PSet 1 Discussion

Conventions:

• Set \(\hbar=1\) throughout.
• For a single spin-\(\tfrac{1}{2}\), \(\ket{+}\) and \(\ket{-}\) denote \(J_z\) eigenkets with eigenvalues \(+\tfrac{1}{2}\) and \(-\tfrac{1}{2}\).
• Coupled (total) angular momentum kets are \(\ket{J,M}\): eigenkets of \(\hat J^2_{\mathrm{tot}}\) and \(\hat J_{z,\mathrm{tot}}\).
• Overall (global) phases of kets are conventional and physically irrelevant.

Reference identities:

Total operators:

\[ \hat{\mathbf J}_{\mathrm{tot}}=\hat{\mathbf J}_1+\hat{\mathbf J}_2 \quad (\text{or } \hat{\mathbf J}_1+\hat{\mathbf J}_2+\hat{\mathbf J}_3). \]

Ladder operator actions on a single \(\ket{j,m}\):

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

Two-spin dot product identity:

\[ \hat{\mathbf J}_{\mathrm{tot}}^2 =(\hat{\mathbf J}_1+\hat{\mathbf J}_2)^2 =\hat{\mathbf J}_1^2+\hat{\mathbf J}_2^2+2\,\hat{\mathbf J}_1\cdot\hat{\mathbf J}_2. \]

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Two spin one-half basics

We are given the normalized state:

\[ \ket{\psi}=\frac{1}{\sqrt{2}}\left(\ket{+-}+i\ket{-+}\right). \]

(a)

Goal: rewrite \(\ket{\psi}\) in the coupled basis \(\{\ket{J,M}\}\).

Key idea: both \(\ket{+-}\) and \(\ket{-+}\) have total \(M=0\), so only the coupled states with \(M=0\) can appear, i.e. \(\ket{1,0}\) and \(\ket{0,0}\).

Step 1: recall the coupled basis (triplet + singlet) for two spins-\(\tfrac{1}{2}\):

\[ \ket{1,0}=\frac{1}{\sqrt{2}}\left(\ket{+-}+\ket{-+}\right),\qquad \ket{0,0}=\frac{1}{\sqrt{2}}\left(\ket{+-}-\ket{-+}\right). \]

Step 2: write an ansatz within the \(M=0\) subspace:

\[ \ket{\psi}=a\ket{1,0}+b\ket{0,0}. \]

Step 3: expand the RHS back into \(\{\ket{+-},\ket{-+}\}\) and match coefficients of the two product kets to solve for \(a,b\).

Checkpoint to do live: you should get two linear equations from matching the amplitudes of \(\ket{+-}\) and \(\ket{-+}\).

Result (one consistent convention):

\[ \textcolor{white}{\boxed{\;\ket{\psi}=\frac{1+i}{2}\ket{1,0}+\frac{1-i}{2}\ket{0,0}\;}} \]

(b)

Goal: outcomes and probabilities when measuring \(\hat J^2_{\mathrm{tot}}\).

Key idea: in the coupled basis,

\[ \hat J^2_{\mathrm{tot}}\ket{J,M}=J(J+1)\ket{J,M}. \]

From (a), only \(J=1\) and \(J=0\) appear, hence possible outcomes are:

\[ J(J+1)=2 \quad \text{or} \quad 0. \]

Probabilities are \(|a|^2\) and \(|b|^2\) from part (a).

Checkpoint: compute \(|1\pm i|^2=2\).

Result:

\[ P(J=1)=\frac{1}{2},\qquad P(J=0)=\frac{1}{2}. \]

(c)

Goal: compute \(\langle \hat{\mathbf J}_1\cdot\hat{\mathbf J}_2\rangle\) in \(\ket{\psi}\).

