {e^2 \over r} = 0 </math>.
:<math>\therefore \mathbf{n} = 0</math>
==2.==
===(1)===
줄 74 ⟶ 450:
:<math>\therefore L_z = m\hbar</math> where <math> m \in \mathbb{N}</math>
==3.==
===(1)===
<math>\psi ( \mathbf{r} ) = A' e^{-\alpha r} Y_0^0</math><br /><br />
Since the spherical harmonics are orthonormal to each other, the wave function only has the <math>|0, 0 \rangle </math> component.<br /><br />
:<math>\therefore P_{lm} =
\begin{cases}
1 & l = m = 0\\
0 & otherwise
\end{cases}
</math>
===(2)===
<math>
\begin{align}
\psi ( \mathbf{r}) & = A \left ( r \sin \theta \cos \phi + r \sin \theta \sin \phi + 2 r \cos \theta \right ) e ^{-\alpha r} \\
& = A \left ( \sin \theta {{e^{i \phi} + e^{-i \phi}} \over 2} - i \sin \theta {{e^{i \phi} - e^{-i \phi}} \over 2} + \cos \theta \right ) r e^{-\alpha r} \\
& = A \left \{ {1 \over 2} \left ( 1 - i \right ) \sin \theta e^{i \phi } + {1 \over 2} \left ( 1 + i \right ) \sin \theta e^{-i \phi } + \cos \theta \right \} r e^{-\alpha r} \\
& = A \left \{ - \sqrt{ 2\pi \over 3 } \left ( 1 - i \right ) Y_1^1 + \sqrt{2\pi \over 3} \left ( 1 + i \right ) Y_1^{\text{-}1} + \sqrt{ 4\pi \over 3} Y_1^0 \right \} r e^{-\alpha r}
\end{align}
</math><br /><br />
<math>\left | \sqrt{2\pi \over 3} \left ( 1-i \right ) \right | ^2 = \left | \sqrt{2\pi \over 3} \left ( 1+i \right ) \right | ^2 = \left | \sqrt{4\pi \over 3} \right | ^2 = {4\pi \over 3} </math><br /><br />
Since the spherical harmonics are orthonormal to each other, the probability of observing <math> | 1, 1 \rangle </math>, <math> | 1, \text{-}1 \rangle </math>, and <math> | 1, 0 \rangle </math> are same and others are impossible.<br /><br />
:<math>\therefore P_{lm} =
\begin{cases}
{1\over 3} & ( l, m ) = ( 1, \text{-}1 ) , \ ( 1, 0 ) , \ or \ ( 1, 1 ) \\
0 & otherwise \\
\end{cases}
</math>
===(3)===
<math>
\begin{align}
\psi(\mathbf{r}) & = A \left ( \sin^2 \theta \sin \phi \cos \phi + \sin \theta \cos \theta \sin \phi + \sin \theta \cos \theta \cos \phi \right ) r^2 e^{-\alpha r^2} \\
& = A \left ( \sin ^2 \theta {{e^{2i\phi} - e^{-2i\phi}}\over 4i} +\sin \theta \cos \theta {{e^{i\phi} - e^{-i\phi}}\over 2i} + \sin \theta \cos \theta {{e^{i\phi} + e^{-i \phi}} \over 2 } \right ) r^2 e^{-\alpha r^2} \\
& = A \left \{ {1 \over 4i} \sin ^2 \theta e^{2i \phi} - {1 \over 4i} \sin ^2 \theta e^{-2i\phi} +{1 \over 2} \left ( 1 - i \right ) \sin \theta \cos \theta e^{i\phi} + {1 \over 2} \left ( 1 + i \right ) e^{-i \phi} \sin \theta \cos \theta \right \} r^2 e^{-\alpha r^2} \\
& = A \left \{ -i \sqrt{ 2\pi \over 15} Y_2^2 + i \sqrt{2\pi \over 15} Y_2^{\text{-}2} - \left ( 1-i \right ) \sqrt{ 2\pi \over 15 } Y_2^1 + \left ( 1+i \right ) \sqrt{2\pi \over 15 }Y_2^{\text{-}1} \right \} r^2 e^{-\alpha r^2}
\end{align}
</math><br /><br />
<math>\left | -i \sqrt{2\pi \over 15} \right | ^2 = \left | i \sqrt{2\pi \over 15} \right | ^2 = {2\pi \over 15}</math> and
<math>\left | - \left ( 1 - i \right ) \sqrt{2\pi \over 15} \right | ^2 = \left | \left ( 1 + i \right ) \sqrt{2\pi \over 15} \right | ^2 = {8\pi \over 15}</math><br /><br />
The spherical harmonics are orthonormal to each other, so
<math>P_{2\text{-}2} : P_{2\text{-}1} : P_{21} : P_{22} = 1 : 4: 4: 1 </math> and <math>P_{2\text{-}2} + P_{2\text{-}1} + P_{21} + P_{22} = 1 </math>.<br /><br />
:<math>\therefore P_{lm} = \begin{cases}
{1 \over 10} & ( l, m) = ( 2, \text{-}2) \ or \ ( 2, 2 ) \\
{2 \over 5} & ( l, m ) = ( 2, \text{-}1 ) \ or \ ( 2, 1 ) \\
0 & otherwise \end{cases}</math>
==4.==
Since <math>\psi(\mathbf{r}) </math> is a energy eigenstate, it satisfies the T.I.S.E.<br />
