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 $\hat{A}^{\dagger}\hat{A} = \hat{A}\hat{A}^{\dagger} = \hat{I}$: \[[\hat{A},\hat{A}^{\dagger}] = \hat{A}\hat{A}^{\dagger} - \hat{A}^{\dagger}\hat{A} = \hat{I} - \hat{I} = 0.\]