# Particle In a finite potential well

1. Oct 24, 2009

### I_am_learning

For a particle in finite potential well we can have several bound states depending on the height of potential well.
Each bound state corresponds to definite energy En.
Then corresponding to Each definite Energy there should be definite Momentum Pn.
Since we have definite momentum---> According to uncertainty principle---We should have no idea about the position of the particle.
BUT, the wave functions for these states don't extend to infinity (thus making infinite uncertainty in position).Instead, They are localized,thus giving us the some idea about where the particle is most likely to be found.

What is the resolution of this apparent contradiction?

2. Oct 24, 2009

### kanato

The resolution is that they don't have definite momentum. What makes you think that there would be a definite momentum? It's only true for the free particle that definite energy means definite momentum (even there, p and -p are degenerate so definite energy still doesn't necessarily mean definite momentum). When you have a potential, then the commutator [p,H] is non-zero, so the momentum and Hamiltonian can't be simultaneously diagonalized, meaning that a state with well defined energy doesn't have well defined momentum.

3. Oct 24, 2009

### I_am_learning

Since they have definite Energy, This energy must be in the form of kinetic Energy. So, they have definite Kinetic Energy thereby having Definite momentum.

Oh! wait,Sorry, that Energy can also be in the form of potential Energy, in this case of finite potential well, since in this case, the wave function can penetrate the wall of potential well to some distance.So you are right that although the energy is definite, the momentum isn't.

But again,however, in the case of infinite potential well, the wave function can't penetrate the wall, so for this case, the apparent contradiction again comes into play, Since for this case, the Energy should always be in the form of Kinetic Energy therefore having Definite Momentum.
Whats the resolution for this case?

4. Oct 24, 2009

### kanato

There still isn't a definite momentum in the infinite square well. The wavefunction is not an eigenstate of the momentum operator:

$$\hat{p}\psi(x) = -i d/dx A \sin(kx) = -i \hbar k A \cos(kx) \neq \hbar k \psi(x)$$

A momentum state eigenstate is $$\psi(x) = e^{ikx}$$, but the eigenstates of the infinite square well are sin(x) and cos(x). These are a combination of complex exponentials, $$2 cos(x) = e^{ikx} + e^{-ikx}$$, so they do not have a definite momentum, instead they are a combination of a momentum state traveling left and one traveling right. The square well states will have <p> = 0, so $$\Delta p = \sqrt {\langle p^2 \rangle} > 0$$.

5. Oct 25, 2009

### Staff: Mentor

Kinetic energy is a scalar quantity. It has magnitude only. Momentum is a vector quantity. It has both magnitude and direction. Fixing the kinetic energy fixes only the magnitude of the momentum.

In a one-dimensional situation, the direction of the momentum is given by its sign, + or -.

6. Oct 25, 2009

### victorphy

Why?I don't understand that wave function can't penetrate the wall can resuilts in definite kinetic energy.

7. Oct 25, 2009

### I_am_learning

If the particle always remains in the 0 potential region (the particle does never penetrate the wall), then all of its energy should be in the kinetic Form. So, for this case, definite Energy Implies Definite Kinetic Energy.

8. Oct 25, 2009

### I_am_learning

So, you mean to say, Although we have definite magnitude of mementum, we don't have definite Derection. So, in overall, We are uncertain about the momentum by large factor. Thats why its Ok (for the uncertainity principle) to have some idea about the particles position.
Is this what you meant to say?

9. Oct 25, 2009

### sweet springs

Hello.

A stationary wave function in and outside of infinite well is wave　train, not a wave of infinite length. Momentum is given by Fourier transform and has continuous distribution. It is similar in finite well case.

http://www.falstad.com/qm1d/directions.html provides applet to show momentum distribution.

Let H=p^2/2m + V= -hbar^2 ∇^2/2m + V0[1 -θ（w+x）+θ（x－w）], then [p,H] = [p,V] = -iV0 hbar/ 2m （∇ (-θ（w+x）+θ（x－w）) -(-θ（w+x）+θ（x－w）) ∇）= -iV0 hbar/ 2m (-δ（w+x）+δ（x－w）)≠0 whether V0 is finite or infinite. That's why momentum cannot have a definite discrete value.

Last edited: Oct 25, 2009
10. Oct 25, 2009

### Staff: Mentor

Correct. If you calculate $\Delta x$ and $\Delta p$ for a "particle in a box" wave function, using the standard definitions in terms of expectation values of x and p:

$$\Delta x = \left< \sqrt{(x - <x>)^2} \right> = \left< x^2 \right> - <x>^2$$

$$\Delta p = \left< \sqrt{(p - <p>)^2} \right> = \left< p^2 \right> - <p>^2$$

you find that they satisfy the HUP:

$$\Delta x \Delta p \ge \frac{\hbar}{2}$$

11. Oct 25, 2009

### I_am_learning

Thanks for that confirmation. It really helped me a lot.
I am still to understand it quantitatively, though.