Bounded operators#

Let $(V, \|\cdot\|_V)$ be a normed space and $(W, \|\cdot\|_W)$ a Banach space. A linear operator $A: V \to W$ is bounded if

equivalently if there exists a constant $C \geq 0$ such that $\|Ax\|_W \leq C\|x\|_V$ for all $x \in V$.

For linear operators, “bounded” and “continuous” are synonymous (this is a standard lemma). We write $\mathcal{B}(\mathcal{H})$ for the space of bounded operators $\mathcal{H} \to \mathcal{H}$.

Unbounded operators#

A linear operator that fails the boundedness condition is called unbounded. The two most important operators in quantum mechanics — position and momentum — are both unbounded on $L^2(\mathbb{R}^n)$:

• Position: $\hat{x}\psi(x) = x\psi(x)$.
• Momentum: $\hat{p}\psi(x) = -i\,\partial_x\psi(x)$.

Why are they unbounded? Position because you can find $L^2$ functions concentrated arbitrarily far from the origin where $\|x\psi\|/\|\psi\|$ blows up. Momentum because differentiation amplifies high-frequency components without bound.

A more elementary way to see that the derivative operator is unbounded: define

Take $f_n(x) = \sin(n\pi x)$, so $\|f_n\|_\infty = 1$ but $\|f_n'\|_\infty = n\pi \to \infty$. The ratio is unbounded.

Important consequence. An unbounded operator cannot be defined on the whole Hilbert space — at least not as a closed linear operator. An everywhere-defined closed operator is automatically bounded (the closed graph theorem), and the special case for symmetric operators is the Hellinger–Toeplitz theorem. It always lives on a proper subspace called its domain. Specifying that domain is part of specifying the operator. Two operators with the same formal expression but different domains are different operators. This is the single most important thing to internalize.

Densely defined operators#

Let $\mathcal{H}$ be a Hilbert space. A linear operator $T: \mathcal{D}_T \to \mathcal{H}$ is densely defined if its domain $\mathcal{D}_T$ is dense in $\mathcal{H}$, i.e. every vector in $\mathcal{H}$ can be approximated arbitrarily well by vectors in $\mathcal{D}_T$.

Density is what allows us to define the adjoint at all. Without it, the adjoint is not even well-defined as a single-valued operator.

The adjoint#

Let $T: \mathcal{D}_T \to \mathcal{H}$ be densely defined. The adjoint $T^*$ is the operator with domain

acting by $T^*\psi := \eta$.

$T^*$ is well-defined. If both $\eta$ and $\tilde\eta$ satisfy $\langle\psi|T\varphi\rangle = \langle\eta|\varphi\rangle = \langle\tilde\eta|\varphi\rangle$ for all $\varphi \in \mathcal{D}_T$, then $\langle\eta - \tilde\eta | \varphi\rangle = 0$ for all $\varphi$ in a dense set, which forces $\eta = \tilde\eta$. (This is exactly where density is used.)

A small but useful proposition: $\ker(T^*) = \operatorname{ran}(T)^\perp$. To see it, $\psi \in \ker(T^*) \iff T^*\psi = 0 \iff \langle\psi|T\varphi\rangle = \langle 0|\varphi\rangle = 0$ for all $\varphi \in \mathcal{D}_T \iff \psi \perp \operatorname{ran}(T)$.

Extensions and a key inclusion lemma#

We say $\tilde T$ is an extension of $T$, written $T \subseteq \tilde T$, if $\mathcal{D}_T \subseteq \mathcal{D}_{\tilde T}$ and $\tilde T\varphi = T\varphi$ for all $\varphi \in \mathcal{D}_T$.

Lemma. If $T \subseteq \tilde T$ are both densely defined, then $\tilde T^* \subseteq T^*$.

