2.1.1 Tensor Product#

Worked solutions for the homework problems in the 2.1.1 Tensor Product lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Identifying entangled states. Three two-qubit states:

A state is a product state if it can be written as \(\vert\psi_A\rangle\otimes\vert\psi_B\rangle\) for some single-qubit kets \(\vert\psi_A\rangle, \vert\psi_B\rangle\); otherwise it is entangled.

(a) For each \(\vert\Psi_i\rangle\), attempt to write it as a product. For the product cases, give explicit \(\vert\psi_A\rangle, \vert\psi_B\rangle\).

(b) For each non-product (entangled) case, argue from the failed factorization equations why no decomposition exists.

(c) The general two-qubit state \(\alpha_{00}\vert 00\rangle + \alpha_{01}\vert 01\rangle + \alpha_{10}\vert 10\rangle + \alpha_{11}\vert 11\rangle\) is a product state if and only if \(\alpha_{00}\alpha_{11} = \alpha_{01}\alpha_{10}\) (the “determinant condition”). Verify this criterion on each of the three states above and confirm it agrees with your conclusions in (a)–(b).

Solution.

(a) Set \(\vert\psi_A\rangle = \alpha_0\vert 0\rangle + \alpha_1\vert 1\rangle\) and \(\vert\psi_B\rangle = \beta_0\vert 0\rangle + \beta_1\vert 1\rangle\). The product expands as

\(\vert\Psi_1\rangle\): amplitudes \((\tfrac{1}{\sqrt 2}, \tfrac{1}{\sqrt 2}, 0, 0)\). We need \(\alpha_0\beta_0 = \alpha_0\beta_1 = \tfrac{1}{\sqrt 2}\) and \(\alpha_1\beta_0 = \alpha_1\beta_1 = 0\). The second pair forces \(\alpha_1 = 0\) (since \(\beta_0, \beta_1\) can’t both vanish without ruining \(\alpha_0\beta_0\)), and the first gives \(\beta_0 = \beta_1\). Normalising with \(\alpha_0 = 1\) and \(\beta_0 = \beta_1 = \tfrac{1}{\sqrt 2}\):

\(\vert\Psi_2\rangle\): amplitudes \((\tfrac{1}{\sqrt 2}, 0, 0, \tfrac{1}{\sqrt 2})\). The factorization requires \(\alpha_0\beta_1 = 0\) AND \(\alpha_1\beta_0 = 0\). The first forces either \(\alpha_0 = 0\) or \(\beta_1 = 0\); either choice immediately kills one of \(\alpha_0\beta_0\) or \(\alpha_1\beta_1\), both of which need to be \(\tfrac{1}{\sqrt 2}\neq 0\). Entangled — no factorization exists.

\(\vert\Psi_3\rangle\): amplitudes \((0, \tfrac{1}{\sqrt 2}, \tfrac{1}{\sqrt 2}, 0)\). The factorization requires \(\alpha_0\beta_0 = 0\) AND \(\alpha_1\beta_1 = 0\) — same obstruction as \(\vert\Psi_2\rangle\). Either choice kills the cross terms. Entangled.

(b) The structural obstruction: a product state’s four amplitudes are determined by only \(2+2 = 4\) complex numbers via two products per amplitude, so the four amplitudes are correlated. A state with \(\alpha_{00} = \alpha_{11} \neq 0\) but \(\alpha_{01} = \alpha_{10} = 0\) (like \(\vert\Psi_2\rangle\)) requires the cross terms to vanish and the diagonal terms to be nonzero — incompatible for any factorization, because the cross terms \(\alpha_0\beta_1, \alpha_1\beta_0\) can vanish only if some single-qubit amplitude vanishes, which then forces the diagonals to vanish too. The “entanglement” is the pattern of zeros: \(\vert\Psi_2\rangle\) and \(\vert\Psi_3\rangle\) have non-product zero/non-zero patterns.

