Editing Self-adjoint operator (section)

== Symmetric vs self-adjoint operators ==
{{see also|Extensions of symmetric operators}}

Although the distinction between a symmetric operator and a (essentially) self-adjoint operator is subtle, it is important since self-adjointness is the hypothesis in the spectral theorem. Here we discuss some concrete examples of the distinction.

=== 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>.

=== Schrödinger operators with singular potentials ===
A more subtle example of the distinction between symmetric and (essentially) self-adjoint operators comes from [[Schrödinger equation|Schrödinger operators]] in quantum mechanics. If the potential energy is singular—particularly if the potential is unbounded below—the associated Schrödinger operator may fail to be essentially self-adjoint. In one dimension, for example, the operator
: <math>\hat{H} := \frac{P^2}{2m} - X^4</math>
is not essentially self-adjoint on the space of smooth, rapidly decaying functions.<ref>{{harvnb|Hall|2013}} Theorem 9.41</ref> In this case, the failure of essential self-adjointness reflects a pathology in the underlying classical system: A classical particle with a <math>-x^4</math> potential escapes to infinity in finite time. This operator does not have a ''unique'' self-adjoint, but it does admit self-adjoint extensions obtained by specifying "boundary conditions at infinity". (Since <math>\hat{H}</math> is a real operator, it commutes with complex conjugation. Thus, the deficiency indices are automatically equal, which is the condition for having a self-adjoint extension.)

In this case, if we initially define <math>\hat{H}</math> on the space of smooth, rapidly decaying functions, the adjoint will be "the same" operator (i.e., given by the same formula) but on the largest possible domain, namely
: <math>\operatorname{Dom}\left(\hat{H}^*\right) = \left\{ \text{twice differentiable functions }f \in L^2(\mathbb{R})\left|\left( -\frac{\hbar^2}{2m}\frac{d^2f}{dx^2} - x^4f(x)\right) \in L^2(\mathbb{R}) \right. \right\}. </math>

It is then possible to show that <math>\hat{H}^*</math> is not a symmetric operator, which certainly implies that <math>\hat{H}</math> is not essentially self-adjoint. Indeed, <math>\hat{H}^*</math> has eigenvectors with pure imaginary eigenvalues,<ref>{{harvnb|Berezin|Shubin|1991}} p. 85</ref><ref>{{harvnb|Hall|2013}} Section 9.10</ref> which is impossible for a symmetric operator. This strange occurrence is possible because of a cancellation between the two terms in <math>\hat{H}^*</math>: There are functions <math>f</math> in the domain of <math>\hat{H}^*</math> for which neither <math>d^2 f/dx^2</math> nor <math>x^4f(x)</math> is separately in <math>L^2(\mathbb{R})</math>, but the combination of them occurring in <math>\hat{H}^*</math> is in <math>L^2(\mathbb{R})</math>. This allows for <math>\hat{H}^*</math> to be nonsymmetric, even though both <math>d^2/dx^2</math> and <math>X^4</math> are symmetric operators. This sort of cancellation does not occur if we replace the repelling potential <math>-x^4</math> with the confining potential <math>x^4</math>.

=== Non-self-adjoint operators in quantum mechanics===
{{see also|Non-Hermitian quantum mechanics}}
In quantum mechanics, observables correspond to self-adjoint operators. By [[Stone's theorem on one-parameter unitary groups]], self-adjoint operators are precisely the infinitesimal generators of unitary groups of [[time evolution]] operators. However, many physical problems are formulated as a time-evolution equation involving differential operators for which the Hamiltonian is only symmetric. In such cases, either the Hamiltonian is essentially self-adjoint, in which case the physical problem has unique solutions or one attempts to find self-adjoint extensions of the Hamiltonian corresponding to different types of boundary conditions or conditions at infinity.

'''Example.''' The one-dimensional Schrödinger operator with the potential <math>V(x) = -(1 + |x|)^\alpha</math>, defined initially on smooth compactly supported functions, is essentially self-adjoint for {{math|0 < ''α'' ≤ 2}} but not for {{math|''α'' > 2}}.{{sfn|Berezin|Shubin|1991|pp=55,86|ps=}}{{sfn|Hall|2013|pp=193-196|ps=}}

The failure of essential self-adjointness for <math>\alpha > 2</math> has a counterpart in the classical dynamics of a particle with potential <math>V(x)</math>: The classical particle escapes to infinity in finite time.<ref>{{harvnb|Hall|2013}} Chapter 2, Exercise 4</ref>

'''Example.''' There is no self-adjoint momentum operator <math>p</math> for a particle moving on a half-line. Nevertheless, the Hamiltonian <math>p^2</math> of a "free" particle on a half-line has several self-adjoint extensions corresponding to different types of boundary conditions. Physically, these boundary conditions are related to reflections of the particle at the origin.{{sfn | Bonneau | Faraut | Valent | 2001 |ps=}}