<math> \left \{ -{\hbar ^2 \over {2\mu}} \left ( {1 \over r^2}{\partial \over {\partial r} }r^2 {\partial \over {\partial r} } + {1 \over {r^2 \sin \theta}}{\partial \over {\partial \theta} } \sin \theta {\partial \over {\partial \theta} } + {1 \over {r^2 \sin ^2 \theta}}{\partial ^2 \over {\partial \phi ^2} } \right ) + V(r) \right \} \psi(\mathbf{r}) = E \psi(\mathbf{r})</math>
<math>\psi(\mathbf{r}) </math> is independent of <math> \theta </math> and <math> \phi </math>.<br />
<math>
\begin{align}
\left ( -{ \hbar ^2 \over {2\mu}} {1 \over r^2}{\partial \over {\partial r} }r^2 {\partial \over {\partial r} } + V(\mathbf{r}) \right ) Ae^{-\alpha r} & = E A e^{-\alpha r} \\
{{\alpha \hbar ^2} \over {2 \mu}}{1\over r^2}{\partial \over \partial r} \left ( r^2 e^{-\alpha r} \right ) + V \left ( r \right ) e^{-\alpha r} & = E e^{-\alpha r} \\
{{\alpha \hbar ^2} \over {2 \mu}}{1\over r^2} \left ( 2r - \alpha r^2 \right ) e^{-\alpha r} + V(r) e^{-\alpha r} & = E e^{-\alpha r} \\
{{\alpha \hbar ^2} \over {2 \mu}}{1\over r^2} \left ( 2r - \alpha r^2 \right ) + V(r) & = E \\
{{\alpha \hbar ^2} \over {\mu r}} - {{\alpha ^2 \hbar ^2} \over {2 \mu}} + V(r) & = E
\end{align}
</math>
<math>
V(r) = - {{\alpha \hbar ^2} \over {\mu r}} + E + {{\alpha ^2 \hbar ^2} \over {2 \mu}} \Rightarrow
\lim_{r \to \infty} V(r) = E + {{\alpha ^2 \hbar ^2} \over {2 \mu}} = 0 </math>
:<math>\therefore V(r) = - {{\alpha \hbar ^2} \over {\mu r}}</math> and <math> E = - {{\alpha ^2 \hbar ^2} \over {2 \mu}} </math>
==5.==
===(1)===
For ground state, <math> n = 1 </math>. Then the only possible <math> (l, m) = (0,0) </math>.<br />
So the corresponding eigenstate is <math>\psi_{1,0,0} (\mathbf{r}) = B R_{10} Y_0^0 = B e^{-r / {a_{0}}}</math>.<br />
To normalize <math>\psi_{1, 0, 0}(\mathbf{r})</math>,<br />
<math>
\begin{align}
\langle \psi_{1, 0, 0} | \psi_{1, 0, 0} \rangle & = \int_0^\infty \int_0^\pi \int_0^{2\pi} \left | \psi_{1, 0, 0} (r, \theta, \phi) \right | ^2 r^2 \sin \theta d\phi d\theta dr \\
& = \int_0^\infty \int_0^\pi \int_0^{2\pi} B^2 e^{-2r / {a_0}} r^2 \sin \theta d\phi d\theta dr \\
& = 2\pi B^2 \left ( \left [ -{{a_0} \over 2} r^2 e^{-2r/{a_0}} \right ] _0 ^\infty + a_0 \int_0^\infty r e^{-2r/{a_0}} dr \right ) \\
& = 2\pi B^2 \left ( \left [ -{{a_0}^2 \over 2} r e^{-2r/{a_0}} \right ] _0 ^\infty + {a_0}^2 \int_0^\infty e^{-2r/{a_0}} dr \right ) \\
& = 2\pi B^2 \left [ -{{a_0}^3 \over 2} r e^{-2r/{a_0}} \right ] _0^\infty = \pi {a_0}^3 B^2 = 1 \Rightarrow B = \left ( {1 \over {\pi {a_0}^3}} \right ) ^ {1/2}
\end{align} </math><br />
By the same logic, <math>A = \left ( {1 \over {\pi {a}^3}} \right ) ^ {1/2}</math>.
<math>
\begin{align}
\langle \psi_{1, 0, 0} | \psi (\mathbf{r}) \rangle & = \int_0^\infty \int_0^\pi \int_0^{2\pi} \psi_{1, 0, 0} ^* (r, \theta, \phi) \psi (r, \theta, \phi) r^2 \sin \theta d\phi d\theta dr \\
& = {1 \over \pi} \int_0^\infty \int_0^\pi \int_0^{2\pi} \left ( aa_0 \right ) ^{-3/2} r^2 \exp \left \{ - \left ( {1 \over a} + {1 \over a_0} \right ) r \right \} \sin \theta d\phi d\theta dr \\
& = 4 \left ( aa_0 \right ) ^{-3/2} \int_0^\infty r^2 \exp \left \{ - \left ( {{a + a_0} \over {aa_0}} \right ) r \right \} dr \\
& = 4 \left ( aa_0 \right ) ^{-3/2} \int_0^\infty \left ( {aa_0 \over {a + a_0}} \right ) ^2 t^2 e^{-t} \left ( {aa_0 \over {a + a_0}} \right ) dt \\
& = 4 \left ( {\sqrt{ aa_0 } \over {a + a_0}} \right ) ^3 \int_0^\infty t^2 e^{-t} dt \\
& = 8 \left ( {\sqrt{ aa_0 } \over {a + a_0}} \right ) ^3
\end{align}</math>.