Step 1: use the identity

\[ \hat{\mathbf J}_1\cdot\hat{\mathbf J}_2 =\frac{1}{2}\left(\hat{\mathbf J}^2_{\mathrm{tot}}-\hat{\mathbf J}_1^2-\hat{\mathbf J}_2^2\right). \]

Step 2: for spin-\(\tfrac{1}{2}\),

\[ \hat{\mathbf J}_1^2=\hat{\mathbf J}_2^2=\frac{3}{4}. \]

Step 3: on a coupled eigenstate, replace \(\hat{\mathbf J}^2_{\mathrm{tot}}\to J(J+1)\) to get the eigenvalues of \(\hat{\mathbf J}_1\cdot\hat{\mathbf J}_2\):

\[ J=1:\;\frac{1}{2}\left(2-\frac{3}{2}\right)=\frac{1}{4},\qquad J=0:\;\frac{1}{2}\left(0-\frac{3}{2}\right)=-\frac{3}{4}. \]

Step 4: take the probability-weighted average using \(P(J=1)=P(J=0)=\tfrac{1}{2}\).

Result:

\[ \textcolor{white}{\boxed{\;\langle \hat{\mathbf J}_1\cdot\hat{\mathbf J}_2\rangle=-\frac{1}{4}\;}} \]

(d)

Goal: reduced density matrix of spin 1 and \(S_{\mathrm{vN}}=-\mathrm{Tr}(\rho_1\log_2\rho_1)\).

Step 1: write \(\ket{\psi}\) emphasizing the bipartition (spin 1) \(\otimes\) (spin 2):

\[ \ket{\psi}=\frac{1}{\sqrt{2}}\ket{+}\ket{-}+\frac{i}{\sqrt{2}}\ket{-}\ket{+}. \]

Step 2: form \(\rho=\ket{\psi}\bra{\psi}\) and trace out spin 2. Orthogonality of \(\ket{+}\) and \(\ket{-}\) kills the off-diagonal terms under the partial trace.

Checkpoint: make sure the cross terms have factors like \(\ket{-}\bra{+}\) on spin 2, which vanish upon tracing.

Result:

\[ \rho_1=\mathrm{Tr}_2(\rho)=\frac{1}{2}\ket{+}\bra{+}+\frac{1}{2}\ket{-}\bra{-}=\frac{\mathbb I_2}{2}. \]

Eigenvalues are \(\lambda_1=\lambda_2=\tfrac{1}{2}\), so

\[ S_{\mathrm{vN}}=-\sum_{k=1}^2 \frac{1}{2}\log_2\left(\frac{1}{2}\right)=1. \]

Interpretation: \(S_{\mathrm{vN}}=1\) bit is the maximum possible entropy for a single qubit, so the state is maximally entangled.

(e)

Goal: after measuring \(\hat J^2_{\mathrm{tot}}\) and obtaining outcome \(2\), find the post-measurement state; then measure \(\hat J_{z,\mathrm{tot}}\).

Key idea: outcome \(2\) means the state is projected into the \(J=1\) subspace.

From (a), the only \(J=1\) component is proportional to \(\ket{1,0}\), so after projection + renormalization:

\[ \textcolor{white}{\boxed{\;\ket{\psi_{\mathrm{post}}}=\ket{1,0}\;}} \]

Then measuring \(\hat J_{z,\mathrm{tot}}\) gives \(M=0\) with probability \(1\).

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Constructing Clebsch–Gordan coefficients

Here \(j_1=1\) and \(j_2=1\). Use the product basis \(\ket{m_1,m_2}\) with \(m_1,m_2\in\{1,0,-1\}\).

(a)

Goal: possible values of \(J\) and \(M\), and dimension check.

Angular momentum addition:

\[ J\in\{|j_1-j_2|,|j_1-j_2|+1,\dots,j_1+j_2\}=\{0,1,2\}. \]

For each \(J\), allowed \(M\) are \(-J,-J+1,\dots,J\).

Dimension counting:

\[ \sum_{J=0}^2(2J+1)=1+3+5=9, \]

and the product basis dimension is

\[ (2j_1+1)(2j_2+1)=3\cdot 3=9. \]

So the coupled basis is complete.

(b)

Goal: compute CG coefficients by explicitly constructing all \(\ket{J,M}\) in terms of \(\ket{m_1,m_2}\), using the “class method” (ladder down the first column, then orthogonality to build the next column).

Phase convention note: within each fixed \(J\) multiplet, you can multiply all \(\ket{J,M}\) by an overall sign (or phase). Pick a consistent convention (e.g., choose the first nonzero coefficient in the top state to be positive real).

Part I: build the \(J=2\) multiplet (the “first column”).

