Eigenvalues, commutators & tensor products

How do small matrices become operators on several spins, and when can a large matrix be divided into smaller problems? This bridge develops the finite-dimensional linear algebra behind those steps. You will check an eigenbasis, distinguish an expectation value from an eigenvalue, construct a tensor-product matrix, and use a commutator to identify invariant sectors. Every example is small enough to calculate by hand before returning to the XXX chain.

Required background. Add and multiply complex numbers, including $i^2=-1$. The entry check repairs conjugation and matrix multiplication. No previous spin-chain calculation is required.

Helpful background. The XXX introduction explains why these operations matter for a finite quantum system.

Complex columns and linear operators

Work in $\mathbb C^d$, for a finite positive integer $d$, with an ordered orthonormal basis $e_1,\ldots,e_d$. A vector is a column; its adjoint is its conjugate transpose. Thus

$$v=\begin{pmatrix}v_1\\v_2\end{pmatrix} \quad\Longrightarrow\quad v^\dagger=\begin{pmatrix}\overline v_1&\overline v_2\end{pmatrix}.$$

The inner product and norm are

$$u^\dagger v=\sum_j\overline u_jv_j, \qquad \|v\|_2=\sqrt{v^\dagger v} =\sqrt{\sum_j|v_j|^2}.$$

For $v\ne0$, the vector $v/\|v\|_2$ has norm one. The zero vector cannot be normalized. Quantum notation $|v\rangle$ denotes the same column, while $\langle v|$ denotes its adjoint. These finite-dimensional conventions agree with Preskill, July 2015 notes, § 2.1, pp. 3–4, equation (2.1), PDF.

A linear operator $A:\mathbb C^d\to\mathbb C^d$ satisfies $A(au+bv)=aAu+bAv$. Its $j$th matrix column is $Ae_j$, expressed in the chosen basis:

$$Ae_j=\sum_i A_{ij}e_i.$$

The second index labels the input basis vector; the first labels an output component. Specifying the basis order is part of specifying a matrix. In a product $ABv$, apply $B$ first, then $A$.

All explicit matrices and eigenvalues below are dimensionless. Multiplying a matrix by an energy scale gives the corresponding energy units when it is used as a Hamiltonian.

Entry check and repair

Let

$$w=\begin{pmatrix}1\\i\end{pmatrix}, \qquad A=\begin{pmatrix}1&-i\\i&1\end{pmatrix}.$$

Calculate $w^\dagger w$, $w^{\mathsf T}w$, and $Aw$. Normalize $w$. What is the first column of $A$, and which input vector produces it?

Repair: conjugate for the norm, use columns for the action

Since $\overline i=-i$,

$$w^\dagger w=1+(-i)i=2, \qquad w^{\mathsf T}w=1+i^2=0.$$

The transpose alone gives the wrong squared norm. The normalized vector is $w/\sqrt2$. Row-by-column multiplication gives

$$Aw=\begin{pmatrix}1+(-i)i\\i+i\end{pmatrix} =\begin{pmatrix}2\\2i\end{pmatrix}=2w.$$

The first column is $(1,i)^{\mathsf T}=Ae_1$, where $e_1=(1,0)^{\mathsf T}$. As a retry, $Ae_2=(-i,1)^{\mathsf T}$ reproduces the second column.

A complete two-dimensional eigenvalue calculation

A matrix is Hermitian when $A^\dagger=A$. Here $A^\dagger$ means transpose the matrix and conjugate every entry; a complex Hermitian matrix need not equal its transpose. The example above is Hermitian because its two off-diagonal entries are complex conjugates.

