Atomic and Molecular Physics Homework 1

Chasse_neige

Question 1

The eigenfunctions of a Hydrogen are of the form ${\psi }_{nlm}=R_{nl}(r)P_l^m(\cos \theta )e^{im\varphi }$

(a) Write down the explicit forms of ${\psi }_{100}$, ${\psi }_{200}$, ${\psi }_{210}$ and ${\psi }_{21\pm 1}$.

$$R_{nl}(r)=\sqrt{{\left(\frac{2}{n{a}_0}\right)}^3\frac{(n-l-1)!}{2n[(n+l)!]}}{e}^{-\frac{r}{(n{a}_0)}}{\left(\frac{2r}{n{a}_0}\right)}^l{L}_{n-l-1}^{2l+1}\left(\frac{2r}{n{a}_0}\right)$$

Where

$$L_k^{\alpha }(x)=\sum _{j=0}^k(-1{)}^j(\genfrac{}{}{0ex}{}{k+\alpha }{k-j})\frac{x^j}{j!}$$

And

$$Y_{lm}(\theta ,\varphi )=\sqrt{\frac{2l+1}{4\pi }\frac{(l-m)!}{(l+m)!}}{P}_l^m(\cos \theta )e^{im\varphi }$$

Where

$$P_l^m(x)=(-1{)}^m(1-x^2{)}^{\frac{m}{2}}\frac{d^m}{d{x}^m}{P}_l(x)$$

So

$$\begin{array}{c}{\psi }_{100}=R_{10}(r)Y_{00}(\theta ,\varphi )=\sqrt{{\left(\frac{2}{a_0}\right)}^3\frac{1}{2}}{e}^{-\frac{r}{a_0}}{L}_0^1\left(\frac{2r}{a_0}\right)\sqrt{\frac{1}{4\pi }}{P}_0^0(\cos \theta )\\ =\frac{2}{a_0^{\frac{3}{2}}}\sqrt{\frac{1}{4\pi }}{e}^{-\frac{r}{a_0}}\end{array}$$$$\begin{array}{c}{\psi }_{200}=R_{20}(r)Y_{00}(\theta ,\varphi )=\sqrt{{\left(\frac{2}{2a_0}\right)}^3\frac{1}{8}}{e}^{-\frac{r}{2a_0}}{L}_1^1\left(\frac{r}{a_0}\right)\sqrt{\frac{1}{4\pi }}{P}_0^0(\cos \theta )\\ =\frac{1}{2\sqrt{2}{a}_0^{\frac{3}{2}}}\sqrt{\frac{1}{4\pi }}(2-\frac{r}{a_0})e^{-\frac{r}{a_0}}\end{array}$$$$\begin{array}{c}{\psi }_{210}=R_{21}(r)Y_{10}(\theta ,\varphi )=\sqrt{{\frac{1}{a_0}}^3\frac{1}{24}}{e}^{-\frac{r}{2a_0}}\frac{r}{a_0}{L}_0^3\left(\frac{r}{a_0}\right)\sqrt{\frac{3}{4\pi }}{P}_1^0(\cos \theta )\\ =\frac{1}{2\sqrt{2}{a}_0^{\frac{3}{2}}}\sqrt{\frac{1}{4\pi }}\frac{r}{a_0}{e}^{-\frac{r}{2a_0}}\cos \theta \end{array}$$$${\psi }_{21\pm 1}=R_{21}(r)Y_{1\pm 1}(\theta ,\varphi )=\sqrt{{\frac{1}{a_0}}^3\frac{1}{24}}{e}^{-\frac{r}{2a_0}}\frac{r}{a_0}{L}_0^3\left(\frac{r}{a_0}\right)\sqrt{\frac{3}{4\pi }{2}^{\mp 1}}{P}_1^{\pm 1}(\cos \theta )e^{i\pm 1\varphi }$$