(c) Compute \(\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10}\) for each:

• \(\vert\Psi_1\rangle\): \((\tfrac{1}{\sqrt 2})(0) - (\tfrac{1}{\sqrt 2})(0) = 0\). Determinant condition satisfied → product. ✓

• \(\vert\Psi_2\rangle\): \((\tfrac{1}{\sqrt 2})(\tfrac{1}{\sqrt 2}) - (0)(0) = \tfrac{1}{2}\). Determinant condition fails → entangled. ✓

• \(\vert\Psi_3\rangle\): \((0)(0) - (\tfrac{1}{\sqrt 2})(\tfrac{1}{\sqrt 2}) = -\tfrac{1}{2}\). Determinant condition fails → entangled. ✓

The determinant condition is the non-zero \(2\times 2\) minor test: a 2-qubit state’s amplitude matrix \(\begin{pmatrix}\alpha_{00} & \alpha_{01} \\ \alpha_{10} & \alpha_{11}\end{pmatrix}\) has rank 1 (i.e. is a product of column-and-row vectors) iff its determinant vanishes. Determinant nonzero = entangled. This rank-1 criterion is the simplest case of the general Schmidt rank classification of bipartite states — a state is product iff its amplitude matrix has rank 1, and the rank measures “how entangled” it is.

2. Operator products via the mixed-product rule. The mixed-product rule states that for tensor-product operators acting on the same composite system,

This is the algebraic backbone of multi-qubit operator manipulation. Apply it concretely.

(a) Compute \((\hat X\otimes\hat Y)(\hat Z\otimes\hat I)\) two ways: (i) using the mixed-product rule and Pauli multiplication (\(\hat X\hat Z = -\mathrm{i}\hat Y\), \(\hat Y\hat I = \hat Y\)); (ii) by explicit matrix multiplication of the two \(4\times 4\) matrices.

(b) Verify the two answers agree.

(c) Use the mixed-product rule to compute \((\hat X\otimes\hat Y)^2\) in two-qubit form. Express your answer as a single Pauli string with sign.

Solution.

(a) (i) Mixed-product rule. Slot 1 holds \(\hat X\hat Z\); slot 2 holds \(\hat Y\hat I\). Using the Pauli multiplication law \(\hat\sigma^i\hat\sigma^j = \delta_{ij}\hat I + \mathrm{i}\epsilon_{ijk}\hat\sigma^k\) (1.1.3),

So

(ii) Explicit matrix multiplication. Build \(\hat X\otimes\hat Y\) and \(\hat Z\otimes\hat I\) as \(4\times 4\) Kronecker products:

Multiplying (column \(k\) of the second matrix is just \(\pm e_k\), so we pick up the corresponding column of the first matrix with sign):

(b) Compare with \(-\mathrm{i}\,\hat Y\otimes\hat Y\). The matrix \(\hat Y\otimes\hat Y\) has corner entries \((-\mathrm{i})(-\mathrm{i}) = -1\) at \((1,4)\) and \((\mathrm{i})(\mathrm{i}) = -1\) at \((4,1)\), and central entries \((-\mathrm{i})(\mathrm{i}) = +1\) at \((2,3)\) and \((\mathrm{i})(-\mathrm{i}) = +1\) at \((3,2)\):

Matches part (ii) exactly. \(\checkmark\)

The mixed-product rule factor-by-factor evaluation is dramatically simpler than the \(4\times 4\) matrix calculation — a \(16\)-entry product becomes two \(2\times 2\) products, each of which is just a Pauli multiplication. This is the practical benefit of working in tensor-product form: the algebra of \(N\)-qubit operators reduces to \(N\) separate single-qubit algebras.

(c) Apply mixed-product rule to the square:

So squaring any “single Pauli on each qubit” string gives \(\hat I\otimes\hat I = \hat I_{\mathrm{2-qubit}}\). This is the natural generalisation of the single-qubit identity \((\hat\sigma^a)^2 = \hat I\) to multi-qubit Pauli strings, and the property that makes Pauli strings ideal eigenoperators of two-level measurements: each has eigenvalues \(\pm 1\), generalising the qubit-level result of 1.1.3 P7.

