Midterm Solution

Problem 1: Angular momentum addition \(\textcolor{red}{\boxed{\text{14 pt}}}\)¶

We have two particles:

Particle 1: \(j_1 = S > 0\), with \(m_1 \in \{-S, -S+1, \dots, S\}\).
Particle 2: \(j_2 = \tfrac{1}{2}\), with \(m_2 \in \{\tfrac{1}{2}, -\tfrac{1}{2}\}\).

Using coupled eigenstates of \(\hat J_{\mathrm{tot}}^2\) and \(\hat J_{\mathrm{tot},z}\), denoted \(\ket{J,M}\), where \(\mathbf{J}_{\mathrm{tot}} = \mathbf{J}_1 + \mathbf{J}_2\),

and the ladder-operator action (from the formula sheet):

\[ J_\pm \ket{j,m} = \hbar\sqrt{j(j+1) - m(m\pm 1)}\,\ket{j,m\pm 1}. \]

(a) Possible values of \(J\) and \(M\)¶

When adding \(j_1=S\) and \(j_2=\tfrac{1}{2}\), the possible total angular momentum quantum numbers are

\[ J \in \left\{S+\tfrac{1}{2},\; S-\tfrac{1}{2}\right\}. \quad\textcolor{red}{\boxed{\text{1 pt for }J}} \]

For a given \(J\), the allowed \(M\) values are

\[ M = -J, -J+1, \dots, J-1, J. \quad\textcolor{red}{\boxed{\text{1 pt for }M}} \]

Equivalently, since \(M=m_1+m_2\), the full set of allowed \(M\) values over the product basis is all integer steps from \(-(S+\tfrac{1}{2})\) to \(S+\tfrac{1}{2}\).

(Consistency check: the product basis has \((2S+1)\cdot 2 = 4S+2\) states; the coupled basis has \((2(S+\tfrac{1}{2})+1) + (2(S-\tfrac{1}{2})+1) = (2S+2) + (2S) = 4S+2\) states.)

(b) Highest-\(J\), highest-\(M\) state in the product basis¶

Here

\[ J_{\max} = S+\tfrac{1}{2},\qquad M_{\max} = J_{\max} = S+\tfrac{1}{2}. \]

To get \(M=S+\tfrac{1}{2}\), we must have \(m_1=S\) and \(m_2=\tfrac{1}{2}\). There is only one product basis state with that \(M\), so (up to a global phase) the coupled state must equal that product state:

\[ \textcolor{red}{\boxed{\text{1 pt for stating that there is only one product state which has the correct }M}} \tag*{} \]

\[ \boxed{\ket{J_{\max},M_{\max}} = \ket{S,S}\otimes\ket{\tfrac{1}{2},\tfrac{1}{2}}.} \quad\textcolor{red}{\boxed{\text{1 pt for correct expression}}} \]

(c) State with \(J=J_{\max}\) and \(M=M_{\max}-1\)¶

We want \(\ket{J_{\max}, M_{\max}-1} = \ket{S+\tfrac{1}{2},\, S-\tfrac{1}{2}}\).

Use the total lowering operator

\[ J_-^{(\mathrm{tot})} = J_{1,-} + J_{2,-}. \]

Apply it to the highest-weight state from part (b):

\[ J_-^{(\mathrm{tot})}\Big(\ket{S,S}\otimes\ket{\tfrac{1}{2},\tfrac{1}{2}}\Big) = \big(J_{1,-}\ket{S,S}\big)\otimes\ket{\tfrac{1}{2},\tfrac{1}{2}} + \ket{S,S}\otimes\big(J_{2,-}\ket{\tfrac{1}{2},\tfrac{1}{2}}\big). \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Using the ladder formula on each factor,

\[ J_{1,-}\ket{S,S} = \hbar\sqrt{S(S+1)-S(S-1)}\,\ket{S,S-1} = \hbar\sqrt{2S}\,\ket{S,S-1}, \]

\[ J_{2,-}\ket{\tfrac{1}{2},\tfrac{1}{2}} = \hbar\sqrt{\tfrac{1}{2}(\tfrac{1}{2}+1)-\tfrac{1}{2}(-\tfrac{1}{2})}\,\ket{\tfrac{1}{2},-\tfrac{1}{2}} = \hbar\,\ket{\tfrac{1}{2},-\tfrac{1}{2}}, \]

so

\[ J_-^{(\mathrm{tot})}\ket{J_{\max},M_{\max}} = \hbar\Big(\sqrt{2S}\,\ket{S,S-1}\otimes\ket{\tfrac{1}{2},\tfrac{1}{2}} + \ket{S,S}\otimes\ket{\tfrac{1}{2},-\tfrac{1}{2}}\Big). \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