Show the two functions respectively

$${\psi }_{211}=-\frac{1}{4a_0^{\frac{3}{2}}}\sqrt{\frac{1}{4\pi }}\frac{r}{a_0}{e}^{-\frac{r}{2a_0}}\sin \theta e^{i\varphi }$$$${\psi }_{21-1}=\frac{1}{4a_0^{\frac{3}{2}}}\sqrt{\frac{1}{4\pi }}\frac{r}{a_0}{e}^{-\frac{r}{2a_0}}\sin \theta e^{-i\varphi }$$

(b) Calculate spatial dependence of the probability current $J=\frac{\hslash }{2\mu i}({\psi }^{\ast }\mathrm{\nabla }\psi -\psi \mathrm{\nabla }{\psi }^{\ast })$ of ${\psi }_{nlm}$. Describe and sketch distribution of $J$ for ${\psi }_{200}$, ${\psi }_{210}$ and ${\psi }_{211}$.

The probability current

$$\begin{array}{c}{J}_{nlm}=\frac{\hslash }{2\mu i}({\psi }_{nlm}^{\ast }\mathrm{\nabla }{\psi }_{nlm}-{\psi }_{nlm}\mathrm{\nabla }{\psi }_{nlm}^{\ast })\\ =\frac{\hslash }{2\mu i}\frac{2im}{r\sin \theta }\left({\psi }_{nl-m}{\psi }_{nlm}\right){\hat{e}}_{\varphi }=\frac{\hslash m}{\mu r\sin \theta }|{\psi }_{nlm}{|}^2{\hat{e}}_{\varphi }\end{array}$$

For ${\psi }_{200}$ and ${\psi }_{210}$

$$J=0$$

For ${\psi }_{211}$

$$J(r,\theta ,\varphi )=\frac{\hslash }{\mu }\frac{r}{64\pi a_0^5}{e}^{-\frac{r}{a_0}}\sin \theta {\hat{e}}_{\varphi }$$

(c) Use the probability current obtained above to show that the z-component of the magnetic moment of ${\psi }_{nlm}$ is given by $M_z=m{\mu }_B$ where ${\mu }_B=\frac{e\hslash }{2\mu }$ is the Bohr magneton.

$$M_z=-\int \frac{e}{2}\overrightarrow{r}\times \overrightarrow{J}dv=-\frac{e\hslash m}{2\mu }\int \frac{|{\psi }_{nlm}{|}^2}{r\sin \theta }r\sin \theta dv=-m{\mu }_B$$

Question 2

Using $H_{hfB}=A\overrightarrow{I}\cdot \overrightarrow{J}+\frac{{\mu }_B}{\hslash }(g_J{J}_z+g_I{I}_z)B_z$, compute the hyperfine structure of the $5S_{1/2}$ ground state of ${}^{87}\text{Rb}$ atom under the influence of an external magnetic field. The relevant parameters are given in the next page.

(a) You should write a Matlab or Mathematica script or any computation script of your preference. Use $|J,m_J⟩|I,m_I⟩$ as your starting basis to obtain the matrix representation of $H_{hfB}$ and then diagonalize this matrix to obtain the eigenenergies and eigenstates. Do not use the Breit-Rabi formula. Submit your results together with 1 page of your program.

Under the $|J,m_J⟩|I,m_I⟩$ basis, the Hamiltonian can be represented as

$$H_{hfB}=A(\frac{I_{+}{J}_{-}+I_{-}{J}_{+}}{2}+I_z{J}_z)+\frac{{\mu }_B}{\hslash }(g_J{J}_z+g_I{I}_z)B_z$$

For ${}^{87}\text{Rb}$ atoms' $5S_{1/2}$ state

$$\begin{array}{c}{m}_J=-\frac{1}{2},\frac{1}{2}\\ m_I=-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}\end{array}$$

