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Ch. 6 Solutions

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Basic

Medium

Advanced

Basic

B-1. Calculation of the Phase Factor for a Stationary State

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Solution strategy: Substitute \(t = \pi\hbar/(E_0 - A)\) into \(C_1(t) = \frac{1}{\sqrt{2}}e^{-iE_{II}t/\hbar}\).

Calculation:

\[C_1(t) = \frac{1}{\sqrt{2}}e^{-iE_{II}t/\hbar} = \frac{1}{\sqrt{2}}e^{-i(E_0 - A)t/\hbar}\]

Substituting \(t = \pi\hbar/(E_0 - A)\), the exponent becomes:

\[-\frac{i(E_0 - A)}{\hbar} \cdot \frac{\pi\hbar}{E_0 - A} = -i\pi\]

Therefore:

\[C_1(t) = \frac{1}{\sqrt{2}}e^{-i\pi} = \frac{1}{\sqrt{2}} \cdot (-1) = -\frac{1}{\sqrt{2}}\]

Final answer:

\[\boxed{C_1(t) = -\frac{1}{\sqrt{2}}}\]

Verification: \(|C_1(t)|^2 = 1/2\), which is consistent with the fact that the probability does not change with time in a stationary state. Also confirmed that \(e^{-i\pi} = \cos\pi + i\sin\pi = -1\).

B-2. Energy Calculation of Tunnel Splitting

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Solution strategy: From \(2A = hf\), calculate \(A = hf/2\) and convert to eV.

Calculation:

\[f = 24{,}000\;\text{MHz} = 2.4 \times 10^{10}\;\text{Hz}\]
\[A = \frac{hf}{2} = \frac{6.626 \times 10^{-34} \times 2.4 \times 10^{10}}{2}\;\text{J}\]
\[A = \frac{15.9024 \times 10^{-24}}{2}\;\text{J} = 7.951 \times 10^{-24}\;\text{J}\]

Converting to eV:

\[A = \frac{7.951 \times 10^{-24}}{1.602 \times 10^{-19}}\;\text{eV} = 4.96 \times 10^{-5}\;\text{eV}\]

Final answer:

\[\boxed{A \approx 4.96 \times 10^{-5}\;\text{eV} \approx 5.0 \times 10^{-5}\;\text{eV}}\]

Verification: The text states that "the energy associated with the inversion motion of nitrogen is on the order of \(10^{-4}\) eV." Since \(2A \approx 10^{-4}\) eV, this is consistent.

B-3. Expansion of the Determinant in the Eigenvalue Equation

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Solution strategy: Expand \(\det(H - EI) = 0\).

Calculation:

\[\det(H - EI) = \det\begin{pmatrix} \alpha - E & \beta \\ \beta^* & \gamma - E \end{pmatrix}\]

Using the formula for the determinant of a \(2 \times 2\) matrix:

\[(\alpha - E)(\gamma - E) - \beta \cdot \beta^* = 0\]

Expanding:

\[\alpha\gamma - \alpha E - \gamma E + E^2 - |\beta|^2 = 0\]

Rearranging:

\[\boxed{E^2 - (\alpha + \gamma)E + (\alpha\gamma - |\beta|^2) = 0}\]

Verification: When \(\beta = 0\), we get \(E^2 - (\alpha + \gamma)E + \alpha\gamma = (E - \alpha)(E - \gamma) = 0\), giving eigenvalues \(E = \alpha, \gamma\) (the eigenvalues of a diagonal matrix), which is correct.

B-4. Verification of Orthogonality of Eigenvectors

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Calculation:

\(\langle I|I\rangle\):

\[\langle I|I\rangle = \frac{1}{2}(\langle 1| - \langle 2|)(|1\rangle - |2\rangle) = \frac{1}{2}(\langle 1|1\rangle - \langle 1|2\rangle - \langle 2|1\rangle + \langle 2|2\rangle)\]
\[= \frac{1}{2}(1 - 0 - 0 + 1) = 1\]

\(\langle II|II\rangle\):

\[\langle II|II\rangle = \frac{1}{2}(\langle 1| + \langle 2|)(|1\rangle + |2\rangle) = \frac{1}{2}(\langle 1|1\rangle + \langle 1|2\rangle + \langle 2|1\rangle + \langle 2|2\rangle)\]
\[= \frac{1}{2}(1 + 0 + 0 + 1) = 1\]

\(\langle I|II\rangle\):

\[\langle I|II\rangle = \frac{1}{2}(\langle 1| - \langle 2|)(|1\rangle + |2\rangle) = \frac{1}{2}(\langle 1|1\rangle + \langle 1|2\rangle - \langle 2|1\rangle - \langle 2|2\rangle)\]
\[= \frac{1}{2}(1 + 0 - 0 - 1) = 0\]

Final answer:

\[\boxed{\langle I|I\rangle = 1, \quad \langle II|II\rangle = 1, \quad \langle I|II\rangle = 0}\]

\(|I\rangle\) and \(|II\rangle\) form an orthonormal set.

