Self-Adjoint Operators

Last time we explored the idea of a unitary operator, $U$, defined by the condition

\begin{equation}\label{004:unitary}U^{\dagger} = U^{-1}.\end{equation}

$U$ acts on a complex vector space, and the condition \eqref{004:unitary} maintains the associated default inner product,

\begin{equation}\label{004:ip}\langle z,z \rangle = \langle Uz,Uz \rangle = |z_{1}|^{2} + |z_{2}|^{2}.\end{equation}

A Reality Condition

Today we'll examine another interesting class of operators, the self-adjoint or self-dual operators. These are defined by the equation

\begin{equation}\label{004:hermitian}H^{\dagger} = H.\end{equation}

If this reminds you of the \textit{reality} condition on the complex numbers,

$$\bar{z} = z,$$

great! Keep that intuition close.

Here's a quick exercise to test your understanding:

Show that \eqref{004:hermitian} implies:

\begin{equation}\label{004:reality} \langle z,Hz \rangle = \langle Hz,z \rangle, \end{equation}

and show that this implies

$$ \langle z,Hz \rangle = \overline{\langle z,Hz \rangle},$$

so that finally,

$$ \langle z,Hz \rangle \in \mathbb{R}.$$

When our liner space is of finite dimension, there is an orthonormal basis for our vector space such that

\begin{equation}\label{004:basis} H\psi_{i} = h_{i}\psi_{i},\end{equation}

where $h_{i}\in\mathbb{R}$.
We can expand $z$ in this basis, so that

$$z = \sum_{i=1}^{N}z^{i}\,\psi_{i},$$

and therefore

$$ Hz = \sum_{i=1}^{N} h_{i} z^{i}\,\psi_{i},$$

so that

$$\langle z,Hz \rangle = \sum_{i=1}^{N} h_{i} |z^{i}|^{2}\,|\psi_{i}|^{2} = \sum_{i=1}^{N} h_{i} |z^{i}|^{2}.$$

Let's return to our familiar example, the vector space $\mathbb{C}^{2}$, and see what the self-adjoint requirement implies.  As we've discussed in recent days, linear transformations on this space amount to $2\times 2$ matrices. The condition \eqref{004:hermitian} constrains these to

\begin{equation}\label{004:matrix}\left(\begin{array}{cc} a & c \\ \bar{c} & b \end{array}\right) \quad a,b\in\mathbb{R},\; c\in\mathbb{C},\end{equation}

which takes the number of real independent degrees of freedom\footnote{That is, four complex numbers of the form $z=a+ib$.} from eight to four.

Matrix multiplication does not commute, but matrix addition does. So we can rewrite \eqref{004:matrix} in a different way,

\begin{equation}\label{004:matrix2} Z = \left(\begin{array}{cc} x_{0} + x_{3} & x_{1} - i x_{2} \\ x_{1} + i x_{2} & x_{0} -  x_{3} \end{array}\right),\end{equation}

where all four $x_{i}\in\mathbb{R}$.

Here's another quick exercise to test your understanding:

Verify that $\det Z$, given in \eqref{004:matrix2} equals

\begin{equation}\label{004:detz}\det Z = x_{0}^{2} + x_{1}^2 + x_{2}^{2} + x_{3}^2,\end{equation}

and is therefore strictly non-negative. This might remind you of an inner product on $\mathbb{R}^{4}$. What does the invertibility of $Z$ imply for this structure?

As the exercise hinted, the form of $Z$ strongly suggests a vector space isomorphism with $\mathbb{R}^{4}$. Taking the structure of \eqref{004:matrix2} seriously, we can expand a generic self-adjoint operator $H$ as,

$$ H = x_{0}\sigma_{0} + x_{1}\sigma_{1} +x_{2}\sigma_{2}+x_{3}\sigma_{3},$$

where

\begin{equation}\label{004:sigmas} \sigma_{0} = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right),\quad  \sigma_{1} = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right),\; \sigma_{2} = \left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right),\; \sigma_{3} = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right). \end{equation}

In other words, the matrices $\sigma_{i}$ take the form of a basis for a vector space.

So the space of self-adjoint operators on $\mathbb{C}^{2}$ itself forms a vector space that is isomorphic to $\mathbb{R}^4$. To complete this discussion, let's explore a possible inner product on the vector space of these matrices and try to replicate \eqref{004:detz}.

We define the inner product on this space of self-adjoint matrices as

\begin{equation}\label{004:ip2} \langle M , N \rangle_{\sigma} = \frac{1}{2}\mathrm{Tr}(MN). \end{equation}

Here's a fun exercise to test your understanding.

Demonstrate by explicit computation that $\langle \sigma_i , \sigma_j \rangle_{\sigma} = \delta_{ij},$ where $\delta_{ij}$ equals one if $i=j$ and zero otherwise. Hence, the $\sigma_i$ form an orthonormal basis for $\mathbb{R}^{4}$.

Based on these results, and \eqref{004:detz} you can verify that

$$\det Z = \langle Z,Z\rangle_{\sigma} = \frac{1}{2}\mathrm{Tr}(ZZ),$$

which is a special property of working in two-complex dimensions.

We will explore the properties of these $\sigma$ matrices in depth in our future discussions, and this accidental relationship to $\mathbb{R}^4$ will play a large role in that discussion.

Here's a final exercise to explore a similar idea.

Define the matrix $J$ by

$$ J = \left(\begin{array}{cc} 0 & 1\\ -1 & 0 \end{array}\right), $$

and consider the space of all matrices such that

$$ M = a\mathbb{1} + bJ, \quad a,b\in\mathbb{R}. $$

Show that $M$ is isomorphic to $\mathbb{R}^{2}$, and that the inner product $\langle\cdot,\cdot\rangle_{\sigma}$ corresponds to the usual, euclidean one. In what way does the space of all such $M$ related to $\mathbb{C}$?

TL;DR

The duality induced by an inner product on a complex vector space can be used to extend the reality condition of the complex numbers to linear operators. These self-adjoint operators have real eigenvalues. This might have been obvious, given that self-adjoint matrices are necessarily square, and square matrices with complex entries are diagonalizable. But using the inner product in the definition ensures these ideas extend to more general spaces. For the specific case of $\mathbb{C}^{2}$, the space of self-adjoint linear operators themselves form a real vector space isomorphic to $\mathbb{R}^{4}$.