On the other hand, within a fixed-\(J\) multiplet,

\[ J_-^{(\mathrm{tot})}\ket{J,M} = \hbar\sqrt{J(J+1) - M(M-1)}\,\ket{J,M-1}. \]

For \(J=J_{\max}\) and \(M=M_{\max}=J_{\max}\), this factor becomes

\[ \sqrt{J(J+1) - J(J-1)} = \sqrt{2J} = \sqrt{2S+1}. \quad\textcolor{red}{\boxed{\text{1 pt, also fine to use normalization}}} \]

Thus

\[ \hbar\sqrt{2S+1}\,\ket{J_{\max},M_{\max}-1} = \hbar\Big(\sqrt{2S}\,\ket{S,S-1}\otimes\ket{\tfrac{1}{2},\tfrac{1}{2}} + \ket{S,S}\otimes\ket{\tfrac{1}{2},-\tfrac{1}{2}}\Big), \]

and dividing by \(\hbar\sqrt{2S+1}\) gives

\[ \boxed{\ket{S+\tfrac{1}{2},\,S-\tfrac{1}{2}} = \sqrt{\frac{2S}{2S+1}}\,\ket{S,S-1}\otimes\ket{\tfrac{1}{2},\tfrac{1}{2}} + \sqrt{\frac{1}{2S+1}}\,\ket{S,S}\otimes\ket{\tfrac{1}{2},-\tfrac{1}{2}}.} \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

(d) Measure particle 1’s \(z\)-angular momentum: outcomes and probabilities¶

The state in part (c) is a superposition of two orthonormal product states with definite \(m_1\).

Term 1 has \(m_1=S-1\) with amplitude \(\sqrt{2S}/\sqrt{2S+1}\).
Term 2 has \(m_1=S\) with amplitude \(1/\sqrt{2S+1}\).

The possible outcomes for measuring \(J_{1,z}\) are

\[ J_{1,z} = \hbar(S-1)\quad\text{or}\quad J_{1,z} = \hbar S, \]

with probabilities given by the squared amplitudes:

\[ \boxed{P(m_1=S-1) = \frac{2S}{2S+1},\qquad P(m_1=S) = \frac{1}{2S+1}.} \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

(e) Entanglement vs \(S\) for the state in part (d)¶

The state from part (c) is

\[ \ket{\psi} = \sqrt{\frac{2S}{2S+1}}\,\underbrace{\ket{S,S-1}}_{\text{particle 1}}\underbrace{\ket{\tfrac{1}{2},\tfrac{1}{2}}}_{\text{particle 2}} + \sqrt{\frac{1}{2S+1}}\,\ket{S,S}\otimes\ket{\tfrac{1}{2},-\tfrac{1}{2}}. \]

This is already in Schmidt form with Schmidt weights

\[ \lambda_1 = \frac{2S}{2S+1},\qquad \lambda_2 = \frac{1}{2S+1}. \]

As \(S\) increases, \(\lambda_2 = 1/(2S+1)\) decreases monotonically to \(0\), so the state becomes less entangled and approaches a product state.

\[ \textcolor{red}{\boxed{ \begin{array}{l} \text{2 pt, fine to just construct reduced density matrix for particle 1,}\\ \text{ then argue that is more pure / more "certain" when S is large} \end{array}} }\tag*{} \]

For example, the von Neumann entanglement entropy is

\[ S_{\mathrm{vN}} = -\lambda_1\ln\lambda_1 - \lambda_2\ln\lambda_2, \]

which decreases with \(S\) and tends to \(0\) as \(S\to\infty\). For large \(S\) (so \(\lambda_2\ll 1\)), it scales like \(S_{\mathrm{vN}}\sim -\lambda_2\ln\lambda_2 \sim (\ln S)/S\) up to constants.

