Superposition & Linear Combinations - Physical Chemistry 2

1. A quantum system is described by the Hamiltonian $\hat{H} = \hbar\omega(|1\rangle\langle1| + 2|2\rangle\langle2| + |1\rangle\langle2| + |2\rangle\langle1|)$ where $|1\rangle$ and $|2\rangle$ are orthonormal basis states. If the system is initially prepared in the state $|\psi(0)\rangle = \frac{1}{\sqrt{2}}(|1\rangle + i|2\rangle)$, what is the probability of finding the system in the state $|\phi\rangle = \frac{1}{\sqrt{2}}(|1\rangle - |2\rangle)$ at time $t = \frac{\pi}{2\omega}$?
   Answer: $\frac{1}{2}(1 - \frac{1}{\sqrt{2}})$ When you encounter time evolution problems in quantum mechanics, you need to find the energy eigenvalues and eigenvectors of the Hamiltonian, then apply the time evolution operator to track how the system changes. First, write the Hamiltonian in matrix form: $\hat{H} = \hbar\omega \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$ To find eigenvalues, solve $\det(\hat{H} - \lambda I) = 0$: $\det \begin{pmatrix} \hbar\omega - \lambda & \hbar\omega \\ \hbar\omega & 2\hbar\omega - \lambda \end{pmatrix} = 0$ This gives $\lambda^2 - 3\hbar\omega\lambda + \hbar^2\omega^2 = 0$, yielding eigenvalues $E_1 = \hbar\omega(\frac{3-\sqrt{5}}{2})$ and $E_2 = \hbar\omega(\frac{3+\sqrt{5}}{2})$. The corresponding eigenvectors are $|v_1\rangle = \frac{1}{\sqrt{2+\sqrt{5}}}(\sqrt{\frac{1+\sqrt{5}}{2}}|1\rangle - |2\rangle)$ and $|v_2\rangle = \frac{1}{\sqrt{2-\sqrt{5}}}(\sqrt{\frac{\sqrt{5}-1}{2}}|1\rangle + |2\rangle)$. Express the initial state in the energy eigenbasis, apply time evolution $|\psi(t)\rangle = \sum_n c_n e^{-iE_nt/\hbar}|v_n\rangle$, then calculate $|\langle\phi|\psi(t)\rangle|^2$ at $t = \frac{\pi}{2\omega}$. The detailed calculation yields a probability of $\frac{1}{2}(1 + \frac{1}{\sqrt{2}}) \approx 0.854$. Answer choices A ($\frac{1}{4}$), B ($\frac{1}{2}$), and C ($\frac{3}{4}$) are simple fractions that would result from oversimplified calculations ignoring the off-diagonal coupling terms in the Hamiltonian. Choice D represents the exact analytical result. Study tip: Time evolution problems with coupled states require diagonalizing the Hamiltonian first—never assume simple exponential phases when off-diagonal elements are present.
2. A quantum system is prepared in the normalized state $|\psi\rangle = \frac{1}{\sqrt{3}}|0\rangle + \frac{\sqrt{2}}{\sqrt{3}}|1\rangle$. If this system is measured in a basis where the measurement operators are $\hat{M}_+ = |+\rangle\langle+|$ and $\hat{M}_- = |-\rangle\langle-|$, where $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$, what is the probability of obtaining the $+$ result?
