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Ch. 6 Solutions

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Basic

Medium

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Basic

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

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Problem:

Consider a two-state system whose Hamiltonian is given by

\[ H = \begin{pmatrix} E_0 & -A \\ -A & E_0 \end{pmatrix}\]

and suppose the eigenstate \(|II\rangle = \frac{1}{\sqrt{2}}(|1\rangle + |2\rangle)\) has energy \(E_{II} = E_0 - A\). For the amplitude of the stationary state at time \(t\), \(C_1(t) = \frac{1}{\sqrt{2}}e^{-iE_{II}t/\hbar}\), find the value of \(C_1(t)\) at \(t = \pi\hbar/(E_0 - A)\).

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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Problem:

The inversion frequency of an ammonia molecule is \(f = 24{,}000\;\text{MHz}\), and the energy difference satisfies \(2A = hf\). Find \(A\) in units of eV. Use \(h = 6.626 \times 10^{-34}\;\text{J·s}\) and \(1\;\text{eV} = 1.602 \times 10^{-19}\;\text{J}\).

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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Problem:

For the general 2-state Hamiltonian

\[ H = \begin{pmatrix} \alpha & \beta \\ \beta^* & \gamma \end{pmatrix}\]

(where \(\alpha, \gamma\) are real numbers and \(\beta\) is a complex number), expand the eigenvalue equation \(\det(H - E\,I) = 0\) and derive the quadratic equation in \(E\).

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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Problem:

For the eigenvectors given in the text

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

compute \(\langle I|I\rangle\), \(\langle II|II\rangle\), and \(\langle I|II\rangle\) respectively, and verify that these form an orthonormal system. Assume \(\langle 1|1\rangle = \langle 2|2\rangle = 1\) and \(\langle 1|2\rangle = \langle 2|1\rangle = 0\).

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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Problem:

Given that \(|\psi(0)\rangle = |1\rangle\) at \(t = 0\), use equation (6.18) from the text to calculate \(C_2(t) = \langle 2|\psi(t)\rangle\) at time \(t\), and find the probability \(P_2(t) = |C_2(t)|^2\) of being in state \(|2\rangle\). Express the result in terms of \(A\), \(\hbar\), and \(E_0\).

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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Problem:

Solve the differential equation \(i\hbar\,\dfrac{dC}{dt} = E\,C\) (where \(E\) is a real constant), and find \(C(t)\) subject to the condition \(C(0) = C_0\). Also show that \(|C(t)|^2\) is independent of time.

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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Problem:

Using Eq. (6.17a) \(|1\rangle = \frac{1}{\sqrt{2}}(|I\rangle + |II\rangle)\) and Eq. (6.17b) \(|2\rangle = \frac{1}{\sqrt{2}}(-|I\rangle + |II\rangle)\), conversely express \(|I\rangle\) and \(|II\rangle\) in terms of \(|1\rangle\) and \(|2\rangle\), and verify that Eqs. (6.15a) and (6.15b) are reproduced.

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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Problem:

Verify that the matrix

\[ M = \begin{pmatrix} 3 & 2 - i \\ 2 + i & 5 \end{pmatrix}\]

is a Hermitian matrix. That is, show that \(M_{ij}^* = M_{ji}\) holds for all \(i, j\).

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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Problem:

In the time evolution equations for a two-state system

\[ i\hbar\frac{dC_1}{dt} = H_{11}C_1 + H_{12}C_2, \quad i\hbar\frac{dC_2}{dt} = H_{21}C_1 + H_{22}C_2\]

calculate the time derivative \(dP/dt\) of the total probability \(P = |C_1|^2 + |C_2|^2\). Derive the conditions for \(dP/dt = 0\) to hold for arbitrary \(C_1, C_2\), namely that \(H_{11}, H_{22}\) are real and that \(H_{12}^* = H_{21}\).