★ 3. The SWAP operator. Define the SWAP operator \(\hat S\) on two qubits by its action on the computational basis:

(a) Write \(\hat S\) as a \(4\times 4\) matrix in the ordered basis \(\{\vert 00\rangle, \vert 01\rangle, \vert 10\rangle, \vert 11\rangle\}\).

(b) Show that \(\hat S^2 = \hat I\) (the SWAP is an involution). Conclude that its eigenvalues are \(\pm 1\).

(c) Identify the symmetric (\(+1\)) and antisymmetric (\(-1\)) eigenspaces of \(\hat S\). Give a basis for each, and state their dimensions.

(d) Show that \(\hat S\) admits the Pauli-string decomposition

(Hint: build each Pauli-string matrix and sum, or use the trace-projection method from 1.1.3 P5 on the 2-qubit Pauli basis.)

(e) Take the SWAP operator itself as the Hamiltonian, \(\hat H = J\hat S\) with coupling \(J > 0\), and prepare the system in the initial state \(\vert\psi(0)\rangle = \vert 01\rangle\). Evolve the state under the Schrödinger equation, \(\vert\psi(t)\rangle = \mathrm{e}^{-\mathrm{i}\hat H t/\hbar}\vert\psi(0)\rangle\), and find \(\vert\psi(t)\rangle\) in closed form. Determine the smallest positive time \(t^*\) at which \(\vert\psi(t^*)\rangle = \vert 10\rangle\) (up to a global phase).

(f) At an arbitrary time \(t\), measure the SWAP operator \(\hat S\) as an observable on the evolved state. List the possible outcomes, the corresponding probabilities, and the post-measurement states. Are the probabilities the same as at \(t = 0\)? Identify the conserved quantity responsible.

Solution.

(a) The action \(\hat S\vert ab\rangle = \vert ba\rangle\) gives: \(\vert 00\rangle\to\vert 00\rangle\), \(\vert 01\rangle\to\vert 10\rangle\), \(\vert 10\rangle\to\vert 01\rangle\), \(\vert 11\rangle\to\vert 11\rangle\). So the matrix (column \(j\) is the image of basis vector \(j\)) is

(b) Applying SWAP twice swaps \(a, b\) then swaps them back: \(\hat S^2\vert ab\rangle = \hat S\vert ba\rangle = \vert ab\rangle\). So \(\hat S^2 = \hat I\). If \(\lambda\) is an eigenvalue of \(\hat S\) with eigenvector \(\vert v\rangle\), applying twice gives \(\lambda^2\vert v\rangle = \vert v\rangle\), so \(\lambda^2 = 1\), \(\lambda = \pm 1\).

(c) Eigenvectors of \(\hat S\) in the computational basis:

• \(\hat S\vert 00\rangle = \vert 00\rangle\): eigenvalue \(+1\).

• \(\hat S\vert 11\rangle = \vert 11\rangle\): eigenvalue \(+1\).

• \(\hat S\,\tfrac{1}{\sqrt 2}(\vert 01\rangle + \vert 10\rangle) = \tfrac{1}{\sqrt 2}(\vert 10\rangle + \vert 01\rangle)\): eigenvalue \(+1\).

• \(\hat S\,\tfrac{1}{\sqrt 2}(\vert 01\rangle - \vert 10\rangle) = \tfrac{1}{\sqrt 2}(\vert 10\rangle - \vert 01\rangle) = -\tfrac{1}{\sqrt 2}(\vert 01\rangle - \vert 10\rangle)\): eigenvalue \(-1\).

The symmetric (\(+1\)) eigenspace is 3-dimensional, spanned by \(\{\vert 00\rangle, \vert 11\rangle, \tfrac{1}{\sqrt 2}(\vert 01\rangle + \vert 10\rangle)\}\). The antisymmetric (\(-1\)) eigenspace is 1-dimensional, spanned by \(\{\tfrac{1}{\sqrt 2}(\vert 01\rangle - \vert 10\rangle)\}\). Dimension count: \(3 + 1 = 4 = \dim(\mathbb C^2\otimes\mathbb C^2)\). ✓

The eigenspaces correspond exactly to the triplet (symmetric, 3-d) and singlet (antisymmetric, 1-d) of two-qubit angular momentum — and they will reappear naturally in 2.1.2 (Symmetrization) as the boson and fermion sectors.