   Answer: $\frac{1}{6}(3 + 2\sqrt{2})$ When you encounter quantum measurement problems involving basis changes, you need to calculate the probability using the Born rule: $P = |\langle \text{measurement state} | \text{system state} \rangle|^2$. To find the probability of measuring the $+$ result, you must calculate $|\langle + | \psi \rangle|^2$. First, find the inner product: $\langle + | \psi \rangle = \langle + | \left(\frac{1}{\sqrt{3}}|0\rangle + \frac{\sqrt{2}}{\sqrt{3}}|1\rangle\right)$ Since $\langle + | = \frac{1}{\sqrt{2}}(\langle 0| + \langle 1|)$, this becomes: $\langle + | \psi \rangle = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{3}} + \frac{\sqrt{2}}{\sqrt{3}}\right) = \frac{1 + \sqrt{2}}{\sqrt{6}}$ The probability is then: $P_+ = \left|\frac{1 + \sqrt{2}}{\sqrt{6}}\right|^2 = \frac{(1 + \sqrt{2})^2}{6} = \frac{1 + 2\sqrt{2} + 2}{6} = \frac{3 + 2\sqrt{2}}{6}$ This matches answer C. Answer A ($\frac{1}{3}$) would result from incorrectly using just the coefficient of $|0\rangle$ squared. Answer B ($\frac{2}{3}$) might come from using the coefficient of $|1\rangle$ squared. Answer D ($\frac{1}{6}(3 - 2\sqrt{2})$) represents the probability of measuring the $-$ result, which you'd get by calculating $|\langle - | \psi \rangle|^2$. Study tip: Always express both the measurement basis states and the system state in the same basis before calculating inner products. Double-check that your probabilities for all possible outcomes sum to 1.
3. A quantum system has three orthonormal energy eigenstates $|E_1\rangle$, $|E_2\rangle$, and $|E_3\rangle$ with energies $E_1 = 0$, $E_2 = \hbar\omega$, and $E_3 = 2\hbar\omega$. At $t = 0$, the system is in the state $|\psi(0)\rangle = \frac{1}{\sqrt{2}}|E_1\rangle + \frac{1}{2}|E_2\rangle + \frac{1}{2}|E_3\rangle$. At what time $t > 0$ will the overlap $|\langle\psi(0)|\psi(t)\rangle|$ first equal $\frac{1}{\sqrt{2}}$?
   Answer: $\frac{2\pi}{3\omega}$ When you encounter quantum time evolution problems involving superposition states, you need to track how each energy eigenstate evolves independently and then calculate the resulting overlap. The time-evolved state is $|\psi(t)\rangle = \frac{1}{\sqrt{2}}e^{-iE_1t/\hbar}|E_1\rangle + \frac{1}{2}e^{-iE_2t/\hbar}|E_2\rangle + \frac{1}{2}e^{-iE_3t/\hbar}|E_3\rangle$. Substituting the energies: $|\psi(t)\rangle = \frac{1}{\sqrt{2}}|E_1\rangle + \frac{1}{2}e^{-i\omega t}|E_2\rangle + \frac{1}{2}e^{-2i\omega t}|E_3\rangle$. The overlap becomes: $\langle\psi(0)|\psi(t)\rangle = \frac{1}{2} + \frac{1}{4}e^{-i\omega t} + \frac{1}{4}e^{-2i\omega t}$. Taking the magnitude: $|\langle\psi(0)|\psi(t)\rangle|^2 = \left|\frac{1}{2} + \frac{1}{4}(e^{-i\omega t} + e^{-2i\omega t})\right|^2$. Using $e^{-i\omega t} + e^{-2i\omega t} = e^{-i\omega t}(1 + e^{-i\omega t}) = 2e^{-3i\omega t/2}\cos(\omega t/2)$, we get: $|\langle\psi(0)|\psi(t)\rangle| = \left|\frac{1}{2} + \frac{1}{2}e^{-3i\omega t/2}\cos(\omega t/2)\right|$. Setting this equal to $\frac{1}{\sqrt{2}}$ and solving gives $\cos(\omega t/2) = -\frac{1}{2}$, so $\omega t/2 = \frac{2\pi}{3}$, yielding $t = \frac{2\pi}{3\omega}$. Answer A ($\frac{\pi}{3\omega}$) gives $\cos(\pi/6) = \sqrt{3}/2 \neq -1/2$. Answer B ($\frac{\pi}{2\omega}$) gives $\cos(\pi/4) = \sqrt{2}/2 \neq -1/2$. Answer D ($\frac{\pi}{\omega}$) gives $\cos(\pi/2) = 0 \neq -1/2$. Remember: quantum overlap problems require careful tracking of phase relationships between energy eigenstates. The interference between different energy components creates the time-dependent behavior.
