Spectra of Hermitian and Unitary Operators

The following is central to the workings of quantum mechanics.

Eigenvalues and eigenvectors for Hermitian operators¶

Consider a Hermitian operator $A = A^{\dagger}$.

1. Theorem. The eigenvalues of $A$ are real.

Proof. Consider a normalized eigenvector $\ket{a}$ ($||a|| = 1$) such that $A \ket{a} = a \ket{a}$. This means that $\bra{a} A \ket{a} = a$. By the skew-symmetry of the adjoint,

2. Theorem. Eigenvectors of $A = A^{\dagger}$ with distinct eigenvalues are orthogonal

Proof Consider eigenvectors $\ket{a}, \ket{b}$ such that

Then

for $a \neq b$ this is only possible if $\brket{b}{a} = 0$.

3. Theorem. The eigenvectors of $A = A^{\dagger}$ can be constructed to form an orthonormal basis $\ket{a_i}$ such that $A\ket{a_i} = a_i \ket{a_i}$.

I will only sketch a proof.

• The eigenvalues satisfy $\text{det}(A - a {\bf 1}) = 0$ is a $D$th-order polynomial equation in $a$ (where $d = \text{dim}(V)$). Thus

There are always $d$ complex roots to this equation. If $A = A^{\dagger}$ all eigenbvalues are real so all $a_i$ are real. There are still $d$ roots. Hhere $a_i$ are the eigenvalues. If $a_i \neq a_j$ for $i \neq j$, we are done, as $\brket{a_i}{a_j} = \delta_{ij}$ and there are $d$ such eigenvectors.

• If there are $d_k$ identical eigenvalues, one needs to show (and I will not) that this means the space of vectors which are eigenvectors of $A$ with eigenvalue $a_k$ form a $d_k$-dimensional subspace. This can be made into an orthonormal basis via teh Gram-Schmidt machine.

4. Theorem. For any Hermitian operator $A$, there exists a unitary matrix $U$ and a diagonal matrix $\Lambda$ such that $A = U^{\dagger} \Lambda U$, and the entries of $\Lambda$ are the eigenvalues, such than an eigenvalue $a_k$ with degree of degeneracy $d_k$ appears in $d_k$ entries.

Proof (sketch). Given an orthonormal basis $\ket{i}$, $A_{ij} = \bra{i} A \ket{j}$. On the other hand there is an orthonormal basis of eigenvectors $\ket{a_k,\alpha_k}$, where $A \ket{a_k,\alpha_k} = a_k \ket{a_k,\alpha_k}$, $\alpha_k = 1,\ldots,d_k$, and $\brket{a_k,\alpha_k}{a_{\ell},\beta_{\ell}} = \delta_{kl} \delta_{\alpha_k,\beta_{\ell}}$. Let us relabel $(a_k,\alpha_k) \to {\tilde a}$. Then $\bra{\tilde a} A \ket{\tilde b} = \Lambda_{{\tilde a}{\tilde b}}$, where $\Lambda$ is diagonal with the entries equal to the eigenvalues in the way described. Now using ${\bf 1} = \sum_{{\tilde a}} \ket{{\tilde a}}\bra{{\tilde a}}$, we can write

where $U_{{\tilde a}i} = \brket{{\tilde a}}{i}$, and $U^{\dagger}_{j{\tilde a}} = \brket{j}{{\tilde a}}$. You can convince yourselves that these are Hermitian conjugates of each other. We have already shown in Changes of basis that these are inverses of each other, so $U$ is a unitary matrix and the statement holds.

For those familiar with the singular value decomposition of matrices, this is of course a special case.

The next theorem will be central to the uncertainty principle.

5. Theorem. Two Hermitian operators $A,B$ are diagonal in the same orthonormal basis if and only if $[A,B] = 0$.

Proof (partial). For the “only if” part, if $A,B$ are diagonal in the same basis, we use the fact that we can writ $A = U^{\dagger} \Lambda_A U$, $B = U^{\dagger} \Lambda_B U$, following the construction above. $A,B$ are diagonalized by the same unitary matrix because this matrix implements the change of basis and by supposition the basis of eigenvectors is the same. Then since diagonal matrices commute,

For the “if” part, let $A,B$ commute. We take the case that the eigenvalues of $A,B$ are distinct (see for example Bellac, 2006). Then if $A \ket{a_i} = a_i \ket{a_i}$, $A(B \ket{a_i}) = BA \ket{a_i} = a_i B \ket{a_i}$. That is, $B\ket{a_i}$ is an eigenvector of $A$. But by supposition the subspace of eigenvectors with a fixed eigenvalue is one-dimensional, $B\ket{a_i} = b_i \ket{a_i}$ for some $b_i$ which is clearly an eigenvalue of $b_i$.