(f) State with \(J=J_{\max}-1\) and largest compatible \(M\)¶

Here

\[ J_{\max}-1 = (S+\tfrac{1}{2}) - 1 = S-\tfrac{1}{2}, \]

and the largest allowed \(M\) in that multiplet is \(M=J=S-\tfrac{1}{2}\). So we want \(\ket{S-\tfrac{1}{2},\,S-\tfrac{1}{2}}\).

In the product basis, the subspace with total \(M=S-\tfrac{1}{2}\) is spanned by

\[ \ket{A} = \ket{S,S-1}\otimes\ket{\tfrac{1}{2},\tfrac{1}{2}},\qquad \ket{B} = \ket{S,S}\otimes\ket{\tfrac{1}{2},-\tfrac{1}{2}}. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

From part (c), the \(J=S+\tfrac{1}{2}\) state in this subspace is

\[ \ket{S+\tfrac{1}{2},\,S-\tfrac{1}{2}} = \frac{\sqrt{2S}\,\ket{A} + \ket{B}}{\sqrt{2S+1}}. \]

The desired \(J=S-\tfrac{1}{2}\) state is the normalized orthogonal vector in the same two-dimensional subspace \(\textcolor{red}{\boxed{\text{1 pt}}}\) (equivalently, the highest-weight state of the lower-\(J\) multiplet, annihilated by \(J_+^{(\mathrm{tot})}\)). A convenient choice is

\[ \boxed{\ket{S-\tfrac{1}{2},\,S-\tfrac{1}{2}} = \frac{\ket{A} - \sqrt{2S}\,\ket{B}}{\sqrt{2S+1}} = \frac{\ket{S,S-1}\otimes\ket{\tfrac{1}{2},\tfrac{1}{2}} - \sqrt{2S}\,\ket{S,S}\otimes\ket{\tfrac{1}{2},-\tfrac{1}{2}}}{\sqrt{2S+1}}} \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

(up to an overall global sign).

Quick check: applying \(J_+^{(\mathrm{tot})}\) to the numerator cancels because \(J_{1,+}\ket{S,S-1} \propto \sqrt{2S}\,\ket{S,S}\) and \(J_{2,+}\ket{\tfrac{1}{2},-\tfrac{1}{2}} \propto \ket{\tfrac{1}{2},\tfrac{1}{2}}\).

Problem 2: Wigner-Eckart theorem \(\textcolor{red}{\boxed{\text{16 pt}}}\)¶

(a) Meaning of each factor on the right-hand side¶

We use the Wigner–Eckart theorem in the form (consistent with Shankar's and Lecture Note's)

\[ \braket{\alpha_2 j_2 m_2|\hat T_k^{q}|\alpha_1 j_1 m_1} = \braket{\alpha_2 j_2\Vert \hat T_k\Vert \alpha_1 j_1}\; \braket{j_2 m_2|k q,\, j_1 m_1}. \]

The factor \(\braket{\alpha_2 j_2\Vert \hat T_k\Vert \alpha_1 j_1}\) is the reduced (double-bar) matrix element. It contains the dynamics (radial integrals / internal structure) and is independent of \(m_1,m_2,q\).

The factor \(\braket{j_2 m_2|k q,\, j_1 m_1}\) is a Clebsch–Gordan coefficient. It contains all angular-momentum geometry and enforces selection rules such as \(m_2=m_1+q\) and \(|j_1-k|\le j_2\le j_1+k\).

\[ \bbox[border:1px solid red; padding:6px]{ \color{red}{ \begin{array}{l} \text{- [1 pt] 2nd factor: CG factor for adding angular momentum }(k,q)\text{ to }(j_1,m_1)\\ \hspace{0.38em}\text{ [0.5 pt}\times\text{3] INDEPENDENT OF form of }T,\ m\ \text{and alphas}\\ \text{- [1.5 pt] 1st factor: depends on form of }T\text{ and alpha, but NO DEPENDENCE on }m_1,\ m_2 \end{array} }} \tag*{} \]

