Editing Self-adjoint operator (section)

=== Boundary conditions ===
In the case where the Hilbert space is a space of functions on a bounded domain, these distinctions have to do with a familiar issue in quantum physics: One cannot define an operator—such as the momentum or Hamiltonian operator—on a bounded domain without specifying ''boundary conditions''. In mathematical terms, choosing the boundary conditions amounts to choosing an appropriate domain for the operator. Consider, for example, the Hilbert space <math>L^2([0, 1])</math> (the space of square-integrable functions on the interval [0,1]). Let us define a momentum operator ''A'' on this space by the usual formula, setting the Planck constant to 1:
: <math>Af = -i\frac{df}{dx}.</math>

We must now specify a domain for ''A'', which amounts to choosing boundary conditions. If we choose
: <math>\operatorname{Dom}(A) = \left\{\text{smooth functions}\right\},</math>

then ''A'' is not symmetric (because the boundary terms in the integration by parts do not vanish).

If we choose 
: <math>\operatorname{Dom}(A) = \left\{\text{smooth functions}\,f \mid f(0) = f(1) = 0\right\},</math>

then using integration by parts, one can easily verify that ''A'' is symmetric. This operator is not essentially self-adjoint,<ref>{{harvnb|Hall|2013}} Proposition 9.27</ref> however, basically because we have specified too many boundary conditions on the domain of ''A'', which makes the domain of the adjoint too big (see also the [[Self-adjoint operator#A symmetric operator that is not essentially self-adjoint|example]] below).

Specifically, with the above choice of domain for ''A'', the domain of the closure <math>A^{\mathrm{cl}}</math> of ''A'' is
: <math>\operatorname{Dom}\left(A^{\mathrm{cl}}\right) = \left\{\text{functions } f \text{ with two derivatives in }L^2 \mid f(0) = f(1) = 0\right\},</math>
whereas the domain of the adjoint <math>A^*</math> of ''A'' is
: <math>\operatorname{Dom}\left(A^*\right) = \left\{\text{functions } f \text{ with two derivatives in }L^2\right\}.</math>

That is to say, the domain of the closure has the same boundary conditions as the domain of ''A'' itself, just a less stringent smoothness assumption. Meanwhile, since there are "too many" boundary conditions on ''A'', there are "too few" (actually, none at all in this case) for <math>A^*</math>. If we compute <math>\langle g, Af\rangle</math> for <math>f \in \operatorname{Dom}(A)</math> using integration by parts, then since <math>f</math> vanishes at both ends of the interval, no boundary conditions on <math>g</math> are needed to cancel out the boundary terms in the integration by parts. Thus, any sufficiently smooth function <math>g</math> is in the domain of <math>A^*</math>, with <math>A^*g = -i\,dg/dx</math>.<ref>{{harvnb|Hall|2013}} Proposition 9.28</ref>

Since the domain of the closure and the domain of the adjoint do not agree, ''A'' is not essentially self-adjoint. After all, a general result says that the domain of the adjoint of <math>A^\mathrm{cl}</math> is the same as the domain of the adjoint of ''A''. Thus, in this case, the domain of the adjoint of <math>A^\mathrm{cl}</math> is bigger than the domain of <math>A^\mathrm{cl}</math> itself, showing that <math>A^\mathrm{cl}</math> is not self-adjoint, which by definition means that ''A'' is not essentially self-adjoint.

The problem with the preceding example is that we imposed too many boundary conditions on the domain of ''A''. A better choice of domain would be to use periodic boundary conditions:
: <math>\operatorname{Dom}(A) = \{\text{smooth functions}\,f \mid f(0) = f(1)\}.</math>

With this domain, ''A'' is essentially self-adjoint.<ref>{{harvnb|Hall|2013}} Example 9.25</ref>

In this case, we can understand the implications of the domain issues for the spectral theorem. If we use the first choice of domain (with no boundary conditions), all functions <math>f_\beta(x) = e^{\beta x}</math> for <math>\beta \in \mathbb C</math> are eigenvectors, with eigenvalues <math>-i \beta</math>, and so the spectrum is the whole complex plane. If we use the second choice of domain (with Dirichlet boundary conditions), ''A'' has no eigenvectors at all. If we use the third choice of domain (with periodic boundary conditions), we can find an orthonormal basis of eigenvectors for ''A'', the functions <math>f_n(x) := e^{2\pi inx}</math>. Thus, in this case finding a domain such that ''A'' is self-adjoint is a compromise: the domain has to be small enough so that ''A'' is symmetric, but large enough so that <math>D(A^*)=D(A)</math>.