Consider $d$-dim Schrodinger equation without internal degree of freedom, that is, we don't consider spin etc.

How to prove that bound state's spectrum must be discrete and scattering state's spectrum must be continuous?

  • $\begingroup$ What is your definition of "bound state" and "scattering state"? Also, it's the spectrum of an operator, not the spectrum of a state. $\endgroup$
    – ACuriousMind
    Sep 26 '16 at 19:46
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/65636/2451 and links therein. $\endgroup$
    – Qmechanic
    Sep 26 '16 at 20:18

The question is not really well-defined in mathematical physics - you first need to specify what you mean by scattering and bound states. Clearly, a bound state should be contained in some compact region for all times and a scattering state should leave every compact region in finite time.

Here is a way to make this more precise: Let's work in $L^2(\mathbb{R}^d)$. Let's consider the operator that is the characteristic function of a ball $B(r):=\{z\in \mathbb{R}^d|\|z\|_2<1\}$ i.e. the operator that maps

$$ L^2(\mathbb{R}^d)\ni \psi\mapsto \begin{cases} \psi(x) &x\in B(r) \\ 0 & else \end{cases} $$

Clearly, this is a bounded linear operator, which is actually also compact. Call this operator $K(r)$, then we could say:

  • The state $\psi$ is a bound state of a self-adjoint operator $A$, if there exists some $r\in \mathbb{R}$ such that $\sup_t\{\|(\operatorname{id}-K(r))e^{iAt}\psi\|\}=0$. This means that we take the state $\psi$, we let it evolve any time and there will be nothing outside of the ball $B(r)$.
  • The state $\psi$ is a scattering state of $A$, if for all $r\in \mathbb{R}$ the state will have essentially left the region $B(r)$ at some time. We can quantify this by taking a Cesaro mean and obtain that $\psi$ is a scattering state if $\lim\limits_{T\to\infty} \frac{1}{T} \int_0^T \|K(r)e^{iAt}\psi\|\,dt=0$. This implies that the state won't spend a large amount of time inside the ball. It could still return and not be a true scattering state, but let's work with this for the moment.

Obviously, that's not enough because a bound state like the bound state from the hydrogen atom will not have a compact support, so it would never be a "bound state" in the definition above. Also, the operator $K(r)$ is somewhat awkward. What you do is to replace these definitions by $\varepsilon$-versions (or a limit of $r\to \infty$) and $K(r)$ by a sequence of compact operators (actually, it's enough to consider operators that are relatively compact with respect to $A$).

Then you have the following theorem known as RAGE theorem that does exactly what you want (I quote the version from Teschl's book "Mathematical Methods in Quantum Mechanics", Theorem 5.7):

Theorem: Let $A$ be self-adjoint and $K_n$ be a sequence of relatively compact operators converging strongly to the identity. Then

$$ \mathcal{H}_c:=\{\psi\in L^2(\mathbb{R}^d)|\lim\limits_{n\to\infty}\lim\limits_{T\to\infty} \frac{1}{T} \int_0^T \|K_ne^{iAt}\psi\|\,dt=0\} \\ \mathcal{H}_{pp}:=\{\psi\in L^2(\mathbb{R}^d)|\lim\limits_{n\to\infty}\sup_t\{\|(\operatorname{id}-K(r))e^{iAt}\psi\|\}=0\}. $$

Here, $\mathcal{H}_c$ is the continuous spectrum and $\mathcal{H}_{pp}$ is the pure point spectrum. Given the specification above, this theorem does exactly what you want.

That doesn't really answer your question how to prove this fact, but you can look up the proof in the book - I don't think I can really give a short and comprehensible sketch of the proof, as I pretty much forgot how it worked.

But let me try to give the most important ingredient to give you an idea where this comes from - namely Fourier analysis:

What we want to know is what the state $\psi$ does when $t\to \infty$, i.e. we want to look at $e^{iAt}\psi$ for large $t$. Using the spectral theorem, for any $\phi,\psi$ you have

$$ \langle \phi|e^{iAt}\psi\rangle=\int e^{i\lambda t} \mu_{\phi,\psi}(\lambda) $$

where $\mu_{\phi,\psi}$ is the Borel measure defined by the positive operator valued measure defined by the spectral projections of $A$ (have a look at the spectral theorem). As you see, that's a Fourier transform of a measure. It turns out that the time average of that Fourier transform is given by adding the pure point parts of the spectrum (Theorem 5.4 in the book and you can derive it using lots of Fourier analysis which I'm not really familiar with).

In a sense, this proves that a scattering state must be inside the continuous spectrum and a bound state must be inside the pure point spectrum.

However - for completeness sake - it's not enough to prove the converse. Not all states in the continuous spectrum are scattering states, because what you really want of a scattering state is that after some time, it'll move along virtually undisturbed to infinity. That's not really what the theorem tells you. In fact, the continuous spectrum can contain states that return or "almost return" to the origin over and over again.

If I recall correctly, one heuristic explanation and construction of such states is the following: Imagine a Hamiltonian with some potential barriers all over the place. There might be a state that passes some barriers but then gets deflected, returns, gets deflected again passes some barriers and gets deflected at a later point, etc. So it'll move to infinity eventually and the state will pass most of its time outside any reasonably sized area, but it's not a true scattering state.

In fact, people often define scattering states to be states in the absolutely continuous spectrum to exclude this behaviour above. What you want to have is that your Hamiltonian doesn't have the weird scattering but non-scattering states - this is related to what is called "asymptotic completeness" in scattering theory. You can then show that the scattering states behave as you want them to do (there is an introduction to scattering theory in the book, but you may also want to consult Barry Simon's overview of rigorous scattering theory).


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