Physical Chemistry 2 Quiz: Superposition And Linear Combinations
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Superposition And Linear CombinationsQuestion 1 of 5

A quantum system is prepared in the normalized state ∣ψ⟩=13∣0⟩+23∣1⟩|\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 M^+=∣+⟩⟨+∣\hat{M}_+ = |+\rangle\langle+| and M^−=∣−⟩⟨−∣\hat{M}_- = |-\rangle\langle-|, where ∣+⟩=12(∣0⟩+∣1⟩)|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) and ∣−⟩=12(∣0⟩−∣1⟩)|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle), what is the probability of obtaining the ++ result?

13\frac{1}{3}
23\frac{2}{3}
16(3+22)\frac{1}{6}(3 + 2\sqrt{2})
16(3−22)\frac{1}{6}(3 - 2\sqrt{2})
12\frac{1}{2}
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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Superposition And Linear Combinations

Practice Superposition And Linear Combinations in Physical Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Superposition And Linear Combinations, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A quantum system is prepared in the normalized state ∣ψ⟩=13∣0⟩+23∣1⟩|\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 M^+=∣+⟩⟨+∣\hat{M}_+ = |+\rangle\langle+| and M^−=∣−⟩⟨−∣\hat{M}_- = |-\rangle\langle-|, where ∣+⟩=12(∣0⟩+∣1⟩)|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) and ∣−⟩=12(∣0⟩−∣1⟩)|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle), what is the probability of obtaining the ++ result?

  1. 13\frac{1}{3}
  2. 23\frac{2}{3}
  3. 16(3+22)\frac{1}{6}(3 + 2\sqrt{2}) (correct answer)
  4. 16(3−22)\frac{1}{6}(3 - 2\sqrt{2})
  5. 12\frac{1}{2}
Explanation: When you encounter quantum measurement problems involving basis changes, you need to calculate the probability using the Born rule: P=∣⟨measurement state∣system state⟩∣2P = |\langle \text{measurement state} | \text{system state} \rangle|^2. To find the probability of measuring the ++ result, you must calculate ∣⟨+∣ψ⟩∣2|\langle + | \psi \rangle|^2. First, find the inner product: ⟨+∣ψ⟩=⟨+∣(13∣0⟩+23∣1⟩)\langle + | \psi \rangle = \langle + | \left(\frac{1}{\sqrt{3}}|0\rangle + \frac{\sqrt{2}}{\sqrt{3}}|1\rangle\right) Since ⟨+∣=12(⟨0∣+⟨1∣)\langle + | = \frac{1}{\sqrt{2}}(\langle 0| + \langle 1|), this becomes: ⟨+∣ψ⟩=12(13+23)=1+26\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+=∣1+26∣2=(1+2)26=1+22+26=3+226P_+ = \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 (13\frac{1}{3}) would result from incorrectly using just the coefficient of ∣0⟩|0\rangle squared. Answer B (23\frac{2}{3}) might come from using the coefficient of ∣1⟩|1\rangle squared. Answer D (16(3−22)\frac{1}{6}(3 - 2\sqrt{2})) represents the probability of measuring the −- result, which you'd get by calculating ∣⟨−∣ψ⟩∣2|\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.

Question 2

A quantum system has three orthonormal energy eigenstates ∣E1⟩|E_1\rangle, ∣E2⟩|E_2\rangle, and ∣E3⟩|E_3\rangle with energies E1=0E_1 = 0, E2=ℏωE_2 = \hbar\omega, and E3=2ℏωE_3 = 2\hbar\omega. At t=0t = 0, the system is in the state ∣ψ(0)⟩=12∣E1⟩+12∣E2⟩+12∣E3⟩|\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>0t > 0 will the overlap ∣⟨ψ(0)∣ψ(t)⟩∣|\langle\psi(0)|\psi(t)\rangle| first equal 12\frac{1}{\sqrt{2}}?

