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Griffiths describes bound and scattering states as follows:

Bound state : $E<V(-\infty $) and $V(+\infty $)

Scattering state: $E>V(-\infty)$ or $V(+\infty)$

Why is that Energy for a system should be greater than the minimum potential in bound states and why is it not a restriction in Scattering states?

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  • $\begingroup$ This isn't a restriction at all - it is purely terminology. $\endgroup$ – Emilio Pisanty Mar 10 '18 at 22:52
  • $\begingroup$ I think it's pretty good terminology though... the bound states are localized to one region of space and the scattering states are basically plane waves. $\endgroup$ – Ryan Thorngren Mar 10 '18 at 22:55
  • $\begingroup$ I am not getting what you are saying. @EmilioPisanty $\endgroup$ – Icchyamoy Mar 10 '18 at 23:00
  • $\begingroup$ @Ryan, Why is this a terminology? I just read the fact that in order for the wave function to normalize in Bound state, from TISE, it can be derived that E >V_min. Why such ani't the case for scattering state? $\endgroup$ – Icchyamoy Mar 10 '18 at 23:02
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Bound states correspond to particles that are localised, that is, found in a restricted region of space. In the one dimensional case, you want the probability density to drop to zero rapidly at $x =\pm \infty$ so that the wavefunction is normalisable to $1$.

On the other hand, for scattering states you would have $e^{\pm ikx}$ behaviour with $k$ real, and you cant normalise such states to one on the open real line.

Consider the Schrodinger equation for a one dimensional potential $$ \left( \ \partial^{2}_{x} +k(x)^2 \ \right) \psi(x) =0 \ , $$ with $$k^2(x) = \frac{2m(E-V(x))}{\hbar^2} \ . $$

Lets look at the asymptotic $x \sim \pm \infty$ (that is, large x) solutions of the differential equation. If $E> V(\pm \infty)$ then $k$ is real there and asymptotically the solutions are like the scattering states mentioned above.

For $E<V(\pm\infty)$, $k$ is complex and you can write $k =i\kappa$ to get solutions that behave like $e^{\pm \kappa x}$ asymptotically. These are the bound states (you want to choose the $\pm$ sign for $x = \mp\infty$).

There are probably technicalities to worry about in the above (heuristic) argument, eg if the asymptotic analysis matches smoothly to actual solutions etc. This should work for generic smooth, bounded potentials. You can find details in older books like "Quantum Mechanics" by Messiah or consult texts on properties of second order differential equations.

Here is one interesting exception to the general claim: It IS POSSIBLE to have bound states (normalised to one) embedded among the scattering states (ie when $E> V(\pm \infty)$) if the potential is of the "Von Neumann-Wigner" type: such potentials oscillate indefinitely even as their magnitude vanishes at infinity. An example is discussed in the Quantum Mechanics book by Ballentine. The unusual oscillatory potential traps the particle that you would have expected to escape. As Ballentine says, its a quantum phenomenon without a classical analog in particle mechanics but may be understood in the wave analogy as due to destructive interference.

So the moral probably is this: Generic potentials do indeed have the properties that Griffiths says they do, but one can construct unusual potentials which exploit technical loopholes to evade the folklore theorem.

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  • $\begingroup$ Clarification: The Von Neumann-Wigner example in Ballentine's book mentioned above is for a three dimensional potential. There might be one dimensional examples but I am not sure. $\endgroup$ – rparwani Mar 21 '18 at 6:33
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In classical mechanics, a state is bound if it can be confined to a compact region of the phase space. Under the assumption that potentials are localised, the condition $E < V(\infty)$ guarantees that the trajectories in phase space can be enclosed in a compact domain. When $E > V(\infty)$, this is no longer possible in general.

Consider the following example: $V:\mathbb R\to\mathbb R$ is the potential of the SHO inside the interval $[-1,1]$, and it is constant otherwise, and such that $V$ is a continuous function:

$$V(x)=\begin{cases}\frac12x^2 & x\in[-1,1]\\\frac12 & x\not\in[-1,1]\end{cases}$$

The Hamiltonian for the system is

$$H(x,p) = \frac12p^2+V(x)$$

and we see that for $E < \frac12 = V(\infty)$, we get circular trajectories in the phase space, all confined inside the square (in fact the circle) of radius $1$. That's because for $E<\frac12$, $x$ and $p$ cannot get larger than $1$ in absolute value. If $E > \frac12$, we can construct a solution for which $p > 0$ at any time, which represents a particle that is travelling from $-\infty$ to $\infty$, interacting (scattering) with the potential at around 0. There is no compact region of the phase space that can contain this trajectory.

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  • $\begingroup$ Can you say why Energy can take any value in Scattering state but not in Bound state? $\endgroup$ – Icchyamoy Mar 10 '18 at 23:13
  • $\begingroup$ The reason is explained above. If the energy is bounded by the maximum of the potential, the state is bounded in phase space. In QM this is like, e.g., a low-energy electron in the ES potential of a nucleus (that was already there or that has been captured). Now you shoot some light to the electron, whose energy jumps above the maximum of the potential, and you now have a free electron. Or, imagine a high energy electron that passes by an ionised atom. The electron comes from $-\infty$, its trajectory is potentially deflected by the atom and continues travelling towards $+\infty$. $\endgroup$ – Phoenix87 Mar 10 '18 at 23:18

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