(d) Verification of the Pauli-string formula. Build each term and sum. Using the matrices computed in Problem 6 and elsewhere:

• \(\hat I\otimes\hat I = \mathrm{diag}(1,1,1,1)\)

• \(\hat X\otimes\hat X\) has \(1\) at the four anti-diagonal positions \((1,4), (2,3), (3,2), (4,1)\)

• \(\hat Y\otimes\hat Y\) has \(-1\) at \((1,4)\) and \((4,1)\), and \(+1\) at \((2,3)\) and \((3,2)\)

• \(\hat Z\otimes\hat Z = \mathrm{diag}(1,-1,-1,1)\)

Sum entry by entry. The diagonal \((1,1) = 1 + 0 + 0 + 1 = 2\). Diagonal \((2,2) = 1 + 0 + 0 + (-1) = 0\). Diagonal \((3,3) = 0\), \((4,4) = 2\). Off-diagonal \((1,4) = 0 + 1 + (-1) + 0 = 0\) and \((4,1) = 0\). Off-diagonal \((2,3) = 0 + 1 + 1 + 0 = 2\) and \((3,2) = 2\). All other entries vanish.

Dividing by \(2\):

This decomposition is the operator-level statement of the identity for the swap of two spin-1/2 particles, \(\hat S = \tfrac{1}{2}(\hat I + \hat\sigma_1\cdot\hat\sigma_2)\) — where \(\hat\sigma_1\cdot\hat\sigma_2 = \sum_a \hat\sigma_1^a\hat\sigma_2^a\). It connects directly to the Heisenberg interaction of Problem 6: a Heisenberg coupling is essentially “twice the SWAP minus the identity,” and its symmetric/antisymmetric eigenstates are exactly the SWAP eigenstates (triplet/singlet).

(e) Exchange dynamics. Decompose the initial state into SWAP eigenstates,

with \(\hat S\vert s\rangle = +\vert s\rangle\) and \(\hat S\vert a\rangle = -\vert a\rangle\). Under \(\hat H = J\hat S\) each eigenstate picks up a phase \(\mathrm{e}^{\mp\mathrm{i}Jt/\hbar}\):

For \(\vert\psi(t)\rangle \propto \vert 10\rangle = \tfrac{1}{\sqrt 2}(\vert s\rangle - \vert a\rangle)\), we need \(\mathrm{e}^{-\mathrm{i}Jt/\hbar} = -\mathrm{e}^{+\mathrm{i}Jt/\hbar}\), i.e. \(\mathrm{e}^{2\mathrm{i}Jt/\hbar} = -1\). The smallest positive solution is

at which \(\vert\psi(t^*)\rangle = -\mathrm{i}\vert 10\rangle\). The two-qubit system therefore performs a coherent SWAP gate in finite time, set by the inverse coupling — a basic primitive of exchange-based quantum gates.

(f) Measurement of SWAP and conservation. The closed-form \(\vert\psi(t)\rangle\) above has unit-modulus amplitudes on each SWAP eigenspace at every \(t\). Measuring \(\hat S\) on the evolved state:

• Outcome \(+1\) with probability \(\tfrac{1}{2}\), post-measurement state \(\vert s\rangle\).

• Outcome \(-1\) with probability \(\tfrac{1}{2}\), post-measurement state \(\vert a\rangle\).

These probabilities equal the \(t = 0\) values: \(\hat H = J\hat S\) commutes trivially with \(\hat S\), so \(\hat S\) is a conserved observable, and the probability distribution of its measurement outcomes on the evolved state is time-independent — even though the state itself oscillates back and forth between \(\vert 01\rangle\) and \(\vert 10\rangle\). This is a direct instance of the general principle that \([\hat H, \hat O] = 0\) implies time-independent statistics for \(\hat O\).

4. Expectation in product states. For a product state \(\vert\psi\rangle = \vert\psi_A\rangle \otimes \vert\psi_B\rangle\), prove that

Explain in one sentence what physical statement this factorization expresses, and why it must fail for entangled states.

