# One-dimensional box of some width with a particle inside

1. Jun 26, 2008

### Domnu

Okay, so let's say we have a one-dimensional box of some width with a particle inside. If the particle is in an eigenstate of the energy, we know its energy exactly. Since the energy is purely kinetic, we also know its momentum. By the uncertainty principle, the position should be infinitely uncertain... but we know that the particle is in the box (hence a finite uncertainty). How does this work?

2. Jun 26, 2008

### Zacku

Hello,

Actually you can only know the norm of the momentum vector this way but you can't have any information about the projections of this vector (and uncertainty prinicple only holds for each projection of the momentum and position vectors).

3. Jun 26, 2008

### Domnu

I don't exactly understand what you meant... could you clarify please? If we have the norm of the momentum vector, don't we essentially have the value of the momentum?

4. Jun 26, 2008

### Zacku

That's not a problem since you can't guess the projection values of the momentum with only the norm. Indeed, the "paradox" you are talking about is not correct since the uncertainty principle refers to the projection values of the momentum and not to its norm value.

Is it more clear this way ?

P.S. : excuse me for my english

5. Jun 26, 2008

### Domnu

Hmm... what exactly do you mean by the "projection values of the momentum?" One thing I did notice was that the energy eigenstates of the particle in a box scenario are not eigenstates of the momentum operator. Is the norm you are speaking of the expected value of the momentum as opposed to the true momentum?

6. Jun 26, 2008

### nughret

One could say the paradox is the fact that the particle's wavefunction is restricted to a box as this requires an infinite potential - impossible.

7. Jun 26, 2008

### gel

You don't know the momentum if a particle is confined to a box. A momentum eigenstate is unchanged by translation (up to a phase change). This can't be true if the particle is confined to a finite region...unless the position of the box itself is uncertain.

8. Jun 26, 2008

### nicksauce

My thoughts (I'm not sure if this has already been mentioned): You know p^2. Thus you know p = -|p|, or |p|, presumably 50% each. Thus <p> = 0, and <p^2> = p^2, so the uncertainty on <p> is p, or (2mE)^(1/2). I work this out and get dx*dp ~= 2.5hbar for the n=1 state.

9. Jun 26, 2008

### gel

Actually, I don't think you do know p^2. The energy will be a function of momentum and the interaction with the sides of the box - not just the kenetic energy term.
For example, you could model the edge of the box as a high potential barrier, and the energy is a function of both position and momentum.

10. Jun 27, 2008

### Zacku

To complete what the others said. The fact that the particle is confined (I don't know if this word is correct) in space implies that the eigen wave functions have to be wave packets of completely free particle solutions ; this is a straightforward result of the Fourier transforms calculus.
You can also see that noting that, as said by gel and nicksauce, the momentum does not commute with the hamiltonian in this system because the potential term depend on the variable of space actually.
Finally, the only measurable states correspond to stationary waves and thus contain both forward and backward plane waves which is equivalent to an indetermination of the sign of the momentum whatever the energy measured.

11. Jun 28, 2008

### per.sundqvist

$$\Delta p=\sqrt{<p>^2-<p^2>},$$
using $$p=\;-i\hbar d/dx, \psi(x)=\sin(\frac{n\pi x}{L})$$