2.1.1 Tensor Product#

Prompts

What is a tensor product of two Hilbert spaces, and how do basis states combine? For a two-particle system where each particle has \(d\) basis states, how many basis states does the combined system have?
How do operators acting on individual particles extend to the full multi-particle Hilbert space? Why do operators on different particles always commute?
What is a Pauli string, and why do Pauli strings form a complete basis for all multi-qubit Hermitian operators?
How does the dimension of the \(N\)-particle Hilbert space grow with \(N\)? What does this exponential growth imply for classical simulation of quantum systems?

Lecture Notes#

Overview#

Chapter 1 described a single qubit. To describe multiple quantum particles—atoms, photons, electrons in a solid—we need a systematic way to combine Hilbert spaces. The tensor product provides this: it constructs the full many-body state space from single-particle spaces, and its dimension grows exponentially with particle number. In this lesson we set up the many-body framework (states, operators, Pauli strings) that will be used throughout the rest of the course.

Many-Body State Space#

Tensor Product of Hilbert Spaces.

Tensor Product

When systems A and B have Hilbert spaces \(\mathcal{H}_A\) and \(\mathcal{H}_B\), the composite system lives in

\[ \mathcal{H}_{AB} = \mathcal{H}_A \otimes \mathcal{H}_B \]

If \(\{\vert a_i\rangle\}\) is a basis for A and \(\{\vert b_j\rangle\}\) a basis for B, then \(\{\vert a_i\rangle \otimes \vert b_j\rangle\}\) is a complete orthonormal basis for \(\mathcal{H}_{AB}\):

(25)#\[ \langle a_i b_j \vert a_{i'} b_{j'} \rangle = \langle a_i \vert a_{i'}\rangle \langle b_j \vert b_{j'}\rangle = \delta_{ii'}\delta_{jj'} \]

Two Qubits.

Each qubit has basis \(\{\vert 0\rangle, \vert 1\rangle\}\). The two-qubit space \(\mathcal{H} = \mathbb{C}^2 \otimes \mathbb{C}^2\) has dimension \(2 \times 2 = 4\), with basis:

\[ \vert 00\rangle,\quad \vert 01\rangle,\quad \vert 10\rangle,\quad \vert 11\rangle \]

where \(\vert ij\rangle \equiv \vert i\rangle_A \otimes \vert j\rangle_B\). Orthonormality: \(\langle i_1 j_1 \vert i_2 j_2\rangle = \delta_{i_1 i_2}\delta_{j_1 j_2}\).

Example: Tensor Product of Two Single-Qubit States

Let

\[ \vert\psi\rangle_A = \alpha\vert 0\rangle + \beta\vert 1\rangle, \qquad \vert\phi\rangle_B = \gamma\vert 0\rangle + \delta\vert 1\rangle. \]

Algebraic approach (distribute over superposition):

\[\begin{split} \begin{split} \vert\psi\rangle_A\otimes\vert\phi\rangle_B &=(\alpha\vert 0\rangle+\beta\vert 1\rangle)\otimes(\gamma\vert 0\rangle+\delta\vert 1\rangle)\\ &=\alpha\gamma\vert 00\rangle+\alpha\delta\vert 01\rangle+\beta\gamma\vert 10\rangle+\beta\delta\vert 11\rangle. \end{split} \end{split}\]

Vector approach (Kronecker product intuition):

\[\begin{split} \vert\psi\rangle_A=\begin{pmatrix}\alpha\\\beta\end{pmatrix}, \qquad \vert\phi\rangle_B=\begin{pmatrix}\gamma\\\delta\end{pmatrix}, \end{split}\]

\[\begin{split} \vert\psi\rangle_A\otimes\vert\phi\rangle_B =\begin{pmatrix}\alpha\,\vert\phi\rangle_B\\\beta\,\vert\phi\rangle_B\end{pmatrix} =\begin{pmatrix}\alpha\begin{pmatrix}\gamma\\\delta\end{pmatrix}\\\beta\begin{pmatrix}\gamma\\\delta\end{pmatrix}\end{pmatrix} =\begin{pmatrix}\alpha\gamma\\\alpha\delta\\\beta\gamma\\\beta\delta\end{pmatrix}, \end{split}\]

which is exactly the coefficient vector of \(\alpha\gamma\vert 00\rangle+\alpha\delta\vert 01\rangle+\beta\gamma\vert 10\rangle+\beta\delta\vert 11\rangle\).

