Von Neumann's Criterion for Self-Adjoint Extensions

Derives Lemma A26.2. Expanding and using \(\braket{Ax}{x}=\braket{x}{Ax}\), which is symmetry, the two cross terms cancel:

the sign of the third term coming from the antilinearity Equation (9.36) of the first slot. Both consequences are immediate. For closedness of the image, let \((A+\ii\mu)x_{n}\longrightarrow u\). Applying Equation (A26.3) to differences shows that both \((x_{n})\) and \((Ax_{n})\) are Cauchy, hence convergent, say \(x_{n}\longrightarrow x\) and \(Ax_{n}\longrightarrow y\); since \(A\) is closed, \(x\in D(A)\) and \(Ax=y\), so \(u=(A+\ii\mu)x\) lies in the image. Equation (A26.4) is then the projection theorem Theorem 16.18 together with Equation (A26.1).

Derives Lemma A26.3. Write \(K_{+}(\mu)=\ker(A^{\dagger}-\ii\mu)\) and suppose \(\dim K_{+}(\nu)<\dim K_{+}(\mu)\) for two positive \(\mu,\nu\) of the same dimension. Then there is a non-zero \(u\in K_{+}(\mu)\cap K_{+}(\nu)^{\perp}\): otherwise the orthogonal projection onto \(K_{+}(\nu)\) would be injective on \(K_{+}(\mu)\), and its adjoint — the projection onto \(K_{+}(\mu)\) restricted to \(K_{+}(\nu)\) — would have dense range in \(K_{+}(\mu)\), forcing \(\dim K_{+}(\mu)\leq\dim K_{+}(\nu)\).

By Equations (A26.1) and (A26.4), \(K_{+}(\nu)^{\perp}=\im(A+\ii\nu)\), so \(u=(A+\ii\nu)x\) for some \(x\in D(A)\); and \(A^{\dagger}u=\ii\mu u\). Hence, using Equation (16.50) and Equation (9.36),

But Equation (A26.3) gives \(\norm{u}\geq\nu\norm{x}\), so \(\abs{\braket{u}{x}}\leq\norm{u}\norm{x}\leq\norm{u}^{2}/\nu\) by Equation (16.1), and Equation (A26.5) forces \(\norm{u}^{2}\leq\abs{\nu-\mu}\,\norm{u}^{2}/\nu\), i.e.\ \(\abs{\nu-\mu}\geq\nu\). So whenever \(\abs{\nu-\mu}<\nu\) no such \(u\) exists and \(\dim K_{+}(\mu)\leq\dim K_{+}(\nu)\); exchanging the roles of \(\mu\) and \(\nu\), the two dimensions are equal whenever \(\abs{\nu-\mu}<\min(\mu,\nu)\). The function \(\mu\longmapsto n_{+}(\mu)\) is therefore locally constant on the connected set \((0,\infty)\), hence constant (Theorem 10.14). The argument for \(K_{-}\) is the same with \(\ii\) replaced by \(-\ii\).

Derives Proposition A26.5. \(A+\ii\mu\) is injective on \(D(A)\) (Lemma A26.2), so every \(u\in\im(A+\ii\mu)\) is \((A+\ii\mu)x\) for exactly one \(x\) and Equation (A26.6) defines \(V_{A}u\) unambiguously; linearity is inherited from that of \(A\). Isometry is the second consequence in Lemma A26.2, \(\norm{V_{A}u}=\norm{(A-\ii\mu)x}=\norm{(A+\ii\mu)x}=\norm{u}\), and the image is \(\im(A-\ii\mu)\) by construction. Both subspaces are closed and are the orthogonal complements of \(K_{\pm}\), by Equation (A26.4).

The two identities Equation (A26.7) are immediate: \((A+\ii\mu)x-(A-\ii\mu)x=2\ii\mu x\) and \((A+\ii\mu)x+(A-\ii\mu)x=2Ax\). The first shows that \((\identity-V_{A})u=0\) forces \(x=0\), hence \(u=0\): injectivity; that \(\im(\identity-V_{A})=D(A)\), since \(x\) runs over \(D(A)\) as \(u\) runs over \(K_{+}^{\perp}\); and, dividing the second identity by the first, Equation (A26.8).