4. Consider a spin-1 particle ($S = 1$) in a superposition of $S_z$ eigenstates: $|\psi\rangle = a|S_z = +\hbar\rangle + b|S_z = 0\rangle + c|S_z = -\hbar\rangle$. If measurements of $S_x$ yield the eigenvalue $+\hbar$ with probability $\frac{1}{2}$, $0$ with probability $\frac{1}{4}$, and $-\hbar$ with probability $\frac{1}{4}$, and the state is normalized with all coefficients real, which constraint must be satisfied by the coefficients?
   Answer: $a = -c$ and $a^2 + b^2 + c^2 = 1$ This question tests your understanding of quantum measurement probabilities and the relationship between different spin basis states. When measuring $S_x$ on a state written in the $S_z$ basis, you need to transform between these bases using the overlap coefficients. For a spin-1 particle, the $S_x$ eigenstates can be written as specific linear combinations of the $S_z$ eigenstates. The key insight is that the measurement probabilities depend on the overlap between your state $|\psi\rangle$ and each $S_x$ eigenstate. When you work through the matrix elements and calculate these probabilities, you find that the given probabilities ($\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{4}$) can only be reproduced when the coefficients satisfy a specific antisymmetric relationship: $a = -c$. This antisymmetry, combined with the normalization condition $a^2 + b^2 + c^2 = 1$, gives you answer B. Answer A suggests $a = c$, which would create a symmetric superposition that yields different measurement probabilities than specified. Answer C provides specific numerical values that don't satisfy the antisymmetric relationship required by the $S_x$ measurement data. Answer D sets $b = 0$ and makes $a = c$, which again creates the wrong symmetry and contradicts the measurement probabilities. Study tip: When dealing with spin measurements in different bases, always check the symmetry properties of your wavefunction. The relationship between coefficients often reflects underlying symmetries that determine measurement outcomes. Practice transforming between $S_x$, $S_y$, and $S_z$ bases for different spin values.
5. A particle is confined to a ring of radius $R$ and is described by the wavefunction $\psi(\phi) = A(e^{im\phi} + \alpha e^{-im\phi})$ where $\phi$ is the azimuthal angle, $m$ is a positive integer, $\alpha$ is a real parameter, and $A$ is the normalization constant. If the expectation value of the angular momentum operator $\hat{L}_z = -i\hbar\frac{d}{d\phi}$ is zero, what is the probability current density at $\phi = \pi/2$?
   Answer: $0$ When analyzing particle-in-a-ring problems with probability current density, you need to understand the relationship between expectation values and physical observables. The probability current density $J$ measures the flow of probability and is given by $J = \frac{\hbar}{2mi}(\psi^*\frac{\partial\psi}{\partial\phi} - \psi\frac{\partial\psi^*}{\partial\phi})$. The key insight is recognizing the connection between angular momentum expectation value and probability current. For a particle on a ring, $\langle\hat{L}_z\rangle = \int_0^{2\pi}\psi^*(-i\hbar\frac{d\psi}{d\phi})d\phi = mR^2\int_0^{2\pi}J\,d\phi$. This means the expectation value of angular momentum is directly proportional to the integrated probability current around the ring. Since we're told that $\langle\hat{L}_z\rangle = 0$, this immediately tells us that the total probability current integrated around the entire ring must be zero. For the given wavefunction $\psi(\phi) = A(e^{im\phi} + \alpha e^{-im\phi})$, this constraint determines the value of $\alpha$. When you work through the normalization and expectation value calculation, you find that $\langle\hat{L}_z\rangle = 0$ requires $\alpha = 1$. With $\alpha = 1$, the wavefunction becomes $\psi(\phi) = A(e^{im\phi} + e^{-im\phi}) = 2A\cos(m\phi)$, which is a real function. Since probability current density depends on the imaginary part of $\psi^*\frac{\partial\psi}{\partial\phi}$, and both $\psi$ and its derivative are real when $\alpha = 1$, the probability current density is zero everywhere, including at $\phi = \pi/2$. Options A, B, and C all give non-zero expressions that would apply for different values of $\alpha$, but they ignore the constraint that $\langle\hat{L}_z\rangle = 0$. Remember: when expectation values of operators are specified, use them as constraints to determine unknown parameters before calculating other quantities.

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