When we have bound states, to my knowledge, we have states that are normalisable and a discrete energy spectrum. However, in the case of scattering states that have a continuous energy spectrum, the states are non-normalisable.

Is there a rule that bounded operators must have normalisable eigenfunctions and discrete eigenvalues? If not, is there an example that violates the above rule?

Is there a connection between normalisability of eigenfunctions and discreteness of eigenvalues?

PS: I know this Phys.SE question is somewhat related, but still couldn't get a complete answer!

  • $\begingroup$ Related: physics.stackexchange.com/q/68639/2451 , physics.stackexchange.com/q/90101/2451 and links therein. $\endgroup$
    – Qmechanic
    Oct 6, 2015 at 7:42
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    $\begingroup$ The first and second paragraphs are talking, in my opinion, about two different things. The questions of the second paragraph are perfectly and clearly answered by @ValterMoretti below. The first paragraph however, talks about bound and scattering states (not operators). A bound state is a state that is localized in space; meaning that either it is of compact support or decays exponentially at infinity. It may be interesting for the OP to know that, for QM operators (elliptic operators), it has been rigorously proved that the eigenvectors decay exponentially, and therefore are bound states. $\endgroup$
    – yuggib
    Oct 6, 2015 at 8:45
  • $\begingroup$ @ValterMoretti No, to Agmon's estimates/method. $\endgroup$
    – yuggib
    Oct 6, 2015 at 8:49
  • $\begingroup$ I think we have already discussed this point. In my opinionb a bound state is nothing but an eigenvector (in proper sense). There is no need to require that the wavefunction decays (esponetially or not) or its support is bounded... $\endgroup$ Oct 6, 2015 at 8:49
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    $\begingroup$ @ValterMoretti I agree it is just a matter of taste... My point of view is that the term "bound" suggests a (spatial) confinement of sorts. I think I am influenced by Landau and Lifshitz ;-) $\endgroup$
    – yuggib
    Oct 6, 2015 at 8:57

2 Answers 2


There is no general rule. However there is a class of bounded self-adjoint operators whose spectrum is made of a bounded set of isolated points (proper eigenvalues) -- except for $0$ at most -- and the eigenspaces associated to these eigenvalues are finite dimensional. They are the so-called compact operators (this class includes classes of operators important in QM, like Hilbert-Schmidt and trace class ones). However there are operators which are not compact but have a pure point spectrum. An example is the Hamiltonian of the harmonic oscillator, whose eigenspaces are also finite dimensional but it is not bounded. The reason why the spectrum has the same features as the one of compact operators is that inverse powers of these operators or the associated resolvent operators are bounded and compact.

Conversely, an example of a bounded operator with pure continuous spectrum is the position operator in $L^2([0,1], dx)$ defined as usual $$(X\psi)(x) = x \psi(x)\quad \forall \psi \in L^2([0,1], dx)\:.$$ It does not admit (proper) eigenvalues. The spectrum is $\sigma(X)= [0,1]$. Since, for a self-adjoint operator (more generally for a normal operator), $$||A||= \sup_{\lambda \in \sigma(A)} |\lambda|$$ you see that $||X||=1$.

NOTE. Regarding your added last point (a connection between normalisability of eigenfunctions and discreteness of eigenvalues) the situation is the following.

If $\lambda\in \sigma(A)$ is an isolated point of the spectrum ($\sigma(A)$) of the self-adjoint operator $A$. Then $\lambda$ is a proper eigenvalue and thus their eigenvectors are proper (normalizable) eigenvectors. So, as you suppose, in the jargon of physicists, "discrete eigenvalues" are proper eigenvalues with normalizable eigenvectors.

The converse is however generally false. You can have points $\lambda$ in a continuous part of $\sigma(A)$ (say, $\lambda \in (a,b)$ with $(a,b) \in \sigma(A)$) which are proper eigenvalues. Even, in a non-separable Hilbert space, it is possible to costruct a self-adjoint operator $A$ such that $\sigma(A)=[0,1]$ and all points of $[0,1]$ are proper eigenvalues with proper eigenvectors. In a separable Hilbert space it is not possible, but one can easily construct an operator whose set of proper eigenvalues is dense in $[0,1]$.

  • $\begingroup$ +1, thanks for difference between different classes. However, I have a doubt, which although my bad have not been put in the question clearly, which is, is the normalisability of eigenstate related to the discreteness of the corresponding eigenvalues ! I believe this answers my question indirectly, but am not able to see through it. $\endgroup$
    – user35952
    Oct 6, 2015 at 9:40
  • $\begingroup$ The point is the following: if $\lambda \in \sigma(A)$ and $\lambda$ is an isolated point then it is a proper eigenvalue and their eigenvectors are proper (normalizable) eigenvectors. The converse is generally false. It is possible, in a non-separable Hilbert space, to costruct a self-adjoint operator $A$ such that $\sigma(A)=[0,1]$ and all points of $[0,1]$ are proper eigenvalues with proper eigenvectors. $\endgroup$ Oct 6, 2015 at 10:01
  • $\begingroup$ Also in separable Hilbert spaces you can have proper eigenvalues immersed in the continuous spectrum... $\endgroup$ Oct 6, 2015 at 10:02
  • $\begingroup$ Alright, although I am not able get across the barrier of mathematical terminology I am trying to understand, I seem to have understood atleast 50% of it. $\endgroup$
    – user35952
    Oct 6, 2015 at 11:03
  • $\begingroup$ I think this is very interesting! I am curious about the counterexample in separable Hilbert space, where eigenvalue of continuous spectrum can have proper eigenfunction. $\endgroup$
    – eigenvalue
    Dec 10, 2020 at 16:02

Preliminaries: If you "limit" your description of quantum mechanics to $L_2$ Hilbert spaces, all your bases will be discrete, both bounded or unbounded. You can have Hilbert spaces of any cardinality, but the one in "standard" quantum mechanics is $L_2$, the space of square integrable functions, which has a countable cardinality, $\aleph_0$. In this case even unbounded solutions, such as the free particle in a box of length $l$, where $l$ can be as large as you like, will be countable.

You might not like the free particle in a box for various reasons, and might want to eliminate the box and make space infinite. Now you cannot longer apply Hilbert space quantum mechanics, because the solutions do not belong to $l_2$.

To fix this we use the dirac delta (a generalized function, or distribution, that does not belong to $L_2$). Initially was a slight of hand that works well in practice but is not mathematically rigorous. Today this has been formalized into what is called rigged Hilbert spaces, that can include distributions, and thus an infinitely continuous number of dimensions, of cardinality $\aleph_1$). Rigged Hilbert spaces are not normally touched in introductions to quantum mechanics, only the sloppy introduction of the delta dirac is done informally.

Answer: After all the preliminaries the answer is short. Continuous solutions appear in rigged Hilbert spaces for both bounded and unbounded states. One set of continuous bounded solutions appear in solid state physics. In the limit of an infinite number of atoms the energy bands for insulators, metals, and everything in between, become continuous (as a result of the "fusion" of the countable number of discrete energy levels that fuse into a continuous band).


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