Inner Product and Hilbert Space
The set of all reasonably well-behaved functions that are possibly subjected to some imposed requirement, on a fixed interval \( [a,b] \), can be regarded to constitute an infinite-dimensional vector space. (Readers may refer to Linear Algebra) A recurring example is \( \mathcal{C}^{n} \), \(n\)-times continuously differentiable functions. For investigations into Sturm-Liouville theory, we are particularly interested in Hilbert spaces, which are (usually infinite-dimensional) complete vector spaces equipped with an inner product that accepts two functions and outputs a scalar. To facilitate, our choice of the inner product will be
\begin{align}
\langle f,g \rangle_w = \int_a^b f(x) w(x) g^*(x) dx \tag{1}
\end{align}
where \( ^* \) denotes complex conjugate and \( w(x) > 0 \) is a positive-definite real weighting function, marked by the subscript. We may omit the integration limits for convenience if there is no confusion. (1) can be seen to satisfy the requirements of an inner product: Conjugate symmetry (\( \langle f, g \rangle = \overline{\langle g,f \rangle} \)), additivity, homogeneity, and positivity. Here we briefly show the positivity part: the inner product (1) of a function with itself
\begin{align}
\langle f,f \rangle_w = \int f(x) w(x) f^*(x) dx = \int w(x) |f(x)|^2 dx > 0 \tag{2}
\end{align}
is always positive, unless \(f(x) \equiv 0 \) is identically zero. Similar to finite-dimensional vector spaces for Linear Algebra, we have the notion of orthogonality: \(f\) and \(g\) are known to be orthogonal if their inner product is zero:
\begin{align}
\langle f,g \rangle_w = \int f(x) w(x) g^*(x) dx = 0 \tag{3}
\end{align}
Orthogonality implies linear independence. The square root of (2) is referred to as the norm, much akin to the length by Pythagoras’ Theorem for a geometric vector:
\begin{align}
\Vert f \Vert_w &= \sqrt{\langle f,f \rangle_w} \\
&= \left(\int f(x) w(x) f^*(x) dx\right)^{\frac{1}{2}} = \left(\int w(x) |f(x)|^2 dx\right)^{\frac{1}{2}} \tag{4}
\end{align}
There are some important observations about inner products in general. First is the Cauchy-Schwarz inequality:
\begin{align}
|\langle f,g \rangle| \leq \Vert f \Vert \Vert g \Vert \tag{5}
\end{align}
The proof is as follows: Consider \(h = f+tg\), where \(t\) is any scalar, \(f,g,h\) are functions in the relevant Hilbert space, then \( \Vert h \Vert^2 = \langle h,h \rangle \geq 0\) by positivity. Also, \( \langle h,h \rangle \) can be written as a quadratic polynomial in \(t\):
\begin{align}
\langle h,h \rangle &= \langle f+tg,f+tg \rangle \\
&= \Vert f \Vert^2 + 2t\langle f,g \rangle + \Vert g \Vert^2 \tag{6}
\end{align}
Since this quantity is always greater than or equal to zero, i.e. the quadratic polynomial has no root or a repeated root, it means that the discriminant must be negative or zero. So,
\begin{align}
\Delta = b^2 − 4ac &\leq 0 \\
(2\langle f,g \rangle)^2 − 4\Vert f \Vert^2\Vert g \Vert^2 &\leq 0 \\
\langle f,g \rangle^2 &\leq \Vert f \Vert^2\Vert g \Vert^2 \tag{7}
\end{align}
which is equivalent to (5). The next one is the triangle inequality:
\begin{align}
\Vert f + g \Vert \leq \Vert f \Vert + \Vert g \Vert \tag{8}
\end{align}
This can be easily proved by squaring (8) itself and using (7), left as an exercise to the curious readers. Finally, for the study of Sturm-Liouville equations, it is encouraged to restrict our attention to separable Hilbert spaces only. “Separable” means that the Hilbert space has a countable Schauder basis \( \{\varphi_j\} = \varphi_1, \varphi_2, \varphi_3, \ldots \) such that any function \( f \) in the Hilbert space can be expressed as a linear combination of
\begin{align}
f = \sum_{j=1}^{\infty} c_j\varphi_j = c_1\varphi_1 + c_2\varphi_2 + c_3\varphi_3 + \cdots \tag{9}
\end{align}
with \(c_j\) being constants. We will soon show that a separable Hilbert space always permits an orthonormal basis in a special way. If it is really the case, i.e. \( \langle \varphi_i,\varphi_j \rangle = 0\) when \( i \neq j\) and \( \langle \varphi_j,\varphi_j \rangle = 1\), then the coefficients in (9) can be computed by taking the inner product with the corresponding basis function (an orthogonal projection):
\begin{align}
\langle f, \varphi_j \rangle &= \langle c_1\varphi_1 + c_2\varphi_2 + c_3\varphi_3 + \cdots , \varphi_j \rangle \\
&= c_1\langle \varphi_1, \varphi_j \rangle + c_2\langle \varphi_2, \varphi_j \rangle + \cdots + c_j\langle \varphi_j, \varphi_j \rangle + \cdots \\
&= (0) + (0) + \cdots + c_j(1) + (0) + \cdots = c_j \tag{10}
\end{align}
We have omitted the rigorous justification (\(L^2\) convergence, see later) from the first to the second line to pass the inner product into the infinite sum.
