Adjoint Boundary Conditions

The operators which are of immediate interest to us are differential operators. Although their actions consist of taking derivatives, their definition is more restrictive. The additional properties they have to satisfy is the consequence of the fact that an operator is a type of mapping, and as such one must always specify not only its fomula (or rule) but also its domain. The domain in our case is a subspace of the given Hilbert space. Thus, to specify uniquely a 2nd order differential operator, one must specify three things:

(i)

The domain $\cal H$ , the Hilbert space, which we shall take to be $L^2(a,b)$ , the space of functions square-integrable on $[a,b]$ .

(ii)

The homogeneous boundary conditions to be satisfied by $u\in \cal H$ .

(iii)

$$L= \alpha (x) \frac{d^2}{dx^2}+\beta (x)\frac{d}{dx}+\gamma (x)$$ , i.e. the ``formula''.

Items (i)-(iii) are referred to collectively as the ``operator $L$ ''. One also should note that (i) and (ii) define a linear subspace $\cal S$ of $\cal H$ as follows: Let $L$ to be the linear map whose image $Lu$ has a well defined inner product, i.e. $\langle v, Lu\rangle =$ finite, for any square-integrable $u$ and $v$ . For the set of continuously differentiable functions, $C^2(a,b)$ , this means that

$$\begin{array}{rcl} L:~~~~~~~~ {\cal{S}}{\cap} C^2(a,b) &\long... ...hcal{H} = L^2(a,b) \\ u &\sim \! \leadsto & L u ~~. \end{array}$$

Here $\mathcal{S}$ is the domain of $L$ , and it is

$${ \mathcal{S}=\{ ~u\in L^2(a,b): ~ \langle v, Lu\rangle=\textrm{finite}~\forall~ v \in L^2(a,b); u~\text{satisfies}\hspace{1.5in} }$$

$$\hspace{1in}\textrm{the given homogeneous boundary conditions at} ~a~ \textrm{and}~ b\}$$

What if $u$ is not a continuously differentiable function? Then its image $Lu$ is not square-integrable, but the inner product $\langle v, Lu\rangle$ is still well-defined because it is finite. For example, if u is a function which has a kink, then $Lu$ would not be defined at that point and $Lu$ would not be square-integrable. Nevertheless, the integral of $\overline v Lu$ would be perfectly finite.

The (Hermitian) adjoint $L^*$ of an operator such as $L$ is defined by the requirement that

$$\langle v,Lu \rangle =\langle L^*v,u \rangle \quad$$

for all $u\in \cal S$ and all $v$ belonging to $\cal S^*$ , the domain of $L^*$ . This is illustrated in the examples below. In compliance with standard notation, we are using $L^*$ , and not $L^H$ to refer to the Hermitian adjoint of the differential operator $L$ . In some physics text books one finds $L^\dagger$ instead.

Example 1. Let $$L=\frac{d}{dx}$$ have as its domain the subspace

$$\mathcal{S} =\{ ~u\in L^2(a,b):~u(a)=2u(b)~~;\langle v,Lu \rangle=\textrm{finite~ whenever}~v \in L^2(a,b)\}~~,$$ (41)

and let the inner product be

$$\langle v,u \rangle = \int _a ^b \overline{v} (x)u(x)~dx~~.$$

FIND the adjoint of this operator.

To do this, one integrates by parts in order to move the operator from the second factor to the first and thereby obtains

$$\langle v,Lu \rangle = \int _a ^b \overline{v} \frac{d}{dx}u~dx$$

$$= \overline{v} (b) u(b) -\overline{v}(a)u(a) - \int _a ^b \frac{d \overline{v}}{dx} u~dx ~~.$$

Using the boundary condition, one obtains

$$\langle v,Lu \rangle = [\overline{v}(b)-2\overline{v}(a)]u(b) +\int_a^b \overline{ (-)\frac{d}{dx}v }~u~dx$$

$$\equiv \langle L^* v, u \rangle ~~.$$

This determines $L^*$ , provided the boundary term vanishes for all $u \in \mathcal{S}$ . This implies that $v$ must satisfy the adjoint boundary condition

$$\overline{v}(b)-2 \overline{v}(a)=0 ~~.$$ (42)

The conclusion is this: the adjoint $L^*$ of $L$ consists of two parts,

(i)

$$L^*= -\frac{d}{dx} \hspace{2in} (\lq\lq the ~formula'')$$

(ii)

the adjoint boundary condition, Eq.(), which determines the domain

$${\mathcal{S}^*}= \{ ~v\in L^2(a,b):~v(a)=\frac{1}{2} v(b)~ ; \langle L^* v,u\rangle<\infty~\forall u \in L^2(a,b)\}~~~~~~~~~~~~~~$$

$$\hspace{.3in} (\lq\lq the~domain'')$$

on which $L^*$ operates.

The expression $L^* =-\frac{d}{dx}$ without the boundary condition is called the formal adjoint of $L$ . If $v$ , and hence $\overline v$ , satisfies the adjoint boundary condition, then the ``formal'' adjoint becomes the adjoint of $L$ . In this case one has

$$\langle v,Lu \rangle =\langle L^* v,u \rangle$$

for all $u \in \mathcal{S}$ and all $v \in \mathcal{S^*}$ .

It is clear that $L$ and its adjoint $L^*$ are different operators: they differ not only in their domain but also in their formula.

Example 2. Consider $$L=i\frac{d}{dx}$$ whose domain is the subspace $\mathcal{S}=\{~u:~u(a)=u(b)~ \}$ .

FIND its adjoint.

Following the familiar procedure, one obtains

$$\langle v,Lu \rangle = i [\overline{v}(b)-\overline{v}(a)]u(b) +\int_a^b \overline{ i\frac{d}{dx}v }~u~dx$$

$$= \langle L^* v, u \rangle ~~.$$

This holds for all $u \in \mathcal{S}$ , provided $v(a)=v(b)$ . It follows that

$$\begin{array}{rcll} L^* &=& \displaystyle i \frac{d}{dx} &(\lq\lq ... ...thcal{S}^*&=& \{ ~v:~v(a)-v(b)=0 ~ \} &(\lq\lq domain'') \end{array}$$

One sees that both the formal adjoint (``the formula'') and its domain are the same as the given operator. This observation motivates the following

Definition. An operator said to be self-adjoint, if both its formula and its domain are the same, i.e.

$$L^*=L$$

and

$$\mathcal{S}^*=\mathcal{S}$$

Reminder: Sometimes we shall mean by $L^*$ only the ``formal'' adjoint of $L$ , at other times we shall mean by $L^*$ the adjoint of $L$ , which includes the boundary conditions. The context will make clear which is which.

Lecture 27