Solution.

Three facts about tensor products are needed.

(i) Bra of a product factorizes. From \(\vert\psi\rangle = \vert\psi_A\rangle\otimes\vert\psi_B\rangle\),

(ii) Mixed-product rule on a product ket:

(iii) Tensor-product inner product factorizes:

Combine: apply (ii) to the ket, (i) to the bra, then (iii):

Physical meaning. In a product state, the joint expectation \(\langle\hat A\otimes\hat B\rangle\) equals the product of the two local expectations \(\langle\hat A\rangle_A\langle\hat B\rangle_B\): the measurements on A and on B are statistically independent, with no correlations between them. There are no joint statistics that cannot be reconstructed from the two single-particle distributions.

For an entangled state this factorization fails: there exist observables \(\hat A,\hat B\) for which \(\langle\hat A\otimes\hat B\rangle \neq \langle\hat A\rangle\langle\hat B\rangle\) on the entangled state. Problem 5(d) computes the explicit case for \(\vert\Phi^+\rangle = (\vert 00\rangle + \vert 11\rangle)/\sqrt 2\): \(\langle\hat Z_1\rangle = \langle\hat Z_2\rangle = 0\) separately, but \(\langle\hat Z_1\hat Z_2\rangle = +1\) — perfect correlation despite individually random measurements. The non-factorization of joint expectation values is the operational signature of entanglement.

5. Single-body and two-body Z measurements. For two qubits, define the single-body observables \(\hat Z_1 = \hat Z\otimes\hat I\) and \(\hat Z_2 = \hat I\otimes\hat Z\), and the two-body observable \(\hat Z_1\hat Z_2 = \hat Z\otimes\hat Z\).

(a) Show \([\hat Z_1, \hat Z_2] = 0\) using the mixed-product rule. State physically why operators on different particles always commute.

(b) Find the four simultaneous eigenstates of \(\hat Z_1\) and \(\hat Z_2\). State their \((\hat Z_1, \hat Z_2)\) eigenvalue pairs.

(c) Show \([\hat Z_1, \hat Z_1\hat Z_2] = 0\). Identify the eigenvalues of \(\hat Z_1\hat Z_2\) and explain physically: \(\hat Z_1\hat Z_2\) measures the parity \((-1)^{n_0+n_1}\) where \(n_i\) is the result of measuring \(\hat Z_i = +1 \to n_i = 0\), \(\hat Z_i = -1 \to n_i = 1\).

(d) For the entangled state \(\vert\Phi^+\rangle = \tfrac{1}{\sqrt 2}(\vert 00\rangle + \vert 11\rangle)\) (from Problem 1(b)), compute \(\langle\hat Z_1\rangle\), \(\langle\hat Z_2\rangle\), and \(\langle\hat Z_1\hat Z_2\rangle\). Interpret the result: individual \(\hat Z\)-measurements are random (\(\langle\hat Z_i\rangle = 0\)), but the parity \(\hat Z_1\hat Z_2\) is deterministic. Why does this contradict the product-state factorization rule of Problem 4?

Solution.

(a) Mixed-product rule:

The two products are equal, so \([\hat Z_1, \hat Z_2] = 0\). Physically, operators carrying \(\hat Z\) on qubit 1 and qubit 2 act on different tensor factors; multiplication in different slots commutes because each slot independently uses \(\hat Z\cdot\hat I = \hat I\cdot\hat Z = \hat Z\).

(b) The four computational basis states are simultaneous eigenstates: \(\hat Z\vert i\rangle = (1-2i)\vert i\rangle\) gives eigenvalue \(+1\) for \(\vert 0\rangle\) and \(-1\) for \(\vert 1\rangle\). So

(c) Since \(\hat Z_1\) and \(\hat Z_2\) commute, \(\hat Z_1\) commutes with any function of them, including their product. Explicitly \([\hat Z_1, \hat Z_1\hat Z_2] = \hat Z_1[\hat Z_1, \hat Z_2] + [\hat Z_1, \hat Z_1]\hat Z_2 = 0\).