Adjoint, Self-adjoint, and Hermitian
With the inner product like (1) defined, we can define the adjoint \( \mathcal{L}^\dagger \) for a given linear operator \( \mathcal{L} \), which is another operator that satisfies
\begin{align}
\langle \mathcal{L}[f],g \rangle_w &= \int w(x) \mathcal{L}[f(x)] \overline{g(x)} dx \\
&= \int w(x) f(x) \overline{(\mathcal{L}^{\dagger}[g(x)])} dx + \text{Boundary Terms} \\
&= \langle f,\mathcal{L}^{\dagger}[g] \rangle_w + \text{Boundary Terms} \tag{11}
\end{align}
for all functions \( f,g \) in the corresponding inner product space \(\mathcal{V}\), and the boundary terms are evaluated at the end-points of the integration interval as a by-product. An overline denotes complex conjugation like the asterisk. The definition of (11) suggests that adjoints are found by the technique of integration by parts.
Adjoints satisfy some common properties: let \( T \) and \( U \) be two operators, then:
- \( (cT)^{\dagger} = c^*T^{\dagger} \), where \(c\) is any complex scalar;
- \( (T^{\dagger})^{\dagger} = T \) (the adjoint of adjoint is itself);
- \( (T \pm U)^{\dagger} = T^{\dagger} \pm U^{\dagger} \);
- \( (TU)^{\dagger} = U^{\dagger}T^{\dagger} \).
We will briefly show the last item here. Using (11) twice, we have
\begin{align}
\langle TU[f], g \rangle = \langle U[f], T^{\dagger}[g] \rangle = \langle f, U^{\dagger}T^{\dagger}[g] \rangle \tag{12}
\end{align}
so we identify that \(TU^{\dagger} = U^{\dagger}T^{\dagger}\).
A linear operator \( \mathcal{L} \) is self-adjoint if its adjoint is equal to itself: \( \mathcal{L}^\dagger = \mathcal{L} \), as outlined in (11). Furthermore, a self-adjoint operator is Hermitian (or called truly self-adjoint) if all boundary terms in (11) vanish.
If the weighting is of unity \( w(x) = 1 \) in (1), then the inner product shall be known as the usual \(L^2\)-inner product, and we will drop the subscript in this case. Now, we are going to show that the Sturm-Liouville operator \( \mathcal{L} = -[\frac{d}{dx}(p(x)\frac{d}{dx}) + q(x)] \) in the last tutorial is self-adjoint with respect to this inner product:
\begin{align}
\langle \mathcal{L}[f], g \rangle &= \int_a^b \mathcal{L}[f(x)]\overline{g(x)} dx \\
&= \int_a^b -[\frac{d}{dx}(p(x)\frac{df}{dx}) + q(x)f(x)]\overline{g(x)} dx \\
&= -\int_a^b \frac{d}{dx}(p(x)\frac{df}{dx})\overline{g(x)} dx -\int_a^b q(x)f(x)\overline{g(x)} dx \\
&= -[p(x)\frac{df}{dx}\overline{g(x)}]_a^b + \int_a^b \frac{df}{dx}p(x)\overline{\frac{dg}{dx}} dx \\
&\quad -\int_a^b q(x)f(x)\overline{g(x)} dx \\
&= -[p(x)\frac{df}{dx}\overline{g(x)}]_a^b + [f(x)p(x)\overline{\frac{dg}{dx}}]_a^b\\
&\quad -\int_a^b f(x) \frac{d}{dx}(p(x) \overline{\frac{dg}{dx}})dx -\int_a^b q(x)f(x)\overline{g(x)} dx \\
&= -[p(x)\frac{df}{dx}\overline{g(x)}]_a^b + [f(x)p(x)\overline{\frac{dg}{dx}}]_a^b\\
&\quad -\int_a^b f(x) [\frac{d}{dx}(p(x)\overline{\frac{dg}{dx}}) + q(x)\overline{g(x)}] dx \\
&= -[p(x)\frac{df}{dx}\overline{g(x)}]_a^b + [f(x)p(x)\overline{\frac{dg}{dx}}]_a^b\\
&\quad +\int_a^b f(x) \overline{(-[\frac{d}{dx}(p(x)\frac{dg}{dx}) + q(x)g(x)])} dx \\
&= -[p(x)\frac{df}{dx}\overline{g(x)}]_a^b + [f(x)p(x)\overline{\frac{dg}{dx}}]_a^b\\
&\quad +\int_a^b f(x) \overline{\mathcal{L}[g(x)]} dx = \text{Boundary Terms} + \langle f, \mathcal{L}[g] \rangle\tag{13}
\end{align}
hence \( \mathcal{L}^{\dagger} = \mathcal{L} \) is self-adjoint where we have integrated by parts twice and use the fact that \(p(x), q(x)\) are real in the second last line. To make the Sturm-Liouville operator really Hermitian, the required overall boundary condition to eliminate the boundary terms is that \( [-(df(x)/dx) p(x) \overline{g(x)} + f(x) p(x) \overline{dg(x)/dx}]_a^b = 0 \) for any pair of \(f,g\). This boundary condition will also have to be obeyed by the eigenfunctions \( \varphi_j \) where \( \mathcal{L}[\varphi_j] = \lambda_j\rho(x)\varphi_j \). However, sometimes \( p(x) \) in \(\mathcal{L}\) may take a form such that at the end-points its value is exactly zero, so that \(\mathcal{L}\) is automatically Hermitian over this natural interval.
