(adsbygoogle = window.adsbygoogle || []).push({}); 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:

[tex]\psi(x, 0) = \frac{1}{10} \forall x \in (-5, 5)[/tex]

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

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

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

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

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

Note that

[tex]\psi(x, 0) = \int_{-\infty}^{\infty} b(k) \phi_k dk[/tex]

[tex]b(k) = \int_{-\infty}^{\infty} \psi(x, 0) \phi_k^* dx[/tex]

imples that

[tex] 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}}[/tex]

Now, we have

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

So,

[tex]\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[/tex]

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

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# Homework Help: Free Protons and Probability

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