General Case: \(N\) Particles.

For \(N\) particles, each with \(d\) single-particle states \(\vert\alpha\rangle\) (\(\alpha = 1, \ldots, d\)), the composite space is \(\mathcal{H}^{\otimes N} = \mathcal{H}_1 \otimes \cdots \otimes \mathcal{H}_N\), with basis:

\[ \vert\alpha_1 \alpha_2 \cdots \alpha_N\rangle = \vert\alpha_1\rangle \otimes \vert\alpha_2\rangle \otimes \cdots \otimes \vert\alpha_N\rangle \]

Here \(\alpha\) is a generic single-particle quantum number: it can label momentum \(\boldsymbol{k}\), position \(\boldsymbol{r}\), spin \(s\), an energy level, or any other set of quantum numbers that specifies a single-particle mode.

Hilbert Space Dimension

(26)#\[ \dim(\mathcal{H}^{\otimes N}) = d^N \]

This grows exponentially with \(N\). For \(N = 300\) qubits: \(2^{300} \approx 10^{90}\) dimensions—more than the number of atoms in the observable universe.

A general state expands as:

(27)#\[ \vert\Psi\rangle = \sum_{\alpha_1, \ldots, \alpha_N} \Psi_{\alpha_1 \cdots \alpha_N}\;\vert\alpha_1 \cdots \alpha_N\rangle \]

with expansion coefficients satisfying \(\sum \vert \Psi_{\alpha_1 \cdots \alpha_N}\vert^2 = 1\).

Many-Body Operators#

Single-Body Operators.

An operator \(\hat{A}\) acting on particle \(i\) alone extends to the full space by tensoring with identity \(\hat{I}\) on all other particles:

(28)#\[ \hat{A}_i = \hat{I}_1 \otimes \cdots \otimes \hat{A} \otimes \cdots \otimes \hat{I}_N \quad (\hat{A}\text{ in position }i) \]

Its action on basis states:

\[ \hat{A}_i \vert\alpha_1 \cdots \alpha_N\rangle = \vert\alpha_1\rangle \cdots (\hat{A}\vert\alpha_i\rangle) \cdots \vert\alpha_N\rangle \]

Example: \(\hat X_1 = \hat X\otimes\hat I\) on Two Qubits

Use the ordered basis \(\{\vert 00\rangle,\vert 01\rangle,\vert 10\rangle,\vert 11\rangle\}\).

Algebraic approach (operator action on kets):

\[\begin{split} \begin{split} \hat X_1\vert 00\rangle&=(\hat X\vert 0\rangle)\otimes(\hat I\vert 0\rangle)=\vert 10\rangle,\\ \hat X_1\vert 01\rangle&=(\hat X\vert 0\rangle)\otimes(\hat I\vert 1\rangle)=\vert 11\rangle,\\ \hat X_1\vert 10\rangle&=(\hat X\vert 1\rangle)\otimes(\hat I\vert 0\rangle)=\vert 00\rangle,\\ \hat X_1\vert 11\rangle&=(\hat X\vert 1\rangle)\otimes(\hat I\vert 1\rangle)=\vert 01\rangle. \end{split} \end{split}\]

So \(\hat X_1\) flips the first qubit and leaves the second unchanged.

Matrix approach (Kronecker product):

\[\begin{split} \hat X\otimes\hat I= \begin{pmatrix}0&1\\1&0\end{pmatrix} \otimes \begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix} 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0 \end{pmatrix}. \end{split}\]

Question. Show similarly that

\[\begin{split} \hat X_2=\hat I\otimes\hat X= \begin{pmatrix} 0&1&0&0\\ 1&0&0&0\\ 0&0&0&1\\ 0&0&1&0 \end{pmatrix}, \end{split}\]

which flips the second qubit while leaving the first unchanged.

