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Consider a potential that on the left of some point $x=x^*>0$ is infinite and on the right of that point it is of the form $$V(x)=-\alpha x^{-3}.$$

I tried to use the WKB method to determine the quantum number $n$ of the last bound state. Since the potential has one rigid wall i use the quantization:

$$\int_{x^*}^bp(x)dx=\int_{x^*}^b\sqrt{2m(E-V(x)}dx=(n+\frac{3}{4})\pi\hbar.$$

I take the limit in which $E\rightarrow 0$ and $b\rightarrow \infty$ where $b$ is the point where the potential energy is equal to the kinetic energy. After doing the integral i get the following result, that the quantum number of the last bound state is:

$$n_{max}=\frac{\sqrt{2x^*ma}}{\pi \hbar}-\frac{3}{4}.$$

To my understanding this is proof that bound states do exist since you can go down a step at a time and figure out the quantum number for the other bound states (given that the mass and $\alpha$ parameter give a sensible $n_{max}$)

I tried another approach to see if i get the same result or at least prove that there are indeed bound states.Using the virial theorem $$2T=-3V$$ for the ground state:

$$E=V+T=\frac{1}{3}T>0$$

so that means no bound states for this potential since the ground state has the lowest energy.

Obviously im doing something wrong since both statements can't be true at the same time, but i can't find what is wrong in my thinking and why these results are indeed incorrect.

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  1. In this answer I won't consider the semiclassical WKB approximation, which is a nice exercise in itself.

  2. Instead, I would like to say a word or two about the virial theorem. The infinite wall at $x=x^{\ast}>0$ effectively means that OP's potential $V(x)$ is not a power law, and therefore OP's argument is not valid.

  3. Nevertheless, it is interesting to instead consider a situation with an attractive power law potential $$\begin{align} V(x)~=~& {\rm sgn}(\alpha) A |x|^{\alpha}, \cr \alpha~\in~&\mathbb{R},\cr A~>~&0. \end{align}\tag{1} $$ If the virial $$ G~:=~xp \tag{2}$$ is bounded in phase space, $$ {\rm const}~>~|G|~\sim~|x| \sqrt{E-V(x)},\tag{3} $$ the classical virial theorem then states that $$\frac{\alpha}{2} \langle V \rangle~\stackrel{(1)}{=}~\langle T \rangle~\geq~0, \tag{4}$$ so that the energy is $$\begin{align} \langle H \rangle ~=~& \langle T \rangle +\langle V \rangle \cr ~\stackrel{(4)}{=}~& (1+\frac{2}{\alpha}) \langle T \rangle .\end{align} \tag{5}$$ There are 5 cases:

    • $\alpha > 0$: The virial (2) is bounded (3). The energy $\langle H \rangle \geq 0$.

    • $\alpha = 0$: Free particle. The energy $\langle H \rangle = \langle T \rangle\geq 0$.

    • $-2 <\alpha < 0$: The virial (2) is bounded (3) if we assume that the energy $\langle H \rangle\equiv E<0$ is negative. Not surprisingly this agrees with eq. (5).

    • $\alpha < -2$: The virial (2) is not bounded (3), so the classical virial theorem does not apply. Hence the naive conclusion $\langle H \rangle \geq 0$ from eq. (5) cannot be trusted. Quantum mechanically, one may rigorously prove that the spectrum of the Hamiltonian $H$ is unbounded from below, i.e. it has no ground state and is unstable, see e.g. the theorem in my Phys.SE answer here.

    • $\alpha = -2$: See e.g. this Phys.SE post.

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  • $\begingroup$ I tried plotting the Virial (3) for different values of $\alpha$ and for $\alpha> 0$, $|G|$ doesn't go to infinity at any limit. But when i plot it for $-2<\alpha< 0$, $|G|$ goes to infinity in the limits, so for these values $|G|$ isn't bounded but the theorem still holds. What am i doing wrong? $\endgroup$ – Μπαμπης Ποζουκιδης Jan 17 at 22:00
  • $\begingroup$ What im trying to say is that according to the plots i made for $|G|$ where $-2\leq \alpha \leq 0$, $|G|$ isn't bounded but you are still stating that $\langle H \rangle \leq 0$, even though the theorem shouldn't apply in this case. $\endgroup$ – Μπαμπης Ποζουκιδης Jan 18 at 12:01
  • $\begingroup$ Thanks. I updated the answer. $\endgroup$ – Qmechanic Jan 18 at 12:24
  • $\begingroup$ Thank you very much!!! I'm gonna continue reading the links you provided to understand these conclusions even better! $\endgroup$ – Μπαμπης Ποζουκιδης Jan 18 at 12:45

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