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In the context of Quantum Field Theory we put restrictions on the potentials we can use. One argument is boundedness. If the potential is unbounded, for example $V(\phi) = \phi^3$, then `the field can always decrease its energy further' by becoming more negative. This results in the no ground state and no bound states.

The analogous argument in the context of Quantum Mechanics is more confusing to me. Consider the example of the hydrogen atom. The potential is not bounded from below and yet we have a discrete spectrum with a unique finite energy ground state.

Another example where boundedness arguments come up is perturbation theory (in Quantum Mechanics). Consider a finite well for a single particle $$V(x)=\Big{\{} \begin{matrix}V_0 \ \ |x| >a \\ 0\ \ \text{otherwise} \end{matrix} $$

If we apply a linear perturbation $\delta V = \lambda x$ then we say that perturbation theory can't converge because the hamiltonian is unbounded and hence has no bound states. And it follows that (this is the point I am not sure if it's true) the corrections to the energy must diverge to $-\infty$. I am not sure what this statement is saying. Is it that the spectrum is just $-\infty$ or is it saying the hamiltonian just doesn't have a spectrum at all (whether thats continuous and/or discrete).

I guess what I'm asking for is a more mathmatical statement about when a single particle hamiltonian has a spectrum.

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    $\begingroup$ Something that might make a good follow up question (i.e., a separate question to this one) is whether the field theory $V(\phi)=k/\phi$ is stable. I believe the answer should be "no" in $3+1$ dimensions, by analogy with your $\phi^3$ example, but "yes" in $0+1$ dimensions, because then I think it will become equivalent to the Hydrogen case. However, I am not 100% sure without thinking on it more. If I'm right, I'm also not sure at which dimension the behavior changes. There are certainly other results that apply in low dimensional quantum theories but break down in higher dimensions. $\endgroup$
    – Andrew
    Nov 29, 2022 at 0:37

3 Answers 3

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The issue can be tackled from a completely mathematical perspective. What you are saying about $\phi^3$ is true at classical level. The classical Hamiltonian functional ${\cal H}(\phi,\partial\phi)$ is not bounded below because we can find field configurations $(\phi_0, \partial \phi_0)$ with $-{\cal H}(\phi_0, \partial \phi_0)$ arbitrarily large, and this gives rise to several pathologies of various nature.

The issue with the hydrogen atom is mathematically different. There you may look for wavefunctions $\psi$ such that $\langle \psi|H\psi\rangle$ is arbitrarily close to $-\infty$.

However, differently from the (classical) Klein-Gordon $\phi^3$, here you also have the constraint $$\langle \psi|\psi\rangle = \int_{\mathbb{R}^3} |\psi(\vec{x})|^2 d^3x =1\:.$$ This constraint prevents us from finding functions with $$-\langle \psi|H\psi\rangle$$ arbitrarily large. The proof is the (well known) computation of the spectrum of the hydrogen atom.

(This is a sort of miracle, and it is one of the reasons why we are here. Since, classically speaking, the hydrogen atom and any system subjected to the attractive Coulomb force is unstable as it is unbounded below. Fortunately for us, our universe is quantum.)

One can also consider the second quantization of $\phi$. In that case, after normal ordering renormalization one sees that it is however possible to find states $\Psi$ of the quantum field where $-\langle \Psi| \hat{\cal H} \Psi\rangle$ is arbitrarily large, in spite of the constraint $\langle \Psi| \Psi\rangle=1$, due to the nature of $\hat{\cal H}$ written in terms of creation and annihilation operators. Maybe there is a straightforward physical reason beyond the evident mathematical difference in the formalisation of the two systems (first and second quantization, respectively) but I do not know, sorry.

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  1. $\phi^3$ model: This is a nice toy model to illustrate some aspects of perturbutative calculations in QFT, but certainly not a fully consistent QFT.

  2. Hydrogen atom with $V(r)=-e^2/r$: The spectrum of the potential term alone is just $(-\infty, 0)$ and, of course, unbounded from below. But the full Hamiltonian (including the kinetic term with spectrum $[0,\infty)$) $H = \vec{p}^2/2m + V(r)$ is bounded from below with the ground state having the lowest energy-eigenvalue $E_1=-me^4/2\hbar^2$. This can be understood as the best compromise of the contribution of the kinetic energy and the potential energy, which are - in contrast to classical mechanics - related by (a generalized version of) the uncertainty relation. As a final result, $H$ has the well known point spectrum $E_n=-m e^4/2 \hbar^2 n^2$ ($n=1,2,\ldots)$ and a continuous spectrum $[0,\infty)$ (corresponding to the scattering states).

  3. Finite well potential: Again, spectrum of the multiplication operator $V(x)$ is just the range of the function $V(x)$ and spectrum of the kinetic energy $P^2/2m$ is $[0,\infty)$. But $H=P^2/2m+V(x)$ has a point spectrum plus a continuous spectrum for $E \ge V_0$.

  4. Hamiltonian $H= P^2/2m + mgX$: Describes e.g. the one-dimensional motion of a particle with mass $m$ in a constant gravitational (or electric $mg \to E$) field. As to be expected, there are no bound states and the spectrum is purely continuous. This has nothing to do with perturbation theory.

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  • $\begingroup$ For 4. the spectrum is $E \in ( -\infty, \infty)$ but has no normalizable eigenstates? $\endgroup$ Nov 28, 2022 at 19:52
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    $\begingroup$ Correct! You can easily find the non-normalizable eigenfunctions in the $p$-representation of the Hamiltonian: $H = p^2/2m +m g i \hbar \partial/ \partial p$. $\endgroup$
    – Hyperon
    Nov 28, 2022 at 20:05
  • $\begingroup$ Ok, I thought so. But now I am confused about the convergence argument. Taking the case of the finite well again, expanding around this hamiltonian, we would expect perturbative treatment of eignestate corrections not to converge (as we start with normalizable states but now with the linear potential they are all unnormalizable) but this doesn't necessary need be the case for the corrections of the energies. $\endgroup$ Nov 28, 2022 at 21:19
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It is hard to give precise necessary and sufficient conditions for the spectrum of the TISE to be bounded from below, but here are some important cases that are definitely unbounded from below, i.e. it has no ground state and is unstable (and hence excluded from OP's sought-for list):

  1. If the potential $V$ has an attractive singular pole of order more than 2, i.e. if $$V({\bf r}) ~\propto~ - |{\bf r}- {\bf r}_0|^{-\alpha}\quad \text{for}\quad {\bf r}~\to~ {\bf r}_0$$ and $\alpha>2$, cf. e.g. eq. (4) in my Phys.SE answer here.

  2. If the preimage of the potential $V$ has the following property $$\exists \epsilon >0~\forall E\in\mathbb{R} ~\exists {\bf r}_0:~~ V^{-1}(]-\infty,E])~\supseteq~ {\rm Ball}({\bf r}_0,\epsilon) .$$

OP's linear perturbation example is prohibit by the 2nd case.

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  • $\begingroup$ So the 2nd condition says, if the spatial region where the potential diverges is unbounded then the spectrum has no bound states? $\endgroup$
    – AfterShave
    Nov 29, 2022 at 1:09
  • $\begingroup$ The 2nd case implies that there's no ground state, yes. $\endgroup$
    – Qmechanic
    Nov 29, 2022 at 6:50

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