Hermitian, anti-Hermitian and unitary operators are the three classes of operators that are used most frequently in quantum mechanics. We can say the following about their eigenvalues.

• The eigenvalues of a Hermitian operator are real:

To prove this, Let $\ket{a}$ be an eigenvector of $\hat{A}$ with eigenvalue $a$ and compute $\sand{a}{\hat{A}}{a}$ in two different ways. First, by acting with $\hat{A}$ to the right, \[\sand{a}{\hat{A}}{a} = \bra{a} \left ( \hat{A} \ket{a}\right ) = a \braket{a}{a} = a.\] Second, using hermiticity and acting with $\hat{A}$ to the left, \[\sand{a}{\hat{A}}{a} = \sand{a}{\hat{A}^{\dagger}}{a} = \left ( \bra{a} \hat{A}^{\dagger} \right ) \ket{a} = a^* \braket{a}{a} = a^*.\] Equating the two expressions gives $a=a^*$, and hence $a$ is real.

• The eigenvalues of an anti-Hermitian operator are imaginary:

Again we let $\ket{a}$ be an eigenvector of $\hat{A}$ with eigenvalue $a$ and compute $\sand{a}{\hat{A}}{a}$ in two different ways. First, by acting with $\hat{A}$ to the right, \[\sand{a}{\hat{A}}{a} = \bra{a} \left ( \hat{A} \ket{a}\right ) = a \braket{a}{a} = a.\] Second, using anti-hermiticity and acting with $\hat{A}$ to the left, \[\sand{a}{\hat{A}}{a} = -\sand{a}{\hat{A}^{\dagger}}{a} = -\left ( \bra{a} \hat{A}^{\dagger} \right ) \ket{a} = -a^* \braket{a}{a} = -a^*.\] Equating the two expressions gives $a=-a^*$, and hence $a$ is pure imaginary.

• The eigenvalues of a unitary operator have modulus $1$:

Again we let $\ket{a}$ be an eigenvector of $\hat{A}$ with eigenvalue $a$, but this time we compute $\sand{a}{\hat{A}^{\dagger}\hat{A}}{a}$ in two different ways. Firstly, by unitarity, $\hat{A}^{\dagger}\hat{A} = \hat{I}$, so \[\sand{a}{\hat{A}^{\dagger}\hat{A}}{a} = \sand{a}{\hat{I}}{a} = \braket{a}{a} = 1.\] Second, we use the eigenvalue equation $\hat{A}\ket{a} = a\ket{a}$, and act with $\hat{A}$ to the right and $\hat{A}^{\dagger}$ to the left. \[\sand{a}{\hat{A}^{\dagger}\hat{A}}{a} = \left ( \bra{a} \hat{A}^{\dagger}\right ) \left ( \hat{A} \ket{a}\right ) = a^* a\braket{a}{a} = a^*a = |a|^2.\] Equating the two expressions gives $|a|^2 = 1$. As a consequence, $a = e^{i\theta}$ for some real-valued angle $\theta$.

Some straightforward properties of eigenvalues and eigenvectors are as follows.

• If $\hat{A}\ket{\psi} = a\ket{\psi}$ then $\hat{A}^{n}\ket{\psi} = a^n \ket{\psi}$.

The case $n=1$ is true by definition, and we can prove the general case by induction. Suppose that $\hat{A}^{n}\ket{\psi} = a^n \ket{\psi}$. Then \[\hat{A}^{n+1}\ket{\psi} = \hat{A}\hat{A}^n \ket{\psi} = \hat{A}a^n\ket{\psi} = a^n \hat{A}\ket{\psi} = a^n a\ket{\psi} = a^{n+1} \ket{\psi}.\]

• If $\hat{A}\ket{\psi} = a\ket{\psi}$ then, for any function $f$, $f(\hat{A})\ket{\psi} = f(a)\ket{\psi}$.

This follows from the fact that functions are represented by their power series, the previous result, and linearity. Suppose the power series representation of $f$ is \[f(z) = \sum_{n=0}^{\infty} c_n z^n.\] Then \begin{align*} f(\hat{A})\ket{\psi} & = \left ( \sum_{n=0}^{\infty} c_n \hat{A}^n \right ) \ket{\psi} \\ & = \sum_{n=0}^{\infty} \left ( c_n \hat{A}^n \ket{\psi} \right ) \qquad\text{by linearity} \\ & = \sum_{n=0}^{\infty} \left ( c_n a^n \ket{\psi} \right )\qquad \text{by the previous result} \\ & = \left ( \sum_{n=0}^{\infty} c_n a^n \right ) \ket{\psi}\qquad \text{by linearity again} \\ & = f(a) \ket{\psi}. \end{align*}

In particular, a special case of this that we use often in quantum mechanics is if $\hat{A}\ket{\psi} = a\ket{\psi}$ then $e^{i\hat{A}} \ket{\psi} = e^{ia}\ket{\psi}$.