Verification: This is consistent with the general theorem that eigenvectors belonging to distinct eigenvalues of a Hermitian matrix are orthogonal.

B-5. Calculating the Time Dependence of Probability

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Solution strategy: Expand the initial state \(|\psi(0)\rangle = |1\rangle\) in terms of energy eigenstates, apply time evolution, and then calculate \(\langle 2|\psi(t)\rangle\).

Calculation:

From Eq. (6.17a), \(|1\rangle = \frac{1}{\sqrt{2}}(|I\rangle + |II\rangle)\), so:

\[|\psi(0)\rangle = |1\rangle = \frac{1}{\sqrt{2}}|I\rangle + \frac{1}{\sqrt{2}}|II\rangle\]

Time evolution:

\[|\psi(t)\rangle = \frac{1}{\sqrt{2}}e^{-iE_I t/\hbar}|I\rangle + \frac{1}{\sqrt{2}}e^{-iE_{II} t/\hbar}|II\rangle\]

$\langle 2|I\rangle = \frac{1}{\sqrt{2}}(\langle 2|1\rangle - \langle 2|2\rangle) \cdot \frac{1}{1} $...

Let us recalculate carefully. From \(|I\rangle = \frac{1}{\sqrt{2}}(|1\rangle - |2\rangle)\), we get \(\langle 2|I\rangle = \frac{1}{\sqrt{2}}(0 - 1) = -\frac{1}{\sqrt{2}}\).

From \(|II\rangle = \frac{1}{\sqrt{2}}(|1\rangle + |2\rangle)\), we get \(\langle 2|II\rangle = \frac{1}{\sqrt{2}}(0 + 1) = \frac{1}{\sqrt{2}}\).

Therefore:

\[C_2(t) = \langle 2|\psi(t)\rangle = \frac{1}{\sqrt{2}}e^{-iE_I t/\hbar}\left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}}e^{-iE_{II} t/\hbar}\cdot\frac{1}{\sqrt{2}}\]
\[= \frac{1}{2}\left(-e^{-iE_I t/\hbar} + e^{-iE_{II} t/\hbar}\right)\]

Substituting \(E_I = E_0 + A\), \(E_{II} = E_0 - A\):

\[C_2(t) = \frac{1}{2}e^{-iE_0 t/\hbar}\left(-e^{-iAt/\hbar} + e^{iAt/\hbar}\right)\]
\[= \frac{1}{2}e^{-iE_0 t/\hbar} \cdot 2i\sin\left(\frac{At}{\hbar}\right) = i\,e^{-iE_0 t/\hbar}\sin\left(\frac{At}{\hbar}\right)\]

Probability:

\[P_2(t) = |C_2(t)|^2 = |i|^2 \cdot |e^{-iE_0 t/\hbar}|^2 \cdot \sin^2\left(\frac{At}{\hbar}\right)\]
\[\boxed{P_2(t) = \sin^2\!\left(\frac{At}{\hbar}\right)}\]

Verification: At \(t = 0\), \(P_2(0) = \sin^2(0) = 0\) (the initial state is \(|1\rangle\), so the probability of being in \(|2\rangle\) is zero). At \(t = \pi\hbar/(2A)\), \(P_2 = 1\) (complete transition to \(|2\rangle\)). This is also consistent with the result from S2.

B-6. Solution of \(i\hbar\,dC/dt = EC\)

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Solution strategy: Solve using separation of variables.

Calculation:

\[i\hbar\frac{dC}{dt} = EC\]
\[\frac{dC}{C} = \frac{E}{i\hbar}dt = -\frac{iE}{\hbar}dt\]

Integrating both sides:

\[\ln C = -\frac{iE}{\hbar}t + \text{const}\]
\[C(t) = C(0)\,e^{-iEt/\hbar}\]

From the initial condition \(C(0) = C_0\):

\[\boxed{C(t) = C_0\,e^{-iEt/\hbar}}\]

Regarding \(|C(t)|^2\):

\[|C(t)|^2 = |C_0|^2 \cdot |e^{-iEt/\hbar}|^2 = |C_0|^2 \cdot 1 = |C_0|^2\]

Since \(E\) is real, \(|e^{-iEt/\hbar}| = 1\), and \(|C(t)|^2\) is independent of time.

Verification: Substituting back into the differential equation: \(i\hbar \cdot C_0 \cdot (-iE/\hbar)e^{-iEt/\hbar} = E \cdot C_0 e^{-iEt/\hbar}\). Since \(i \cdot (-i) = 1\), the left-hand side \(= EC_0 e^{-iEt/\hbar}\) = right-hand side. ✓

B-7. Inverse of Basis Transformation

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Solution strategy: Solve equations (6.17a) and (6.17b) as a system of simultaneous equations.