An eigenvector is a nonzero vector $v$ satisfying $Av=\lambda v$. For a Hermitian matrix its eigenvalue is real: the scalar $v^\dagger Av$ equals its own complex conjugate, and

$$\lambda=\frac{v^\dagger Av}{v^\dagger v}.$$

To find the eigenvalues of our $A$, compute

$$\det(A-\lambda I) =(1-\lambda)^2-(-i)i =\lambda(\lambda-2).$$

The eigenvalues are $0$ and $2$. Solving each two-row equation and normalizing gives

$$v_0=\frac1{\sqrt2}\begin{pmatrix}1\\-i\end{pmatrix}, \qquad v_2=\frac1{\sqrt2}\begin{pmatrix}1\\i\end{pmatrix}, \qquad Av_0=0,\quad Av_2=2v_2.$$

The subscripts label eigenvalues. In particular, $v_0$ is a unit vector with eigenvalue zero, not the zero vector. Direct checks give

$$v_0^\dagger v_0=v_2^\dagger v_2=1, \qquad v_0^\dagger v_2=\frac{1+i^2}{2}=0.$$

Two orthonormal vectors in a two-dimensional space form a complete basis. More explicitly, their outer products satisfy

$$v_0v_0^\dagger+v_2v_2^\dagger=I, \qquad A=0\,v_0v_0^\dagger+2v_2v_2^\dagger.$$

An outer product $vv^\dagger$ is a matrix, while $v^\dagger v$ is a scalar. For normalized $v$, the matrix $vv^\dagger$ projects a vector onto the direction $v$. The displayed identity is the spectral resolution of this particular matrix. The finite-dimensional Hermitian spectral theorem supplies an orthonormal eigenbasis in general; see Preskill, § 2.1, pp. 4–5, equations (2.3)–(2.6), PDF.

The resolution also proves positive semidefiniteness, meaning $z^\dagger Az\geq0$ for every $z$:

$$z^\dagger Az=2|v_2^\dagger z|^2 =|z_1-iz_2|^2\geq0.$$

It is not positive definite, since the nonzero vector $v_0$ gives zero. Hermiticity by itself does not require nonnegative eigenvalues; the first exercise tests that distinction.

An expectation value is not an eigenvalue test

For a normalized vector $v$, the expectation value of $A$ is $v^\dagger Av$. At $v=e_1$ it is $1$, even though $1$ is not an eigenvalue of $A$. Indeed,

$$Ae_1-1e_1=\begin{pmatrix}0\\i\end{pmatrix}, \qquad \|Ae_1-1e_1\|_2=1.$$

For a proposed pair $(\lambda,v)$ with $v\ne0$, define the normalized eigenvector residual

$$r(A,v,\lambda)=\frac{\|Av-\lambda v\|_2}{\|v\|_2}.$$

It is exactly zero precisely when $v$ is an eigenvector with the proposed eigenvalue. Dividing by $\|v\|_2$ prevents shrinking a wrong vector from making its test look better. An expectation value describes an average in a state; it does not verify the full vector equation. Numerical residuals also require a stated precision and an appropriate operator scale; this page’s checks are exact.

Tensor products need an ordered basis

Take two labeled spaces $V_1=\mathbb C^2$ and $V_2=\mathbb C^2$. In each, use

$$|0\rangle=\begin{pmatrix}1\\0\end{pmatrix}, \qquad |1\rangle=\begin{pmatrix}0\\1\end{pmatrix}.$$

These are basis labels. For spin chains they can represent $|\uparrow\rangle$ and $|\downarrow\rangle$, respectively. The space $V_1\otimes V_2$ has dimension $2\times2=4$. Fix its basis order as follows:

Column position	Basis vector	Factor 1	Factor 2
1	$\lvert00\rangle$	$\lvert0\rangle$	$\lvert0\rangle$
2	$\lvert01\rangle$	$\lvert0\rangle$	$\lvert1\rangle$
3	$\lvert10\rangle$	$\lvert1\rangle$	$\lvert0\rangle$
4	$\lvert11\rangle$	$\lvert1\rangle$	$\lvert1\rangle$

Here $|a b\rangle=|a\rangle\otimes|b\rangle$, and the second label varies fastest in this order. Bilinearity gives, for local columns $u=(u_0,u_1)^{\mathsf T}$ and $v=(v_0,v_1)^{\mathsf T}$,

$$u\otimes v= \begin{pmatrix}u_0v_0\\u_0v_1\\u_1v_0\\u_1v_1\end{pmatrix}.$$

General vectors in the four-dimensional space are linear combinations of these basis vectors; they need not be a single product $u\otimes v$. For example, the normalized vector $(|00\rangle+i|11\rangle)/\sqrt2$ used below must be acted on term by term. No factorization assumption is needed.