The set of eigenvectors of $\hat{A}$ sharing the same eigenvalue $a$ forms a subspace of the Hilbert space, i.e. if $\hat{A}\ket{\psi} = a\ket{\psi}$ and $\hat{A}\ket{\phi} = a\ket{\phi}$ then \[\hat{A}\left ( c \ket{\psi} + d\ket{\phi} \right ) = a\left ( c \ket{\psi} + d\ket{\phi} \right ),\] for any scalars $c$ and $d$.

This property is a straightforward consequence of linearity. A subspace formed by eigenvectors sharing the same eigenvalue of $\hat{A}$ is called an eigenspace of $\hat{A}$.

Note that, since eigenvectors with a given eigenvalue form a subspace, if $\ket{\psi}$ is an eigenvector of $\hat{A}$ with eigenvalue $a$ then so is $c\ket{\psi}$ for any scalar $c$. For this reason, we can always choose to work with normalized eigenvectors such that $\|\psi \| = 1$. If you have an unnormalized eigenvector you can just multiply it by whatever scalar is needed to make it normalized. From now on, we will do this and by eigenvector I will mean normalized eigenvector unless otherwise stated. Note that this does not completely eliminate the ambiguity because if $\ket{\psi}$ is a normalized eigenvector of $\hat{A}$ with eigenvalue $a$ then so is $e^{i\theta}\ket{\psi}$ for any phase angle $\theta$.

An eigenvalue $a$ of $\hat{A}$ is called nondegenerate if the corresponding eigenspace is one-dimensional, i.e. up to multiplication by a scalar there is a unique vector such that $\hat{A}\ket{\psi} = a\ket{\psi}$. If the dimension of the eigenspace is $\geq 2$ then the eigenvalue is called degenerate. The operator $\hat{A}$ itself is called nondegenerate if all of its eigenspaces are nondegenerate and is otherwise called degenerate. Nondegenerate operators are much easier to deal with, but unfortunately we do often have to deal with degenerate operators in quantum mechanics.

For a nondegenerate eigenvalue, any orthonormal basis for the eigenspace just consists of a single vector, which is unique up to multiplication by a phase $e^{i\theta}$. We will label the basis vector corresponding to eigenvalue $a$ as $\ket{a}$. This shows another advantage of Dirac notation. Since the ket symbol $\ket{}$ indicates that we are dealing with a vector, we use whatever label we like inside the ket to describe the vector. The symbol $\ket{a}$ should be read as “the normalized eigenvector corresponding to eigenvalue $a$”.

For a degenerate eigenvalue, we need more than one vector to form a basis for the eigenspace, so we are going to need another label in addition to $a$. We can construct an orthonormal basis for the eigenspace and label the vectors $\ket{a,1}, \ket{a_2}, \cdots$ where we should read $\ket{a,j}$ as “the $j^{\text{th}}$ vector in an orthonormal basis for the eigenspace corresponding to eigenvalue $a$.

Since the eigenspace corresponding to an eigenvalue $a$ is a subspace, we can form the projection operator $\hat{P}_a$ onto that subspace. This is defined as follows. For all eigenvectors $\ket{\psi}$ with eigenvalue $a$, $\hat{P}_a\ket{\psi} = \ket{\psi}$ and for any vector $\ket{\phi}$ that is orthogonal to all of these vectors $\hat{P}_a\ket{\phi} = 0$.

More explicitly, if $a$ is nondegenerate then \[\hat{P}_a = \proj{a}.\] If $a$ is degenerate then we construct an orthonormal basis $\ket{a,1}, \ket{a,2}, \cdots$ for the eigenspace and then \[\hat{P}_a = \sum_j \proj{a,j}.\] These projection operators act like the identity operator on the eigenspace and the zero operator on the orthogonal complement.

An operator $\hat{A}$ is normal if $[\hat{A},\hat{A}^{\dagger}] = 0$. The relevance of normal operators is that they are completely characterized by their eigenvalues and the projectors onto the corresponding eigenspaces. This makes them simple to deal with. We will discuss this in the next section.