Calculation:

Given equations: - (6.17a): \(|1\rangle = \frac{1}{\sqrt{2}}(|I\rangle + |II\rangle)\) - (6.17b): \(|2\rangle = \frac{1}{\sqrt{2}}(-|I\rangle + |II\rangle)\)

(6.17a) \(-\) (6.17b):

\[|1\rangle - |2\rangle = \frac{1}{\sqrt{2}}(|I\rangle + |II\rangle) - \frac{1}{\sqrt{2}}(-|I\rangle + |II\rangle) = \frac{1}{\sqrt{2}}(2|I\rangle) = \sqrt{2}\,|I\rangle\]
\[|I\rangle = \frac{1}{\sqrt{2}}(|1\rangle - |2\rangle)\]

(6.17a) \(+\) (6.17b):

\[|1\rangle + |2\rangle = \frac{1}{\sqrt{2}}(|I\rangle + |II\rangle) + \frac{1}{\sqrt{2}}(-|I\rangle + |II\rangle) = \frac{1}{\sqrt{2}}(2|II\rangle) = \sqrt{2}\,|II\rangle\]
\[|II\rangle = \frac{1}{\sqrt{2}}(|1\rangle + |2\rangle)\]

Final answer:

\[\boxed{|I\rangle = \frac{1}{\sqrt{2}}(|1\rangle - |2\rangle), \quad |II\rangle = \frac{1}{\sqrt{2}}(|1\rangle + |2\rangle)}\]

This is consistent with equations (6.15a) and (6.15b). ✓

B-8. Verification of Hermiticity

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Calculation:

Verify that \(M_{ij}^* = M_{ji}\) for each element of matrix \(M\).

  • \(M_{11} = 3\): \(M_{11}^* = 3 = M_{11}\) ✓ (real)
  • \(M_{22} = 5\): \(M_{22}^* = 5 = M_{22}\) ✓ (real)
  • \(M_{12} = 2 - i\): \(M_{12}^* = 2 + i = M_{21}\) ✓
  • \(M_{21} = 2 + i\): \(M_{21}^* = 2 - i = M_{12}\) ✓

Final Answer: Since \(M_{ij}^* = M_{ji}\) holds for all elements, \(M\) is a Hermitian matrix. \(\square\)

Verification: The eigenvalues of a Hermitian matrix should be real. \(\text{tr}(M) = 8\), \(\det(M) = 15 - 5 = 10\). The eigenvalues are \(E = 4 \pm \sqrt{6}\) from \(E^2 - 8E + 10 = 0\). Indeed real. ✓

Medium

M-1. Derivation of Hermiticity from Probability Conservation

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Strategy: Compute \(dP/dt = d(|C_1|^2 + |C_2|^2)/dt\) and find the conditions under which it is identically zero.

Calculation:

\[\frac{d|C_1|^2}{dt} = C_1^*\frac{dC_1}{dt} + C_1\frac{dC_1^*}{dt}\]

From Equation (6.2a):

\[\frac{dC_1}{dt} = \frac{1}{i\hbar}(H_{11}C_1 + H_{12}C_2) = -\frac{i}{\hbar}(H_{11}C_1 + H_{12}C_2)\]

Its complex conjugate:

\[\frac{dC_1^*}{dt} = \frac{i}{\hbar}(H_{11}^*C_1^* + H_{12}^*C_2^*)\]

Therefore:

\[\frac{d|C_1|^2}{dt} = C_1^*\left[-\frac{i}{\hbar}(H_{11}C_1 + H_{12}C_2)\right] + C_1\left[\frac{i}{\hbar}(H_{11}^*C_1^* + H_{12}^*C_2^*)\right]\]
\[= \frac{i}{\hbar}\left[-(H_{11}|C_1|^2 + H_{12}C_1^*C_2) + (H_{11}^*|C_1|^2 + H_{12}^*C_1 C_2^*)\right]\]
\[= \frac{i}{\hbar}\left[(H_{11}^* - H_{11})|C_1|^2 + H_{12}^*C_1 C_2^* - H_{12}C_1^*C_2\right]\]

Similarly for \(|C_2|^2\):

\[\frac{d|C_2|^2}{dt} = \frac{i}{\hbar}\left[(H_{22}^* - H_{22})|C_2|^2 + H_{21}^*C_2 C_1^* - H_{21}C_2^*C_1\right]\]

Summing:

\[\frac{dP}{dt} = \frac{i}{\hbar}\Big[(H_{11}^* - H_{11})|C_1|^2 + (H_{22}^* - H_{22})|C_2|^2 + (H_{12}^* - H_{21})C_1 C_2^* + (H_{21}^* - H_{12})C_1^* C_2\Big]\]

For \(dP/dt = 0\) to hold for arbitrary \(C_1, C_2\), each coefficient must independently vanish.

