Discreteness of bound vs unbound states

[Mayan Fung]

I observe that all bound states have discrete energy levels, eg. particle in a box, hydrogen atoms. But unbound states always have a continuous energy spectrum. For example, for the case of a finite potential well, when $E<V_0$, we have discrete energy for the bound states. When $E>V_0$, the energy spectrum is continuous. Many books only solve the Schrödinger Equation to show that we have some constraints so that the allowed energy can only take certain value.

Apart from the maths, is there any physical interpretation for this fact? Is there any theory explaining some characteristics of the bound and unbound states?

 

[DrClaude]

It is a question of boundary conditions. When the wave function can extend to infinity, the spectrum becomes continuous.

 

[phyzguy]

One way to look at it is to assume the unbound states are in a box. The spacing between the states is a function of the size of the box. By moving the walls of the box to infinity, the spacing between the states can be made arbitrarily small.

 

[Mayan Fung]

Thanks! That's a good way of thinking!

 

[Keith_McClary]

a finite potential well, when $E<V_0$, we have discrete energy for the bound states. When $E>V_0$, the energy spectrum is continuous.

This is true if $V$ drops off rapidly enough at $\infty$ but is not true in general.

 

[strangerep]

I don't think discreteness is a question of boundary conditions, nor whether the wave function can extend to infinity. Consider an isolated hydrogen atom: there's nothing stopping the electron's wave function from extending to radial infinity, yet the bound states are discrete.

Here's my take on the subject...

The relevant Hamiltonian for the H-atom is $$H ~=~ K+V ~\equiv~ \frac{p^2}{2\mu} ~-~ \frac{Y}{r} ~,$$where $K,V$ correspond to kinetic and potential energies respectively, $\mu$ is the reduced mass, $p^2$ is the squared spatial momentum, $r$ is the radius (from the center of mass), $Y := ke^2$ and $e$ is the elementary electron charge with $k$ being a constant.

The distinction between unbound and bound states corresponds to the sign of $H$'s eigenvalues. Heuristically, $K>V$ (resp. $K<V$) corresponds to unbound (resp. bound) states. (The case $K=V$ is actually a distinct 3rd case which I'll skip over here.)

The reason for this correspondence can be traced back to the dynamical symmetry groups applicable in both cases. (A dynamical symmetry group maps solutions of the equations of motion among themselves.) The dynamical symmetry Lie algebra can be found by finding conserved quantities and constructing the commutator algebra among these quantities. For the H-atom, the basic conserved quantities are the Hamiltonian $H$, the angular momentum $L_i$, and the Laplace-Runge-Lenz (LRL) vector $A_i$. The crucial thing here is that the $A_i$ satisfy a commutation relation among themselves of the form $$[A_i, A_j] ~=~ 2\mu H i \hbar \varepsilon_{jk\ell} L_\ell ~~~~~~ (1)~.$$ Normally, in an ordinary Lie algebra you would not expect to see such a product of generators on the RHS of the commutator, so this is a hint that a more careful analysis is needed.

$H$ is a (self-adjoint) physical observable (corresponding to energy) so its eigenvalues $E$ are real and the corresponding eigenvectors span the Hilbert space. There is no other operator in the algebra which can invert the sign of $H$, hence we may conclude that the Hilbert space can be partitioned usefully into 3 disjoint subspaces corresponding to the sign of the $H$ eigenvalues. The disjointness of these subspaces is crucial here. Moreover, since $L_i$ and $A_i$ commute with $H$, the action of $L_i$ and $A_i$ preserves these subspaces.

The next step is to define a scaled LRL vector $$A'_i ~:=~ \frac{A_i}{\sqrt{2\mu |H|}} ~~~~ (2) ~,$$ which satisfies a modified commutation relation $$[A'_i, A'_j] ~=~ - \text{sign}(H) i \hbar \varepsilon_{jk\ell} L_\ell ~~~~~~ (1)~.$$ The different signs on the RHS mean that the Lie algebras applicable in the 3 subspaces are totally different. For $E<0$ we have the compact $SO(4)$, but for $E>0$ we have a noncompact group ($SO(3,1)?$). The former gives discrete energy eigenvalues (by an essentially similar analysis to how ordinary $SO(3)$ gives discrete angular momentum eigenvalues). This doesn't happen with the noncompact group and the positive energy eigenvalues turn out to be continuous.

In summary, the gross structure of the complete energy spectrum is all about the relative sizes of kinetic energy (usually a function of momentum) and potential energy (usually a function of position), plus the fact that position and momentum don't commute, and then how these parts of the Hamiltonian $H=K+V$ fit into the overall dynamical group for the (class of) systems under consideration.

HTH.

 

[Mayan Fung]

@strangerep Thank you for your detailed explanation. That is a bit too advanced for me. I will spend some time figuring out the details.

 

[strangerep]

[...] That is a bit too advanced for me. I will spend some time figuring out the details.[...]

Have you studied (and/or do you understand) how discrete eigenvalues arise for quantum angular momentum? That's a simpler case, concentrating on angular momentum and the SO(3) rotation group only, without the extra complications of energy, etc.

If you haven't studied it, I suggest careful reading of section 7.1 of Ballentine.