(b) Reduced matrix element for the angular momentum operator¶

The angular momentum operator defines a rank-\(k=1\) spherical tensor. Use the \(q=0\) component, which corresponds to \(J_z\) in the standard spherical basis. The Wigner–Eckart theorem gives

\[ \braket{\alpha_2 j_2 m_2|J_z|\alpha_1 j_1 m_1} = \braket{\alpha_2 j_2\Vert \hat J_1\Vert \alpha_1 j_1}\; \braket{j_2 m_2|1\,0,\, j_1 m_1}. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

We are told the Clebsch–Gordan coefficient

\[ \braket{j\,j|1\,0,\, j\,j}=\lambda. \]

Choose \(j_2=j_1=j\) and \(m_2=m_1=j\) \(\;\textcolor{red}{\boxed{\text{1 pt}}}\). Then the left-hand side is

\[ \braket{\alpha_2 j j|J_z|\alpha_1 j j} = \hbar j\,\delta_{\alpha_2\alpha_1}. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Equating both sides gives

\[ \hbar j\,\delta_{\alpha_2\alpha_1} = \braket{\alpha_2 j\Vert \hat J_1\Vert \alpha_1 j}\;\lambda, \]

so

\[ \boxed{ \braket{\alpha_2 j_2\Vert \hat J_1\Vert \alpha_1 j_1} = \delta_{j_2 j_1}\,\delta_{\alpha_2\alpha_1}\,\frac{\hbar j_1}{\lambda}. } \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

(c) Show that \(\hat Q=\vec J\cdot\vec B\) is a \(k=0\) tensor operator¶

If \(\vec B\) is a vector operator, then under a rotation \(R\) with unitary \(U(R)\) we have

\[ U(R)\,J_i\,U^\dagger(R)=\sum_j R_{ij} J_j, \qquad U(R)\,B_i\,U^\dagger(R)=\sum_j R_{ij} B_j. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Consider \(\hat Q=\sum_i J_i B_i\). Under rotation,

\[ U(R)\,\hat Q\,U^\dagger(R) = \sum_i \Big(U(R)J_iU^\dagger(R)\Big)\Big(U(R)B_iU^\dagger(R)\Big) = \sum_{i,j,k} R_{ij}R_{ik}\,J_j B_k. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Using orthogonality of \(R\) (i.e. \(\sum_i R_{ij}R_{ik}=\delta_{jk}\)),

\[ U(R)\,\hat Q\,U^\dagger(R) = \sum_{j,k}\delta_{jk} J_j B_k = \sum_j J_j B_j = \hat Q. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

So \(\hat Q\) is invariant under rotations (a scalar operator). Therefore \(\hat Q\) transforms as a rank-\(k=0\) spherical tensor operator.

(d) Counting nonzero matrix elements for the \(n=3\) Stark problem¶

In a weak uniform electric field along \(z\), the perturbation is proportional to the position operator \(z\), i.e. \(\hat W \propto z\). The operator \(z\) is the \(q=0\) component of a rank-\(k=1\) tensor operator, so the angular selection rules inside a fixed-\(n\) manifold are

\[ \Delta m = 0, \qquad \Delta \ell = \pm 1. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

For \(n=3\) (ignoring spin), the degenerate subspace has \(3^2=9\) states: \(\quad\textcolor{red}{\boxed{\text{1 pt}}}\)

\(\ell=0\): one state (\(m=0\)),
\(\ell=1\): three states (\(m=-1,0,1\)),
\(\ell=2\): five states (\(m=-2,-1,0,1,2\)).

Within this \(9\times 9\) matrix, only \(\ell\leftrightarrow \ell\pm 1\) with the same \(m\) can be nonzero, and all diagonal entries vanish by parity.

The allowed couplings are: \(\quad\textcolor{red}{\boxed{\text{1 pt}}}\)

\(\ket{3,\ell=0,m=0}\leftrightarrow \ket{3,\ell=1,m=0}\) (one pair),
\(\ket{3,\ell=1,m}\leftrightarrow \ket{3,\ell=2,m}\) for \(m=-1,0,1\) (three pairs).