Derives Lemma A26.6. An isometry preserves inner products, by the polarization identity Equation (16.3) applied on its domain. Let \(Vz=z\) and let \(u\in K_{+}^{\perp}\) be arbitrary. Then

using \(Vz=z\) in the third step. By Equation (A26.7), \((\identity-V)u=(\identity-V_{A})u\) ranges over \(2\ii\mu\,D(A)\), which is dense because \(A\) is densely defined. Hence \(z\perp\mathcal{H}\) and \(z=0\).

Derives Lemma A26.7. \(B\) is well defined and single valued because \(\identity-V\) is injective, and linear because \(V\) is; its domain is dense by hypothesis.

Symmetry. Let \(z,z'\in D(V)\) and put \(x=(\identity-V)z\), \(y=(\identity-V)z'\), so \(Bx=\ii\mu(\identity+V)z\) and \(By=\ii\mu(\identity+V)z'\). Using \(\braket{Vz}{Vz'}=\braket{z}{z'}\) and the antilinearity of the first slot, which turns the prefactor \(\ii\mu\) into \(-\ii\mu\),

while, the prefactor now sitting in the second slot,

So \(\braket{Bx}{y}=\braket{x}{By}\) for all \(x,y\in D(B)\).

Its Cayley transform is \(V\). From Equation (A26.9),

so \(\im(B+\ii\mu)=D(V)\), \(\im(B-\ii\mu)=\im V\), and \(V_{B}\) sends \(2\ii\mu z\) to \(2\ii\mu Vz\): that is, \(V_{B}=V\).

Extension. If \(V\supseteq V_{A}\) then, by Equation (A26.7), \(D(B)=\im(\identity-V)\supseteq \im(\identity-V_{A})=D(A)\), and on \(D(A)\) the prescription Equation (A26.9) is Equation (A26.7) itself, so \(B\) agrees with \(A\) there.

Derives Proposition A26.8. Suppose \(A=A^{\dagger}\). Then \(K_{\pm}=\ker(A^{\dagger}\mp\ii\mu)=\ker(A\mp\ii\mu)=\set{0}\) by Equation (A26.3), so \(n_{\pm}=0\) and, by Equation (A26.4), \(V_{A}\) is an isometry of \(\mathcal{H}\) onto \(\mathcal{H}\): unitary.

Conversely suppose \(n_{+}=n_{-}=0\), i.e.\ \(\im(A\pm\ii\mu)=\mathcal{H}\). Let \(y\in D(A^{\dagger})\). There is \(x\in D(A)\) with \((A-\ii\mu)x=(A^{\dagger}-\ii\mu)y\), and since \(A\subseteq A^{\dagger}\) this reads \((A^{\dagger}-\ii\mu)(y-x)=0\), i.e. \(y-x\in K_{+}=\set{0}\). Hence \(y=x\in D(A)\), so \(D(A^{\dagger})=D(A)\) and \(A\) is self-adjoint. This is part (1) of Theorem A26.1.

Proof of Theorem A26.1, parts (2) and (3). Derives Theorem A26.1. From an extension to a unitary. Let \(B\supseteq A\) be self-adjoint. By Proposition A26.8 its Cayley transform \(V_{B}\) is unitary on \(\mathcal{H}\), and it extends \(V_{A}\): indeed \(\im(A+\ii\mu)\subseteq\im(B+\ii\mu)\) and, on that subspace, \(V_{B}(A+\ii\mu)x=(B-\ii\mu)x=(A-\ii\mu)x\). A unitary carrying the closed subspace \(K_{+}^{\perp}\) onto \(K_{-}^{\perp}\) carries the orthogonal complement onto the orthogonal complement, so \(W=V_{B}|_{K_{+}}\) is a unitary map \(K_{+}\longrightarrow K_{-}\).

From a unitary to an extension. Conversely let \(W\!:K_{+}\longrightarrow K_{-}\) be unitary and define, using the decompositions Equation (A26.4),

Both summands on the right are orthogonal — \(V_{A}u\in K_{-}^{\perp}\) and \(Wk\in K_{-}\) — so \(\norm{V_{W}(u+k)}^{2} =\norm{u}^{2}+\norm{k}^{2}\) and \(V_{W}\) is an isometry of \(\mathcal{H}\); its image is \(K_{-}^{\perp}\oplus K_{-}=\mathcal{H}\), so \(V_{W}\) is unitary. By Lemma A26.6, \(\identity-V_{W}\) is injective, and its image is dense because

where the middle equality holds because for a unitary \(V_{W}\) the relations \(V_{W}z=z\) and \(V_{W}^{\dagger}z=z\) are equivalent. So Lemma A26.7 applies: it produces a densely defined symmetric \(A_{W}\supseteq A\) whose Cayley transform is the unitary \(V_{W}\), and \(A_{W}\) is therefore self-adjoint by Proposition A26.8.