Proof. Let $\psi \in \mathcal{D}_{\tilde T^*}$, so there exists $\eta \in \mathcal{H}$ with $\langle\psi|\tilde T\beta\rangle = \langle\eta|\beta\rangle$ for all $\beta \in \mathcal{D}_{\tilde T}$. Since $\mathcal{D}_T \subseteq \mathcal{D}_{\tilde T}$ and $\tilde T = T$ on $\mathcal{D}_T$, this gives $\langle\psi|T\alpha\rangle = \langle\eta|\alpha\rangle$ for all $\alpha \in \mathcal{D}_T$. Hence $\psi \in \mathcal{D}_{T^*}$ and $T^*\psi = \eta = \tilde T^*\psi$. $\square$

Notice the direction-reversal: a bigger operator has a smaller adjoint. This is the crucial structural fact.

Symmetric vs. self-adjoint#

A densely defined operator $T$ is symmetric if

It is self-adjoint if $T = T^*$ — meaning both the actions agree and the domains agree:

1. $\mathcal{D}_T = \mathcal{D}_{T^*}$,
2. $T\varphi = T^*\varphi$ for all $\varphi \in \mathcal{D}_T$.

A linguistic warning. In physics, the word “Hermitian” is used inconsistently: sometimes as a synonym for symmetric, sometimes as a synonym for self-adjoint. Statements like “observables correspond to Hermitian operators” gloss over a real distinction that becomes invisible only in finite dimensions. I’ll avoid the word entirely.

If $T$ is symmetric, then $T \subseteq T^*$. (For $\psi \in \mathcal{D}_T$, set $\eta := T\psi$; symmetry gives $\langle\psi|T\alpha\rangle = \langle\eta|\alpha\rangle$ for all $\alpha \in \mathcal{D}_T$, so $\psi \in \mathcal{D}_{T^*}$ with $T^*\psi = T\psi$.) The whole game is to ask whether the inclusion is equality.

A maximality property#

Self-adjoint operators have no proper symmetric extensions. Concretely: if $T$ is self-adjoint, $\tilde T$ is symmetric, and $T \subseteq \tilde T$, then $T = \tilde T$.

Proof. From $T \subseteq \tilde T$, the lemma gives $\tilde T^* \subseteq T^*$. Symmetric means $\tilde T \subseteq \tilde T^*$, so $\tilde T \subseteq \tilde T^* \subseteq T^* = T$. Combined with $T \subseteq \tilde T$, we get $T = \tilde T$. $\square$

This is why self-adjointness is the right condition for observables. Spectral theorem guarantees self-adjoint operators have real spectra and a sensible functional calculus; symmetric operators in general do not. And the maximality property tells us that a self-adjoint operator is “as big as it can be” — there’s no room to extend it further while keeping the symmetric property.

Closures#

A densely defined operator $T$ is closable if its adjoint $T^*$ is itself densely defined. Equivalently (and more usefully), $T$ has a smallest closed extension, which we call the closure $\overline T$. It satisfies $\overline T = T^{**}$.

Symmetric operators are always closable. If $T$ is symmetric, then $T \subseteq T^*$ implies $\mathcal{D}_T \subseteq \mathcal{D}_{T^*}$. Since $\mathcal{D}_T$ is dense, so is $\mathcal{D}_{T^*}$, hence $T$ is closable. So we always have access to $\overline T = T^{**}$ for symmetric operators.

For symmetric $T$, the closure sits between $T$ and $T^*$:

Essentially self-adjoint operators#

A symmetric operator $T$ is essentially self-adjoint (e.s.a.) if its closure $\overline T = T^{**}$ is self-adjoint.

This is weaker than self-adjointness. A self-adjoint operator is automatically e.s.a.: if $T = T^*$, then taking adjoints gives $T^* = T^{**}$, so $T = T^{**}$, so $\overline T = T$ is self-adjoint.

Theorem. If $T$ is essentially self-adjoint, then $\overline T$ is the unique self-adjoint extension of $T$.

Proof. $\overline T$ is a self-adjoint extension by hypothesis. For uniqueness, suppose $S$ is any other self-adjoint extension, $T \subseteq S$. Self-adjoint operators are closed (since $S = S^*$ and adjoints are always closed), so $\overline T \subseteq S$ (the closure is the smallest closed extension). But $\overline T$ is itself self-adjoint, and a self-adjoint operator has no proper symmetric extensions — so $\overline T = S$. $\square$

The strategic upshot. When we want to elevate a symmetric operator to a genuine observable, we do not need it to already be self-adjoint. It suffices that it be essentially self-adjoint. Then the closure $\overline T$ is the canonical observable, picked out uniquely.