Operators on Different Particles Commute

(29)#\[ [\hat{A}_i, \hat{B}_j] = 0 \quad \text{for } i \neq j \]

This is because they act on different tensor factors.

Two-Body Operators.

An operator coupling particles \(i\) and \(j\) acts nontrivially on both tensor factors:

\[ \hat{V}_{ij} = \hat{I}_1 \otimes \cdots \otimes \hat{V} \otimes \cdots \otimes \hat{I}_N \quad (\hat{V}\text{ acts on positions }i,j) \]

Example: Constructing \(\hat{X}_1\hat{X}_2\)

Problem. For two qubits, compute \(\hat{X}_1\hat{X}_2\) in matrix form and verify it matches \(\hat{X}\otimes\hat{X}\).

Solution. First write

\[ \hat{X}_1 = \hat{X}\otimes\hat{I}, \qquad \hat{X}_2 = \hat{I}\otimes\hat{X}. \]

Using the computational basis \(\{\vert 00\rangle,\vert 01\rangle,\vert 10\rangle,\vert 11\rangle\}\),

\[\begin{split} \hat{X}_1= \begin{pmatrix} 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0 \end{pmatrix}, \qquad \hat{X}_2= \begin{pmatrix} 0&1&0&0\\ 1&0&0&0\\ 0&0&0&1\\ 0&0&1&0 \end{pmatrix}. \end{split}\]

Now multiply the two matrices:

\[\begin{split} \hat{X}_1\hat{X}_2= \begin{pmatrix} 0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0 \end{pmatrix}. \end{split}\]

If we construct \(\hat{X}\otimes\hat{X}\) directly,

\[\begin{split} \hat{X}\otimes\hat{X} = \begin{pmatrix} 0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0 \end{pmatrix}, \end{split}\]

which is exactly the same matrix. This works because tensor products and matrix products satisfy the mixed-product rule

\[ (\hat{A}\otimes\hat{B})(\hat{C}\otimes\hat{D})=(\hat{A}\hat{C})\otimes(\hat{B}\hat{D}), \]

so here \((\hat{X}\otimes\hat{I})(\hat{I}\otimes\hat{X})=\hat{X}\otimes\hat{X}\).

General Operator Expansion.

Any operator on the \(N\)-particle space can be expanded in the basis of ket-bra products:

\[ \hat{O} = \sum_{\{\alpha\},\{\beta\}} O_{\alpha_1\cdots\alpha_N,\,\beta_1\cdots\beta_N}\;\vert\alpha_1\cdots\alpha_N\rangle\langle\beta_1\cdots\beta_N\vert \]

For multi-qubit systems, a more natural basis is given by Pauli strings.

Pauli Strings.

Pauli String Expansion

Any Hermitian operator on \(N\) qubits can be written as:

(30)#\[ \hat{O} = \sum_{\{s_i\}} c_{\{s_i\}} \bigotimes_{i=1}^N \hat{\sigma}_i^{s_i}, \qquad \hat{\sigma}_i^{s_i} \in \{\hat{I},\, \hat{X},\, \hat{Y},\, \hat{Z}\} \]

with \(c_{\{s_i\}} \in \mathbb{R}\). There are \(4^N\) Pauli strings; they form an orthogonal basis for all \(2^N \times 2^N\) Hermitian matrices.

Each Pauli string \(\bigotimes_i \hat{\sigma}_i^{s_i}\) is itself a tensor product of single-qubit operators. The decomposition is unique because the \(4^N\) Pauli strings are orthogonal under the trace inner product \(\langle \hat{A}, \hat{B}\rangle = \mathrm{Tr}(\hat{A}^\dagger \hat{B})\).