  1. π3ω\frac{\pi}{3\omega}
  2. π2ω\frac{\pi}{2\omega}
  3. 2π3ω\frac{2\pi}{3\omega} (correct answer)
  4. πω\frac{\pi}{\omega}
  5. 4π3ω\frac{4\pi}{3\omega}
Explanation: 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 ∣ψ(t)⟩=12e−iE1t/ℏ∣E1⟩+12e−iE2t/ℏ∣E2⟩+12e−iE3t/ℏ∣E3⟩|\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: ∣ψ(t)⟩=12∣E1⟩+12e−iωt∣E2⟩+12e−2iωt∣E3⟩|\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: ⟨ψ(0)∣ψ(t)⟩=12+14e−iωt+14e−2iωt\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: ∣⟨ψ(0)∣ψ(t)⟩∣2=∣12+14(e−iωt+e−2iωt)∣2|\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−iωt+e−2iωt=e−iωt(1+e−iωt)=2e−3iωt/2cos⁡(ωt/2)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: ∣⟨ψ(0)∣ψ(t)⟩∣=∣12+12e−3iωt/2cos⁡(ωt/2)∣|\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 12\frac{1}{\sqrt{2}} and solving gives cos⁡(ωt/2)=−12\cos(\omega t/2) = -\frac{1}{2}, so ωt/2=2π3\omega t/2 = \frac{2\pi}{3}, yielding t=2π3ωt = \frac{2\pi}{3\omega}. Answer A (π3ω\frac{\pi}{3\omega}) gives cos⁡(π/6)=3/2≠−1/2\cos(\pi/6) = \sqrt{3}/2 \neq -1/2. Answer B (π2ω\frac{\pi}{2\omega}) gives cos⁡(π/4)=2/2≠−1/2\cos(\pi/4) = \sqrt{2}/2 \neq -1/2. Answer D (πω\frac{\pi}{\omega}) gives cos⁡(π/2)=0≠−1/2\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.

Question 3

A particle is confined to a ring of radius RR and is described by the wavefunction ψ(ϕ)=A(eimϕ+αe−imϕ)\psi(\phi) = A(e^{im\phi} + \alpha e^{-im\phi}) where ϕ\phi is the azimuthal angle, mm is a positive integer, α\alpha is a real parameter, and AA is the normalization constant. If the expectation value of the angular momentum operator L^z=−iℏddϕ\hat{L}_z = -i\hbar\frac{d}{d\phi} is zero, what is the probability current density at ϕ=π/2\phi = \pi/2?

  1. ℏα(1−α2)mR2(1+α2)\frac{\hbar\alpha(1-\alpha^2)}{mR^2(1+\alpha^2)}
  2. 2ℏαmR2(1+α2)\frac{2\hbar\alpha}{mR^2(1+\alpha^2)}
  3. ℏ(1−α2)mR2(1+α2)\frac{\hbar(1-\alpha^2)}{mR^2(1+\alpha^2)}
  4. 00 (correct answer)
  5. ℏαmR2(1+α2)\frac{\hbar\alpha}{mR^2(1+\alpha^2)}
Explanation: 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 JJ measures the flow of probability and is given by J=ℏ2mi(ψ∗∂ψ∂ϕ−ψ∂ψ∗∂ϕ)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, ⟨L^z⟩=∫02πψ∗(−iℏdψdϕ)dϕ=mR2∫02πJ dϕ\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 ⟨L^z⟩=0\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 ψ(ϕ)=A(eimϕ+αe−imϕ)\psi(\phi) = A(e^{im\phi} + \alpha e^{-im\phi}), this constraint determines the value of α\alpha. When you work through the normalization and expectation value calculation, you find that ⟨L^z⟩=0\langle\hat{L}_z\rangle = 0 requires α=1\alpha = 1. With α=1\alpha = 1, the wavefunction becomes ψ(ϕ)=A(eimϕ+e−imϕ)=2Acos⁡(mϕ)\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 ψ∗∂ψ∂ϕ\psi^*\frac{\partial\psi}{\partial\phi}, and both ψ\psi and its derivative are real when α=1\alpha = 1, the probability current density is zero everywhere, including at ϕ=π/2\phi = \pi/2. Options A, B, and C all give non-zero expressions that would apply for different values of α\alpha, but they ignore the constraint that ⟨L^z⟩=0\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.

Question 4

Consider a spin-1 particle (S=1S = 1) in a superposition of SzS_z eigenstates: ∣ψ⟩=a∣Sz=+ℏ⟩+b∣Sz=0⟩+c∣Sz=−ℏ⟩|\psi\rangle = a|S_z = +\hbar\rangle + b|S_z = 0\rangle + c|S_z = -\hbar\rangle. If measurements of SxS_x yield the eigenvalue +ℏ+\hbar with probability 12\frac{1}{2}, 00 with probability 14\frac{1}{4}, and −ℏ-\hbar with probability 14\frac{1}{4}, and the state is normalized with all coefficients real, which constraint must be satisfied by the coefficients?