When essential self-adjointness fails, life gets harder: the symmetric operator may admit many self-adjoint extensions (or even none), and choosing one becomes a piece of physical input. The technology to count and parametrize these extensions is deficiency index theory, due to von Neumann — beyond our scope here, but coming up explicitly when we look at the interval case.

Step 1. $\hat{P}_j^{\text{int}}$ is symmetric#

For $\psi, \varphi \in \mathcal{D}_P$, integrate by parts:

The boundary term vanishes because $\psi(0) = \psi(2\pi) = 0$. The remaining integral is

So $\hat{P}_j^{\text{int}}$ is symmetric. ✓

Step 2. The adjoint $(\hat{P}_j^{\text{int}})^*$ has a bigger domain#

Now we ask: for which $\psi \in L^2$ does there exist $\eta \in L^2$ with

The strategy: pick any $N \in AC([0,2\pi])$ with $N' = \eta$ a.e. (such an $N$ exists because every $L^2$ function on a compact interval has an absolutely continuous antiderivative). Substituting $\eta = N'$ and integrating the right-hand side by parts,

and the boundary term vanishes because $\varphi \in \mathcal{D}_P$ vanishes at the endpoints. So $(\star)$ becomes

Equivalently, $\psi - iN \perp \{\varphi' \mid \varphi \in \mathcal{D}_P\}$ in $L^2$.

Now we identify this orthogonal complement. The set $\{\varphi' \mid \varphi \in \mathcal{D}_P\}$ consists exactly of those continuous functions $\xi$ on $[0, 2\pi]$ satisfying $\int_0^{2\pi} \xi(x)\,dx = 0$ — that is, $\xi \perp \mathbf{1}$, where $\mathbf{1}$ is the constant function. (One direction: if $\xi = \varphi'$ with $\varphi(0) = \varphi(2\pi) = 0$, then $\int_0^{2\pi}\xi\,dx = \varphi(2\pi) - \varphi(0) = 0$. Conversely, given such a $\xi$, the antiderivative $\varphi(x) = \int_0^x \xi(y)\,dy$ vanishes at both endpoints and lies in $C^1$.)

So the closure (in $L^2$) of $\{\varphi' \mid \varphi \in \mathcal{D}_P\}$ is $\{\mathbf{1}\}^\perp$, and its orthogonal complement is therefore $\{\mathbf{1}\}^{\perp\perp} = \overline{\operatorname{span}\{\mathbf{1}\}} = \mathbb{C}\cdot\mathbf{1}$, the space of constant functions. Hence

Since $N \in AC$, this gives $\psi \in AC([0, 2\pi])$ with no boundary conditions whatsoever. The requirement $\hat{P}_j^*\psi = \eta = N' = -i\psi' \in L^2$ then forces $\psi' \in L^2$, i.e. $\psi \in H^1([0, 2\pi])$. Conversely, every $H^1$ function arises this way, and so

The adjoint is the same differential expression, but acting on a much larger domain — $H^1$ functions with no boundary conditions, versus $C^1$ functions vanishing at both endpoints.

In particular, $\hat{P}_j^{\text{int}} \subsetneq (\hat{P}_j^{\text{int}})^*$, so $\hat{P}_j^{\text{int}}$ is not self-adjoint.

Step 3. The closure $\hat{P}_j^{**}$ is also too small#

Maybe the closure rescues us? Let’s compute $\mathcal{D}_{P^{**}}$ directly.

$\psi \in \mathcal{D}_{P^{**}}$ means: there exists $\zeta \in L^2$ such that $\langle\psi | \hat P_j^* \varphi\rangle = \langle \zeta | \varphi\rangle$ for all $\varphi \in \mathcal{D}_{P^*} = H^1$. Since $\hat P_j^{**} \subseteq \hat P_j^*$ and both act as $-i\partial_x$, $\zeta = -i\psi'$, and the question reduces to whether the appropriate boundary terms vanish.