Example: Pauli String Decomposition

Problem. Decompose the projector \(\hat{P} = \vert 00\rangle\langle 00\vert\) into Pauli strings.

Solution. Use \(\vert 0\rangle\langle 0\vert = \tfrac{1}{2}(\hat{I} + \hat{Z})\):

\[ \vert 00\rangle\langle 00\vert = \vert 0\rangle\langle 0\vert \otimes \vert 0\rangle\langle 0\vert = \tfrac{1}{4}(\hat{I} + \hat{Z}) \otimes (\hat{I} + \hat{Z}) = \tfrac{1}{4}(\hat{I}\hat{I} + \hat{I}\hat{Z} + \hat{Z}\hat{I} + \hat{Z}\hat{Z}) \]

Four Pauli strings, each with real coefficient \(\tfrac{1}{4}\).

Discussion: Exponential Complexity

The number of Pauli strings grows as \(4^N\), which is exponential in \(N\). This means specifying a general \(N\)-qubit Hamiltonian requires exponentially many parameters. In practice, physical Hamiltonians are local (each term couples only a few nearby qubits), so only polynomially many Pauli strings have nonzero coefficients. What are the implications for classical simulation of quantum systems?

Poll: Hilbert space of two qubits

Two qubits live in \(\mathcal{H}_A \otimes \mathcal{H}_B\) where each space is 2-dimensional. What is the dimension of the combined space?

(A) 2 (sum of dimensions).

(B) 4 (product of dimensions).

(D) Infinite (tensor products are always infinite-dimensional).

Summary#

Tensor product: \(\mathcal{H}^{\otimes N} = \mathcal{H}_1 \otimes \cdots \otimes \mathcal{H}_N\) combines single-particle spaces; basis states \(\vert\alpha_1 \cdots \alpha_N\rangle\) are products of single-particle states; dimension \(d^N\) grows exponentially.
Many-body states: General state \(\vert\Psi\rangle = \sum_{\{\alpha_i\}} \Psi_{\alpha_1 \cdots \alpha_N} \vert\alpha_1 \cdots \alpha_N\rangle\) requires \(d^N\) coefficients; entangled states do not factor into products.
Single-body operators: \(\hat{A}_i = \hat{I} \otimes \cdots \otimes \hat{A} \otimes \cdots \otimes \hat{I}\) acts only on particle \(i\); operators on different particles commute.
Pauli strings: Any Hermitian operator on \(N\) qubits expands as \(\hat{O} = \sum_{\{s_i\}} c_{\{s_i\}} \bigotimes_i \hat{\sigma}_i^{s_i}\) with \(4^N\) orthogonal Pauli basis elements (exponential complexity for general operators).

Homework#

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

\[ \vert\Psi_1\rangle = \tfrac{1}{\sqrt 2}\bigl(\vert 00\rangle + \vert 01\rangle\bigr),\qquad \vert\Psi_2\rangle = \tfrac{1}{\sqrt 2}\bigl(\vert 00\rangle + \vert 11\rangle\bigr),\qquad \vert\Psi_3\rangle = \tfrac{1}{\sqrt 2}\bigl(\vert 01\rangle + \vert 10\rangle\bigr). \]

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).

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

\[ (\hat A\otimes\hat B)(\hat C\otimes\hat D) = (\hat A\hat C)\otimes(\hat B\hat D). \]

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.

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

\[ \hat S\vert ab\rangle = \vert ba\rangle \quad\text{for all } a, b \in \{0, 1\}. \]

(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

\[ \hat S = \tfrac{1}{2}\bigl(\hat I\otimes\hat I + \hat X\otimes\hat X + \hat Y\otimes\hat Y + \hat Z\otimes\hat Z\bigr). \]

(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.

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

\[ \langle\psi\vert\,(\hat A\otimes\hat B)\,\vert\psi\rangle = \langle\psi_A\vert\hat A\vert\psi_A\rangle\,\langle\psi_B\vert\hat B\vert\psi_B\rangle. \]

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

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?

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?

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?

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.