Coefficient of \(|C_1|^2\): \(H_{11}^* - H_{11} = 0\) → \(H_{11}^* = H_{11}\) → \(H_{11}\) is real

Coefficient of \(|C_2|^2\): \(H_{22}^* - H_{22} = 0\) → \(H_{22}^* = H_{22}\) → \(H_{22}\) is real

Coefficient of \(C_1 C_2^*\): \(H_{12}^* - H_{21} = 0\) → \(H_{12}^* = H_{21}\)

Coefficient of \(C_1^* C_2\): \(H_{21}^* - H_{12} = 0\) → \(H_{21}^* = H_{12}\) (equivalent to the condition above)

Final Answer:

\[\boxed{H_{11}, H_{22} \in \mathbb{R}, \quad H_{12}^* = H_{21}}\]

This is precisely the condition for the Hamiltonian matrix to be Hermitian: \(H^\dagger = H\). \(\square\)

Verification: For a Hermitian matrix, we confirm that every term in the expression for \(dP/dt\) vanishes. Since \(H_{11}^* = H_{11}\), the first term vanishes; since \(H_{22}^* = H_{22}\), the second term vanishes; since \(H_{12}^* = H_{21}\), the third and fourth terms give \((H_{21} - H_{21})C_1C_2^* + (H_{12} - H_{12})C_1^*C_2 = 0\). ✓

M-2. Derivation of Rabi Oscillations

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Solution strategy: Expand the initial state in energy eigenstates, apply time evolution, then compute the projection onto each basis state.

Calculation:

Initial state \(|\psi(0)\rangle = |1\rangle = \frac{1}{\sqrt{2}}(|I\rangle + |II\rangle)\)

Time evolution:

\[|\psi(t)\rangle = \frac{1}{\sqrt{2}}e^{-iE_I t/\hbar}|I\rangle + \frac{1}{\sqrt{2}}e^{-iE_{II} t/\hbar}|II\rangle\]

Calculation of \(C_1(t) = \langle 1|\psi(t)\rangle\):

From \(\langle 1|I\rangle = \frac{1}{\sqrt{2}}\), \(\langle 1|II\rangle = \frac{1}{\sqrt{2}}\):

\[C_1(t) = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}e^{-iE_I t/\hbar} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}e^{-iE_{II} t/\hbar}\]
\[= \frac{1}{2}\left(e^{-i(E_0+A)t/\hbar} + e^{-i(E_0-A)t/\hbar}\right)\]
\[= \frac{1}{2}e^{-iE_0 t/\hbar}\left(e^{-iAt/\hbar} + e^{iAt/\hbar}\right) = e^{-iE_0 t/\hbar}\cos\!\left(\frac{At}{\hbar}\right)\]
\[P_1(t) = |C_1(t)|^2 = \cos^2\!\left(\frac{At}{\hbar}\right)\]

Calculation of \(C_2(t) = \langle 2|\psi(t)\rangle\):

From \(\langle 2|I\rangle = -\frac{1}{\sqrt{2}}\), \(\langle 2|II\rangle = \frac{1}{\sqrt{2}}\):

\[C_2(t) = \frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{\sqrt{2}}\right)e^{-iE_I t/\hbar} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}e^{-iE_{II} t/\hbar}\]
\[= \frac{1}{2}e^{-iE_0 t/\hbar}\left(-e^{-iAt/\hbar} + e^{iAt/\hbar}\right) = i\,e^{-iE_0 t/\hbar}\sin\!\left(\frac{At}{\hbar}\right)\]
\[P_2(t) = |C_2(t)|^2 = \sin^2\!\left(\frac{At}{\hbar}\right)\]

Verification of probability conservation:

\[P_1(t) + P_2(t) = \cos^2\!\left(\frac{At}{\hbar}\right) + \sin^2\!\left(\frac{At}{\hbar}\right) = 1 \quad \checkmark\]

Oscillation period:

Complete transition \(P_2(t) = 1\) occurs when \(At/\hbar = \pi/2\), i.e., \(t = \pi\hbar/(2A)\). The period for the system to complete one full cycle \(|1\rangle \to |2\rangle \to |1\rangle\) is:

\[\boxed{T = \frac{\pi\hbar}{A}}\]

Final answer:

\[\boxed{P_1(t) = \cos^2\!\left(\frac{At}{\hbar}\right), \quad P_2(t) = \sin^2\!\left(\frac{At}{\hbar}\right), \quad T = \frac{\pi\hbar}{A}}\]

Verification: Defining the angular frequency \(\omega_R = 2A/\hbar\), the period is \(T = 2\pi/\omega_R = \pi\hbar/A\). Also, from \(2A = hf\) we get \(T = 1/(2f)\)... actually, the angular frequency of oscillation of \(P_2\) is \(2A/\hbar\) (since the argument of \(\sin^2\) is \(At/\hbar\), and the period of \(\sin^2\theta\) is \(\pi\), so one full period corresponds to \(At/\hbar = \pi\), giving \(t = \pi\hbar/A\)). ✓

M-3. Energy Levels of an Ammonia Molecule in an Electric Field

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Solution strategy: Solve the eigenvalue equation \(\det(H - EI) = 0\).