Step 1: start from the stretched state (unique product state at max \(M\)):

\[ \ket{2,2}=\ket{1,1}. \]

General ladder-operator action (for any \(j\)):

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

Step 2: apply \(\hat J_{-,\mathrm{tot}}=\hat J_{-,1}+\hat J_{-,2}\) to obtain \(\ket{2,1}\) (then normalize).

For \(j=1\),

\[ \hat J_-\ket{1,1}=\sqrt{2}\ket{1,0}. \]

So

\[ \hat J_{-,\mathrm{tot}}\ket{1,1} =\sqrt{2}\ket{1,0}+\sqrt{2}\ket{0,1}. \]

Normalize to get

\[ \ket{2,1}=\frac{1}{\sqrt{2}}\left(\ket{1,0}+\ket{0,1}\right). \]

Step 3: apply \(\hat J_{-,\mathrm{tot}}\) again to get \(\ket{2,0}\).

Do it explicitly (good board exercise): apply \(\hat J_{-,1}\) and \(\hat J_{-,2}\) to each term in \(\ket{2,1}\), collect like terms, then normalize.

Result:

\[ \ket{2,0}=\frac{1}{\sqrt{6}}\left(\ket{1,-1}+2\ket{0,0}+\ket{-1,1}\right). \]

Step 4: continue laddering down (or use symmetry) to get

\[ \ket{2,-1}=\frac{1}{\sqrt{2}}\left(\ket{-1,0}+\ket{0,-1}\right),\qquad \ket{2,-2}=\ket{-1,-1}. \]

Part II: build the \(J=1\) multiplet (the “second column”) by orthogonality.

Step 5: focus on the \(M=1\) subspace. The product kets with \(M=1\) are \(\ket{1,0}\) and \(\ket{0,1}\), so any \(M=1\) state is

\[ \alpha\ket{1,0}+\beta\ket{0,1}. \]

We already have the \(J=2\) state in this subspace:

\[ \ket{2,1}\propto \ket{1,0}+\ket{0,1}. \]

Impose orthogonality to \(\ket{2,1}\) to get \(\alpha+\beta=0\), then normalize:

\[ \ket{1,1}=\frac{1}{\sqrt{2}}\left(\ket{1,0}-\ket{0,1}\right). \]

Step 6: ladder down within the \(J=1\) multiplet:

\[ \ket{1,0}=\frac{1}{\sqrt{2}}\left(\ket{1,-1}-\ket{-1,1}\right), \]

\[ \ket{1,-1}=\frac{1}{\sqrt{2}}\left(\ket{0,-1}-\ket{-1,0}\right). \]

Part III: build the \(J=0\) state as the remaining orthogonal vector at \(M=0\).

Step 7: the \(M=0\) product subspace is spanned by

\[ \ket{1,-1},\quad \ket{0,0},\quad \ket{-1,1}. \]

We already have two orthonormal states in this subspace: \(\ket{2,0}\) and \(\ket{1,0}\). The \(J=0\) state must be orthogonal to both. Solve the linear system (two orthogonality constraints) and then normalize.

Result (one standard convention):

\[ \ket{0,0}=\frac{1}{\sqrt{3}}\left(\ket{1,-1}-\ket{0,0}+\ket{-1,1}\right). \]

At this point, the CG coefficients are read off directly as the amplitudes of each product ket in the coupled ket expansions.

(c)

Goal: which coupled states are most entangled, and how close to maximum?

Setup: this is a bipartite system of two spin-1’s (two qutrits). For a pure state, entanglement can be quantified by the von Neumann entropy of the reduced density matrix, \(S_{\mathrm{vN}}(\rho_1)\).

Maximum possible entanglement for two qutrits occurs when

\[ \rho_1=\frac{\mathbb I_3}{3}, \]

which gives

\[ S_{\max}=\log_2 3. \]

Fast evaluation strategy:

• If the state has the form of a “Schmidt-like” sum with equal magnitudes on three orthogonal \(\ket{m_1}\) labels, it is maximally entangled.
• If it has only two product terms with equal magnitude, then the reduced density matrix has eigenvalues \(\{1/2,1/2,0\}\) and \(S_{\mathrm{vN}}=1\).

