Chapter 11 Quiz: Tensor Products and Composite Systems

Question 1. If $\mathcal{H}_A$ has dimension 3 and $\mathcal{H}_B$ has dimension 4, what is the dimension of $\mathcal{H}_A \otimes \mathcal{H}_B$?

(a) 7 (b) 12 (c) 64 (d) 81

Answer: (b). The dimension of a tensor product space is $d_A \times d_B = 3 \times 4 = 12$.

Question 2. Which of the following is a valid notation for the tensor product $|\psi\rangle_A \otimes |\phi\rangle_B$?

(a) $|\psi\rangle|\phi\rangle$ (b) $|\psi, \phi\rangle$ (c) $|\psi\phi\rangle$ (d) All of the above

Answer: (d). All three are standard shorthand notations for the tensor product.

Question 3. The state $|\Psi\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |01\rangle)$ is:

(a) Entangled (b) Separable (c) Neither entangled nor separable (d) Undefined

Answer: (b). We can factor: $|\Psi\rangle = |0\rangle \otimes \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = |0\rangle|+\rangle$. The coefficient matrix is $C = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}$, which has rank 1.

Question 4. The state $|\Psi\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$ is:

(a) A product state in a different basis (b) Entangled with Schmidt rank 2 (c) Separable because it has only two terms (d) Not a valid quantum state

Answer: (b). The coefficient matrix $C = \frac{1}{\sqrt{2}}\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$ has rank 2 (determinant $= -1/2 \neq 0$), so the state is entangled.

Question 5. What is the Kronecker product of $|0\rangle = \begin{pmatrix}1\\0\end{pmatrix}$ and $|1\rangle = \begin{pmatrix}0\\1\end{pmatrix}$?

(a) $\begin{pmatrix}0\\1\\0\\0\end{pmatrix}$ (b) $\begin{pmatrix}1\\0\\0\\1\end{pmatrix}$ (c) $\begin{pmatrix}0\\0\\1\\0\end{pmatrix}$ (d) $\begin{pmatrix}0\\1\\1\\0\end{pmatrix}$

Answer: (a). $|0\rangle \otimes |1\rangle = \begin{pmatrix}1 \cdot \begin{pmatrix}0\\1\end{pmatrix} \\ 0 \cdot \begin{pmatrix}0\\1\end{pmatrix}\end{pmatrix} = \begin{pmatrix}0\\1\\0\\0\end{pmatrix}$.

Question 6. The Schmidt decomposition of a bipartite pure state:

(a) Is a double sum over two indices (b) Is a single sum with orthonormal Schmidt bases for each subsystem (c) Only exists for entangled states (d) Requires the state to be in the computational basis

Answer: (b). The key feature of the Schmidt decomposition is that it reduces to a single sum $\sum_k \lambda_k |a_k\rangle|b_k\rangle$ with orthonormal bases, which is why it is so powerful.

Question 7. A bipartite pure state has Schmidt coefficients $\lambda_1 = 1$ and all others zero. This state is:

(a) Maximally entangled (b) Partially entangled (c) Separable (a product state) (d) A mixed state

Answer: (c). Schmidt rank 1 (only one nonzero coefficient) means the state is a product state.

Question 8. The reduced density matrix $\hat{\rho}_A = \text{Tr}_B(|\Phi^+\rangle\langle\Phi^+|)$ for the Bell state $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ is:

(a) $|0\rangle\langle 0|$ (b) $|+\rangle\langle +|$ (c) $\frac{1}{2}\hat{I}$ (d) $\frac{1}{2}(|0\rangle\langle 0| + |0\rangle\langle 1|)$

Answer: (c). The maximally entangled Bell state produces the maximally mixed reduced state $\hat{\rho}_A = \frac{1}{2}|0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle 1| = \frac{1}{2}\hat{I}$.

Question 9. The partial trace operation is the quantum analogue of:

(a) Taking the determinant of a matrix (b) Marginalizing a joint probability distribution (c) Diagonalizing a Hamiltonian (d) Projecting onto an eigenstate

Answer: (b). Just as marginalizing a joint distribution $P(a,b)$ over $b$ gives $P(a) = \sum_b P(a,b)$, the partial trace over $B$ gives the effective state of $A$ alone.

Question 10. How many Bell states form a complete orthonormal basis for the two-qubit Hilbert space?

(a) 2 (b) 3 (c) 4 (d) 8

Answer: (c). The four Bell states ($|\Phi^+\rangle, |\Phi^-\rangle, |\Psi^+\rangle, |\Psi^-\rangle$) form a complete orthonormal basis for $\mathbb{C}^2 \otimes \mathbb{C}^2$.