Thus there are \(1+3=4\) independent off-diagonal couplings; by Hermiticity each produces two nonzero matrix entries, so the perturbation matrix has

\[ \boxed{8\ \text{nonzero entries in the }9\times 9\text{ matrix}.} \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

For the reduced (double-bar) matrix elements: by Wigner–Eckart, all \(m\)-dependence is carried by Clebsch–Gordan coefficients, so we only need one reduced matrix element for each distinct \((\ell,\ell')\) channel within \(n=3\):

\(\braket{3,\ell=1\Vert z\Vert 3,\ell=0}\),
\(\braket{3,\ell=2\Vert z\Vert 3,\ell=1}\).

Hence,

\[ \boxed{2\ \text{distinct reduced matrix elements are needed.}} \quad\textcolor{red}{\boxed{\text{2 pt}}} \]

Problem 3: Non-degenerate perturbation theory \(\textcolor{red}{\boxed{\text{8 pt}}}\)¶

We consider

\[ \hat H=\hat H_0+\hat W, \]

where the eigenstates and eigenvalues of \(\hat H_0\) are known:

\[ \hat H_0\ket{n}=E_n^{(0)}\ket{n}, \qquad \braket{m|n}=\delta_{mn}. \]

To organize perturbation theory, introduce a dimensionless parameter \(\lambda\) and write

\[ \hat H(\lambda)=\hat H_0+\lambda \hat W, \]

and expand the exact eigenpair connected to \(\ket{n}\) as

\[ \ket{\psi_n(\lambda)}=\ket{n}+\lambda\ket{n^{(1)}}+O(\lambda^2), \qquad E_n(\lambda)=E_n^{(0)}+\lambda E_n^{(1)}+O(\lambda^2). \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

We will use the standard phase/normalization choice (intermediate normalization)

\[ \braket{n|\psi_n(\lambda)}=1 \quad\Rightarrow\quad \braket{n|n^{(1)}}=0. \]

(a) First-order shift in the ground-state energy¶

Start from the time-independent Schrödinger equation

\[ \hat H(\lambda)\ket{\psi_1(\lambda)}=E_1(\lambda)\ket{\psi_1(\lambda)}. \]

Insert the expansions and collect terms order-by-order in \(\lambda\).

Order \(\lambda^0\) gives

\[ \hat H_0\ket{1}=E_1^{(0)}\ket{1}. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Order \(\lambda^1\) gives

\[ \hat H_0\ket{1^{(1)}}+\hat W\ket{1} = E_1^{(0)}\ket{1^{(1)}}+E_1^{(1)}\ket{1}. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Rearrange:

\[ (\hat H_0-E_1^{(0)})\ket{1^{(1)}}+\hat W\ket{1}=E_1^{(1)}\ket{1}. \]

Now eft-multiply by \(\bra{1}\) \(\;\textcolor{red}{\boxed{\text{1 pt}}}\). Since \(\bra{1}(\hat H_0-E_1^{(0)})=0\), we obtain

\[ E_1^{(1)}=\bra{1}\hat W\ket{1}. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

(b) First-order shift in the ground-state wavefunction¶

Start again from the order-\(\lambda^1\) equation

\[ (\hat H_0-E_1^{(0)})\ket{1^{(1)}}+\hat W\ket{1}=E_1^{(1)}\ket{1}. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Project onto an excited unperturbed eigenstate \(\;\textcolor{red}{\boxed{\text{1 pt}}}\) \(\bra{m}\) with \(m\neq 1\):

\[ \bra{m}(\hat H_0-E_1^{(0)})\ket{1^{(1)}}+\bra{m}\hat W\ket{1} = E_1^{(1)}\braket{m|1}. \]

The right-hand side vanishes because \(\braket{m|1}=0\) for \(m\neq 1\). Using \(\hat H_0\ket{m}=E_m^{(0)}\ket{m}\) gives

\[ (E_m^{(0)}-E_1^{(0)})\braket{m|1^{(1)}}+\bra{m}\hat W\ket{1}=0, \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

so

\[ \braket{m|1^{(1)}}=\frac{\bra{m}\hat W\ket{1}}{E_1^{(0)}-E_m^{(0)}} \quad (m\neq 1). \]

