hw #4

Q1: 7.12

For an in-depth discussion of block diagonal matrices see notes for week 4.

a)

The unperturbed hamiltonian matrix is obtained by setting $\epsilon\to 0$ gives us a diagonal matrix. An eigenbasis of any diagonal matrix $A_{ij}$ can be given by the standard basis $\hat{e}_n$ whose $m^{\text{th}}$ coefficient is $\delta_{nm}$ and eigenvalue is $A_{nn}$.

b)

Recognizing the matrix as block diagonal, we reduce our problem to find the eigenvectors / eigenvalues of the blocks $\begin{bmatrix}1-\epsilon\end{bmatrix}$ and $\begin{bmatrix} 1 & \epsilon \\ \epsilon & 2 \end{bmatrix}$. The first matrix obviously has one eigenvalue equal to $1 - \epsilon$ and an eigenvector of $\begin{bmatrix}1\end{bmatrix}$. For the second we solve the characteristic polynomial: $$ (1-\lambda)(2-\lambda) - \epsilon^2 = 0 $$ which has solutions $$ \lambda = \frac{3}{2} \pm \frac{1}{2}\sqrt{1 + 4 \epsilon^2} \approx \frac{3}{2} \pm \frac{1}{2}\left( 1 + \frac{1}{2} 4 \epsilon^2 \right) = \frac{1}{2} \left( 3 \pm \left( 1 + 2 \epsilon^2 \right) \right) = \begin{array}{l} 2 + \epsilon^2 \\ 1 - \epsilon^2 \end{array} $$ with corresponding (non-normalized) eigenvectors $\begin{bmatrix} \epsilon \\ \frac{1}{2} \left( 1 \pm \sqrt{1 + 4\epsilon^2} \right) \end{bmatrix}$

c)

The rules of non-degenerate perturbation theory stipulate that given any eigenbasis $\vec{v}_n$ of $H^o$ with eigenvalues $E^o_n$ we can compute the first order perturbations $\Delta E^{(1)}_n$ to the energy levels according to the rule: $$ \Delta E^{(1)}_n = \vec{v}_n^{\intercal} V' \vec{v}_n $$ where $V'$ is the perturbation matrix, and the second order perturbations $\Delta E^{(2)}_n$ from: $$ \Delta E^{(2)}_n = \sum_{m \neq n}\frac{|\vec{v}_m^{\intercal} V' \vec{v}_n|^2}{E^o_n-E^o_m} $$ Using our $\hat{e}_n$ as such an eigenbasis and obtaining the pertubation matrix $\begin{bmatrix} -\epsilon & 0 & 0 \\ 0 & 0 & \epsilon \\ 0 & \epsilon & 0 \end{bmatrix}$ from the difference $H(\epsilon) - H(\epsilon \to 0)$, we find first order corrections given by $-\epsilon, 0, 0$. For the second order corrections we note that the correction for the eigenstate $\hat{e}_1$ is zero since $V'_{1j}=0$ for $j\neq 1$. To second order the corrections to the second and third eigenstates will be equal and opposite since they only contain off diagonal elements amongst themselves, so that: $$ \Delta E^{(2)}_3 = -\Delta E^{(2)}_2 = \frac{\epsilon^2}{2-1} = \epsilon^2 $$ These are in perfect agreement with the second-order expansion of the exact eigenvalues from part a).

d)

The rules of degenerate perturbation theory require us to first diagonalize the degenerate subspaces of eigenvectors. The $E^o = 2$ energy level is non-degenerate so its first-order correction is the same as obtained in part b), namely zero. The $E^o = 1$ energy level is however degenerate and spanned by the eigenvectors $\hat{e}_1$ and $\hat{e}_2$. The perturbation matrix in this subspace is $\begin{bmatrix} -\epsilon & 0 \\ 0 & 0 \end{bmatrix}$. This matrix is already in diagonal form (they should include a fourth sense in which Jupiter has brought us good fortune!), so the original $\hat{e}_n$ all happen to be "good eigenstates" as the term is applied to degenerate perturbation theory. The first order corrections are thus all the same as in part c), and our work is done.

Q2: 7.15

A good thing to have memorized is the schrodinger equation for the hydrogen atom in atomic units: $$ -\frac{1}{2}\psi'' - \frac{1}{r} \psi = \varepsilon_n \psi $$ Wheren $\varepsilon_n \equiv - \frac{1}{2} \frac{1}{n^2}$, with energy in units of the hartree ($\equiv 2 |E_1| \approx 27.2\text{eV}$) and the length in units of the bohr radius $a_o$. So our task is equivalent to finding the negative of the expectation value of the potential energy, which by the virial theorem is simply $2 \varepsilon_n$, i.e. $$ \ev{\frac{1}{r}} = - \ev{v(r)} = - 2 \varepsilon_n = - 2 ( - \frac{1}{2} \frac{1}{n^2} ) = \frac{1}{n^2} $$ See how by working in natural units we avoid all the messy constants. Note the answer here is in units of inverse bohr radii.

Q3: 7.20

Note that the 3 on the RHS of 7.58 accounts for the 3 in the first term in the sum on the RHS of 7.68, so it only remains to show, in units of $\frac{n E_n^2}{mc^2}$, that the remaining term in the fine structure formula is $-\frac{2}{j+1/2}$.

Take the case where $j = l + 1 / 2$. The relevant term in the relativistic correction is then $$ -\frac{2}{j} = \frac{1/2 - 2j^2}{(j-1/2)j(j+1/2)}$$ and the corresponding term for the spin orbit correction is $$ \frac{j(j+1) - (j-1/2)(j+1/2) - 3/4}{(j-1/2)j(j+1/2)} = \frac{j-1/2}{(j-1/2)j(j+1/2)} $$ putting the two together we get $$ \frac{1/2 - 2j^2 + j-1/2}{(j-1/2)j(j+1/2)} = -2 \frac{j(j-1/2)}{(j-1/2)j(j+1/2)} = -\frac{2}{j+1/2} $$ we will trust Griffiths that the $j = l - 1/2$ case works similarly.