Properties of Hermitian Operators
There are two important properties about Hermitian operators. The first one is
Properties 1. (Reality of Eigenvalues) The eigenvalues of any Hermitian operator \( \mathcal{L}^\dagger = \mathcal{L} \) are real.
The proof is as follows (we shall do it for any inner product with some weight \(w\)). Let \( \varphi \) be an eigenfunction (weighted by a real positive function \(\rho > 0\)) of \( \mathcal{L} \), then
\begin{align}
\langle \mathcal{L}[\varphi], \varphi \rangle_w &= \langle \lambda\rho \varphi, \varphi \rangle_w \\
&= \lambda \langle \rho \varphi, \varphi \rangle_w \tag{14}
\end{align}
but also, since \( \mathcal{L}^\dagger = \mathcal{L} \) is Hermitian, and by conjugate symmetry:
\begin{align}
\langle \mathcal{L}[\varphi], \varphi \rangle_w = \langle \varphi, \mathcal{L}^\dagger[\varphi] \rangle_w &= \langle \varphi, \mathcal{L}[\varphi] \rangle_w \\
&= \langle \varphi, \lambda \rho \varphi \rangle_w \\
&= \overline{\langle \lambda \rho \varphi, \varphi \rangle}_w \\
&= \lambda^* \overline{\langle \rho \varphi, \varphi \rangle_w} \tag{15}
\end{align}
It is not hard to show that in general, the weighted norm \( \langle \rho \varphi, \varphi \rangle_w\) is real and positive-definite for any non-zero eigenfunction \( \varphi \neq 0 \) under a suitable inner product and \(\rho > 0\). To be simple, we will prove that particularly for the form (1):
\begin{align}
\langle \rho \varphi, \varphi \rangle_w &= \int_a^b \rho(x) \varphi(x) w(x) \varphi^*(x) dx = \langle \varphi, \varphi \rangle_{\rho w} \\
&= \int_a^b (\rho(x)^{\frac{1}{2}} w(x)^{\frac{1}{2}})^2 |\varphi(x)|^2 dx \\
&= \int_a^b |\rho(x)^{\frac{1}{2}} w(x)^{\frac{1}{2}}\varphi(x)|^2 dx > 0 \tag{16}
\end{align}
Hence by (14), (15), and the remark:
\begin{align}
\lambda \langle \rho \varphi, \varphi \rangle_w = \lambda^* \overline{\langle \rho \varphi, \varphi \rangle_w} &= \lambda^* \langle \rho \varphi, \varphi \rangle_w \tag{17}
\end{align}
Furthermore, \( \varphi \neq 0 \) and \( \langle \rho \varphi, \varphi \rangle_w \neq 0 \) are both non-zero, so this implies that \( \lambda = \lambda^* \), i.e. the eigenvalue is real.
The second result is
Properties 2. (Orthogonality between Eigenfunctions) Any two eigenfunctions of a Hermitian operator \( \mathcal{L}^\dagger = \mathcal{L} \) corresponding to two distinct eigenvalues are always orthogonal to each other (with respect to the weighting function after the eigenvalue).