Expanding \(\ket{1^{(1)}}\) in the \(\{\ket{m}\}\) basis and using \(\braket{1|1^{(1)}}=0\) yields

\[ \boxed{ \ket{1^{(1)}} = \sum_{m\neq 1} \ket{m}\, \frac{\bra{m}\hat W\ket{1}}{E_1^{(0)}-E_m^{(0)}}. } \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Thus, to first order,

\[ \ket{\psi_1} = \ket{1} + \sum_{m\neq 1} \ket{m}\, \frac{\bra{m}\hat W\ket{1}}{E_1^{(0)}-E_m^{(0)}} +O(\hat W^2). \]

(c) Upper bound on the magnitude of the second-order energy shift (Serves as additional reading)¶

Consider the 2nd order energy shift of a general energy eigenstate \(n\) (and again assume no degeneracies). Starting with the general expression for this 2nd order shift, show that the magnitude of this shift can be upper bounded by an expression involving the uncertainty of \(\hat{W}\) in the unperturbed energy level \(n\). What other quantity appears in this bound?

For a non-degenerate level \(n\), the second-order energy shift is

\[ E_n^{(2)} = \sum_{m\neq n} \frac{\left|\bra{m}\hat W\ket{n}\right|^2}{E_n^{(0)}-E_m^{(0)}}. \]

Define the minimum level spacing from level \(n\) to any other unperturbed level:

\[ \Delta_n \equiv \min_{m\neq n}\left|E_m^{(0)}-E_n^{(0)}\right|. \]

Then

\[ \left|E_n^{(2)}\right| \le \sum_{m\neq n} \frac{\left|\bra{m}\hat W\ket{n}\right|^2}{\left|E_n^{(0)}-E_m^{(0)}\right|} \le \frac{1}{\Delta_n} \sum_{m\neq n} \left|\bra{m}\hat W\ket{n}\right|^2. \]

Use completeness to rewrite the numerator sum:

\[ \bra{n}\hat W^2\ket{n} = \sum_m \bra{n}\hat W\ket{m}\bra{m}\hat W\ket{n} = \sum_m \left|\bra{m}\hat W\ket{n}\right|^2. \]

Hence

\[ \sum_{m\neq n}\left|\bra{m}\hat W\ket{n}\right|^2 = \bra{n}\hat W^2\ket{n}-\left|\bra{n}\hat W\ket{n}\right|^2 \equiv (\Delta W)_n^2, \]

where \((\Delta W)_n\) is the uncertainty of \(\hat W\) in the unperturbed eigenstate \(\ket{n}\).

Therefore,

\[ \boxed{ \left|E_n^{(2)}\right| \le \frac{(\Delta W)_n^2}{\Delta_n}. } \]

The “other quantity” appearing in the bound is \(\Delta_n\), i.e. the minimum unperturbed energy spacing (spectral gap) between level \(n\) and all other levels.

Problem 4: Degenerate perturbation theory \(\textcolor{red}{\boxed{\text{8 pt}}}\)¶

We have two harmonic oscillators with lowering operators \(\hat a,\hat b\) and

\[ \hat H=\hat H_0+\hat W, \qquad \hat H_0=\omega_a\left(\hat a^\dagger \hat a+\hat b^\dagger \hat b\right), \qquad \hat W=g\left(\hat a^\dagger \hat b+\hat b^\dagger \hat a\right). \]

Let \(\ket{n_a,n_b}\) denote the simultaneous eigenkets of \(\hat a^\dagger\hat a\) and \(\hat b^\dagger\hat b\):

\[ \hat a^\dagger \hat a\ket{n_a,n_b}=n_a\ket{n_a,n_b}, \qquad \hat b^\dagger \hat b\ket{n_a,n_b}=n_b\ket{n_a,n_b}. \]

We use the ladder action

\[ \hat a\ket{n_a,n_b}=\sqrt{n_a}\ket{n_a-1,n_b}, \quad \hat a^\dagger\ket{n_a,n_b}=\sqrt{n_a+1}\ket{n_a+1,n_b}, \]

\[ \hat b\ket{n_a,n_b}=\sqrt{n_b}\ket{n_a,n_b-1}, \quad \hat b^\dagger\ket{n_a,n_b}=\sqrt{n_b+1}\ket{n_a,n_b+1}. \]

(a) Lowest three energy eigenvalues of \(\hat H_0\) and degeneracies¶

Since

\[ \hat H_0\ket{n_a,n_b}=\omega_a(n_a+n_b)\ket{n_a,n_b}, \quad\textcolor{red}{\boxed{\text{0.5 pt}}} \]

the unperturbed energies depend only on the total excitation number \(N=n_a+n_b\).

