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  1. An operator $A$ is said to be self-adjoint 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?

  2. Why are the operators in quantum mechanics in general unbounded?

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Related question by OP: physics.stackexchange.com/q/99542/2451 –  Qmechanic Mar 7 at 18:32
    
I cannot understand why you say that "for the free particle momentum operator $\hat{p}$ these inner products does not exist". Instead they exist and the momentum operator is self-adjoint (and it has no eigenvalues, since its spectrum is real but purely continuous). The operators of QM are generally not defined on the whole Hilbert space, but only on a dense domain and cannot be extended on a larger domain remaining symmetric. There is an important theorem which proves that they are bounded if and only if the domain is the whole Hilbert space. So they are not bounded, in general. –  Valter Moretti Mar 7 at 18:45
    
$-i\hbar\frac{d}{dx}e^{ikx}=\hbar k e^{ikx}$. So, eigenvalues are real, as k is real. For $\psi(x)=e^{ikx}$, the inner product $\int\limits_{-\infty}^{\infty}\psi^* \hat p \psi dx=\int\limits_{-\infty}^{\infty}e^{-ikx} (\hbar k) e^{ikx} dx=\infty$. So diverges. –  Roopam Mar 7 at 18:54
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Well, $-i\hbar d/dx$ is not the momentum operator; the momentum operator is the unique self-adjoint extension of $-i\hbar d/dx$ initially defined on ${\cal S}(\mathbb R)$. Secondly, $e^{ikx}$ does not belong to the Hilbert space (and no divergences arise) so it is not an eigenvector of $-id/dx$ and $k$ is not an eigenvalue. The point is: if one wants to really understand these issues he/she should give up with intuitive-formal discussions and should use the appropriate mathematical tools. Otherwise everything remains vague and undefined. –  Valter Moretti Mar 7 at 19:00
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To be more explicit, I think your question is worth and deserves an answer, but the answer cannot be given without the necessary mathematical rigor in the definitions at least. –  Valter Moretti Mar 7 at 19:05

1 Answer 1

up vote 10 down vote accepted

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 unique operator, the adjoint $A^\dagger$ 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 self-adjoint 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 symmetric. It is clear that the adjoint of $A$, in this case, is an extension of $A$ itself.

A symmetric operator is essentially selfadjoint 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 $(A^\dagger)^\dagger$).

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.

The spectrum of $p$ is continuous and coincides with the whole real line.

Passing to Fourier-Plancherel transform, the operator $p:= \left(-i\frac{d}{dx}\right)^\dagger$ turns out to coincide with the multiplicative operator $k \cdot$.

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:

a (densely-defined) self-adjoint operator $A :D(A) \to \cal H$ is bounded if and only if $D(A)= \cal H$

and almost all operators of QM, for various reasons, are not 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.

ADDENDUM. 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.

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 possible values of the observable. So if the observable takes an unbounded set of values, the spectrum $\sigma(A)$ must be an unbounded subset of $\mathbb R$.

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.

So, if an observable, like $p$ or $x$ or the angular momentum, takes an unbounded set of values, the self-adjoint representing it has necessarily to be unbounded (and therefore defined in a proper dense subspace of the Hilbert space).

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Amazing answer. Concise, clear, correct and complete. –  Emilio Pisanty Mar 8 at 11:21
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Although I'm a mathematician, and although this answer is an excellent summary of the relevant mathematics, I think it's worthwhile to add a physical reason why operators like momentum ought to be unbounded. If, for example, the operator $p$ were bounded, then this would entail an absolute upper bound $\Vert p \Vert$ for all possible measurements of the momentum of this particle. That doesn't conform to our understanding (classical or quantum) of how momentum works. Analogous comments apply to position operators, orbital angular momentum, etc. –  Andreas Blass Mar 8 at 17:33
    
@Andreas Blass Yes the reasons you gave are very good (the spectrum is not bounded so the observable cannot be bounded). However, there is another, physical, reason for unboundedness of $p$. It is a standard result that there are not couples of operators both bounded and satisfying the canonical relation commutations on a common domain as $x$ and $p$ do. Since $x$ and $p$ are transformed into each other through a unitary map (Fourier-Plancherel transform) none of them can be bounded. –  Valter Moretti Mar 8 at 18:06
    
@V.Moretti: Actually its not the spectrum but the numerical range what is of interest here. –  Freeze_S May 29 at 11:55
    
@Freeze_S I do not understand your remark. What I wrote above (last comment) is that the operator representing the observable is unbounded if and only if its spectrum, i.e. the set of possible values of the observable, is unbounded. What is wrong/imprecise with it? –  Valter Moretti May 29 at 12:19

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