QM: Fate of a particle of an arbitrary energy in the particle in a box problem

[shanu_bhaiya]

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.

Ψ_n = √(2/L) sin(nπx/L)
E_n = (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:
Ψ_n = √(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.

 

[JoAuSc]

The electron will leave the electron gun with a distribution of energies centered around the value you mentioned. When it gets trapped it will, for a split second, have the same wavefunction it had before the gun was replaced with a wall. In the pic I've attached, I have the electron wavefunction leaving the electron gun at three different times. Note that in the third drawing, just after it's trapped, the wavefunction hasn't changed much, and that it's not one of the infinite square well's eigenfunctions. It is in a superposition of states, each state with a different energy. The expectation value of the energy before and after being trapped should be similar, though, but I'm not an expert. Hopefully this clears things up.

 

[nasu]

The eigenstates (psi_n with energy En) you describe are not the only states possible.
Any combination of them is also possible (Schrodinger equation is linear).
Besides, any arbitrary state can be written as a linear combination of these eigenstates (they form a base).
They are like the normal modes of a string. The string does not need to oscillate in a normal mode.

However, these are solutions of the stationary Schrodinger equation. If you have a time dependent system, as you describe, you need to use the time dependent equation. Even more, the potential changes in time..
What you can say is that AFTER the system becomes again stationary, the state will be some combination of the eigenstates. Not necessarily the initial state.

In the stationary state, you cannot have energy less than the ground state.
(Like in a string with fixed ends and length L you cannot have standing waves longer than 2L. You may excite any non-standing wave but in the end (after some time) it either "die" or excite some standing wave modes)

 

[shanu_bhaiya]

Thanks a lot to both of you for replying.

The electron will leave the electron gun with a distribution of energies centered around the value you mentioned.

I didn't understand this clearly. I'm sure that my gun will shot the e^- with E = ((0.2)πħ/L)²/2m. And if you talk of Uncertainty Principle, then I can safely modify my problem by saying that the electron leaves the gun with an energy between ((0.1)πħ/L)²/2m to ((0.8)πħ/L)²/2m. But you see that the problem remains same. Or, does your statement mean that it's initial state or Ψ before hitting the wall first time will be a composition of Ψ_n I mentioned?

It is in a superposition of states, each state with a different energy. The expectation value of the energy before and after being trapped should be similar.

Thanks for the picture. But, it is impossible that the expectation value of allowed energies can be equal to E = ((0.2)πħ/L)²/2m, because it is less than all the allowed E_n.

The eigenstates (psi_n with energy En) you describe are not the only states possible.
Any combination of them is also possible (Schrodinger equation is linear).

I agree with Ψ, but what will happen with E, because it is even less than all the allowed E_n. The two options are:
1. It will remain same, which is not possible.
2. It will change, which must violate conservation of energy.

In the stationary state, you cannot have energy less than the ground state. (Like in a string with fixed ends and length L you cannot have standing waves longer than 2L. You may excite any non-standing wave but in the end (after some time) it either "die" or excite some standing wave modes)

So, option 2 must be true, i.e. it will change to one of the E_n. But since it's initial energy was already less than all the E_n, then from where it will gain the energy? Assume potential inside the well to be zero always.

 

[Dick]

It doesn't violate conservation of energy. The wavefunction of the electron penetrates the region where you are going to erect the wall, if the electron has a definite energy it's not localized. The act of erecting the wall can bump the electron into a higher energy state.

 

[shanu_bhaiya]

Thanks a lot for replying.

It doesn't violate conservation of energy.

Do you want to say that the electron will gain the energy from walls? As a wall, imagine a capacitor, a negatively charged one. Now according to what you said the projected or test electron must gain the energy from the interaction between the outer electrons on the capacitor and the test-electron; but how can it be possible because the electrons on the capacitor can be imagined to be not moving at all and so the test electron must leave with the kinetic energy with which it came, so after any number of hits to the wall, it's kinetic energy must not change. So how it bumps to higher E_n without breaking the law of conservation of energy?

The wave function of the electron penetrates the region where you are going to erect the wall, if the electron has a definite energy it's not localized. The act of erecting the wall can bump the electron into a higher energy state.

I really couldn't understand what you want to say, can you please be more descriptive?

 

[Dick]

The electron 'interacts' with the wall. You know that this isn't classical physics, right? The final state the electron winds up in isn't even deterministic. You have an initial state, a 'big disturbance' and then a choice of final states to settle into. In QM there often isn't a much more detailed picture of what exactly happens in between the initial state and the final state than that.

 

[shanu_bhaiya]

Thanks for clearing the matter, any more suggestions are still welcome. It seems now there is a long time until I start getting the feel of Quantum Mechanics, although I've just begun.