The eigenvalue of \(\hat Z_1\hat Z_2\) on \(\vert ab\rangle\) is the product of the individual eigenvalues: \(+1\) for \(\vert 00\rangle\) and \(\vert 11\rangle\) (both qubits the same), \(-1\) for \(\vert 01\rangle\) and \(\vert 10\rangle\) (qubits opposite). So \(\hat Z_1\hat Z_2\) has eigenvalues \(\{+1, +1, -1, -1\}\) — two doubly-degenerate sectors.

This is the parity observable: it returns \(+1\) when both qubits are in the same state (\(\vert 00\rangle\) or \(\vert 11\rangle\), \(n_0 + n_1\) even) and \(-1\) when they differ (\(\vert 01\rangle\) or \(\vert 10\rangle\), \(n_0 + n_1\) odd). Parity is a single bit of correlation information that the joint measurement extracts; it does not specify which of the two same-parity states the system is in.

(d) On \(\vert\Phi^+\rangle = \tfrac{1}{\sqrt 2}(\vert 00\rangle + \vert 11\rangle)\), the \(\hat Z_1\) outcomes are \(+1\) (from \(\vert 00\rangle\)) and \(-1\) (from \(\vert 11\rangle\)) with probability \(\tfrac{1}{2}\) each. So

But the parity is always \(+1\) on \(\vert\Phi^+\rangle\): both \(\vert 00\rangle\) and \(\vert 11\rangle\) have \(\hat Z_1\hat Z_2 = +1\). So

Contradiction with product-state factorization. If \(\vert\Phi^+\rangle\) were a product state, Problem 4 would force \(\langle\hat Z_1\hat Z_2\rangle = \langle\hat Z_1\rangle\langle\hat Z_2\rangle = 0\cdot 0 = 0\). The actual value is \(+1\) — a glaring violation of factorization. This is the operational signature of entanglement: the two qubits’ individual measurements are random, but their correlation is deterministic. The joint expectation value cannot be recovered from the local expectations alone, because the state carries information about the correlation that no single-particle description can capture. This is the foundation of every entangled-state-based protocol in quantum information, and it shows up here directly from the lecture’s tensor-product machinery.

6. Heisenberg interaction matrix. The Heisenberg interaction is \(\hat{H} = J(\hat{X}\otimes\hat{X} + \hat{Y}\otimes\hat{Y} + \hat{Z}\otimes\hat{Z})\).

(a) Write \(\hat{H}\) as a \(4\times 4\) matrix.

(b) How many Pauli strings appear?

Solution.

(a) Build each two-qubit Pauli string as a Kronecker product in the ordered basis \(\{\vert 00\rangle, \vert 01\rangle, \vert 10\rangle, \vert 11\rangle\}\). With \(\hat X = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\), \(\hat Y = \begin{pmatrix}0 & -\mathrm{i} \\ \mathrm{i} & 0\end{pmatrix}\), \(\hat Z = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\):

(For \(\hat Y\otimes\hat Y\) the corner entries are \((-\mathrm{i})(-\mathrm{i}) = -1\) and \((\mathrm{i})(\mathrm{i}) = -1\), while the central entries are \((-\mathrm{i})(\mathrm{i}) = +1\).) Adding the three and multiplying by \(J\):

The off-diagonal \(2J\) in the central block comes from \(\hat X\otimes\hat X\) and \(\hat Y\otimes\hat Y\) (each contributes \(+1\) there); in the corner entries those same two strings cancel (\(+1\) from \(\hat X\otimes\hat X\), \(-1\) from \(\hat Y\otimes\hat Y\)). Diagonalising the central block \(\begin{pmatrix}-1 & 2 \\ 2 & -1\end{pmatrix}\) gives eigenvalues \(-1 \pm 2 \in \{+1, -3\}\), so \(\hat H\) has spectrum \(J\{+1, +1, +1, -3\}\): a three-fold-degenerate triplet at energy \(+J\) (the states \(\vert 00\rangle\), \(\vert 11\rangle\), and \((\vert 01\rangle + \vert 10\rangle)/\sqrt 2\)) and a singlet at \(-3J\) (the antisymmetric \((\vert 01\rangle - \vert 10\rangle)/\sqrt 2\)). The triplet/singlet structure is exactly the SWAP eigenspace structure of Problem 3 — this is the origin of magnetic ordering in the Heisenberg model.