For operators $C$ on $V_1$ and $D$ on $V_2$, define

$$(C\otimes D)(u\otimes v)=(Cu)\otimes(Dv),$$

and extend by linearity. This factor-wise rule, together with the product inner product, is given in Preskill, § 2.1, p. 7, equations (2.13)–(2.14), PDF.

Derive the Kronecker matrix rule

Write $C|a\rangle=\sum_b C_{ba}|b\rangle$ and $D|\mu\rangle=\sum_\nu D_{\nu\mu}|\nu\rangle$. Applying both operators gives

$$(C\otimes D)|a\mu\rangle =\sum_{b,\nu}C_{ba}D_{\nu\mu}|b\nu\rangle.$$

Thus the entry in row $(b,\nu)$ and column $(a,\mu)$ is $C_{ba}D_{\nu\mu}$. Grouping the rows and columns by their first label gives the Kronecker product matrix

$$C\otimes D= \begin{pmatrix} C_{00}D&C_{01}D\\ C_{10}D&C_{11}D \end{pmatrix}.$$

Each entry shown here is a $2\times2$ block. This is a $4\times4$ matrix, not the ordinary $2\times2$ product $CD$.

For a concrete column check, define the Pauli matrices

$$X=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad Y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\qquad Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$

Then $(X\otimes Z)|01\rangle=-|11\rangle$, so column 2 of $X\otimes Z$ is $(0,0,0,-1)^{\mathsf T}$. Recovering a column from a basis input is a useful way to catch a reversed factor order. With $\hbar=1$, physical spin operators are $S^x=X/2$, $S^y=Y/2$, $S^z=Z/2$, as in Preskill, § 2.2.1, p. 11, equations (2.28)–(2.29), PDF.

Commutators depend on which space is acted on

For square matrices on the same space, define the commutator

$$[A,B]=AB-BA.$$

It measures the difference between applying the operators in the two possible orders. A statement $[A,B]=0$ is an operator identity: it must hold on every vector. Agreement on a single input does not establish it.

The tensor action rule proves

$$(C\otimes D)(E\otimes F)=CE\otimes DF$$

by applying both sides to $u\otimes v$ and extending linearly. In particular, operators acting on separate factors commute:

$$(C\otimes I)(I\otimes D)=C\otimes D =(I\otimes D)(C\otimes I).$$

This remains true even if the local matrices $C$ and $D$, considered on a single copy of $\mathbb C^2$, do not commute. When both act on the first factor instead,

$$[C\otimes I,D\otimes I]=[C,D]\otimes I.$$

Sharing a factor does not force noncommutation, but the separate-factor argument no longer settles the question. For $N$ sites, the notation $C_n$ means

$$C_n=I\otimes\cdots\otimes C\otimes\cdots\otimes I,$$

with $C$ in position $n$; it acts on a space of dimension $2^N$. This is the local-operator embedding used in Faddeev 1996, § 2, equations (6), (14)–(15), PDF. The label $n$ specifies a tensor factor, not a matrix row.

Why a commuting operator identifies invariant sectors

Let $H$ and $Q$ be Hermitian matrices on the same finite-dimensional space, and suppose $[H,Q]=0$. For an eigenvalue $q$ of $Q$, define its eigenspace

$$\mathcal E_q=\{v:Qv=qv\}.$$

It includes the zero vector and may have dimension larger than one. If $v\in\mathcal E_q$, then

$$Q(Hv)=H(Qv)=qHv.$$

Therefore $Hv\in\mathcal E_q$: $H$ preserves the whole eigenspace. Choosing a basis grouped by these eigenspaces makes $H$ block diagonal, so each block can be solved separately. If $H$ generates time evolution, its powers and hence its exponential preserve the same spaces. The commutation condition is the conservation condition discussed in Preskill, § 2.2.1, p. 10, equation (2.23), PDF.