For now, we note that all the important types of operators used in quantum mechanics are normal. For example:

• Hermitian operators $\hat{A}^{\dagger} = \hat{A}$: \[[\hat{A},\hat{A}^{\dagger}] = [\hat{A},\hat{A}] = 0\]
• Anti-Hermitian operators $\hat{A}^{\dagger} = -\hat{A}$: \[[\hat{A},\hat{A}^{\dagger}] = [\hat{A},-\hat{A}] = - [\hat{A},\hat{A}]= 0\]
• Unitary Operators $\hat{A}^{\dagger}\hat{A} = \hat{A}\hat{A}^{\dagger} = \hat{I}$: \[[\hat{A},\hat{A}^{\dagger}] = \hat{A}\hat{A}^{\dagger} - \hat{A}^{\dagger}\hat{A} = \hat{I} - \hat{I} = 0.\]

The set of eigenvalues of an operator is called its spectrum. Note, there is a connection between this mathematics terminology and the physicist's notion of the spectrum of an atom, i.e. the set of frequencies of electromagnetic radiation that it can emit and absorb. The eigenvalues of the Hamiltonian operator in quantum mechanics are the possible energies of the system, so, for the Hamiltonian operator, the mathematician's spectrum is this set of possible energies. But radiation is emitted and absorbed when the system makes a transition between its possible energy levels, so the set of possible energies of an absorbed or emitted photon are the magnitudes of the differences between the eigenvalues of the Hamiltonain. Therefore, the physicists spectrum is the set of magnitudes of differences between the eigenvalues. The mathematician's term spectrum is clearly inspired by the physicist's, although they are not quite the same thing.

The spectral theorem for an operator with a discrete set of eigenvalues $a_1,a_2,\cdots$ is as follows.

Theorem Let $a_1,a_2,\cdots$ be the (distinct) eigenvalues of an operator $\hat{A}$. Then, $\hat{A}$ is a normal operator if an only if all of the following hold:

1. Its eigenspaces are orthogonal: $\hat{P}_{a_j}\hat{P}_{a_k} = \delta_{jk}\hat{P}_{a_j}$.
2. Its eigenspaces are complete: $\sum_j \hat{P}_{a_j} = \hat{I}$.
3. The spectral decomposition holds: $\hat{A} = \sum_j a_j \hat{P}_{a_j}$.

Note that it is possible that one of the eigenvalues $a_j = 0$, in which case the term $a_j P_{a_j} = 0$ may be omitted from the spectral decomposition. However, the completness relation $\sum_j \hat{P}_{a_j} = \hat{I}$ only holds if we include the projector onto the eigenspace with eigenvalue zero.

Note, if the operator $\hat{A}$ has a continuous spectrum on an interval $a_{\mathrm{min}} < a < a_{\mathrm{max}}$ then we would have to write orthogonality as $\hat{P}_a\hat{P}_{a'} = \delta(a - a') \hat{P}_a$, completeness as \[\int_{a_{\mathrm{min}}}^{a_{\mathrm{max}}} \hat{P}_{a} \, \mathrm{d}a = \hat{I},\] and the spectral decomposition as \[\hat{A} = \int_{a_{\mathrm{min}}}^{a_{\mathrm{max}}} a \hat{P}_{a} \, \mathrm{d}a.\] Even more generally, $\hat{A}$ might have a discrete set $a_1,a_2$ of eigenvalues within some interval as well as a continuous set $a_{\mathrm{min}} < a < a_{\mathrm{max}}$ in a disjoint interval, and then we would have to write the spectral decomposition as \[\hat{A} = \sum_j a_j \hat{P}_{a_j} + \int_{a_{\mathrm{min}}}^{a_{\mathrm{max}}} a \hat{P}_{a} \, \mathrm{d}a.\]

The proof of the most general version of the spectral theorem is well beyond the scope of this course. It can be found in many advanced mathematics textbooks such as

• Henry Helson, "The Spectral Theorem", pp. 23--41 (Springer, 1986)

In these notes, we will prove the finite-dimensional case. That proof is fairly straightforward, but involved, so it has its own page. On this page, we will prove a small part of the theorem.

Theorem The eigenvectors corresponding to distinct eigenvalues of a Hermitian operator $\hat{A}$ are orthogonal.

Proof Let $a \neq b$ be distinct eigenvalues of $\hat{A}$ with eigenvectors $\ket{a}$ and $\ket{b}$. Then we can calculate $\sand{b}{\hat{A}}{a}$ in two different ways. First, acting with $\hat{A}$ on the right, \[\sand{b}{\hat{A}}{a} = \bra{b} \left ( \hat{A} \ket{a}\right ) = a \braket{b}{a}.\] Second, using hermiticity and acting with $\hat{A}$ to the left, \[\sand{b}{\hat{A}}{a} = \sand{b}{\hat{A}^{\dagger}}{a} = \left ( \bra{b}\hat{A}^{\dagger} \right ) \ket{a} = b^* \braket{b}{a} = b\braket{b}{a},\] where the last step follows because the eigenvalues of a Hermitian operator are real.

Equating the two expressions gives $a \braket{b}{a} = b\braket{b}{a}$, which we can rearrange to \[(a-b) \braket{b}{a} = 0.\] In order to satisfy this equation, it must be the case that either $(a-b) = 0$ or $\braket{b}{a} = 0$, but we have assumed that $a$ and $b$ are distinct eigenvalues, so $(a-b) \neq 0$ and hence $\braket{b}{a} = 0$.