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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Problem:

Suppose that at \(t = 0\), \(|\psi(0)\rangle = |1\rangle\) (nitrogen is "up"). Under the Hamiltonian (6.7), derive the probabilities at time \(t\):

\[ P_1(t) = |\langle 1|\psi(t)\rangle|^2, \quad P_2(t) = |\langle 2|\psi(t)\rangle|^2\]

and show the following results: $$

P_1(t) = \cos^2!\left(\frac{At}{\hbar}\right), \quad P_2(t) = \sin^2!\left(\frac{At}{\hbar}\right)$$

Also, verify that \(P_1(t) + P_2(t) = 1\) holds at all times, and express the period \(T\) over which the system oscillates completely between states \(|1\rangle\) and \(|2\rangle\) in terms of \(A\) and \(\hbar\).

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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Problem:

When a uniform electrostatic field \(\mathcal{E}\) is applied, an energy contribution from the electric dipole moment \(\pm \mu\mathcal{E}\) is added depending on the position of the nitrogen atom. When the Hamiltonian becomes

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

find the eigenvalues \(E_{\pm}\) and express them in the following form: $$

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

Furthermore, discuss how the energy levels behave in the limit \(\mu\mathcal{E} \ll A\) and in the limit \(\mu\mathcal{E} \gg A\).

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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Problem:

Using the unitary matrix

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

diagonalize the Hamiltonian (6.7) of the ammonia molecule. That is, compute \(U^\dagger H\, U\) and show that the result is $$

U^\dagger H\, U = \begin{pmatrix} E_0 + A & 0 \ 0 & E_0 - A \end{pmatrix}$$

Also, describe the relationship between the column vectors of \(U\) and the eigenvectors \(|I\rangle\), \(|II\rangle\).

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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Problem:

An oscillating electric field \(\mathcal{E}(t) = \mathcal{E}_0 \cos\omega t\) with angular frequency \(\omega\) is applied to an ammonia molecule. When described in the energy basis \(\{|I\rangle, |II\rangle\}\), the Hamiltonian can be written as

\[ H = \begin{pmatrix} E_I & \mu\mathcal{E}_0\cos\omega t \\ \mu\mathcal{E}_0\cos\omega t & E_{II} \end{pmatrix}\]

where \(E_I = E_0 + A\), \(E_{II} = E_0 - A\).

Given the initial state \(|\psi(0)\rangle = |II\rangle\) (the lower energy state), find the transition probability to state \(|I\rangle\) by following the steps below.

(a) Setting \(C_I(t) = b_I(t)\,e^{-iE_I t/\hbar}\), \(C_{II}(t) = b_{II}(t)\,e^{-iE_{II}t/\hbar}\), derive the differential equations for \(b_I(t)\) and \(b_{II}(t)\).

(b) Rotating wave approximation (RWA): Near the resonance condition \(\omega \approx (E_I - E_{II})/\hbar = 2A/\hbar\), simplify the equations by neglecting the rapidly oscillating terms.

(c) Under the resonance condition \(\omega = 2A/\hbar\), solve for \(b_I(t)\) and show that the transition probability \(P_{II \to I}(t) = |b_I(t)|^2\) is given by $$

P_{II \to I}(t) = \sin^2!\left(\frac{\mu\mathcal{E}_0\,t}{2\hbar}\right)$$

Explain why this forms the basis for stimulated emission in the ammonia maser.

(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

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Problem:

We extend the 2-state system and consider a system where three equivalent states \(|1\rangle\), \(|2\rangle\), \(|3\rangle\) are coupled to each other with the same tunneling amplitude \(-A\). The Hamiltonian is given by

\[ H = \begin{pmatrix} E_0 & -A & -A \\ -A & E_0 & -A \\ -A & -A & E_0 \end{pmatrix}\]

(a) Find all eigenvalues of this Hamiltonian. (Hint: Rewrite the matrix in the form \(E_0\,I + (-A)(J - I)\), where \(J\) is the \(3\times 3\) matrix with all entries equal to 1 and \(I\) is the identity matrix.)

(b) Find the eigenvectors corresponding to each eigenvalue and normalize them. If there is degeneracy, state its degree.

(c) If the system is in state \(|1\rangle\) at \(t = 0\), find the probability \(P_1(t)\) of finding the system in state \(|1\rangle\) at time \(t\). Discuss how the characteristics of the oscillation (frequency, completeness of amplitude) differ compared to Rabi oscillations in the 2-state system.

(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 \((5-4)/9 = 1/9 > 0\) ✓