Apply to the coupled kets:

• The singlet \(\ket{0,0}\) has three terms of equal magnitude and uses all three \(m_1\) values exactly once. Tracing out spin 2 gives

\[ \rho_1=\frac{\mathbb I_3}{3}\quad\Rightarrow\quad S_{\mathrm{vN}}=\log_2 3. \]

Conclusion:

\[ \textcolor{white}{\boxed{\;\text{The most entangled coupled state is } \ket{0,0},\ \text{and it reaches the maximum } \log_2 3.\;}} \]

(Other coupled kets generally have smaller entropy; e.g. two-term superpositions give \(S_{\mathrm{vN}}=1\).)

(d)

Goal: determine which coupled states are even/odd under exchange of spins 1 and 2.

Key general rule (from lecture / Shankar):

\[ \hat P_{12}\ket{J,M}=(-1)^{j_1+j_2-J}\ket{J,M}. \]

Here \(j_1=j_2=1\), so the phase is \((-1)^{2-J}=(-1)^J\).

Therefore:

\[ \textcolor{white}{\boxed{\;\text{Even (symmetric): }J=2,0\qquad \text{Odd (antisymmetric): }J=1.\;}} \]

Quick verification: swap \(m_1\leftrightarrow m_2\) in the explicit expansions from (b) and check the sign.

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Multiple spins and a cavity mode

We have three spins-\(\tfrac{1}{2}\) coupled to a cavity. Define

\[ \hat{\mathbf J}_{\mathrm{tot}}=\hat{\mathbf J}_1+\hat{\mathbf J}_2+\hat{\mathbf J}_3. \]

Hamiltonian:

\[ \hat H=\omega_c\hat a^\dagger\hat a+\omega_{at}\hat J_{z,\mathrm{tot}} +g\left(\hat a^\dagger \hat J_{-,\mathrm{tot}}+\hat J_{+,\mathrm{tot}}\hat a\right). \]

(a)

Goal: possible \(J_{\mathrm{tot}}\) values and naive state count.

Classical “vector addition” expectation:

• Maximum: all aligned \(\Rightarrow J_{\mathrm{tot}}=\tfrac{3}{2}\).
• Minimum: partial cancellation \(\Rightarrow J_{\mathrm{tot}}=\tfrac{1}{2}\).

So

\[ J_{\mathrm{tot}}\in\left\{\frac{3}{2},\frac{1}{2}\right\}. \]

Naive count:

\[ (2\cdot\tfrac{3}{2}+1)+(2\cdot\tfrac{1}{2}+1)=4+2=6, \]

but the full product space has dimension \(2^3=8\).

(b)

Goal: explain the degeneracy and identify which \(\ket{J,M}\) are degenerate.

Core explanation: the total-\(J=\tfrac{1}{2}\) subspace must appear twice (two orthogonal \(J=\tfrac{1}{2}\) doublets).

Clean coupling-order argument:

\[ \left(\tfrac{1}{2}\right)\otimes\left(\tfrac{1}{2}\right)=0\oplus 1, \]

then

\[ (0\oplus 1)\otimes\left(\tfrac{1}{2}\right) =\left(\tfrac{1}{2}\right)\oplus\left(\tfrac{1}{2}\oplus\tfrac{3}{2}\right). \]

So \(J=\tfrac{1}{2}\) appears twice (two orthogonal doublets), while \(J=\tfrac{3}{2}\) appears once (one quartet). Concretely, for each \(M=\pm\tfrac{1}{2}\) there are two orthogonal states with the same \((J,M)\) labels; an additional label is the intermediate \(j_{12}\in\{0,1\}\).

(c)

Goal: explicitly construct the coupled kets in terms of product kets.

Part I: the fully symmetric \(J=\tfrac{3}{2}\) quartet.

Start from the stretched state and ladder down:

\[ \ket{\tfrac{3}{2},\tfrac{3}{2}}=\ket{+++}, \]

\[ \ket{\tfrac{3}{2},\tfrac{1}{2}} =\frac{1}{\sqrt{3}}\left(\ket{++-}+\ket{+-+}+\ket{-++}\right), \]

\[ \ket{\tfrac{3}{2},-\tfrac{1}{2}} =\frac{1}{\sqrt{3}}\left(\ket{--+}+\ket{-+-}+\ket{+--}\right), \]

\[ \ket{\tfrac{3}{2},-\tfrac{3}{2}}=\ket{---}. \]

Part II: one \(J=\tfrac{1}{2}\) doublet from \(j_{12}=0\) (singlet of spins 1 and 2).