For $\psi \in H^1$ (which we already know is necessary, since $\mathcal{D}_{P^{**}} \subseteq \mathcal{D}_{P^*} = H^1$) and $\varphi \in H^1$, integration by parts gives

Requiring this to vanish for all $\varphi \in H^1$ — and $H^1$ functions can take arbitrary independent values at $0$ and $2\pi$ — forces $\psi(0) = 0$ and $\psi(2\pi) = 0$.

So

which is strictly smaller than $\mathcal{D}_{P^*} = H^1([0, 2\pi])$. Hence $\hat{P}_j^{**} \subsetneq \hat{P}_j^*$, and $\hat{P}_j^{\text{int}}$ is not essentially self-adjoint.

What’s really going on#

This is not a defect of our choice of “starting domain” — it reflects a real physical ambiguity. There is in fact a one-parameter family of self-adjoint extensions of $\hat{P}_j^{\text{int}}$, parametrized by $\theta \in [0, 2\pi)$, with domains

Different choices of $\theta$ correspond to physically different theories (think: a particle on an interval threaded by a magnetic flux, or a “twisted” boundary condition). The deficiency-index analysis confirms that this is the full story: the deficiency indices of $\hat P_j^{\text{int}}$ are $(1, 1)$, giving exactly a $U(1)$-worth of self-adjoint extensions.

The lesson: when an operator is symmetric but not essentially self-adjoint, the physics of which observable to use is not determined by the formal expression alone. You have to add a choice of boundary condition, and that choice is part of defining the system.

Step 1. Symmetric#

For $\psi, \varphi$ with $\psi(0) = \psi(2\pi)$ and $\varphi(0) = \varphi(2\pi)$, the boundary term in integration by parts is

using both boundary conditions. So $\hat P_j^{\text{circ}}$ is symmetric. ✓

Step 2. The adjoint#

We can shortcut the computation by using the inclusion $\hat{P}_j^{\text{int}} \subsetneq \hat{P}_j^{\text{circ}}$. By the inclusion lemma, $(\hat{P}_j^{\text{circ}})^* \subseteq (\hat{P}_j^{\text{int}})^* = \hat{P}_j^*|_{H^1}$. So $\mathcal{D}_{(P^{\text{circ}})^*} \subseteq H^1([0, 2\pi])$ and the action is still $-i\partial_x$.

To pin down the boundary conditions: $\psi \in \mathcal{D}_{(P^{\text{circ}})^*}$ requires the boundary term to vanish for all $\varphi$ with $\varphi(0) = \varphi(2\pi)$:

Since $\varphi(0)$ is arbitrary, this forces $\psi(2\pi) = \psi(0)$. So

The adjoint inherits the same periodic boundary condition, just on a bigger function space.

Step 3. Not self-adjoint, but…#

Comparing,

So $\hat{P}_j^{\text{circ}}$ is not self-adjoint either — but only because of the regularity gap $C^1 \subsetneq H^1$, not because of any boundary-condition mismatch.

Step 4. Essentially self-adjoint!#

Let’s compute the closure. For $\psi \in \mathcal{D}_{P^{**}}$ and $\varphi \in \mathcal{D}_{P^*} = \{H^1\text{ with periodic BC}\}$, the boundary term is

(using $\varphi(2\pi) = \varphi(0)$). For this to vanish for all such $\varphi$, we need $\psi(0) = \psi(2\pi)$ — but no further constraint, because $\varphi(0)$ is the only free quantity. So

Therefore $\hat{P}_j^{**} = \hat{P}_j^*$, and $\hat{P}_j^{\text{circ}}$ is essentially self-adjoint.

Conclusion#

The momentum operator on the circle has a unique self-adjoint realization: take the closure of the naive operator on $C^1$-with-periodic-BC, and you land on the operator $-i\partial_x$ with domain $\{\psi \in H^1 \mid \psi(0) = \psi(2\pi)\}$.

This is the operator whose eigenfunctions are $e^{inx}$ with eigenvalues $n \in \mathbb{Z}$, which is exactly the spectrum of momentum on $S^1$ that you’d compute formally in any QM textbook. The point of the whole apparatus is to justify that calculation by showing that the operator we’re diagonalizing actually exists as a genuine observable.