Calculation:

\[\det\begin{pmatrix} E_0 + \mu\mathcal{E} - E & -A \\ -A & E_0 - \mu\mathcal{E} - E \end{pmatrix} = 0\]

Substituting \(\lambda = E - E_0\):

\[(\mu\mathcal{E} - \lambda)(-\mu\mathcal{E} - \lambda) - A^2 = 0\]
\[-(\mu\mathcal{E})^2 + \lambda^2 - A^2 = 0\]
\[\lambda^2 = A^2 + (\mu\mathcal{E})^2\]
\[\lambda = \pm\sqrt{A^2 + (\mu\mathcal{E})^2}\]

Therefore:

\[\boxed{E_{\pm} = E_0 \pm \sqrt{A^2 + (\mu\mathcal{E})^2}}\]

Discussion of limiting cases:

Case \(\mu\mathcal{E} \ll A\) (weak field limit):

\[\sqrt{A^2 + (\mu\mathcal{E})^2} = A\sqrt{1 + \left(\frac{\mu\mathcal{E}}{A}\right)^2} \approx A\left(1 + \frac{(\mu\mathcal{E})^2}{2A^2}\right)\]
\[E_{\pm} \approx E_0 \pm A \pm \frac{(\mu\mathcal{E})^2}{2A}\]

This is a quadratic shift from the zero-field levels \(E_0 \pm A\) (quadratic Stark effect). The energy levels shift slightly in proportion to the square of the electric field.

Case \(\mu\mathcal{E} \gg A\) (strong field limit):

\[\sqrt{A^2 + (\mu\mathcal{E})^2} \approx \mu\mathcal{E}\sqrt{1 + \left(\frac{A}{\mu\mathcal{E}}\right)^2} \approx \mu\mathcal{E}\left(1 + \frac{A^2}{2(\mu\mathcal{E})^2}\right)\]
\[E_{\pm} \approx E_0 \pm \mu\mathcal{E}\]

Tunneling becomes negligible, and the system approaches the classical picture where the nitrogen atom is localized either "above" or "below." The energy splitting is linearly proportional to the electric field (linear Stark effect).

Verification: When \(\mathcal{E} = 0\), we get \(E_{\pm} = E_0 \pm A\), which agrees with the result in the text (6.13). ✓

M-4. Diagonalization of the Hamiltonian and Change of Matrix Representation

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Strategy: Directly compute \(U^\dagger H U\).

Calculation:

\[U = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}, \quad H = \begin{pmatrix} E_0 & -A \\ -A & E_0 \end{pmatrix}\]

Since \(U\) is a real matrix, \(U^\dagger = U^T\):

\[U^\dagger = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\]

Step 1: Compute \(U^\dagger H\):

\[U^\dagger H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} E_0 & -A \\ -A & E_0 \end{pmatrix}\]
\[= \frac{1}{\sqrt{2}}\begin{pmatrix} E_0 - (-A) & -A - E_0 \\ E_0 + (-A) & -A + E_0 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} E_0 + A & -A - E_0 \\ E_0 - A & E_0 - A \end{pmatrix}\]

Redoing the calculation carefully:

\((1,1)\) entry: \(1 \cdot E_0 + (-1)(-A) = E_0 + A\)

\((1,2)\) entry: \(1 \cdot (-A) + (-1) \cdot E_0 = -A - E_0\)

\((2,1)\) entry: \(1 \cdot E_0 + 1 \cdot (-A) = E_0 - A\)

\((2,2)\) entry: \(1 \cdot (-A) + 1 \cdot E_0 = E_0 - A\)

\[U^\dagger H = \frac{1}{\sqrt{2}}\begin{pmatrix} E_0 + A & -(E_0 + A) \\ E_0 - A & E_0 - A \end{pmatrix}\]

Step 2: Compute \((U^\dagger H)U\):

\[U^\dagger H U = \frac{1}{\sqrt{2}}\begin{pmatrix} E_0 + A & -(E_0 + A) \\ E_0 - A & E_0 - A \end{pmatrix} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}\]
\[= \frac{1}{2}\begin{pmatrix} (E_0+A)\cdot 1 + (-(E_0+A))\cdot(-1) & (E_0+A)\cdot 1 + (-(E_0+A))\cdot 1 \\ (E_0-A)\cdot 1 + (E_0-A)\cdot(-1) & (E_0-A)\cdot 1 + (E_0-A)\cdot 1 \end{pmatrix}\]
\[= \frac{1}{2}\begin{pmatrix} (E_0+A) + (E_0+A) & (E_0+A) - (E_0+A) \\ (E_0-A) - (E_0-A) & (E_0-A) + (E_0-A) \end{pmatrix}\]
\[= \frac{1}{2}\begin{pmatrix} 2(E_0+A) & 0 \\ 0 & 2(E_0-A) \end{pmatrix}\]
\[\boxed{U^\dagger H U = \begin{pmatrix} E_0 + A & 0 \\ 0 & E_0 - A \end{pmatrix}}\]

Relationship between the column vectors of \(U\) and the eigenvectors:

The first column of \(U\) is \(\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}\). This corresponds to the component representation of \(|I\rangle = \frac{1}{\sqrt{2}}(|1\rangle - |2\rangle)\) in the \(\{|1\rangle, |2\rangle\}\) basis. The \((1,1)\) entry after diagonalization is \(E_I = E_0 + A\).