First define the singlet of spins 1 and 2:

\[ \ket{0,0}_{12}=\frac{1}{\sqrt{2}}\left(\ket{+-}-\ket{-+}\right)_{12}. \]

Then tensor with spin 3:

\[ \ket{\tfrac{1}{2},\tfrac{1}{2}}_{j_{12}=0}=\ket{0,0}_{12}\ket{+}_3 =\frac{1}{\sqrt{2}}\left(\ket{+-+}-\ket{-++}\right), \]

\[ \ket{\tfrac{1}{2},-\tfrac{1}{2}}_{j_{12}=0}=\ket{0,0}_{12}\ket{-}_3 =\frac{1}{\sqrt{2}}\left(\ket{+--}-\ket{-+-}\right). \]

Part III: the other \(J=\tfrac{1}{2}\) doublet from \(j_{12}=1\) (triplet of spins 1 and 2).

Practical construction at \(M=\tfrac{1}{2}\):

• The subspace is spanned by \(\{\ket{++-},\ket{+-+},\ket{-++}\}\).
• One vector is the symmetric \(\ket{\tfrac{3}{2},\tfrac{1}{2}}\).
• One vector is already used by the singlet-derived state \(\ket{\tfrac{1}{2},\tfrac{1}{2}}_{j_{12}=0}\).
• The remaining orthonormal vector (orthogonal to both) is the second \(J=\tfrac{1}{2}\) state.

A convenient normalized choice is:

\[ \ket{\tfrac{1}{2},\tfrac{1}{2}}_{j_{12}=1} =\frac{1}{\sqrt{6}}\left(\ket{+-+}+\ket{-++}-2\ket{++-}\right), \]

\[ \ket{\tfrac{1}{2},-\tfrac{1}{2}}_{j_{12}=1} =\frac{1}{\sqrt{6}}\left(\ket{+--}+\ket{-+-}-2\ket{--+}\right). \]

Checkpoint: verify orthonormality and confirm these are orthogonal to the \(J=\tfrac{3}{2}\) states.

(d)

Goal: interpret the Hamiltonian structure.

Talking points to emphasize:

• The interaction exchanges a photon with a collective spin flip.
• Because only total-spin operators \(\hat J_{\pm,\mathrm{tot}}\) appear, the Hamiltonian does not mix different \(J\) sectors.

(e)

Goal: show \([\hat H,\hat J^2_{\mathrm{tot}}]=0\) but \([\hat H,\hat J_{z,\mathrm{tot}}]\neq 0\).

Key commutators:

\[ [\hat J^2_{\mathrm{tot}},\hat J_{z,\mathrm{tot}}]=0,\qquad [\hat J^2_{\mathrm{tot}},\hat J_{\pm,\mathrm{tot}}]=0, \]

and cavity operators commute with spin operators.

Therefore,

\[ [\hat H,\hat J^2_{\mathrm{tot}}]=0. \]

For \(\hat J_{z,\mathrm{tot}}\), use

\[ [\hat J_{z,\mathrm{tot}},\hat J_{-,\mathrm{tot}}]=-\hat J_{-,\mathrm{tot}},\qquad [\hat J_{z,\mathrm{tot}},\hat J_{+,\mathrm{tot}}]=+\hat J_{+,\mathrm{tot}}. \]

You should obtain

\[ [\hat H,\hat J_{z,\mathrm{tot}}] =g\left(\hat a^\dagger \hat J_{-,\mathrm{tot}}-\hat J_{+,\mathrm{tot}}\hat a\right)\neq 0. \]

(f)

Goal: show \([\hat H,\hat Q]=0\) where

\[ \hat Q=\hat a^\dagger\hat a+\hat J_{z,\mathrm{tot}}. \]

Oscillator commutators:

\[ [\hat a^\dagger\hat a,\hat a^\dagger]=\hat a^\dagger,\qquad [\hat a^\dagger\hat a,\hat a]=-\hat a. \]

Sketch of the cancellation to show live:

• \(\hat a^\dagger\hat J_-\) increases photon number by 1 and decreases \(J_z\) by 1.
• \(\hat J_+\hat a\) decreases photon number by 1 and increases \(J_z\) by 1.