The formula Equation (A26.2). Write \(z=u+k\) with \(u=(A+\ii\mu)x\), \(x\in D(A)\). Then, by Equations (A26.7) and (A26.12),

so, dividing by \(2\ii\mu\) and renaming \(k/(2\ii\mu)\) as \(k\) — which is a bijection of the subspace \(K_{+}\), and \(W\) is linear — Equation (A26.9) becomes exactly Equation (A26.2). The sum there is direct: if \(x+k-Wk=0\) then applying \(\identity-V_{W}\) backwards, i.e. using injectivity on \(2\ii\mu x=(\identity-V_{W})u\) and \(k-Wk=(\identity-V_{W})k\), gives \(u+k=0\) with \(u\perp k\), hence \(u=k=0\).

The correspondence is bijective. The two constructions are inverse to each other: from \(A_{W}\) one recovers \(V_{A_{W}}=V_{W}\) (Lemma A26.7) and hence \(W=V_{W}|_{K_{+}}\); and from a self-adjoint \(B\supseteq A\) one gets \(W=V_{B}|_{K_{+}}\), whose associated operator has Cayley transform \(V_{B}\) and therefore equals \(B\), since \(B\) is determined by \(V_{B}\) through Equation (A26.8).

Counting. A unitary map \(K_{+}\longrightarrow K_{-}\) exists if and only if the two spaces have orthonormal bases of the same cardinality — given such bases, the induced map is unitary; conversely a unitary carries a basis to a basis — that is, if and only if \(n_{+}=n_{-}\). If \(n_{+}\neq n_{-}\) there is none, and by the first paragraph a self-adjoint extension would produce one: this is part (3). When \(n_{+}=n_{-}=n<\infty\), fixing orthonormal bases of \(K_{\pm}\) identifies the unitary maps with the matrix group \(\U(n)\), a transitive and free action of \(\U(n)\) on the set of extensions.

Derives Corollary A26.9. Indices. This is the computation of Example 16.81: the equations \(P^{\dagger}\psi=\pm\ii\mu\psi\) have the one-dimensional solution spaces spanned by

both square integrable on the compact interval, so \(n_{+}=n_{-}=1\) and Theorem A26.1 gives a family of extensions parametrised by \(\U(1)\). Normalising, \(N_{+}^{-2}=\int_{0}^{L}\ee^{-2\mu x/\hbar}\dd x =\bigl(\hbar/2\mu\bigr)\left(1-q^{2}\right)\) and \(N_{-}^{-2}=\bigl(\hbar/2\mu\bigr)\left(q^{-2}-1\right)\), whence

Each \(P_{\theta}\) is self-adjoint. Symmetry on \(D_{\theta}\) is Equation (A26.13): for \(\varphi,\psi\in D_{\theta}\), \(\varphi(L)^{\ast}\psi(L) =\ee^{-\ii\theta}\ee^{\ii\theta}\varphi(0)^{\ast}\psi(0) =\varphi(0)^{\ast}\psi(0)\) and the right-hand side vanishes. For the converse inclusion let \(y\in D(P_{\theta}^{\dagger})\). Since \(D_{0}\subseteq D_{\theta}\), \(y\) belongs to the maximal domain \(D(P^{\dagger})\) and \(P_{\theta}^{\dagger}y=P^{\dagger}y\); so Equation (A26.13) with \(\varphi=y\) must vanish for every \(\psi\in D_{\theta}\):

Choosing \(\psi(x)=1+(\ee^{\ii\theta}-1)x/L\), which lies in \(D_{\theta}\) and has \(\psi(0)=1\), gives \(y(L)^{\ast}\ee^{\ii\theta}=y(0)^{\ast}\), i.e.\ \(y(L)=\ee^{\ii\theta}y(0)\) and \(y\in D_{\theta}\). Hence \(D(P_{\theta}^{\dagger})=D_{\theta}\) and \(P_{\theta}\) is self-adjoint.