The second column of \(U\) is \(\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}\). This corresponds to the component representation of \(|II\rangle = \frac{1}{\sqrt{2}}(|1\rangle + |2\rangle)\). The \((2,2)\) entry after diagonalization is \(E_{II} = E_0 - A\).

In other words, each column vector of \(U\) is an eigenvector of the Hamiltonian, and the diagonal entries of the diagonalized matrix are the corresponding eigenvalues.

Verification: Confirm that \(U\) is unitary. \(U^\dagger U = \frac{1}{2}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = I\). ✓

Advanced

A-1. Oscillating Electric Field as a Time-Dependent Perturbation and Transition Probability

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(a) Derivation of the Differential Equations for \(b_I(t)\), \(b_{II}(t)\)

Solution strategy: Substitute \(C_I(t) = b_I(t)e^{-iE_I t/\hbar}\), \(C_{II}(t) = b_{II}(t)e^{-iE_{II}t/\hbar}\) into Equation (6.3).

Calculation:

The time evolution equations in the energy basis \(\{|I\rangle, |II\rangle\}\):

\[i\hbar\frac{dC_I}{dt} = E_I C_I + \mu\mathcal{E}_0\cos\omega t \cdot C_{II}\]
\[i\hbar\frac{dC_{II}}{dt} = \mu\mathcal{E}_0\cos\omega t \cdot C_I + E_{II} C_{II}\]

Substituting \(C_I = b_I e^{-iE_I t/\hbar}\). Left-hand side:

\[i\hbar\left(\dot{b}_I e^{-iE_I t/\hbar} + b_I \cdot \left(-\frac{iE_I}{\hbar}\right)e^{-iE_I t/\hbar}\right) = i\hbar\dot{b}_I e^{-iE_I t/\hbar} + E_I b_I e^{-iE_I t/\hbar}\]

Right-hand side:

\[E_I b_I e^{-iE_I t/\hbar} + \mu\mathcal{E}_0\cos\omega t \cdot b_{II} e^{-iE_{II}t/\hbar}\]

The terms \(E_I b_I e^{-iE_I t/\hbar}\) cancel on both sides:

\[i\hbar\dot{b}_I = \mu\mathcal{E}_0\cos\omega t \cdot b_{II}\, e^{-i(E_{II}-E_I)t/\hbar}\]

Defining \(\omega_0 \equiv (E_I - E_{II})/\hbar = 2A/\hbar\), so that \(E_{II} - E_I = -\hbar\omega_0\):

\[\boxed{i\hbar\dot{b}_I = \mu\mathcal{E}_0\cos\omega t \cdot b_{II}\, e^{i\omega_0 t}}\]

Similarly:

\[\boxed{i\hbar\dot{b}_{II} = \mu\mathcal{E}_0\cos\omega t \cdot b_I\, e^{-i\omega_0 t}}\]

(b) Rotating Wave Approximation (RWA)

Substituting \(\cos\omega t = \frac{1}{2}(e^{i\omega t} + e^{-i\omega t})\):

\[i\hbar\dot{b}_I = \frac{\mu\mathcal{E}_0}{2}\,b_{II}\left(e^{i(\omega + \omega_0)t} + e^{-i(\omega - \omega_0)t}\right)\]
\[i\hbar\dot{b}_{II} = \frac{\mu\mathcal{E}_0}{2}\,b_I\left(e^{i(\omega - \omega_0)t} + e^{-i(\omega + \omega_0)t}\right)\]

Under the resonance condition \(\omega \approx \omega_0\): - \(e^{\pm i(\omega - \omega_0)t}\) varies slowly (\(\omega - \omega_0 \approx 0\)) - \(e^{\pm i(\omega + \omega_0)t}\) oscillates rapidly (\(\omega + \omega_0 \approx 2\omega_0\))

The rapidly oscillating terms average to approximately zero over time and are neglected (rotating wave approximation):

\[\boxed{i\hbar\dot{b}_I = \frac{\mu\mathcal{E}_0}{2}\,b_{II}\,e^{-i(\omega - \omega_0)t}}\]
\[\boxed{i\hbar\dot{b}_{II} = \frac{\mu\mathcal{E}_0}{2}\,b_I\,e^{i(\omega - \omega_0)t}}\]

(c) Solution at Resonance

When \(\omega = \omega_0\), all exponential factors become unity:

\[i\hbar\dot{b}_I = \frac{\mu\mathcal{E}_0}{2}\,b_{II}\]
\[i\hbar\dot{b}_{II} = \frac{\mu\mathcal{E}_0}{2}\,b_I\]

Defining \(\Omega_R \equiv \mu\mathcal{E}_0/(2\hbar)\) (Rabi frequency):

\[\dot{b}_I = -i\Omega_R\, b_{II}, \quad \dot{b}_{II} = -i\Omega_R\, b_I\]

We derive a second-order differential equation for \(b_I\). Differentiating the first equation:

\[\ddot{b}_I = -i\Omega_R\,\dot{b}_{II} = -i\Omega_R \cdot (-i\Omega_R\, b_I) = -\Omega_R^2\, b_I\]

This is a simple harmonic oscillator equation with the general solution:

\[b_I(t) = \alpha\cos(\Omega_R t) + \beta\sin(\Omega_R t)\]

Initial conditions: \(b_I(0) = 0\) (the initial state is \(|II\rangle\), so \(C_I(0) = 0\)), \(b_{II}(0) = 1\).