(b) Three Pauli strings appear: \(\hat X\otimes\hat X\), \(\hat Y\otimes\hat Y\), and \(\hat Z\otimes\hat Z\), each with the real coefficient \(J\). Every term is a genuine two-qubit Pauli string — a tensor product of single-qubit Paulis, here with no identity factor. For comparison, a general Hermitian operator on two qubits needs up to \(4^2 = 16\) Pauli strings; the Heisenberg Hamiltonian uses only \(3\) because it is highly constrained: isotropic (the three axes enter symmetrically), traceless, and purely an interaction (no single-body terms of the form \(\hat I\otimes\hat\sigma\) or \(\hat\sigma\otimes\hat I\)).

7. Pauli string decomposition. Decompose \(\vert 00\rangle\langle 00\vert\) into Pauli strings \(\hat\sigma_1^{s_1}\otimes\hat\sigma_2^{s_2}\) with \(\hat\sigma^s\in\{\hat I, \hat X, \hat Y, \hat Z\}\).

(a) Use the single-qubit identity \(\vert 0\rangle\langle 0\vert = \tfrac{1}{2}(\hat I + \hat Z)\) and tensor-product factorization to expand \(\vert 00\rangle\langle 00\vert\) in the 16-term Pauli basis. How many strings have non-zero coefficient?

(b) Verify (a) independently using the trace-projection formula \(c_{s_1 s_2} = \tfrac{1}{4}\operatorname{Tr}\!\bigl[(\hat\sigma^{s_1}\otimes\hat\sigma^{s_2})\,\vert 00\rangle\langle 00\vert\bigr]\) (from 1.1.3 P5, generalised to two qubits).

(c) Generalise: what are the non-zero Pauli-string coefficients for a 2-qubit projector \(\vert ij\rangle\langle ij\vert\) in general?

Solution.

(a) Start from \(\vert 0\rangle\langle 0\vert = \tfrac{1}{2}(\hat I + \hat Z)\), verifiable by direct matrix sum: \(\tfrac{1}{2}(\hat I + \hat Z) = \tfrac{1}{2}\bigl(\begin{smallmatrix}1 & 0 \\ 0 & 1\end{smallmatrix}\bigr) + \tfrac{1}{2}\bigl(\begin{smallmatrix}1 & 0 \\ 0 & -1\end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix}1 & 0 \\ 0 & 0\end{smallmatrix}\bigr) = \vert 0\rangle\langle 0\vert\).

The two-qubit projector factorizes:

Expanding by bilinearity of \(\otimes\):

Exactly four Pauli strings have non-zero coefficient: \(\hat I\otimes\hat I\), \(\hat I\otimes\hat Z\), \(\hat Z\otimes\hat I\), \(\hat Z\otimes\hat Z\), each with coefficient \(\tfrac{1}{4}\). All 12 strings containing \(\hat X\) or \(\hat Y\) have coefficient zero.

(b) Apply the trace-projection formula. Using \(\operatorname{Tr}(\vert 00\rangle\langle 00\vert\cdot\hat O) = \langle 00\vert\hat O\vert 00\rangle\),

The diagonal matrix element \(\langle 0\vert\hat\sigma\vert 0\rangle\) equals \(1\) for \(\hat\sigma\in\{\hat I, \hat Z\}\) (both diagonal with \(1\) in the upper-left), and \(0\) for \(\hat\sigma\in\{\hat X, \hat Y\}\) (zero diagonal entries in the Z basis). So \(c_{s_1 s_2}\) is non-zero only when both \(\hat\sigma^{s_1}\) and \(\hat\sigma^{s_2}\) are in \(\{\hat I, \hat Z\}\), yielding the four coefficients \(\tfrac{1}{4}\cdot 1\cdot 1 = \tfrac{1}{4}\) — matching (a). \(\checkmark\)