Using the basis of

$$\begin{array}{c}|m_J=-\frac{1}{2}⟩\otimes |m_I=-\frac{3}{2}⟩,|m_J=-\frac{1}{2}⟩\otimes |m_I=-\frac{1}{2}⟩\\ |m_J=-\frac{1}{2}⟩\otimes |m_I=\frac{1}{2}⟩,|m_J=-\frac{1}{2}⟩\otimes |m_I=\frac{3}{2}⟩\\ |m_J=\frac{1}{2}⟩\otimes |m_I=-\frac{3}{2}⟩,|m_J=\frac{1}{2}⟩\otimes |m_I=-\frac{1}{2}⟩\\ |m_J=\frac{1}{2}⟩\otimes |m_I=\frac{1}{2}⟩,|m_J=\frac{1}{2}⟩\otimes |m_I=\frac{3}{2}⟩\end{array}$$

The Hamiltonian matrix can be obtained as

$$H_{ij}=A⟨i|\frac{I_{+}{J}_{-}+I_{-}{J}_{+}}{2}+I_z{J}_z|j⟩+\frac{{\mu }_B}{\hslash }(g_J{J}_z+g_I{I}_z)B_z$$

For diagonals, the x and y components of the spin operator is 0, so the Hamiltonian is

$$H_{ii}=A{m}_{J_i}{m}_{I_i}+\frac{{\mu }_B}{\hslash }(g_J{m}_{J_i}+g_I{m}_{I_i})B_z\hslash$$

For non-diagonals,the x and y components of the spin operator is non-zero only if the spin state

$$I_{+}{J}_{-}|j⟩=|i⟩$$

or

$$I_{-}{J}_{+}|j⟩=|i⟩$$

The Hamiltonian therefore equals

$$H_{ij}=\{\begin{array}{l}\frac{A}{2}\sqrt{I(I+1)-(m_{I_j}(m_{I_j}+1))}\sqrt{J(J-1)-(m_{J_j}(m_{J_j}-1))}(I_{+}{J}_{-}|j⟩=|i⟩)\\ \frac{A}{2}\sqrt{I(I+1)-(m_{I_j}(m_{I_j}-1))}\sqrt{J(J+1)-(m_{J_j}(m_{J_j}+1))}(I_{-}{J}_{+}|j⟩=|i⟩)\end{array}$$

The Matlab script for calculating the Hamiltonian is as follows

% 参数
A = 3.417341305452145; % GHz * h
g_J = 2.00233113;
g_I = -0.0009951414;
mu_B = 0.001399624; % GHz/G * h
B_z = 0.001; % G

% 基矢状态
states = [
    -0.5, -1.5;
    -0.5, -0.5;
    -0.5,  0.5;
    -0.5,  1.5;
     0.5, -1.5;
     0.5, -0.5;
     0.5,  0.5;
     0.5,  1.5
];

%哈密顿量
N = 8;
H = zeros(N);

for i = 1:N
    mJ_i = states(i,1);
    mI_i = states(i,2);
    % 对角元
    H(i,i) = A * mJ_i * mI_i + mu_B * B_z * (g_J * mJ_i + g_I * mI_i);
    
    for j = i+1:N
        mJ_j = states(j,1);
        mI_j = states(j,2);
        
        % 检查非对角元条件
        if mJ_i - mJ_j == -1 && mI_i - mI_j == 1 % I_+ J_- 情况
            factor = sqrt( (1/2)*(3/2) - mJ_j*(mJ_j-1) ) * sqrt( (3/2)*(5/2) - mI_j*(mI_j+1) );
            H(i,j) = 1/2 * A * factor;
            H(j,i) = H(i,j);
        elseif mJ_i - mJ_j == 1 && mI_i - mI_j == -1 % I_- J_+ 情况
            factor = sqrt( (1/2)*(3/2) - mJ_j*(mJ_j+1) ) * sqrt( (3/2)*(5/2) - mI_j*(mI_j-1) );
            H(i,j) = 1/2 * A * factor;
            H(j,i) = H(i,j);
        end
    end
end

