Self-adjoint and unbounded operators in QM - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-18T00:45:21Z https://physics.stackexchange.com/feeds/question/102476 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/102476 20 Self-adjoint and unbounded operators in QM SRS https://physics.stackexchange.com/users/36793 2014-03-07T18:19:37Z 2014-03-08T18:40:37Z <ol> <li><p>An operator $A$ is said to be <a href="http://en.wikipedia.org/wiki/Self-adjoint_operator">self-adjoint</a> if $(\chi,A\psi)=(A\chi,\psi)$ for $\psi, \chi \in D_A$ and $D_A=D_{A^\dagger}$. But for the free particle momentum operator $\hat{p}$ these inner products does not exist, however its eigenvalues are real. So, is $\hat{p}$ a self-adjoint operator? </p></li> <li><p>Why are the operators in quantum mechanics in general <a href="http://en.wikipedia.org/wiki/Unbounded_operator">unbounded</a>?</p></li> </ol> https://physics.stackexchange.com/questions/102476/-/102485#102485 27 Answer by Valter Moretti for Self-adjoint and unbounded operators in QM Valter Moretti https://physics.stackexchange.com/users/35354 2014-03-07T19:37:22Z 2014-03-08T18:40:37Z <p>If $\cal H$ is a complex Hilbert space, and $A :D(A) \to \cal H$ is linear with $D(A)\subset \cal H$ dense subspace, there is a <em>unique</em> operator, the <strong>adjoint</strong> $A^\dagger$ of $A$ satisfying (this is its definition) $$\langle A^\dagger \psi| \phi \rangle = \langle \psi | A \phi \rangle\quad \forall \phi \in D(A)\:,\forall \psi \in D(A^\dagger)$$ with: $$D(A^\dagger) := \{ \phi \in {\cal H}\:|\: \exists \phi_1 \in {\cal H} \mbox{ with} \: \langle \phi_1 |\psi \rangle = \langle \phi | A \psi\rangle \:\: \forall \psi \in D(A)\}$$ The above densely-defined operator $A$ is said to be <strong>self-adjoint</strong> if $A= A^\dagger$. A densely-defined operator satisfying $$\langle A \psi| \phi \rangle = \langle \psi | A \phi \rangle\quad \forall \psi,\phi \in D(A)$$ is said to be <strong>symmetric</strong>. It is clear that the adjoint of $A$, in this case, is an <em>extension</em> of $A$ itself. </p> <p>A symmetric operator is <strong>essentially selfadjoint</strong> if $A^\dagger$ is self-adjoint. It is possible to prove that it is equivalent to say that $A$ admits a unique self-adjoint extension (given by $(A^\dagger)^\dagger$).</p> <p>The operator $-i\frac{d}{dx}$ with domain given by Schwartz' space ${\cal S}(\mathbb R)$ (but everything follows holds also if the initial domain is $C_0^\infty(\mathbb R)$) is symmetric and essentially self-adjoint. Both $-i\frac{d}{dx}$ and the true momentum operator $p:= \left(-i\frac{d}{dx}\right)^\dagger$ are not bounded. Both operators do not admit eigenvalues and eigenvectors. </p> <p>The spectrum of $p$ is continuous and coincides with the whole real line.</p> <p>Passing to Fourier-Plancherel transform, the operator $p:= \left(-i\frac{d}{dx}\right)^\dagger$ turns out to coincide with the multiplicative operator $k \cdot$.</p> <p>Concerning the issue of unboundedness of most self-adjoint quantum operators, the point is that a celebrated theorem (one of the possible versions of Hellinger–Toeplitz theorem) establishes that:</p> <p><em>a (densely-defined) self-adjoint operator $A :D(A) \to \cal H$ is bounded if and only if $D(A)= \cal H$</em></p> <p>and almost all operators of QM, for various reasons, are <strong>not</strong> defined in the whole Hilbert space (unless the space is finite dimensional). These operators are not initially defined on the whole Hilbert space because they usually are differential operators. Differential operators need some degree of regularity to be applied on a function, whereas the generic element of a $L^2$ space is incredibly non-regular (it is defined up to zero-measure sets). The subsequent extension to self-adjoint operators exploits a weaker notion of derivative (weak derivative in the sense of Sobolev) but the so-obtained larger domain is however very small with respect to the whole $L^2$ space.</p> <p><strong>ADDENDUM.</strong> In view of a remarkable Andreas Blass' comment, I think it is worth stressing a further physical reason for unboundedness of some self-adjoint operators representing observables in QM.</p> <p>First of all the spectrum $\sigma(A)$ of a self-adjoint observable represented by a self-adjoint operator $A$ has the physical meaning of the set of all <strong>possible values</strong> of the observable. So if the observable takes an <strong>unbounded set of values</strong>, the spectrum $\sigma(A)$ must be an <strong>unbounded</strong> subset of $\mathbb R$.</p> <p>Secondly, if $A$ is a self-adjoint operator (more generally a normal operator), the important result holds true: $$||A|| = \sup\{ |\lambda| \:|\: \lambda \in \sigma(A)\}$$ including the unbounded cases where both sides are $+\infty$ simultaneously.</p> <p>So, if an observable, like $p$ or $x$ or the angular momentum, takes an <strong>unbounded set of values</strong>, the self-adjoint representing it has necessarily to be <strong>unbounded</strong> (and therefore defined in a proper dense subspace of the Hilbert space).</p>