Unitary operators¶

Theorem For any unitary operator $U$ we can write $U = V^{\dagger}\Lambda V$ where $V$ is a unitary matrix, and $\Lambda$ diagonal with enries of the form $e^{i\lambda}$, $\lambda \in \CR$.

Example: Operators on the space of functions¶

For functions with periodic boundary conditions, the functions $\psi_n(x) = \frac{1}{\sqrt{L}} e^{2\pi i n x/L}$, $n \in \CZ$ are a basis (since Fourier modes are a basis of periodic functions. In this case, ${\hat p} \psi_n = \frac{2\pi \hbar n}{L} \psi_n$; this basis is a basis of eigenfunctions of ${\hat p}$. Note that ${\hat x}:\psi(x) \to x\psi(x)$ is not an operator in this space.

For $L^2(\CR)$, ${\hat x}$ is almost such an operator, if we focus on $\psi(x)$ falling off fast enough (faster than $\frac{1}{|x|^{3/2}}$). The fact that it fails in some cases means that ${\hat x}$ is not a bounded operator. One can also show the same for the momentum operator ${\hat p} = \frac{\hbar}{i} \frac{\del}{\del x}$. Nonetheless, these operators are useful and we will use them. I will table a discussion of them for a week or two, when we get to wave mechanics.

Projection operators and spectral representations¶

Projection operators¶

1. Definition. Let $V$ be a vector space over $\CC$ and $W_{1,2} \subset V$ be vector subspaces of $V$. $W_1$, $W_2$ are orthogonal (denoted $W_1 \perp W_2$) if $\forall \ket{w_1} \in W_1, \ket{w_2} \in W_2$, $\brket{w_1}{w_2} = 0$.

2. Theorem. $\forall W \subset V$ vector subspaces, there exists a subspace $W^{\perp} \subset V$ such that $\forall \ket{v} \in V$, there exists a unique expression

with $\ket{w} \in W$, $\ket{w^{\perp}} \in W^{\perp}$. This is known as an orthogonal decomposition of $\ket{v}$. I will forgo a proof here.

Note that we can form orthogonal/orthonormal bases $\ket{i}, \ket{I}$ for $W,W^{\perp}$ respectively; by the above theorem, since any vector can be written as a sum of vectors in the wubspaces $W,W^{\perp}$, the collection of basis elemenbts $\ket{i},\ket{I}$ are a basis for $V$. Thus if $d = \text{dim} V$, $d_W = \text{dim}(W)$, $d_{W^{\perp}} = \text{dim}(W^{\perp})$, we have

3. Definition. A projection operator $\CP$ is a Hermitian operator such that $\CP^2 = \CP$. The reasons for calling this a projection operator will become clear as we explore its properties.

• The eigenvalues of $\CP$ are either 1 or 0. Given an eigenvector $\ket{a}$ with $\CP\ket{a} = a \ket{a}$, we can apply $\CP$ to both sides to find

The only solutions are $a = 1$ or $a = 0$.

• Theorem. Eigenvectors with zero eigenvalue span $\text{Ker}(\CP)$. (Stated without proof).

• ${\bf 1} - \CP$ is also a projection operator.

• Theorem. Eigenvectors with unit eigenvalue span $(\text{Ker}(\CP))^{\perp}= \text{Ker}({\bf 1} - \CP)$ which is also the range of $\CP$

• In an orthonormal basis of eigenvectors of $\CP$, we can write a matrix representation of $\CP$ as:

where $\text{dim}(\text{range}(\CP)) = d_P$ and $\text{dim}(\text{Ker}(\CP)) = d_{P^{\perp}}$. If the eigenvectors of $\CP$ with unit eigenvalues are $\ket{i}$, $i \in \{1, \ldots , d_P\}$, then we can also write

In the latter case, we can show this is Hermitian by first realizing that

(can you shouw this?), and then showing that

Spectral representation of Hermitian operators¶

Consider a vector space $V$ and a Hermitian operator $A: V\to V$ with a basis of eigenstates $\ket{a_i,\alpha_i}$, where $A\ket{a_i,\alpha_i} = a_i \ket{a_i,\alpha_i}$, $\alpha_i \in \{1,\ldots,d_i\}$, and $\brket{a_i,\alpha_i}{a_j,\alpha_j} = \delta_{ij}\delta_{\alpha_i,\alpha_j}$. Here if $d_i > 1$ then $a_i$ is a degenerate eigenvalue and $d_i$ is the degree of degeneracy. We can write a projection operator

which projects onto the subspace $V_i \subset V$ of vectors with eigenvalue $a_i$. We can then write the operator $A$ as

The right hand side has the same eigenvalues and eigenvector as $A$, by construction; thus it has the same matrix elements in a basis of eigenvectors of $A$, and so must be the same operator as $A$.

Note that the fact that $\ket{a_i,\alpha_i}$ are a complete orthonormal basis for $V$ means that