# Free Protons and Probability

1. Jun 27, 2008

### Domnu

At t = 0, 10^5 noninteracting protons are known to be on a line segment 10 cm long. It is equally probable to find any proton at any point on this segment. How many protons remain on the segment at t = 10 s?

Attempt at Solution
We can create a wavefunction for this scenario:

$$\psi(x, 0) = \frac{1}{10} \forall x \in (-5, 5)$$

Everywhere else, we can let $$\phi = 0$$. Now, we need to try to find $$|\psi(x, 10)|^2$$ where $$x$$ ranges from -5 to 5. To do this, we first construct $$|\psi(x, t)|^2$$. Since

$$\psi(x, t) = \exp \left(-\frac{i\hat{H}t}{\hbar}\right) \psi(x, 0)$$

we want to have $$\psi(x, 0)$$ in terms of the eigenstates of the Hamiltonian operator so that we can have

$$\psi(x, t) = \exp \left(-\frac{iE_n t}{\hbar}\right) \psi(x, 0) = \exp (-i \omega_n t)$$

However, since the eigenstates of the momentum operator are eigenstates of the Hamiltonian operator, we can find $$\psi(x, 0)$$ in terms of the eigenstates of the momentum operator. We will now do this.

Note that

$$\psi(x, 0) = \int_{-\infty}^{\infty} b(k) \phi_k dk$$

$$b(k) = \int_{-\infty}^{\infty} \psi(x, 0) \phi_k^* dx$$

imples that

$$b(k) = \int_{-5}^{5} \frac{1}{10} \cdot \frac{1}{\sqrt{2\pi}} \cdot e^{-ikx} dx = \frac{\sin 5k}{5k\sqrt{2\pi}}$$

Now, we have

$$\psi(x, 0) = \int_{-\infty}^{\infty} \frac{\sin 5k}{5k\sqrt{2\pi}} \phi_k dk$$

So,

$$\psi(x, t) = \int_{-\infty}^{\infty} e^{-i \omega t}\frac{\sin 5k}{5k\sqrt{2\pi}} \cdot \frac{1}{\sqrt{2\pi}}\cdot e^{ikx} dk$$

How do I proceed to integrate from here? It turns out to be extremely ugly...

2. Jun 28, 2008

### Domnu

Actually, I think I got the coefficients $$b(k)$$ incorrectly. Would the bounds actually go from - to + infinity? The protons are actually free, right? In that case, we'd have

$$b(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-ikx} dx = \frac{\sqrt{2\pi}}{10} \delta (k)$$

Is this correct?

3. Jul 3, 2008

### Gokul43201

Staff Emeritus
No, $\psi(x,0)=0$, outside (-5,5).

4. Jul 3, 2008

### Domnu

If that were the case, then wouldn't the protons always be in the region? Or would the wavefunction eventually evolve so that it would spread out?