  1. a=ca = c and a2+b2+c2=1a^2 + b^2 + c^2 = 1
  2. a=−ca = -c and a2+b2+c2=1a^2 + b^2 + c^2 = 1 (correct answer)
  3. a2=c2=14a^2 = c^2 = \frac{1}{4} and b2=12b^2 = \frac{1}{2}
  4. a=12a = \frac{1}{2}, b=0b = 0, c=12c = \frac{1}{2}
  5. a2+c2=34a^2 + c^2 = \frac{3}{4} and b=0b = 0
Explanation: This question tests your understanding of quantum measurement probabilities and the relationship between different spin basis states. When measuring SxS_x on a state written in the SzS_z basis, you need to transform between these bases using the overlap coefficients. For a spin-1 particle, the SxS_x eigenstates can be written as specific linear combinations of the SzS_z eigenstates. The key insight is that the measurement probabilities depend on the overlap between your state ∣ψ⟩|\psi\rangle and each SxS_x eigenstate. When you work through the matrix elements and calculate these probabilities, you find that the given probabilities (12\frac{1}{2}, 14\frac{1}{4}, 14\frac{1}{4}) can only be reproduced when the coefficients satisfy a specific antisymmetric relationship: a=−ca = -c. This antisymmetry, combined with the normalization condition a2+b2+c2=1a^2 + b^2 + c^2 = 1, gives you answer B. Answer A suggests a=ca = 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 SxS_x measurement data. Answer D sets b=0b = 0 and makes a=ca = 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 SxS_x, SyS_y, and SzS_z bases for different spin values.

Question 5

A quantum system is described by the Hamiltonian H^=ℏω(∣1⟩⟨1∣+2∣2⟩⟨2∣+∣1⟩⟨2∣+∣2⟩⟨1∣)\hat{H} = \hbar\omega(|1\rangle\langle1| + 2|2\rangle\langle2| + |1\rangle\langle2| + |2\rangle\langle1|) where ∣1⟩|1\rangle and ∣2⟩|2\rangle are orthonormal basis states. If the system is initially prepared in the state ∣ψ(0)⟩=12(∣1⟩+i∣2⟩)|\psi(0)\rangle = \frac{1}{\sqrt{2}}(|1\rangle + i|2\rangle), what is the probability of finding the system in the state ∣ϕ⟩=12(∣1⟩−∣2⟩)|\phi\rangle = \frac{1}{\sqrt{2}}(|1\rangle - |2\rangle) at time t=π2ωt = \frac{\pi}{2\omega}?

  1. 14\frac{1}{4}
  2. 12\frac{1}{2}
  3. 34\frac{3}{4}
  4. 12(1+12)\frac{1}{2}(1 + \frac{1}{\sqrt{2}})
  5. 12(1−12)\frac{1}{2}(1 - \frac{1}{\sqrt{2}}) (correct answer)
Explanation: 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: To find eigenvalues, solve det⁡(H^−λI)=0\det(\hat{H} - \lambda I) = 0: det⁡(ℏω−λℏωℏω2ℏω−λ)=0\det \begin{pmatrix} \hbar\omega - \lambda & \hbar\omega \\ \hbar\omega & 2\hbar\omega - \lambda \end{pmatrix} = 0 This gives λ2−3ℏωλ+ℏ2ω2=0\lambda^2 - 3\hbar\omega\lambda + \hbar^2\omega^2 = 0, yielding eigenvalues E1=ℏω(3−52)E_1 = \hbar\omega(\frac{3-\sqrt{5}}{2}) and E2=ℏω(3+52)E_2 = \hbar\omega(\frac{3+\sqrt{5}}{2}). The corresponding eigenvectors are ∣v1⟩=12+5(1+52∣1⟩−∣2⟩)|v_1\rangle = \frac{1}{\sqrt{2+\sqrt{5}}}(\sqrt{\frac{1+\sqrt{5}}{2}}|1\rangle - |2\rangle) and ∣v2⟩=12−5(5−12∣1⟩+∣2⟩)|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 ∣ψ(t)⟩=∑ncne−iEnt/ℏ∣vn⟩|\psi(t)\rangle = \sum_n c_n e^{-iE_nt/\hbar}|v_n\rangle, then calculate ∣⟨ϕ∣ψ(t)⟩∣2|\langle\phi|\psi(t)\rangle|^2 at t=π2ωt = \frac{\pi}{2\omega}. The detailed calculation yields a probability of 12(1+12)≈0.854\frac{1}{2}(1 + \frac{1}{\sqrt{2}}) \approx 0.854. Answer choices A (14\frac{1}{4}), B (12\frac{1}{2}), and C (34\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.