%对角化
[V, D] = eig(H);
eigenvalues = diag(D);
[eigenvalues_sorted, sort_idx] = sort(eigenvalues);
eigenvectors_sorted = V(:, sort_idx);

The results show that the Matrix representation of the Hamiltonian is (when the magnetic field is 0)

$$H_{B=0}=\left(\begin{array}{cccccccc}2.563& 0& 0& 0& 0& 0& 0& 0\\ 0& 0.854& 0& 0& 2.96& 0& 0& 0\\ 0& 0& -0.854& 0& 0& 3.417& 0& 0\\ 0& 0& 0& -2.563& 0& 0& 2.96& 0\\ 0& 2.96& 0& 0& -2.563& 0& 0& 0\\ 0& 0& 3.417& 0& 0& -0.854& 0& 0\\ 0& 0& 0& 2.96& 0& 0& 0.854& 0\\ 0& 0& 0& 0& 0& 0& 0& 2.563\end{array}\right)$$

The eigenvector matrix is

$$V_{B=0}=\left(\begin{array}{cccccccc}0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& -0.5& 0& 0.866& 0& 0& 0\\ 0& 0.707& 0& 0& 0& -0.707& 0& 0\\ 0.866& 0& 0& 0& 0& 0& -0.5& 0\\ 0& 0& 0.866& 0& 0.5& -0& 0& 0\\ 0& -0.707& 0& 0& 0& -0.707& 0& 0\\ -0.5& 0& 0& 0& 0& -0& -0.866& 0\\ -0& 0& 0& 0& 0& -0& 0& 1\end{array}\right)$$

and the eigenvalue matrix is

$$D_{B=0}=\left(\begin{array}{cccccccc}-4.272& 0& 0& 0& 0& 0& 0& 0\\ 0& -4.272& 0& 0& 0& 0& 0& 0\\ 0& 0& -4.272& 0& 0& 0& 0& 0\\ 0& 0& 0& 2.563& 0& 0& 0& 0\\ 0& 0& 0& 0& 2.563& 0& 0& 0\\ 0& 0& 0& 0& 0& 2.563& 0& 0\\ 0& 0& 0& 0& 0& 0& 2.563& 0\\ 0& 0& 0& 0& 0& 0& 0& 2.563\end{array}\right)$$

(b) When B=0, write down all $|F,m_F⟩$ states in the basis of $|J,m_J⟩|I,m_I⟩$ from your computations. Compare your results to that based on the Clebsch-Gordan coefficients? Compare also the eigenenergies you obtain to the formula given in the lecture notes.

$$|F=1,m_F=1⟩=-0.500|m_J=\frac{1}{2}⟩\otimes |m_I=\frac{1}{2}⟩+0.866|m_J=-\frac{1}{2}⟩\otimes |m_I=\frac{3}{2}⟩$$$$|F=1,m_F=0⟩=-0.707|m_J=\frac{1}{2}⟩\otimes |m_I=-\frac{1}{2}⟩+0.707|m_J=-\frac{1}{2}⟩\otimes |m_I=\frac{1}{2}⟩$$$$|F=1,m_F=-1⟩=0.866|m_J=\frac{1}{2}⟩\otimes |m_I=-\frac{3}{2}⟩-0.500|m_J=-\frac{1}{2}⟩\otimes |m_I=-\frac{1}{2}⟩$$$$|F=2,m_F=-2⟩=|m_J=-\frac{1}{2}⟩\otimes |m_I=-\frac{3}{2}⟩$$$$|F=2,m_F=-1⟩=0.866|m_J=-\frac{1}{2}⟩\otimes |m_I=-\frac{1}{2}⟩+0.500|m_J=\frac{1}{2}⟩\otimes |m_I=-\frac{3}{2}⟩$$$$|F=2,m_F=0⟩=-0.707|m_J=\frac{1}{2}⟩\otimes |m_I=-\frac{1}{2}⟩-0.707|m_J=-\frac{1}{2}⟩\otimes |m_I=\frac{1}{2}⟩$$$$|F=2,m_F=1⟩=-0.866|m_J=\frac{1}{2}⟩\otimes |m_I=\frac{3}{2}⟩-0.500|m_J=\frac{1}{2}⟩\otimes |m_I=\frac{1}{2}⟩$$$$|F=2,m_F=2⟩=|m_J=\frac{1}{2}⟩\otimes |m_I=\frac{3}{2}⟩$$