:cf.) <math> \int_0^\infty x^2 e^{-x} dx = - \left [ x^2 e^{-x} \right ] _0 ^\infty + 2 \int_0^\infty x e^{-x} dx
= - \left [ x e^{-x} \right ] _0^\infty + 2 \int_0^\infty e^{-x}dx = 2
</math>
:<math>\therefore P = \left \| \langle \psi_{1, 0, 0} | \psi (\mathbf{r}) \rangle \right \| ^2 = 64 \left ( {\sqrt{ aa_0 } \over {a + a_0}} \right ) ^6 </math>
===(2)===
Reduced mass of <math> ^3 \text{H}</math> is <math> {{\left ( m_p + 2m_n \right ) m_e} \over {m_p + 2m_n + m_e}} = 9.10773026 \times 10^{\text{-}31} \, \text{kg} </math> so its bohr radius is <math>a_1 = 5.29273186\times 10^{\text{-}11} \, \text{m}</math>.
Reduced mass of <math> ^3 \text{He}^+</math> is <math> {{\left ( 2m_p + m_n \right ) m_e} \over {2m_p + m_n + m_e}} = 9.10772950 \times 10^{\text{-}31} \, \text{kg} </math> so its bohr radius is <math>a_2 = 5.29273230\times 10^{\text{-}11} \, \text{m}</math>.
From the result of '''(1)''', <math> 1-P= 5.18\times10^{-15}</math>.
==6.==
<math>
\begin{align}
H = \sqrt{ \mathbf{p}^2 c^2 + \mu^2 c^4} + V(\mathbf{r} ) & \Rightarrow \left ( H -V(\mathbf{r} ) \right ) ^2 = \mathbf{p}^2 c^2 + \mu^2 c^4 \\
& \Rightarrow \left ( H -V( \mathbf{r} ) \right ) ^2 | \psi ( \mathbf{r} ) \rangle = \left ( \mathbf{p}^2 c^2 + \mu^2 c^4 \right ) | \psi (\mathbf{r} ) \rangle \\
\end{align}</math>
The potential is rotationally invariant. The Hamiltonian operator is only the function of <math>\hat{\mathbf{R}}^2 </math> and <math>\hat{\mathbf{P}}^2 </math>.
So the Hamiltonian and Angular momentum operators mutually commute, which means that <math>| l, m \rangle </math> are eigenstates.
<math> \Rightarrow \psi _E (\mathbf{r}) = R_{El} Y_l^m </math>
<math>
\begin{align}
\left ( E - V(r) \right ) ^2 R_{El} Y_l^m & = -{{\hbar^2 c^2} \over {2\mu}} \nabla ^2 R_{El} Y_l^m + \mu ^2 c^4 R_{El} Y_l ^m
= -{{\hbar^2 c^2} \over {2\mu}} \left [ { 1\over r^2} {\partial \over { \partial r}} \left ( r^2 {\partial \over {\partial r}} \right ) - {1\over {\hbar^2 r^2}} \hat{\mathbf{L}}^2 \right ] R_{El} Y_l^m + \mu^2 c^4 R_{El} Y_l^m \\
& = -{{\hbar^2 c^2} \over {2\mu}} \left [ { 1\over r^2} {\partial \over { \partial r}} \left ( r^2 {\partial \over {\partial r}} \right ) - {{l \left ( l + 1 \right ) }\over r^2}\right ] R_{El} Y_l^m + \mu^2 c^4 R_{El} Y_l^m
\end{align} </math>
Dividing each side with <math>Y_l^m</math> we get,
<math>
\begin{align}
\left ( E - V(r) \right ) ^2 R_{El}
& = -{{\hbar^2 c^2} \over {2\mu}} \left [ { 1\over r^2} {d \over { dr}} \left ( r^2 {d \over {dr}} \right ) - {{l \left ( l + 1 \right ) }\over r^2}\right ] R_{El} + \mu^2 c^4 R_{El}
\end{align} </math>
Let <math> U_{El} = r R_{El} </math>.
<math>\left ( E - V(r) \right ) ^2 U_{El} = \left ( E + {e^2 \over {4 \pi \epsilon_0 r}} \right ) ^2 U_{El} = \left [ -{{\hbar^2 c^2} \over {2\mu}}{d^2\over {dr^2}} + {{l \left ( l + 1 \right ) \hbar^2 c^2} \over {2\mu r^2}} \right ] U_{El} + \mu ^2 c^4 U_{El} </math>
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