# Eigenfunction & Uncertainty

1. Feb 29, 2004

### SilentSilver

According to my book, uncertainty Q = 0 (where Q is an observable) is true when the state function is an eigenfunction.

The energy eigenfunction for a particle in a 1-D box with infinitely high walls is sin(n*pi*x/a). This implies that the linear momentum, p, is known with zero uncertainty. By the uncertainty principle, the position, x, should have infinite uncertainty. This should mean an eigenfunction whose absolute value squared is a constant.

But the eigenfunction above (sin[]) doesn't meet that requirement.

I'd really appreciate it if someone could help me out here Thanks!

2. Mar 1, 2004

### turin

As far as I know, the expectation value of momentum is zero in this case, and the uncertainty in momentum would be:

&Delta;p = &radic;{<p2> - <p>2}
= &radic;{2m<E> - 0}
/= 0.

Last edited: Mar 1, 2004
3. Mar 1, 2004

### jeff

We can see that in fact the opposite is true - that p is completely unknown - by writing any wavefunction &psi;(x) as a fourier transform &psi;(x) &prop; &int; dp &psi;(p)eip&sdot;x showing that all values of momentum p are needed to reconstruct the wavefunction at a given position x.

4. Mar 1, 2004

### arivero

Jeff, the values are used with different weight for each p, so it is not completely undetermined.

My vote goes to Turin; explicitly we have
$$<p>= -i n {\pi \over a} \int_0^a sin(n \pi x/a) cos (n \pi x/a) dx =0$$

The particle is in a box, so x can not be spread across all the line. It is always between 0 and a, for every state. In consequence it is not strange that p can not be completely determined.

The usual expresions are only valid taking as configuration space the infinite real line.

In fact even self-adjointness becomes tricky; instead of a unique way, in the box there is a four-paramitrized way to extend the hermitian hamiltonian operator to get a self-adjoint one.

Last edited: Mar 1, 2004
5. Mar 1, 2004

### jeff

Hmm, this is interesting. For some reason I didn't notice that I'd left out the first sentence of my intended post, which was:

The momentum is given by the energy eigenvalue, but consider an unbound particle. We can see that in fact the opposite is true - that p is completely unknown - by writing any wavefunction &psi;(x) as a fourier transform &psi;(x) &prop; &int; dp &psi;(p)eip&sdot;x showing that all values of momentum p are needed to reconstruct the wavefunction at a given position x.

So you're right, we disagree.

6. Mar 1, 2004

### turin

Jeff,
What are you talking about? I think I'm missing something in your arguement. The wavefunction is in an energy eigenstate, not a position eigenstate. So why is the value of &psi; at some particular point x significant?

7. Mar 1, 2004

### jeff

Only my initial remark, that the momentum is given by the energy eigenvalue, was in reference to the particle bound by the potential well. Expanding on this a bit, that the particle is in an energy eigenstate means thats it has a specific energy, the eigenvalue of the state under the action of the hamiltonian. On the other hand, we don't know where the in the box the particle is, i.e., the eigenfunction gives the probability of measuring it to be at position x. The wavefunction &psi;(x) in my latter remark was for a different type of system, an unbound one.

8. Mar 1, 2004

### Tom Mattson

Staff Emeritus
It is not: E=p2/2m. For a given E, p can be anywhere in the range

[-(2mE)1/2,+(2mE)1/2].

edit: typo

9. Mar 1, 2004

### jeff

Thanks tom. Momentum does have direction. I should have said "up to sign". Bad jeff. Bad, bad jeff. (I've assumed that by "anywhere in the range [-(2mE)1/2,+(2mE)1/2]" you mean just those two values).

10. Mar 2, 2004

### arivero

Lets go for the energy in the box, then. E=p^2/2m; let me put h=1 and m=1 too, ok? Also, the eigenfuntion is not normalized, just that sine. Well, usual stuff
$$p \to -i {d \over dx} ; E \to - \frac12 {d^2\over dx^2}$$
then
$$<E>=({n\pi\over a})^2 \int_0^a sin^2(n\pi x /a) dx$$
So for a particle bound between absorbing walls, <E> is positive but <p> is zero.

Note that it is different for a particle bound between mirror walls, ie if we ask $$\psi'(a)=0$$ instead of $$\psi(a)=0$$. Further insights in papers by Seba and also by Carreau.

11. Mar 2, 2004

### Tom Mattson

Staff Emeritus
Actually, I meant the continuous range. Bad Tom. Bad, bad Tom.

12. Mar 3, 2004

### jeff

The solutions of schrodinger's equation for the 1D infinite potential well are given, as is well known to students having taken introductory courses in quantum mechanics, by the eigenstates

&psi;n(x) = (2/a)&frac12;sin(knx)

of the hamiltonian with energy eigenvalues &epsilon;n = kn&sup2;/2m in which the allowed momenta are kn = &pi;n/a for n = 1,2... In other words, the particle in one of these states &psi;n(x) has definite momentum kn and energy &epsilon;n. In particular, note that &psi;0(x) = 0, so vanishing momentum and energy are not even allowed. Please consult any quantum mechanics textbook.

Last edited: Mar 3, 2004
13. Mar 3, 2004

### turin

Not quite. The nth energy eigenstate in the position basis is

&psi;n(x) = (2/a)&frac12;sin(knx), 0 < x < a
&psi;n(x) = 0, otherwise.

This is actually an important distinction as it shows that the state is a bound state and not a free particle plane wave.

It would seem that way, but check again. Go back to that post where you have the Fourier transform. This is actually the idea that you must apply. From it, you will realize that the energy eigenstate in k-space is

|&phi;n(k)| ~ 1/(a2k2 - n2pi2)

Are you saying that even if the wavefunction vanishes everywhere in the position basis, it can still have non-trivial momentum?