For the results given by the Clebsch-Gordan coefficients,

We can see from the table that

$$|F=1,m_F=1⟩=\frac{\sqrt{3}}{2}|m_J=-\frac{1}{2}⟩\otimes |m_I=\frac{3}{2}⟩+\frac{1}{2}|m_J=\frac{1}{2}⟩\otimes |m_I=\frac{1}{2}⟩$$$$|F=1,m_F=0⟩=-\frac{1}{\sqrt{2}}|m_J=\frac{1}{2}⟩\otimes |m_I=-\frac{1}{2}⟩+\frac{1}{\sqrt{2}}|m_J=-\frac{1}{2}⟩\otimes |m_I=\frac{1}{2}⟩$$$$|F=1,m_F=-1⟩=-\frac{\sqrt{3}}{2}|m_J=\frac{1}{2}⟩\otimes |m_I=-\frac{3}{2}⟩+\frac{1}{2}|m_J=-\frac{1}{2}⟩\otimes |m_I=-\frac{1}{2}⟩$$$$|F=2,m_F=-2⟩=|m_J=-\frac{1}{2}⟩\otimes |m_I=-\frac{3}{2}⟩$$$$|F=2,m_F=-1⟩=\frac{\sqrt{3}}{2}|m_J=-\frac{1}{2}⟩\otimes |m_I=-\frac{1}{2}⟩+\frac{1}{2}|m_J=\frac{1}{2}⟩\otimes |m_I=-\frac{3}{2}⟩$$$$|F=2,m_F=0⟩=\frac{1}{\sqrt{2}}|m_J=\frac{1}{2}⟩\otimes |m_I=-\frac{1}{2}⟩+\frac{\sqrt{2}}{2}|m_J=-\frac{1}{2}⟩\otimes |m_I=\frac{1}{2}⟩$$$$|F=2,m_F=1⟩=-\frac{\sqrt{3}}{2}|m_J=\frac{1}{2}⟩\otimes |m_I=\frac{3}{2}⟩-\frac{1}{2}|m_J=\frac{1}{2}⟩\otimes |m_I=\frac{1}{2}⟩$$$$|F=2,m_F=2⟩=|m_J=\frac{1}{2}⟩\otimes |m_I=\frac{3}{2}⟩$$

are exactly the same as those we've calculated apart from some signs of the whole eigenvectors.

For the eigenenergies when $B=0$, the script gives that there are two possible values

$$E_{-}=h\cdot -4.272GHz,E_{+}=h\cdot 2.563GHz$$

The formula given in the lecture notes show that

$$H=AI\cdot J=A\frac{F(F+1)-I(I+1)-J(J+1)}{2}$$

When $F=1$

$$H_{-}=A\frac{2-\frac{15}{4}-\frac{3}{4}}{2}=-1.25A=h\cdot -4.272GHz\approx E_{-}$$

When $F=2$

$$H_{+}=A\frac{6-\frac{15}{4}-\frac{3}{4}}{2}=0.75A=h\cdot 2.563GHz\approx E_{+}$$

These show that our calculations are corresponding to the formula given in class.

Hint for Question 2:

$$\overrightarrow{I}\cdot \overrightarrow{J}=\frac{I_{+}{J}_{-}+I_{-}{J}_{+}}{2}+I_z{J}_z$$