- #1
Domnu
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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...
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...
