Tensor Products and Composite Systems
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You should be comfortable with the definition of a linear operator, the concept of an inner product space, and the spectral theorem for self-adjoint operators as established in previous lectures.
Why this matters
Tensor products allow us to describe composite systems where the state of the whole is not simply the sum of its parts.
In classical mechanics, the state of two particles is described by a single point in a combined configuration space. In quantum mechanics and linear algebra, the state space of a composite system is constructed via the tensor product of the individual state spaces. This construction is fundamental because it preserves the linear structure while allowing for correlations (entanglement) that cannot be captured by simple Cartesian products. Understanding how operators act on these tensor product spaces is essential for analyzing coupled oscillators, spin systems, and any multi-component physical system.
The Idea of a Composite Space
Imagine you have two independent systems, System A and System B. If System A can be in any of several states and System B can be in any of several states, the combined system can be in a state that specifies the state of A and the state of B simultaneously. In linear algebra, if \( V \) is the space of states for A and \( W \) is the space for B, the space of states for the combined system is not just the set of pairs \( (v, w) \), but a new vector space constructed from them, denoted \( V \otimes W \). The key intuition is that a basis for the combined space is formed by taking every possible pair of basis vectors from \( V \) and \( W \). If \( V \) has dimension \( m \) and \( W \) has dimension \( n \), the combined space has dimension \( m \times n \). This multiplicative growth in dimension is the hallmark of tensor products and is why composite quantum systems become complex so quickly.
Definition and Basis Construction
Let \( V \) be a vector space with basis \( \{v_1, \dots, v_m\} \) and \( W \) be a vector space with basis \( \{w_1, \dots, w_n\} \). The tensor product space \( V \otimes W \) is defined such that it has a basis consisting of the \( mn \) vectors \( \{v_i \otimes w_j \mid 1 \le i \le m, 1 \le j \le n\} \). Any vector in \( V \otimes W \) can be written as a linear combination of these basis elements. The tensor product operation \( \otimes \) is bilinear, meaning \( (a v_1 + b v_2) \otimes w = a(v_1 \otimes w) + b(v_2 \otimes w) \) and \( v \otimes (c w_1 + d w_2) = c(v \otimes w_1) + d(v \otimes w_2) \). This linearity allows us to decompose complex states into simpler components.
For example, if \( V \) is a 2-dimensional space (like a spin-1/2 particle) and \( W \) is a 3-dimensional space (like a spin-1 particle), the composite space is 6-dimensional. The basis vectors are \( v_1 \otimes w_1, v_1 \otimes w_2, v_1 \otimes w_3, v_2 \otimes w_1, v_2 \otimes w_2, v_2 \otimes w_3 \).
Operators on Tensor Product Spaces
When we have operators acting on \( V \) and \( W \), we often want to define operators on the composite space \( V \otimes W \). The most natural way is the tensor product of operators. If \( \hat{A} \) is an operator on \( V \) and \( \hat{B} \) is an operator on \( W \), then \( \hat{A} \otimes \hat{B} \) is an operator on \( V \otimes W \) defined by its action on basis vectors: \( (\hat{A} \otimes \hat{B})(v_i \otimes w_j) = (\hat{A}v_i) \otimes (\hat{B}w_j) \). This operator acts on the first component with \( \hat{A} \) and the second component with \( \hat{B} \) simultaneously. It is crucial to distinguish this from operators that act on only one part of the system. For instance, \( \hat{A} \otimes \hat{I}_W \) acts on \( V \) via \( \hat{A} \) and leaves \( W \) unchanged, while \( \hat{I}_V \otimes \hat{B} \) acts on \( W \) via \( \hat{B} \) and leaves \( V \) unchanged.
These operators commute if \( \hat{A} \) and \( \hat{B} \) act on different spaces, which is a key property for simultaneous diagonalization in composite systems.
Inner Products and Norms
To make \( V \otimes W \) a proper inner product space, we define the inner product on the basis vectors. If \( \langle v_i | v_j \rangle \) is the inner product in \( V \) and \( \langle w_k | w_l \rangle \) is the inner product in \( W \), the inner product in \( V \otimes W \) is defined by extending bilinearly from the basis: \( \langle v_i \otimes w_k | v_j \otimes w_l \rangle = \langle v_i | v_j \rangle \langle w_k | w_l \rangle \). This ensures that if the bases of \( V \) and \( W \) are orthonormal, the basis of \( V \otimes W \) is also orthonormal. The norm of a vector \( \psi = \sum_{i,j} c_{ij} (v_i \otimes w_j) \) is given by \( \|\psi\|^2 = \sum_{i,j} |c_{ij}|^2 \). This structure is essential for probability interpretation in quantum mechanics, where the norm squared represents the total probability.
The derivations
Spectral Theorem for Self-Adjoint Operators
Let \( \hat{A} \) be a self-adjoint operator on a finite-dimensional complex inner-product space \( V \) with \( \dim V = n \), so that \( \hat{A}=\hat{A}^{\dagger} \). Then \( \hat{A} \) has real eigenvalues, its eigenvectors belonging to distinct eigenvalues are orthogonal, and \( V \) possesses an orthonormal basis \( \{|e_1\rangle,\dots,|e_n\rangle\} \) consisting entirely of eigenvectors of \( \hat{A} \). Equivalently, the unitary \( \hat{U} \) whose columns are those eigenvectors diagonalizes \( \hat{A} \): \( \hat{U}^{\dagger}\hat{A}\hat{U}=\hat{\Lambda}=\mathrm{diag}(\lambda_1,\dots,\lambda_n) \), and \( \hat{A}=\sum_{k}\lambda_k\,|e_k\rangle\langle e_k| \).