Within a degenerate eigenspace of $Q$, $H$ can still mix vectors. Commutation does not make every prechosen eigenvector of $Q$ an eigenvector of $H$. The restriction of Hermitian $H$ to each invariant eigenspace can itself be diagonalized; doing this supplies a common orthonormal eigenbasis. The transfer exercise exhibits the distinction explicitly.

In the XXX chain, $Q$ can be total magnetization. Its eigenspaces are the sectors with a fixed number of down spins. This linear-algebra argument explains the reduction used by the short-chain lesson; symmetry sectors alone do not establish quantum integrability.

Exercises

Guided: shift and rescale the spectrum

Using the worked matrix $A$, set $B=3I-2A$. Find its explicit matrix, normalized eigenvectors and eigenvalues without solving a new characteristic equation. Is $B$ Hermitian? Is it positive semidefinite? Explain separately what an identity shift and a negative rescaling do to eigenvalues and eigenvectors.

Hint

If $Av=\lambda v$, compute $(3I-2A)v$. To disprove positive semidefiniteness, one vector with a negative quadratic form is sufficient.

Solution

Direct subtraction gives

$$B=\begin{pmatrix}1&2i\\-2i&1\end{pmatrix}.$$

For the same normalized vectors as before,

$$Bv_0=3v_0,\qquad Bv_2=-v_2.$$

Thus the eigenvalues are $3,-1$. The off-diagonal entries are complex conjugates, so $B$ is Hermitian, but $v_2^\dagger Bv_2=-1$ shows it is not positive semidefinite.

More generally, $cI+dA$ retains every eigenvector of $A$ and changes its eigenvalue to $c+d\lambda$. An identity shift adds the same constant to all eigenvalues; a negative real rescaling reverses their ordering. Neither requires changing the eigenvectors. This is the matrix reason an energy offset can change the vacuum energy without changing states.

Independent: separate factors and shared factors

In the ordered basis $(|00\rangle,|01\rangle,|10\rangle,|11\rangle)$, construct $X\otimes I$, $I\otimes Z$, and $Z\otimes I$. Apply each to

$$|\psi\rangle=\frac{|00\rangle+i|11\rangle}{\sqrt2}.$$

Then calculate $[X\otimes I,I\otimes Z]$ and $[X\otimes I,Z\otimes I]$ as operator identities. Two of the three individual actions on $|\psi\rangle$ coincide. Does that make their operators equal? Find a basis input that answers the question.

Hint

$X$ interchanges the local basis states; $Z$ leaves $|0\rangle$ unchanged and negates $|1\rangle$. Find the four columns from those rules. For the shared-factor commutator, multiply the two local matrices first.

Solution

The matrices are

$$X\otimes I= \begin{pmatrix} 0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0 \end{pmatrix},$$ $$I\otimes Z=\operatorname{diag}(1,-1,1,-1), \qquad Z\otimes I=\operatorname{diag}(1,1,-1,-1).$$

Acting term by term or multiplying these matrices gives

$$\begin{aligned} (X\otimes I)|\psi\rangle &=\frac{|10\rangle+i|01\rangle}{\sqrt2},\\ (I\otimes Z)|\psi\rangle &=\frac{|00\rangle-i|11\rangle}{\sqrt2},\\ (Z\otimes I)|\psi\rangle &=\frac{|00\rangle-i|11\rangle}{\sqrt2}. \end{aligned}$$

The tensor multiplication rule proves

$$[X\otimes I,I\otimes Z]=0.$$

For the other pair, $XZ=-iY$ and $ZX=iY$, so

$$[X\otimes I,Z\otimes I]=-2iY\otimes I\ne0.$$

As an independent sign check, this last commutator sends $|\psi\rangle$ to $\sqrt2(|10\rangle-i|01\rangle)$, a nonzero vector of norm $2$.

Finally, $I\otimes Z$ and $Z\otimes I$ are different despite their equal action on $|\psi\rangle$: on $|01\rangle$ their outputs are $-|01\rangle$ and $+|01\rangle$, respectively. One successful input test cannot prove equality of operators.

