Particle in Infinite Square Well

[AgingStudent49]

TL;DR

Comparison of Free vs. Fixed end boundary conditions for a particle in an infinite square well

In thinking about the particle in an infinite square well, it the commonly espoused boundary conditions of ψ(0) = 0 and ψ(L) = 0 seem somewhat arbitrary. What in nature forces the wave function to vanish at the boundaries? If the particle can't escape and there is no energy loss, why not impose free end boundary conditions where the wave is reflected from each boundary? In working through this, the time-independent wave function with free-end boundary conditions is:

$ψ(x) = sqrt(2/L) cos(2nπx/L)$​

and the energy levels are:

$En = 4 (n^2 π^2 ħ^2)/(2m L^2)$​

(Notice that the energy levels are 4 times greater that those resulting from the fixed-end boundary conditions.).

Comments are encouraged.

 

[PeroK]

TL;DR Summary: Comparison of Free vs. Fixed end boundary conditions for a particle in an infinite square well

In thinking about the particle in an infinite square well, it the commonly espoused boundary conditions of ψ(0) = 0 and ψ(L) = 0 seem somewhat arbitrary. What in nature forces the wave function to vanish at the boundaries?

It's not so much arbitrary as idealized. The infinite square well can be seen as the limit of a finite square well with a large potential boundary. In that case, continuity of the wave function imposes the boundary conditions.

 

[bob012345]

TL;DR Summary: Comparison of Free vs. Fixed end boundary conditions for a particle in an infinite square well

In thinking about the particle in an infinite square well, it the commonly espoused boundary conditions of ψ(0) = 0 and ψ(L) = 0 seem somewhat arbitrary. What in nature forces the wave function to vanish at the boundaries? If the particle can't escape and there is no energy loss, why not impose free end boundary conditions where the wave is reflected from each boundary? In working through this, the time-independent wave function with free-end boundary conditions is:

$ψ(x) = sqrt(2/L) cos(2nπx/L)$​

and the energy levels are:

$En = 4 (n^2 π^2 ħ^2)/(2m L^2)$​

(Notice that the energy levels are 4 times greater that those resulting from the fixed-end boundary conditions.).




Comments are encouraged.

I’m not sure what you mean by a “free end boundary condition”. How is that different from a classical particle?

 

[snorkack]

The thing is that infinite energy requires infinite second derivative of Ψ for any nonzero Ψ.
You can have nonzero Ψ with infinite second derivative for a point - as is the case at nucleus. The function passes through an angle and its first derivative has discontinuous jump.
But you cannot have infinite second derivative for an extended volume of space, like everywhere outside the potential well. Therefore the only allowed option is for Ψ to be zero there.

 

[gaiussheh]

Well, you don't deduce the boundary from the potential. It is, in fact, the opposite. You know that you are dealing with a confined particle such that it has zero probability of appearing outside the well. You know that the wave function must be continuous. Thus you construct something like $\psi(0)=\psi(L)=0$. Then you think - "okay, the only reasonable interpretation is to think of an infinite potential outside of the box."

 

[Albertus Magnus]

I think that the nearest thing to "free end" type boundary conditions (like those for transverse vibrations on a string) for a quantum particle, where there is reflection from the "walls" is the finite square well problem. In that case, the particle does indeed reflect from the sides, however, it also "tunnels" through each of the sides as well. Thats the thing about quantum scattering problems, one always gets non-zero amplitudes for both transmission and reflection. The only way to totally confine a quantum particle is to use the $\psi=0$ on some surface type of boundary conditions.

 

[Albertus Magnus]

The thing is that infinite energy requires infinite second derivative of Ψ for any nonzero Ψ.
You can have nonzero Ψ with infinite second derivative for a point - as is the case at nucleus. The function passes through an angle and its first derivative has discontinuous jump.
But you cannot have infinite second derivative for an extended volume of space, like everywhere outside the potential well. Therefore the only allowed option is for Ψ to be zero there.

Yeah, this answer pins it down exactly as to why you can't have $\psi\neq 0$ for some region where the potential is infinite.