Question 11. The state $|\Psi^-\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$ is special because:

(a) It is the only non-entangled Bell state (b) It is the singlet state with total spin $S = 0$ (c) It has different Schmidt coefficients from the other Bell states (d) It cannot be created with a CNOT gate

Answer: (b). $|\Psi^-\rangle$ is the unique spin-singlet state, invariant under rotation of the spin quantization axis. It has the same Schmidt coefficients ($1/\sqrt{2}, 1/\sqrt{2}$) as all other Bell states.

Question 12. The coefficient matrix $C$ for a two-qubit state has entries $C_{ij} = \alpha_{ij}$. The state is entangled if and only if:

(a) $C$ is unitary (b) $C$ is Hermitian (c) $\det(C) \neq 0$ (equivalently, $\text{rank}(C) > 1$) (d) $\text{Tr}(C) = 1$

Answer: (c). A pure bipartite state is entangled iff its coefficient matrix has rank greater than 1, which for a $2 \times 2$ matrix is equivalent to having nonzero determinant.

Question 13. For 10 qubits, the dimension of the composite Hilbert space is:

(a) 20 (b) 100 (c) 1,024 (d) $10^{10}$

Answer: (c). $2^{10} = 1024$.

Question 14. If a pure state $|\Psi\rangle_{AB}$ has entanglement entropy $S = 0$, then:

(a) The state is maximally entangled (b) The state is separable (c) The state is a mixed state (d) The partial trace is undefined

Answer: (b). $S = 0$ means the reduced density matrix is pure, which happens if and only if the composite state is a product state.

Question 15. The maximum entanglement entropy for two qubits is:

(a) 0 (b) $\frac{1}{2}$ (c) 1 (d) 2

Answer: (c). For two qubits, $S_{\max} = \log_2(\min(d_A, d_B)) = \log_2(2) = 1$ ebit.

Question 16. The operator $\hat{\sigma}_z \otimes \hat{I}$ acting on the two-qubit state $|10\rangle$ gives:

(a) $|10\rangle$ (b) $-|10\rangle$ (c) $|00\rangle$ (d) $|11\rangle$

Answer: (b). $(\hat{\sigma}_z \otimes \hat{I})|10\rangle = (\hat{\sigma}_z|1\rangle) \otimes (\hat{I}|0\rangle) = (-|1\rangle) \otimes |0\rangle = -|10\rangle$.

Question 17. Which statement about the Schmidt decomposition is FALSE?

(a) The Schmidt coefficients are unique (up to ordering) (b) The Schmidt bases are always unique (c) The Schmidt rank equals the rank of the coefficient matrix (d) The Schmidt coefficients are the singular values of the coefficient matrix

Answer: (b). The Schmidt bases are not unique when two or more Schmidt coefficients are equal (degenerate). In that case, any unitary rotation within the degenerate subspace produces an equally valid Schmidt basis.

Question 18. Alice and Bob share the Bell state $|\Phi^+\rangle$. Alice measures her qubit in the $\{|0\rangle, |1\rangle\}$ basis and obtains $|0\rangle$. What is the post-measurement state of Bob's qubit?

(a) $|0\rangle$ (b) $|1\rangle$ (c) $|+\rangle$ (d) $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$

Answer: (a). In $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$, if Alice gets $|0\rangle$, Bob's qubit collapses to $|0\rangle$ (the outcomes are perfectly correlated).

Question 19. The EPR argument concluded that quantum mechanics is:

(a) Correct and complete (b) Incomplete — hidden variables must exist (c) Incorrect — entanglement is impossible (d) Non-local — signals travel faster than light

Answer: (b). EPR argued that if both $S_z$ and $S_x$ are "elements of reality" (since either can be predicted with certainty via distant measurement), but QM says they cannot both have definite values simultaneously, then QM must be incomplete.

Question 20. A composite system $AB$ is in a pure entangled state. Which of the following is TRUE?

(a) Both $\hat{\rho}_A$ and $\hat{\rho}_B$ are pure states (b) Both $\hat{\rho}_A$ and $\hat{\rho}_B$ are mixed states (c) $\hat{\rho}_A$ is pure but $\hat{\rho}_B$ is mixed (d) The purity of the reduced states depends on the specific entangled state

Answer: (b). For any pure entangled state, the reduced density matrices of both subsystems are mixed. This is because the Schmidt decomposition has more than one nonzero coefficient, giving $\hat{\rho}_A = \sum_k \lambda_k^2 |a_k\rangle\langle a_k|$ with at least two terms — a mixed state. By Corollary 11.1, $\hat{\rho}_A$ and $\hat{\rho}_B$ share the same nonzero eigenvalues, so both are mixed.

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