Transfer: break an invariant sector

In $\mathbb C^3$, use the ordered basis $(e_1,e_2,e_3)$ and matrices

$$H=\begin{pmatrix}1&-i&0\\i&1&0\\0&0&3\end{pmatrix}, \qquad Q=\operatorname{diag}(1,1,-1).$$

Verify $[H,Q]=0$, identify the two $Q$-sectors, and find the complete eigenbasis of $H$. Is the displayed basis already a common eigenbasis of both operators?

Now let $\epsilon$ be real and define

$$H_\epsilon=H+\epsilon(e_1e_3^\dagger+e_3e_1^\dagger).$$

Compute $[H_\epsilon,Q]$. For $\epsilon\ne0$, show explicitly that the old sectors are no longer invariant. Which properties survive? You do not need to solve the perturbed cubic eigenvalue equation.

Hint

For diagonal $Q$ with entries $q_j$, $[H,Q]_{ij}=(q_j-q_i)H_{ij}$. Reuse the worked eigenvectors in the upper $2\times2$ block. To test the perturbation, inspect $H_\epsilon e_1$.

Solution

The only off-diagonal entries of $H$ join basis states with the same $Q$-eigenvalue, so its commutator with $Q$ vanishes. The sectors are

$$\mathcal E_{+1}=\operatorname{span}\{e_1,e_2\}, \qquad \mathcal E_{-1}=\operatorname{span}\{e_3\}.$$

The normalized eigenvectors and corresponding $H$-eigenvalues are

$$\frac{e_1-ie_2}{\sqrt2}:0,\qquad \frac{e_1+ie_2}{\sqrt2}:2,\qquad e_3:3.$$

They form a common orthonormal eigenbasis of $H$ and $Q$. The original displayed basis is an eigenbasis of $Q$, but not of $H$: for example, $He_1=e_1+ie_2$ is not proportional to $e_1$. Degeneracy permits mixing inside $\mathcal E_{+1}$.

The perturbed commutator is

$$[H_\epsilon,Q] =\begin{pmatrix}0&0&-2\epsilon\\0&0&0\\2\epsilon&0&0\end{pmatrix}.$$

For nonzero $\epsilon$,

$$H_\epsilon e_1=e_1+ie_2+\epsilon e_3 \notin\mathcal E_{+1}, \qquad H_\epsilon e_3=\epsilon e_1+3e_3 \notin\mathcal E_{-1}.$$

The old sector decomposition therefore fails. The matrix remains Hermitian for real $\epsilon$, so its eigenvalues are real and it has an orthonormal eigenbasis, but the old three eigenvectors no longer solve its eigenvalue problem. A broken symmetry label does not imply that the matrix ceases to define a valid finite quantum problem.

Return to the spin chain

Use these calculations in Construct a small spin-chain Hamiltonian: the site labels specify tensor factors, matrix columns encode actions on basis states, and magnetization conservation selects invariant blocks. That lesson uses these operations to derive the physical bond interaction.

For the algebraic route, continue to Check a rational R-matrix. Its permutation operators act on specified tensor factors and compose from right to left, exactly as the matrices do here. The larger argument will establish commuting transfer matrices; basic Hermiticity, finite diagonalization or a single symmetry reduction is not that integrability argument.

These results apply to finite-dimensional complex spaces. No assertion about domains of unbounded operators or completeness of an infinite-dimensional eigenfunction expansion is being used.

References

• Faddeev, L. D. How Algebraic Bethe Ansatz works for integrable model. Les Houches lecture notes, 1996. arXiv:hep-th/9605187v1, 26 May 1996. Version record. Open PDF, § 2, equations (6), (14)–(15).
• Preskill, John. Lecture Notes for Ph219/CS219: Quantum Information, Chapter 2, “Foundations I: States and Ensembles.” California Institute of Technology, updated July 2015. Author’s PDF, § 2.1, pp. 3–7, equations (2.1)–(2.6) and (2.13)–(2.14); § 2.2.1, pp. 10–11, equations (2.23), (2.28)–(2.29).