 

[vanhees71]

For a more intuitive understanding it's also good to look at the analogous problem in classical mechanics. If you have a central potential, using angular-momentum conservation you can reduce the problem to an effective 1D problem for the equation of motion of the radial coordinate $r=|\vec{x}|$. Usually the potential is defined such that $V(r) \rightarrow 0$ for $r \rightarrow \infty$ and the effective potential for the 1D motion is given by
$$V_{\text{eff}}(r)=V(r) + \frac{\ell^2}{2m r^2}.$$
If, as for, e.g., the attractive Coulomb potential (hydrogen(-like) atom) the effective potential has a minimum there are bound solution for the equation of motion since $E=m \dot{r}^2/2+V_{\text{eff}}(r) \geq V_{\text{eff}}(r)$ and thus there are two classical turning points if the energy $E<0$. For $E \geq 0$ the classical motion is not bound.

In the quantum case you have the very same effective Hamiltonian for the radial wave function (looking for the simultaneous eigenstates of $\hat{H}$, $\hat{\vec{L}}^2$ nad $\hat{L}_z$), leading to the wave function
$$u_E(\vec{x})=\frac{1}{r} R_{E,\ell}(r) \text{Y}_{\ell m}(\vartheta,\varphi)$$
with the Schrödinger equation for the radial part
$$\hat{H}_{\text{eff}} R_{E,\ell}=E u_{E,\ell}(r)$$
with the effective 1D Hamiltonian
$$\hat{H}_{\text{eff}}=-\frac{\hbar^2}{2m} \partial_r^2 + \frac{\hbar^2 \ell (\ell+1)}{2m r^2} + V(r).$$
In order that that $\hat{H}_{\text{eff}}$ is self-adjoint on the Hilbert space $\text{L}^2(\mathbb{R}_+)$ you need a boundary condition at $r=0$, which leads to $u_{E,\ell}(r) \rightarrow 0$ for $r \rightarrow 0$. At infinity a proper $\text{L}^2$ function must of course go to 0 sufficiently quickly such that $u_{E,\ell}$ is normalizable to 1, and this happens only for $E<0$ as known from the Schrödinger-equation theory in 1D, i.e., when there is a "classical forbidden region". Of course in quantum mechanics the probability for the particle to be in this classical forbidden region is non-zero (tunnel effect) but at least you get a normalizable wave function. Usually it's going somehow exponentially to 0 $u_{E,\ell}(r) \simeq \tilde{u}_{E,\ell}(r) \exp(-k r)$ with $k>0$. Then you find that only for a discrete set of energies this asymptotic condition is fulfilled.

For $E>0$ you don't have an exponentially decaying solution at infinity but rather some function going like $\exp(\pm \mathrm{i} k r)$ with $k>0$. Then the energy eigenvalues build a continuous part of the spectrum of the effective Hamiltonian, and the wave functions are only "normalizable to a Dirac-$\delta$ distribution, i.e., $\langle u_{E,\ell}|u_{E',\ell'} \rangle=\delta(E-E') \delta_{\ell \ell'}$.

 

[stevendaryl]

This thread showed me that I really don't completely understand unbound states.

If we view $H \psi = E \psi$ as purely a differential equation to be solved, then we can classify the solutions in three different types:

1. Solutions where $\psi$ is square-integrable.
2. Solutions where $\psi$ is not square-integrable, but $|\psi|^2$ is bounded.
3. Solutions where $|\psi|$ is unbounded.

Examples of the three types:

1. $\psi(x) = e^{-Kx}$ in the region $x > 0$, where $K > 0$.
2. $\psi(x) = e^{iKx}$ in the region $x > 0$, where $K$ is real.
3. $\psi(x) = e^{+Kx}$ in the region $x > 0$, where $K > 0$.

For the examples that I know of (harmonic oscillator potential, coulomb potential, square well potential, delta-function potential), there are discretely many solutions of type 1, while continuously many solutions of types 2 and 3. I don't know if that's the case with every potential function.

 

[stevendaryl]

This thread showed me that I really don't completely understand unbound states.

If we view $H \psi = E \psi$ as purely a differential equation to be solved, then we can classify the solutions in three different types:

1. Solutions where $\psi$ is square-integrable.
2. Solutions where $\psi$ is not square-integrable, but $|\psi|^2$ is bounded.
3. Solutions where $|\psi|$ is unbounded.

Examples of the three types:

1. $\psi(x) = e^{-Kx}$ in the region $x > 0$, where $K > 0$.
2. $\psi(x) = e^{iKx}$ in the region $x > 0$, where $K$ is real.
3. $\psi(x) = e^{+Kx}$ in the region $x > 0$, where $K > 0$.

For the examples that I know of (harmonic oscillator potential, coulomb potential, square well potential, delta-function potential), there are discretely many solutions of type 1, while continuously many solutions of types 2 and 3. I don't know if that's the case with every potential function.

Correction: The harmonic oscillator potential has no solutions of type 2, I don't think.

 

[vanhees71]

I think so, but maybe there are some odd potentials where the wave functions behave differently. Of course there are potentials, where there is no continuous part, i.e., only bound states if the potential goes to infinity at $x \pm \infty$ (the standard example is the harmonic oscillator).

Already the Coulomb potential is special because it's long-ranged, and you have infinitely many bound states (the hydrogen bound states with $E_n \propto 1/n^2$, $n \in \mathbb{N}$) with a limiting point $E=0$ (which belongs to the continuous spectrum), and the scattering solutions with $E>0$ (continuous spectrum) are modified in their asymptotic behavior compared to the standard scattering solutions ("incoming plane wave + scattered spherical wave").

It's worth reading about the 1D case in Messiah's QM book, where everything is nicely discussed (using the adequate math of differential equations).