Lowest level: \(N=0\) only gives \(\ket{0,0}\), so

\[ E_1=0, \qquad \text{degeneracy}=1. \quad\textcolor{red}{\boxed{\text{0.5 pt}}} \]

Next level: \(N=1\) gives \(\ket{1,0}\) and \(\ket{0,1}\), so

\[ E_2=\omega_a, \qquad \text{degeneracy}=2. \quad\textcolor{red}{\boxed{\text{0.5 pt}}} \]

Third level: \(N=2\) gives \(\ket{2,0},\ket{1,1},\ket{0,2}\), so

\[ E_3=2\omega_a, \qquad \text{degeneracy}=3. \quad\textcolor{red}{\boxed{\text{0.5 pt}}} \]

(b) First-order energy shifts of the \(E_2\) manifold¶

The \(E_2\) subspace is spanned by

\[ \ket{1,0},\qquad \ket{0,1}. \]

In degenerate perturbation theory, the first-order corrected energies are obtained by diagonalizing the perturbation \(\hat W\) restricted to this degenerate subspace, i.e. the \(2\times 2\) matrix with elements

\[ W_{ij}=\bra{i}\hat W\ket{j}, \qquad \ket{i},\ket{j}\in\{\ket{1,0},\ket{0,1}\}. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Compute the action of \(\hat W\) on the basis states.

First,

\[ \hat W\ket{1,0} = g\left(\hat a^\dagger \hat b+\hat b^\dagger \hat a\right)\ket{1,0}. \]

Since \(\hat b\ket{1,0}=0\), we have \(\hat a^\dagger \hat b\ket{1,0}=0\). For the other term,

\[ \hat b^\dagger \hat a\ket{1,0} = \hat b^\dagger\left(\sqrt{1}\ket{0,0}\right) = \sqrt{1}\sqrt{0+1}\ket{0,1} = \ket{0,1}. \]

Hence,

\[ \hat W\ket{1,0}=g\ket{0,1}. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Similarly,

\[ \hat W\ket{0,1} = g\left(\hat a^\dagger \hat b+\hat b^\dagger \hat a\right)\ket{0,1}. \]

Now \(\hat a\ket{0,1}=0\), so \(\hat b^\dagger \hat a\ket{0,1}=0\). For the first term,

\[ \hat a^\dagger \hat b\ket{0,1} = \hat a^\dagger\left(\sqrt{1}\ket{0,0}\right) = \sqrt{1}\sqrt{0+1}\ket{1,0} = \ket{1,0}, \]

so

\[ \hat W\ket{0,1}=g\ket{1,0}. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Therefore, in the ordered basis \(\{\ket{1,0},\ket{0,1}\}\),

\[ W = \begin{pmatrix} \bra{1,0}\hat W\ket{1,0} & \bra{1,0}\hat W\ket{0,1} \\ \bra{0,1}\hat W\ket{1,0} & \bra{0,1}\hat W\ket{0,1} \end{pmatrix} = \begin{pmatrix} 0 & g\\ g & 0 \end{pmatrix}. \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Diagonalizing this matrix gives eigenvalues \(\pm g\) with normalized eigenvectors

\[ \ket{\pm}=\frac{1}{\sqrt{2}}\left(\ket{1,0}\pm\ket{0,1}\right). \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

Thus the first-order energy shifts in the \(E_2\) manifold are

\[ \Delta E_{2,\pm}^{(1)}=\pm g, \quad\textcolor{red}{\boxed{\text{1 pt}}} \]

so the two first-order-corrected energies are

\[ \boxed{ E_{2,\pm}=E_2+\Delta E_{2,\pm}^{(1)}=\omega_a\pm g, } \]

with corresponding (zeroth-order) eigenstates \(\ket{\pm}\) within the degenerate subspace.