From \(b_I(0) = 0\): \(\alpha = 0\).

From \(\dot{b}_I(0) = -i\Omega_R\, b_{II}(0) = -i\Omega_R\): \(\beta\Omega_R = -i\Omega_R\), giving \(\beta = -i\).

\[b_I(t) = -i\sin(\Omega_R t)\]

Transition probability:

\[P_{II \to I}(t) = |b_I(t)|^2 = |-i\sin(\Omega_R t)|^2 = \sin^2(\Omega_R t)\]
\[\boxed{P_{II \to I}(t) = \sin^2\!\left(\frac{\mu\mathcal{E}_0\,t}{2\hbar}\right)}\]

Connection to the ammonia maser:

In the ammonia maser, a state selector directs only molecules in the high-energy state \(|I\rangle\) into a resonant cavity. When the electromagnetic field frequency in the cavity satisfies the resonance condition \(\omega = 2A/\hbar\), molecules undergo transitions from \(|I\rangle\) to \(|II\rangle\), emitting microwave photons with energy equal to the energy difference \(2A\). This is stimulated emission. The result above shows that a complete transition (\(P = 1\)) occurs at the appropriate interaction time \(t = \pi\hbar/(\mu\mathcal{E}_0)\), which forms the basis for efficient maser operation.

Verification: - At \(t = 0\): \(P_{II \to I} = 0\) (initial state is \(|II\rangle\)) ✓ - Confirming \(b_{II}(t) = \cos(\Omega_R t)\): \(\dot{b}_{II} = -\Omega_R\sin(\Omega_R t) = -i\Omega_R \cdot (-i\sin(\Omega_R t)) = -i\Omega_R b_I\) ✓ - Probability conservation: \(|b_I|^2 + |b_{II}|^2 = \sin^2(\Omega_R t) + \cos^2(\Omega_R t) = 1\) ✓

A-2. Extension to a 3-State System: Generalized Quantum Oscillations

→ Back to problem

(a) Calculating the Eigenvalues

Solution strategy: Rewrite \(H\) in terms of \(J\) (the all-ones matrix) and \(I\) (the identity matrix).

Calculation:

Let \(J\) be the \(3\times 3\) matrix with all entries equal to 1:

\[J = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}\]

Rewrite \(H\):

\[H = \begin{pmatrix} E_0 & -A & -A \\ -A & E_0 & -A \\ -A & -A & E_0 \end{pmatrix} = E_0 I + (-A)(J - I) = (E_0 + A)I - AJ\]

Find the eigenvalues of \(J\). Since \(J\) is a rank-1 matrix: - Eigenvalue \(3\): eigenvector \(\frac{1}{\sqrt{3}}(1, 1, 1)^T\) (non-degenerate) - Eigenvalue \(0\): any vector orthogonal to \((1, 1, 1)^T\) (2-fold degenerate)

Since \(H = (E_0 + A)I - AJ\), the eigenvalue of \(H\) corresponding to eigenvalue \(j\) of \(J\) is:

\[E = (E_0 + A) - Aj\]
  • For \(j = 3\): \(E = E_0 + A - 3A = E_0 - 2A\)
  • For \(j = 0\): \(E = E_0 + A - 0 = E_0 + A\) (2-fold degenerate)

Final answer:

\[\boxed{E_1 = E_0 - 2A \quad (\text{non-degenerate}), \quad E_2 = E_0 + A \quad (\text{2-fold degenerate})}\]

(b) Eigenvectors

Eigenvector corresponding to \(E = E_0 - 2A\):

The vector corresponding to eigenvalue 3 of \(J\):

\[|s\rangle = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\]

Eigenvectors corresponding to \(E = E_0 + A\) (2-fold degenerate):

As an orthonormal basis for the 2-dimensional space orthogonal to \((1, 1, 1)^T\), we can choose, for example:

\[|d_1\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \quad |d_2\rangle = \frac{1}{\sqrt{6}}\begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}\]

Verification: \(\langle d_1|d_2\rangle = \frac{1}{\sqrt{12}}(1 \cdot 1 + (-1)\cdot 1 + 0 \cdot (-2)) = 0\) ✓

\(\langle s|d_1\rangle = \frac{1}{\sqrt{6}}(1 - 1 + 0) = 0\) ✓

Direct calculation of \(H|s\rangle\):

\[H|s\rangle = \frac{1}{\sqrt{3}}\begin{pmatrix} E_0 - A - A \\ -A + E_0 - A \\ -A - A + E_0 \end{pmatrix} = \frac{1}{\sqrt{3}}\begin{pmatrix} E_0 - 2A \\ E_0 - 2A \\ E_0 - 2A \end{pmatrix} = (E_0 - 2A)|s\rangle \quad \checkmark\]

(c) Calculating the Probability \(P_1(t)\)

Solution strategy: Expand \(|1\rangle\) in terms of energy eigenstates and apply time evolution.