We start by denoting the two (weighted) eigenfunctions as \( \varphi_1, \varphi_2 \) with the respective eigenvalues \( \lambda_1, \lambda_2 \), then
\begin{align}
\langle \mathcal{L}[\varphi_1], \varphi_2 \rangle_w &= \langle \lambda_1\rho\varphi_1, \varphi_2 \rangle_w \\
&= \lambda_1 \langle \rho\varphi_1, \varphi_2 \rangle_w \tag{18}
\end{align}
On the other hand,
\begin{align}
\langle \mathcal{L}[\varphi_1], \varphi_2 \rangle_w = \langle \varphi_1, \mathcal{L}^\dagger[\varphi_2] \rangle_w &= \langle \varphi_1, \mathcal{L}[\varphi_2] \rangle_w \\
&= \langle \varphi_1, \lambda_2\rho\varphi_2 \rangle_w \\
&= \lambda_2^* \langle \varphi_1, \rho\varphi_2 \rangle_w = \lambda_2 \langle \varphi_1, \rho\varphi_2 \rangle_w \tag{19}
\end{align}
where we have used the fact that the Hermitian eigenvalue is real as in Properties 1. Again, it can be shown that for the inner product (1), \( \langle \rho\varphi_1, \varphi_2 \rangle_w = \int w\rho\varphi_1\varphi_2^* dx = \int w\varphi_1(\rho\varphi_2)^* dx = \langle \varphi_1, \rho\varphi_2 \rangle_w \). So (18) and (19) together mean that
\begin{align}
\lambda_1 \langle \rho\varphi_1, \varphi_2 \rangle_w &= \lambda_2 \langle \rho\varphi_1, \varphi_2 \rangle_w \\
(\lambda_1-\lambda_2) \langle \rho\varphi_1, \varphi_2 \rangle_w &= 0 \tag{20}
\end{align}
but it is supposed that the two eigenvalues are distinct \( \lambda_1 \neq \lambda_2 \), hence it must be that \( \langle \rho\varphi_1, \varphi_2 \rangle_w = \langle \varphi_1, \varphi_2 \rangle_{\rho w} = 0 \) and the eigenfunctions \(\varphi_1, \varphi_2 \) are orthogonal with respect to the combined weighting \( \rho w \) (this means that we can absorb either one of \(\rho, w\) into a single weight and the conclusion remains the same). In integral form, we have \( \int w(x)\rho(x)\varphi_1\varphi_2^* dx = 0 \). With this, we can go one step further and divide the orthogonal eigenfunctions of the Hermitian operator by their norms to make them have unit length and hence become orthonormal.
A final emphasis is that, since Sturm-Liouville operators are Hermitian (as long as the boundary conditions are fulfilled), the two nice properties above also cover them.
Exercise
Show that \( \mathcal{L} = -i\hbar\frac{d}{dx} \), (\(i\) is the imaginary number) famously the momentum operator in Quantum Mechanics, is a self-adjoint operator with respect to the \(L^2\)-inner product over the entire \(x\)-axis. Find the form of its eigenfunctions.
Answer
Integrating by parts to get
\begin{align}
\langle \mathcal{L}[f], g \rangle &= \int_{-\infty}^{\infty} \mathcal{L}[f(x)] \overline{g(x)} dx \\
&= \int_{-\infty}^{\infty} -i\hbar\frac{d}{dx}(f(x)) \overline{g(x)} dx \\
&= [-i\hbar f(x)\overline{g(x)}]_{-\infty}^{\infty} -\int_{-\infty}^{\infty} f(x) (-i\hbar\frac{d}{dx}(\overline{g(x)})) dx \\
&= \int_{-\infty}^{\infty} f(x) (i\hbar\frac{d}{dx}(\overline{g(x)})) dx + [-i\hbar f(x)\overline{g(x)}]_{-\infty}^{\infty} \\
&= \int_{-\infty}^{\infty} f(x) \overline{(-i\hbar\frac{d}{dx}(g(x)))} dx + [-i\hbar f(x)\overline{g(x)}]_{-\infty}^{\infty} \\
&= \int_{-\infty}^{\infty} f(x) \overline{\mathcal{L}[g(x)]} dx + [-i\hbar f(x)\overline{g(x)}]_{-\infty}^{\infty} \\
\end{align}
So we see that \( \mathcal{L}^\dagger = \mathcal{L} = -i\hbar\frac{d}{dx}\) is self-adjoint. The eigenfunctions \( \varphi_m \) of \( \mathcal{L} \) can be found from solving the first-order ODE:
\begin{align}
\mathcal{L}[\varphi_m] = -i\hbar\frac{d\varphi_m}{dx} &= m\varphi_m \\
-\int i\hbar\frac{d\varphi_m}{\varphi_m} &= \int m dx \\
-i\hbar \ln |\varphi_m| &= m x + C \\
\varphi_m &= Ke^{\frac{i}{\hbar}m x}
\end{align}
where \( K = \pm e^{\frac{i}{\hbar}C} \) is any scaling constant. Notice that the eigenvalue \( m \) can take any value ranging from \(-\infty\) to \(\infty\) as there is no boundary condition being enforced. In this case, the eigenvalues form a continuous spectrum, and we note that the inner product space generated by those eigenfunctions, which are uncountable in this problem, is not separable.
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