- #1
shanu_bhaiya
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QM: Fate of a particle of an arbitrary energy in "the particle in a box" problem
1. The Doubt
Consider the situation of a particle in a box with infinite potential beyond the walls.
Ψn = √(2/L) sin(nπx/L)
En = (nπħ/L)2/2m
My doubt is that what will happen, i.e. what will be the state of a single particle when left in such a box with initial kinetic energy E = ((0.2)πħ/L)2/2m.
The process of leaving the particle goes as follows (in one dimension):
First an electron-gun is placed in front of a wall and then e- is fired from it and suddenly after firing the e-, the gun is replaced by another wall.
Note that the length of the wall is L and the gun was placed at origin.
2. The attempt at a solution
I tried to calculate it and assumed that it's state will be:
Ψn = √(2/L) sin((0.2)πx/L)
It does obey the Schrodinger's Equation but then it doesn't obey the the boundary conditions and has a finite probability at x=L. So my assumption was wrong.
1. The Doubt
Consider the situation of a particle in a box with infinite potential beyond the walls.
Ψn = √(2/L) sin(nπx/L)
En = (nπħ/L)2/2m
My doubt is that what will happen, i.e. what will be the state of a single particle when left in such a box with initial kinetic energy E = ((0.2)πħ/L)2/2m.
The process of leaving the particle goes as follows (in one dimension):
First an electron-gun is placed in front of a wall and then e- is fired from it and suddenly after firing the e-, the gun is replaced by another wall.
Note that the length of the wall is L and the gun was placed at origin.
2. The attempt at a solution
I tried to calculate it and assumed that it's state will be:
Ψn = √(2/L) sin((0.2)πx/L)
It does obey the Schrodinger's Equation but then it doesn't obey the the boundary conditions and has a finite probability at x=L. So my assumption was wrong.