Calculation:

Expand \(|1\rangle\) in the eigenbasis:

\[\langle s|1\rangle = \frac{1}{\sqrt{3}}, \quad \langle d_1|1\rangle = \frac{1}{\sqrt{2}}, \quad \langle d_2|1\rangle = \frac{1}{\sqrt{6}}\]

Therefore:

\[|1\rangle = \frac{1}{\sqrt{3}}|s\rangle + \frac{1}{\sqrt{2}}|d_1\rangle + \frac{1}{\sqrt{6}}|d_2\rangle\]

Verification: sum of squared coefficients \(= \frac{1}{3} + \frac{1}{2} + \frac{1}{6} = \frac{2+3+1}{6} = 1\) ✓

Time evolution:

\[|\psi(t)\rangle = \frac{1}{\sqrt{3}}e^{-i(E_0-2A)t/\hbar}|s\rangle + \frac{1}{\sqrt{2}}e^{-i(E_0+A)t/\hbar}|d_1\rangle + \frac{1}{\sqrt{6}}e^{-i(E_0+A)t/\hbar}|d_2\rangle\]

Calculate \(\langle 1|\psi(t)\rangle\):

\[C_1(t) = \frac{1}{3}e^{-i(E_0-2A)t/\hbar} + \frac{1}{2}e^{-i(E_0+A)t/\hbar} + \frac{1}{6}e^{-i(E_0+A)t/\hbar}\]
\[= \frac{1}{3}e^{-i(E_0-2A)t/\hbar} + \frac{2}{3}e^{-i(E_0+A)t/\hbar}\]

Factoring out the common phase:

\[C_1(t) = e^{-iE_0 t/\hbar}\left(\frac{1}{3}e^{2iAt/\hbar} + \frac{2}{3}e^{-iAt/\hbar}\right)\]

Probability:

\[P_1(t) = \left|\frac{1}{3}e^{2iAt/\hbar} + \frac{2}{3}e^{-iAt/\hbar}\right|^2\]

Setting \(\phi = At/\hbar\):

\[P_1(t) = \left|\frac{1}{3}e^{2i\phi} + \frac{2}{3}e^{-i\phi}\right|^2\]

Expanding:

\[= \frac{1}{9}|e^{2i\phi}|^2 + \frac{4}{9}|e^{-i\phi}|^2 + \frac{2}{3}\cdot\frac{2}{3}\text{Re}(e^{2i\phi}\cdot e^{i\phi})\]
\[= \frac{1}{9} + \frac{4}{9} + \frac{4}{9}\text{Re}(e^{3i\phi})\]
\[= \frac{5}{9} + \frac{4}{9}\cos(3\phi)\]
\[\boxed{P_1(t) = \frac{1}{9}\left(5 + 4\cos\frac{3At}{\hbar}\right)}\]

Comparison with the two-state system:

Feature Two-state system Three-state system
Oscillation frequency \(A/\hbar\) (argument of \(\cos^2\) in \(P_1\)) \(3A/\hbar\)
Maximum transition probability \(P_1 = 0\) (complete transition) \(P_1 = 1/9\) (minimum value)
Completeness of oscillation Complete (\(P_1\): 1→0→1) Incomplete (\(P_1\): 1→1/9→1)

Discussion:

  1. Frequency: In the three-state system, the angular frequency of oscillation is \(3A/\hbar\), which differs from \(2A/\hbar\) in the two-state system (the angular frequency of oscillation of \(P_1 = \cos^2(At/\hbar)\)).

  2. Completeness of amplitude: In the two-state system, \(P_1(t)\) oscillates completely from 0 to 1 (complete Rabi oscillation). In contrast, in the three-state system the minimum value of \(P_1(t)\) is \(1/9\) (when \(\cos(3At/\hbar) = -1\)), so the system never completely leaves state \(|1\rangle\). This is because the initial state \(|1\rangle\) has weight \(2/3\) in the degenerate eigenspace, and that portion evolves with the same phase, so the interference with the non-degenerate part (weight \(1/3\)) only partially cancels.

  3. Physical interpretation: In the three-state system, since there are two possible "destinations," quantum interference becomes more complex, and simple complete oscillation does not occur.

Verification: - \(t = 0\): \(P_1(0) = \frac{1}{9}(5 + 4) = 1\) ✓ - Time average of \(P_1\): \(\overline{P_1} = 5/9\). Meanwhile, the sum of squared projection probabilities onto each eigenstate \(= (1/3)^2 + (2/3)^2 = 1/9 + 4/9 = 5/9\) ✓ (consistent with the general formula for the long-time average) - Check that \(P_1(t) \geq 0\): minimum value \((5-4)/9 = 1/9 > 0\) ✓