Does anyone know why in quantum mechanics the second statement is always true?

"When the spectrum of an operator $A$ has a continuous part, we associate a bra $\langle a|$ and a ket $|a \rangle$ to each element $a$ of the continuous spectrum of $A$. Obviously, the bras $\langle a|$ and kets $|a \rangle$ are not in the Hilbert space."

  • 1
    $\begingroup$ Hilbert spaces in QM belong to $l_2$, the space of square integrable functions. It has an countable infinite number of dimensions. If the spectrum is continuous then the dimension of the space is also continuous, thus those vectors , or bras and kets are no longer in hilbert space. You can generalize hilbert space to include them though, see en.wikipedia.org/wiki/Rigged_Hilbert_space $\endgroup$ – user126422 Oct 4 '16 at 18:37
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/90101/2451 , physics.stackexchange.com/q/68639/2451 and links therein. $\endgroup$ – Qmechanic Oct 4 '16 at 18:38
  • $\begingroup$ Because if kets or bras were in the Hilbert space the corresponding points would by definition be in the discrete or residual spectrum, not in the continuous one. There is no such thing as "continuous dimension", but the original Hilbert space can be extended to include elements that can be interpreted as "eigenvectors" for continuous spectrum. This extended ("rigged") space is as countably dimensional however as the original was. An example is extending $L^2$ to $H^{-1}$ to include $\delta$ functions, which are "kets" of the position operator. $\endgroup$ – Conifold Oct 4 '16 at 19:59

Because by definition the eigenvalues of an operator $A$ are part of the point spectrum $\sigma_p(A)$. For self-adjoint operators, the continuous spectrum is the complement $\sigma_c(A)=\sigma(A) \setminus \sigma_p(A)$.

Therefore if, in any sense $Af=af$ for some $a \in \sigma_c(A)$, $f$ cannot be an eigenvector. For this reason it cannot belong to the Hilbert space.

As a matter of fact, the identity $Af=af$ where $a \in \sigma_c(A)$ holds in a different sense than the standard one, a distributional sense if the Hilbert space is $L^2(\mathbb R, d^nx)$.

It is worth noticing that the point spectrum, in spite of its name, may be a continuous set, all $\mathbb R$ for instance. In this case, however, the Hilbert space would not be separable. A famous theorem by Stone and von Neumann proves that the Hilbert space of a particle (irreducible representation of Weyl group) must be separable necessarily. For this reason Hilbert spaces of non-relativistic elementary systems in QM are separable and point spectra are at most countable.

  • $\begingroup$ For your last sentence "For this reason Hilbert spaces of non-relativistic elementary systems in QM are separable and point spectra are at most countable." Is this point spectra of self-adjoint operators? Also, is $\sigma(A)$ at the top ($\sigma_{c} = \sigma \setminus \sigma_{p}$) defined as $\sigma := \{ \lambda|~ A - \lambda I ~\text{ not invertible} \}$? $\endgroup$ – Alex Oct 4 '16 at 22:59
  • $\begingroup$ Regarding your first question, my statement holds for every normal operator, like self-adjoint anti self-adjoint, unitary... Regarding the second one, since self-adjoint operators are closed, $a\not \in \sigma(A)$ if and only if $A-aI$ has image given by the whole Hilbert space, it is invertible and the inverse operator is bounded. $\endgroup$ – Valter Moretti Oct 5 '16 at 6:16
  • $\begingroup$ Okay thanks. If you have a chance please see MSE my spectral theory post. $\endgroup$ – Alex Oct 5 '16 at 13:10
  • $\begingroup$ @ValterMoretti Nice answer. I would like to confirm my understanding of your answer for this question please. As I understand, the state of any system is represented by a vector $| \psi \rangle$ in a Hilbert space. The type of Hilbert space depends on the observable in question. So for position and momentum observables, we can consider the operators acting on the Hilbert space $L^2(\mathbb{R})$, but for spin $\frac{1}{2}$, we consider the Hilbert space which is spanned by vectors $\endgroup$ – user100411 Nov 7 '16 at 13:58
  • $\begingroup$ \begin{equation} \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \end{equation} and \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} but clearly for finite dimensional Hilbert space, separability holds vacuously, so the result still holds. More generally, by the Stone and von Neumann which you referred to it is true that for any operator corresponding to an observable, the Hilbert space in question is separable, hence the point spectrum is at most countable for any observable Is my understanding correct? Thanks for your time. $\endgroup$ – user100411 Nov 7 '16 at 13:58

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