Simultaneous Diagonalization of Commuting Operators
Let \(\hat{A}\) and \(\hat{B}\) be self-adjoint operators on a finite-dimensional complex inner-product space \(V\), \(\dim V = n\), and suppose they commute, \([\hat{A},\hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A} = 0\). Then \(V\) possesses a single orthonormal basis \(\{\lvert e_1\rangle,\dots,\lvert e_n\rangle\}\) whose every vector is simultaneously an eigenvector of \(\hat{A}\) and of \(\hat{B}\), so that one unitary \(U\) brings both to diagonal form: \(U^{\dagger}\hat{A}U = \operatorname{diag}(a_1,\dots,a_n)\) and \(U^{\dagger}\hat{B}U = \operatorname{diag}(b_1,\dots,b_n)\). The converse also holds: two self-adjoint operators sharing an orthonormal eigenbasis necessarily commute.
A common misconception is that every vector in \( V \otimes W \) can be written as a single tensor product \( v \otimes w \). This is false; such vectors are called "separable" or "product states." Most vectors in the tensor product space are "entangled" and cannot be factored into a simple product of a vector from \( V \) and a vector from \( W \). For example, if \( V \) and \( W \) are both 2-dimensional, the vector \( \frac{1}{\sqrt{2}}(v_1 \otimes w_1 + v_2 \otimes w_2) \) is entangled. Recognizing this distinction is critical for understanding the richness of composite systems. C
Worked example
Problem. Let \( V \) be a 2-dimensional complex vector space with orthonormal basis \( \{|1\rangle, |2\rangle\} \) and \( W \) be a 2-dimensional complex vector space with orthonormal basis \( \{|a\rangle, |b\rangle\} \). Consider the operator \( \hat{A} \) on \( V \) defined by \( \hat{A}|1\rangle = |1\rangle \) and \( \hat{A}|2\rangle = -|2\rangle \), and the operator \( \hat{B} \) on \( W \) defined by \( \hat{B}|a\rangle = |b\rangle \) and \( \hat{B}|b\rangle = |a\rangle \). Find the eigenvalues and eigenvectors of the operator \( \hat{C} = \hat{A} \otimes \hat{I}_W \) on \( V \otimes W \).
- Identify the basis of \( V \otimes W \): \( \{|1\rangle \otimes |a\rangle, |1\rangle \otimes |b\rangle, |2\rangle \otimes |a\rangle, |2\rangle \otimes |b\rangle\} \).
- Determine the action of \( \hat{C} = \hat{A} \otimes \hat{I}_W \) on each basis vector. Since \( \hat{I}_W \) leaves \( W \) unchanged, \( \hat{C} \) acts only on the \( V \) part.
- Calculate \( \hat{C}(|1\rangle \otimes |a\rangle) = (\hat{A}|1\rangle) \otimes |a\rangle = |1\rangle \otimes |a\rangle \). Thus, \( |1\rangle \otimes |a\rangle \) is an eigenvector with eigenvalue \( 1 \).
- Calculate \( \hat{C}(|1\rangle \otimes |b\rangle) = (\hat{A}|1\rangle) \otimes |b\rangle = |1\rangle \otimes |b\rangle \). Thus, \( |1\rangle \otimes |b\rangle \) is an eigenvector with eigenvalue \( 1 \).
- Calculate \( \hat{C}(|2\rangle \otimes |a\rangle) = (\hat{A}|2\rangle) \otimes |a\rangle = -|2\rangle \otimes |a\rangle \). Thus, \( |2\rangle \otimes |a\rangle \) is an eigenvector with eigenvalue \( -1 \).
- Calculate \( \hat{C}(|2\rangle \otimes |b\rangle) = (\hat{A}|2\rangle) \otimes |b\rangle = -|2\rangle \otimes |b\rangle \). Thus, \( |2\rangle \otimes |b\rangle \) is an eigenvector with eigenvalue \( -1 \).
Units check. The eigenvalues are dimensionless scalars, consistent with the definition of the operator \( \hat{A} \) which maps basis vectors to themselves or their negatives. The eigenvectors are dimensionless state vectors in the normalized basis.
Three quick tests
- If \( \dim(V) = 3 \) and \( \dim(W) = 4 \), what is the dimension of \( V \otimes W \)?
Answer
12 - Does the operator \( \hat{A} \otimes \hat{I} \) commute with \( \hat{I} \otimes \hat{B} \)?
Answer
Yes, they act on different subspaces and thus commute. - Is the vector \( \frac{1}{\sqrt{2}}(|1\rangle \otimes |a\rangle + |2\rangle \otimes |b\rangle) \) a product state?
Answer
No, it is an entangled state.
Problem set for PU-104 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Derivations cited: D-082 · D-084 · Threads: symmetry, waves, matter