Thursday 31 October 2019

quantum mechanics - Bound states, scattering states and infinite potentials


I am doing my first semester of Quantum Mechanics and we're using Griffith's Introduction to Quantum Mechanics. As he is introducing the Dirac delta function potential he explains bound and scattering states, and I understand that a system is considered bound if the energy of the system is less than the potential at infinity, that is


\begin{align} \text{Bound state: } E &< \lim_{|x| \to \infty} V(x)\\ \text{Scattering state: } E &> \lim_{|x| \to \infty} V(x). \end{align}


That makes sense, and he then continues by saying that this implies that $E<0$ for bound states and $E>0$ for scattering states, as you can always add a constant to the potential energy to make it zero at infinity.


He also explains how the solution to the Schrödinger equation for bound states is a discrete linear combination, and the solution for scattering states is an integral which cannot be normalized, and therefore does not exist. He then continues with the Dirac delta function potential unabated.


The problem I have is how to reconcile this with the previous chapter, in which he treated the harmonic oscilator - a bound system - and found the energy levels to be


$$E_n = \hbar \omega \left(\frac{1}{2}+n\right),$$


which is positive even though the potential goes to infinity at infinity. I suppose you could "subtract infinity" and get an infinitely negative energy (and 0 potential at infinity), but that's a bit weird at best. Part 1 of the question: Is that all there is to it? "Subtract infinity" and then the second inequality ($E<0$) works?


Part 2 of the qustion: since infinite potentials are just approximations and do not really exist (or do they?), how can bound states ever exist (Griffith remarks that finite potentials can be overcome by tunneling)? Additionally, scattering states also do not exists as their wavefunctions are non-normalizable. So the conclusion is that nothing really exists according to Quantum Mechanics... which can't be right, surely?



Answer




The distinction to be made here is that, for the quantum harmonic oscillator system, there are no unbound states, only bound states thus, there is no benefit to insisting the states have negative energy, no reason to 'subtract infinity' in order to zero the potential at infinity.


However, in systems that permit both bound and unbound states, it is reasonable to zero the potential at infinity for the same reason that we do this classically.


For example, in the classical central force problem, there is a state in which particle can 'escape to infinity' where it will have zero kinetic energy (more precisely, the kinetic energy of the particle asymptotically approaches zero). If we set the potential energy to be zero at infinity, then the total energy 'at infinity' is zero. Thus, the particle with zero total energy 'sits on the boundary' between those particles with not enough energy to 'reach' infinity and those that do.


But, for the classical harmonic oscillator potential, no particle can escape to infinity. The kinetic energy of the particle will periodically and instantaneously be zero. In this case, it is reasonable that the state where the total energy is always equal to the potential energy (the state where the kinetic energy is always zero) be the zero total energy state; all other states having positive total energy.



So the conclusion is that nothing really exists according to Quantum Mechanics... which can't be right, surely?



That's not remotely the correct conclusion to draw. One might conclude instead that


(1) The conception of bound state must be modified in the passage from classical mechanics to quantum mechanics and


(2) the physical (normalizable) unbound states are not eigenstates of the Hamiltonian, i.e., the physical unbound states are not states of definite energy but are, instead, a distribution of energy eigenstates, e.g., a wavepacket.



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