# Particle in a box?

1. Mar 5, 2006

### pivoxa15

If we have the case of an infinite potential well, the particle is assumed to have 0 potential energy when inside the well at all times. Why is that? Does it assume that the box itself is 0 net charge? This is because any non zero net charge and the electron will either be repelled or attracted meaning positive potential energy for some time.

Or does it imply either infinite negative or positive charge on both sides? If that is the case than why is it still 0 potential evergy at all points inside the well?

2. Mar 6, 2006

### Staff: Mentor

Try it yourself: assume that $V = V_0$ inside the box (where $V_0 \ne 0$). What are the stationary-state solutions of the Schrödinger equation for this case, and their energies?

3. Mar 6, 2006

### pivoxa15

Assuming potential energy V = a, the total energy levels = Energy level assuming 0 potential + a. In other words, ((hbar)(n)(pie))^2/(2ml^2) + a.

So including a potential just added that amount of energy to the total energy of the particle for each quantum state.

However, that still does not tell me why we shouldn't include a potential energy for the particle in the well. I guess I just don't have a physical understanding of why there is 0 potential energy for the particle anywhere in the well. When it reaches one end of the well and is repelled than surely it should have positive potential energy when that happens? Or is the trick that since each end has infinite potential, whenever the particle amass some potential, it is instanteously canceld from the well on the other end. So it is the infinity that is making things unintuitive?

Last edited: Mar 6, 2006
4. Mar 6, 2006

### Ratzinger

The particle moves in a potential, it 'gains' potential energy when it is in a region with a higher V. So it has potential energy at the boundaries.

5. Mar 6, 2006

### Staff: Mentor

Keep in mind that in physics, the "zero point" of potential energy is basically arbitrary. All that matters in terms of physically observable outcomes are differences in potential energy.

With the particle in a box, regardless of what you choose as the potential energy inside the box, a state with a given n is the same "height" above that "floor", and that's what matters, physically. The wave function comes out to be the same regardless of the value of a in your energy-level formula:

$$\psi(x) = \sqrt{\frac{2}{L}} \sin \left( n \pi \frac {x}{L} \right)$$

So the expectation values of position, momentum, and quantities derived from them, are the same regardless of the value of a. It's purely a matter of convenience, which value you choose for a.

6. Mar 6, 2006

### pivoxa15

A constant, non zero potential inside the box is not realistic. How about potential V varies as x as it should because when the electron is nearer to the walls or boundaries of the box, it should be repelled. Hence in the middle, its potential should be lowest and closer to the edges, it should be higher. so V=(x-L/2)^2 where L is the length of the box.

This is the point I was trying to make which was that potential energy of the electron should be non zero near the walls. The problem is that now I do not know how to solve this equation as V is not a constant anymore. What would the solution look like? Wouldn't this be more realistic?

7. Mar 16, 2006

### the iron maiden

If it was an infinate potential well, then any charge that the electron possesses can be equalled by the potential in the well ( since it is infinate ) meaning that there would be no difference in potential of the electron and the well - in other words no potential, so the electron wouldn't repel from the walls as a result of its charge, but just simply by interacting with the walls of the " box ". much like a classical case of linear momentum. the energy levels of the electron would be constant due to the conservation of linear momentum.

8. Mar 16, 2006

### pivoxa15

I have not assumed an infinite potential well. I am considering a realistic case with Potential V=(x-L/2)^2 which means that the walls are slightly negatively charged hence the electron will be repelled as it travels near the wall.

Hence the physical system is now a quantum oscillator. I suspect this is a more realistic model of the electron made out of walls that are slightly negatively charged.

9. Mar 16, 2006

### masudr

First
then,
Anyway, I'll ignore that for now.

---

If you are talking about a quadratic potential then you do indeed have a harmonic oscillator (a constant potential would give the "particle in a box").

Note that if

$$V=\frac{1}{2}mw^2x^2+V_0$$

then we would still have the usual quantum harmonic oscillator.

At this stage, considering the repelling charge of electrons at the edge of the box is not actually going to work. The reason is because you must justify the potential you claim will work. We can work out classically the potential a series of static charges will produce. However, taking a full quantum mechanical description is obviously going to be more complicated.

In your original post you considered a particle in an infinite square well, but asked why the potential in the well had to be zero. Well, as has been pointed out potential is only defined up to a constant, so this is irrelevant.

When considering a quadratic potential, you say you cannot solve the equation. Well, you can find the energy eigenvalues of a quantum harmonic oscillator, and this problem is tackled in almost every standard QM text. If you can't afford that, try http://en.wikipedia.org/wiki/Harmonic_oscillators.