(c) For a general \(\vert ij\rangle\langle ij\vert\) with \(i, j\in\{0, 1\}\), the factorization is \(\vert i\rangle\langle i\vert\otimes\vert j\rangle\langle j\vert\) with \(\vert k\rangle\langle k\vert = \tfrac{1}{2}(\hat I + (-1)^k\hat Z) = \tfrac{1}{2}(\hat I + \langle\hat Z\rangle_k\hat Z)\). So

Always four non-zero Pauli strings (the diagonal \(\{I, Z\}\otimes\{I, Z\}\) block), with \(\pm\tfrac{1}{4}\) coefficients determined by the signs \((-1)^i, (-1)^j\). Computational-basis projectors always live in the diagonal \(\{I, Z\}^{\otimes 2}\) sector — a general principle: the Pauli-string expansion of any operator diagonal in the computational basis is supported entirely on tensor products of \(\hat I\) and \(\hat Z\).

8. Hilbert-space parametrization. For \(N\) qubits, how many independent real parameters specify

(a) a general normalized state in \((\mathbb{C}^2)^{\otimes N}\), and

(b) a normalized product state \(\vert\psi_1\rangle\otimes\cdots\otimes\vert\psi_N\rangle\)?

(c) What does the difference between (a) and (b) represent physically? Tabulate the two counts for \(N = 1, 2, 3, 10\) to illustrate the scaling.

Solution.

(a) A general \(N\)-qubit state lives in \((\mathbb{C}^2)^{\otimes N} = \mathbb{C}^{2^N}\), so it is specified by \(2^N\) complex amplitudes — that is, \(2\cdot 2^N = 2^{N+1}\) real numbers. Two constraints cut this count:

• Normalization \(\sum\vert\Psi_{\alpha_1\cdots\alpha_N}\vert^2 = 1\) removes \(1\) real parameter.

• The global phase is unobservable, removing \(1\) more.

Hence a general normalized \(N\)-qubit state carries

(This is the \(n\)-level count \(2n - 2\) from 1.1.2 Problem 8, evaluated at \(n = 2^N\). Check \(N = 1\): \(2^2 - 2 = 2\), the two Bloch-sphere angles \((\theta, \varphi)\).)

(b) A product state \(\vert\psi_1\rangle\otimes\cdots\otimes\vert\psi_N\rangle\) is fixed by choosing each single-qubit factor independently. Each factor is a normalized qubit state — a Bloch-sphere point — carrying \(2^{1+1} - 2 = 2\) real parameters. With \(N\) independent factors:

(Phase subtlety: each factor carries its own phase freedom, but a rephasing \(\vert\psi_k\rangle\to\mathrm{e}^{\mathrm{i}\phi_k}\vert\psi_k\rangle\) multiplies the whole product by \(\mathrm{e}^{\mathrm{i}(\phi_1 + \cdots + \phi_N)}\) — a single global phase. So \(N\) per-factor phases collapse into one unphysical global phase; counting each factor as a Bloch-sphere point already accounts for this correctly.)

(c) The difference

is the number of parameters a general state carries beyond anything a product state can reach. These “extra” parameters describe entanglement — correlations not attributable to any single particle. As \(N\) grows, product-state parameters scale linearly (\(2N\)), while the full count scales exponentially (\(2^{N+1} - 2\)); entangled states overwhelmingly dominate the Hilbert space, and product states become an exponentially thin slice.

For a single qubit (\(N=1\)), there is no second particle to entangle with, so \(\Delta = 0\). From \(N = 2\) onward, the entanglement count strictly exceeds the product count, and at \(N = 10\) the entanglement degrees of freedom outnumber product-state parameters by a factor of \(\sim 100\). This exponential separation is the central computational resource of quantum information — a classical description of the state would require listing all \(2^{N+1} - 2\) parameters, whereas a product state only \(2N\). Quantum dynamics in the entangled sector is precisely the dynamics that cannot be efficiently simulated by classical bookkeeping of single-particle states.