So the sum \(\hat a^\dagger\hat a+\hat J_z\) is conserved.

Result:

\[ \textcolor{white}{\boxed{\;[\hat H,\hat Q]=0.\;}} \]

(g)

Goal: energy eigenstates for 0 and 1 total excitations; express eigenkets in terms of total angular momentum states.

It is convenient to define a nonnegative excitation number (constant shift of \(\hat Q\)):

\[ \hat N_{\mathrm{exc}}=\hat a^\dagger\hat a+\left(\hat J_{z,\mathrm{tot}}+\frac{3}{2}\right). \]

Zero excitation sector (\(N_{\mathrm{exc}}=0\)):

\[ \ket{0}\otimes\ket{\tfrac{3}{2},-\tfrac{3}{2}} \]

with energy

\[ E_0=-\frac{3}{2}\omega_{at}. \]

One excitation sector (\(N_{\mathrm{exc}}=1\)): one cavity-coupled subspace plus two uncoupled (“dark”) atomic states.

Cavity-coupled subspace (in \(J=\tfrac{3}{2}\)), basis:

\[ \ket{e_{\mathrm{ph}}}=\ket{1}\otimes\ket{\tfrac{3}{2},-\tfrac{3}{2}},\qquad \ket{e_{\mathrm{at}}}=\ket{0}\otimes\ket{\tfrac{3}{2},-\tfrac{1}{2}}. \]

Diagonal energies:

\[ E_{\mathrm{ph}}=\omega_c-\frac{3}{2}\omega_{at},\qquad E_{\mathrm{at}}=-\frac{1}{2}\omega_{at}. \]

Coupling matrix element uses

\[ \hat J_+\ket{J,M}=\sqrt{J(J+1)-M(M+1)}\,\ket{J,M+1}, \]

so from \(J=\tfrac{3}{2}\), \(M=-\tfrac{3}{2}\) to \(M=-\tfrac{1}{2}\) the factor is \(\sqrt{3}\), giving off-diagonal coupling \(g\sqrt{3}\).

Define detuning \(\delta=\omega_c-\omega_{at}\). After diagonalizing the \(2\times 2\) block, eigenenergies:

\[ E_{\pm}=\left(\frac{\omega_c}{2}-\omega_{at}\right)\pm\sqrt{\left(\frac{\delta}{2}\right)^2+3g^2}. \]

Corresponding dressed eigenkets can be written as:

\[ \ket{+}=\cos\theta\,\ket{e_{\mathrm{ph}}}+\sin\theta\,\ket{e_{\mathrm{at}}},\qquad \ket{-}=-\sin\theta\,\ket{e_{\mathrm{ph}}}+\cos\theta\,\ket{e_{\mathrm{at}}}, \]

with

\[ \tan(2\theta)=\frac{2g\sqrt{3}}{\delta}. \]

Uncoupled atomic states (in \(J=\tfrac{1}{2}\)):

With one spin excitation and zero photons, there are three product states \(\{\ket{+--},\ket{-+-},\ket{--+}\}\). In the coupled basis these split into one symmetric \(J=\tfrac{3}{2},M=-\tfrac{1}{2}\) state (which is the only one coupled to the cavity via the collective operator \(\hat J^\pm_{\mathrm{tot}}\)) plus two orthogonal \(J=\tfrac{1}{2},M=-\tfrac{1}{2}\) states. Because \(\hat H\) conserves \(\hat J^2_{\mathrm{tot}}\), those two \(J=\tfrac{1}{2}\) states do not couple to the photon state; “dark” here just means uncoupled from the cavity interaction.

\[ \ket{0}\otimes\ket{\tfrac{1}{2},-\tfrac{1}{2}}_{j_{12}=0},\qquad \ket{0}\otimes\ket{\tfrac{1}{2},-\tfrac{1}{2}}_{j_{12}=1} \]

are eigenstates with energy

\[ E_{\mathrm{dark}}=-\frac{1}{2}\omega_{at}. \]

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