There are no others. Let \(B\) be a self-adjoint extension of \(P\). Then \(P\subseteq B=B^{\dagger}\subseteq P^{\dagger}\), so \(D(B)\) lies in the maximal domain and Equation (A26.13) vanishes on \(D(B)\times D(B)\). Taking \(\varphi=\psi\) gives \(\abs{\psi(L)}^{2}=\abs{\psi(0)}^{2}\) for every \(\psi\in D(B)\): the image \(S\subseteq\C^{2}\) of the boundary map \(\psi\longmapsto(\psi(0),\psi(L))\) is a subspace isotropic for the hermitian form \(\abs{b}^{2}-\abs{a}^{2}\), which has signature \((1,1)\), so \(\dim S\leq1\). If \(\dim S=1\), say \(S=\C\,(1,\ee^{\ii\theta})\), then \(D(B)\subseteq D_{\theta}\), i.e. \(B\subseteq P_{\theta}\); taking adjoints reverses the inclusion, so \(P_{\theta}=P_{\theta}^{\dagger}\subseteq B^{\dagger}=B\) and \(B=P_{\theta}\). If \(S=\set{0}\) then \(D(B)\) is contained in \(D_{\theta}\) for every \(\theta\), so \(B\subseteq P_{\theta}\) and the same argument gives \(B=P_{\theta}\) — contradicting \(S=\set{0}\), since the domain of \(P_{\theta}\) contains \(\psi(x) =1+(\ee^{\ii\theta}-1)x/L\), whose boundary values are not both zero. So every self-adjoint extension is one of the \(P_{\theta}\).

Matching the parameters. Let \(W k_{+}=\ee^{\ii\alpha}k_{-}\) and let \(A\) be the closure of \(P\), a closed symmetric operator with the same adjoint and the same deficiency subspaces — the defining condition Equation (16.50) for \(A^{\dagger}\) extends from \(D_{0}\) to its graph closure by continuity of the inner product. By Equation (A26.2), every element of \(D(A_{W})\) is \(\psi=\psi_{0}+c\left(k_{+}-\ee^{\ii\alpha}k_{-}\right)\) with \(\psi_{0}\in D(A)\) and \(c\in\C\). Every \(\psi_{0}\in D(A)\) vanishes at both endpoints: if \(\psi_{n}\in D_{0}\) converges to \(\psi_{0}\) in the graph norm then, for absolutely continuous \(\phi\) on \([0,L]\),

— average \(\phi(x)=\phi(y)+\int_{y}^{x}\phi'\) over \(y\) and use Equation (16.1) — so graph convergence implies uniform convergence and \(\psi_{0}(0)=\lim\psi_{n}(0)=0\), likewise at \(L\). Therefore the boundary values of \(\psi\) come from the second term alone: by Equations (A26.15) and (A26.16),

using \(k_{+}(L)=N_{+}q\) and \(k_{-}(L)=N_{-}q^{-1}=N_{+}\). The denominator \(1-\ee^{\ii\alpha}q\) never vanishes because \(q<1\), and

so the ratio \(\psi(L)/\psi(0)\) is a phase, which is Equation (A26.14). Hence \(D(A_{W})\subseteq D_{\theta}\) with that \(\theta\), and since both operators are self-adjoint, \(A_{W}=P_{\theta}\) by the adjoint argument used above. Finally \(z\longmapsto(q-z)/(1-qz)\) is a Möbius transformation with \(\abs{q}<1\); Equation (A26.18) shows it maps the unit circle into itself, and it is injective, so it is a bijection of the circle: as \(\alpha\) runs once around, so does \(\theta\), and the \(\U(1)\) of Theorem A26.1 is exactly the circle of boundary conditions Equation (16.55).

Derives Corollary A26.10. The solutions of \(P^{\dagger}\psi=\pm\ii\mu\psi\) are again \(\psi(x)=C\ee^{\mp\mu x/\hbar}\); on the half-line the decaying one is square integrable, with \(\norm{\psi}^{2} =\abs{C}^{2}\hbar/(2\mu)\), and the growing one is not, so \(n_{+}=1\) and \(n_{-}=0\) (Example 16.82). By part (3) of Theorem A26.1 — there is no unitary map from a one-dimensional space onto \(\set{0}\) — there is no self-adjoint extension. By Proposition A25.11 there is therefore no strongly continuous unitary group whose generator is a momentum on the half-line, which is the statement that probability cannot be conserved under translations that push it off the end of the half-line (Remark 16.83).