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Bounded operator - definition?

+ 3 like - 0 dislike
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As mentioned also in Bounded and Unbounded Operator, an operator $A$ is said to be bounded, if $$\|Af\|\leq k \|f\|,$$ where the constant $k$ does not depend on the choice of $f$ (let us consider a map to the same Banach space).

However, in a mathematical physics text I came across a definition: a symmetric operator $B$ is said to be bounded from below if there $\exists$ a constant $c$ such that $$\langle\psi|B\psi\rangle\geq c\|\psi\|^2$$ for all $\psi$ in the domain of $B$.

Both definitions are logical (in the second one we can imagine $B$ being the Hamiltonian, than the system energy is bounded from below and hence the system is stable).

The only think that bothers me is when we rewrite the first definition into a similar form to the second one (we assume the norm comes from an inner product), namely:

$$\langle Af, Af\rangle \leq k\|f\|^2,$$ we get something quite different on the left-hand side, so the same words (bounded operator) refer to different things. Any hints how I can clarify this to myself?

This post imported from StackExchange Physics at 2015-02-13 11:48 (UTC), posted by SE-user wondering
asked Feb 12, 2015 in Mathematics by wondering (30 points) [ no revision ]
retagged Feb 13, 2015
@Sofia I returned one of the edits, as it was not a typo. I can explain in case it is not clear.

This post imported from StackExchange Physics at 2015-02-13 11:48 (UTC), posted by SE-user wondering

3 Answers

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Actually, the conditions $$\langle x|A x \rangle \geq a \langle x |x \rangle$$ and $$\langle x|A x \rangle \leq b \langle x |x \rangle$$ with $a,b \in \mathbb R$ fixed and all $x\in D(A)$ (the domain of $A$) refer to boundedness (resp. from below or from above) of  the quadratic form associated to the linear operator $A$. This operator can always be supposed to be Hermitian since the anti-Hermitian part does not play any role in  $\langle x|A x \rangle$.

However, making stronger the hypotheses and assuming that $D(A)\subset {\cal H}$ is a dense linear manifold in the Hilbert space ${\cal H}$ and that $A=A^*$, namely,  $A$ is self-adjoint, then the (both from below and from above) boundedness condition of the quadratic form associated to $A$ is equivalent to that of the operator $A$ itself.

Indeed, if the operator $A$ is bounded, it must be $D(A)={\cal H}$ (well-known property of self-adjoint operators) and 

$$| \langle x |A x \rangle | \leq ||x|| ||Ax|| \leq ||x|| ||A|| ||x|| = ||A|| ||x||^2$$ 

so that the quadratic form is  bounded both from below and from above with $a = -||A||$ and $b=||A||$.

If, conversely the quadratic form is bounded, the spectral theorem immediately implies that

$$\lambda - a \geq 0\quad\mbox{and}\quad  b- \lambda \geq 0\quad \mbox{if $\lambda \in \sigma(A)$,}$$

so that the spectrum $\sigma(A)$ is included in the bounded interval $[a,b]$. The spectral radius formula, in turn, implies that 

$$||A|| \leq \max\{|a|, |b|\} < + \infty\:.$$

The two notions of boundedness, for operators representing observables in QM, are therefore equivalent.  This is particularly remarkable because the boundedness of the quadratic form of $A$ is equivalent to the boundedness of the spectrum (see above) of $A$, which means that the values attained by the observable $A$ form a bounded set. This equivalence shows in particular that most observables (think of position or momentum) cannot be represented by bounded operators (since the attained values are arbitrarily large) and this is one of the reasons why QM is technically complicated mathematically speaking, the reason being physical however!

answered Feb 20, 2015 by Valter Moretti (2,025 points) [ no revision ]
+ 1 like - 0 dislike

When one says that an operator is bounded, you can think of it as being bounded from above. This is different from being bounded from below. An operator can be bounded (from above) and bounded from below, or perhaps just bounded, or just bounded from below.

Observe that $(Af,Af)$ and $(\psi,B\psi)$ are slightly different: the former is always positive for any operator $A$, not necessarily symmetric, while the latter can be negative. If $c$ turns out to be non-negative then the operator $B$ is positive and, if it is also bounded (from above) its spectrum is contained in $[c,\Vert B\Vert]$.

This post imported from StackExchange Physics at 2015-02-13 11:48 (UTC), posted by SE-user Phoenix87
answered Feb 12, 2015 by Phoenix87 (20 points) [ no revision ]
Can you tell me why $\hat B$ is required to be symmetric? And, symmetric in which sense, i.e. with respect to what? Should it be equal to its transposed? Should it be also real ?

This post imported from StackExchange Physics at 2015-02-13 11:48 (UTC), posted by SE-user Sofia
symmetric here is a synonym of Hermitian, i.e. an operator that looks like a self-adjoint operator, but it is not defined everywhere (but it is usually densely defined)

This post imported from StackExchange Physics at 2015-02-13 11:48 (UTC), posted by SE-user Phoenix87
I understand. Your explanation is very clear.

This post imported from StackExchange Physics at 2015-02-13 11:48 (UTC), posted by SE-user Sofia
+ 1 like - 0 dislike

TL;DR: The property bounded, bounded from above, and bounded from below are different things, cf. Wikipedia.

In detail, consider a densely defined symmetric linear operator $A:D\subseteq H \to H$ in a complex Hilbert space $H$. Let $$\langle A \rangle_{\psi}~:=~ \frac{\langle \psi, A\psi\rangle}{||\psi||^2}$$ for $\psi\in D\backslash\{0\}$.

  1. That $A$ is bounded from below means that $$\exists C\in \mathbb{R}~ \forall \psi\in D\backslash\{0\}: ~~ \langle A \rangle_{\psi}~\geq ~C. $$

  2. That $A$ is bounded from above means that $$\exists C\in \mathbb{R}~ \forall \psi\in D\backslash\{0\}: ~~ \langle A \rangle_{\psi}~\leq ~C. $$

  3. That $A$ is bounded means that $$\exists C\geq 0~ \forall \psi\in D\backslash\{0\}: ~~ ||A \psi||~\leq ~C||\psi||, $$ which is equivalent to $A^{\dagger}A$ $(=A^2)$ being bounded from above, which in turn is equivalent to $A$ being bounded from both above and below.

This post imported from StackExchange Physics at 2015-02-13 11:48 (UTC), posted by SE-user Qmechanic
answered Feb 12, 2015 by Qmechanic (2,790 points) [ no revision ]
also relative boundedness: planetmath.org/katorellichtheorem

This post imported from StackExchange Physics at 2015-02-13 11:48 (UTC), posted by SE-user Phoenix87
I don't see how this answers the question. This answer (comment?..) doesn't even talk about two definitions of boundedness, am I missing something?

This post imported from StackExchange Physics at 2015-02-13 11:48 (UTC), posted by SE-user Ruslan
I updated the answer.

This post imported from StackExchange Physics at 2015-02-13 11:48 (UTC), posted by SE-user Qmechanic

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