Question

For a proton in the ground state of a 1-dimensional infinite square well, what is the probability of finding the proton in the central 2% of the well?

Answer 1

Answer:

The probability of finding the proton at the central 2% of the well is almost exactly 4%

Explanation:

If we solve Schrödinger's equation for the infinite square well, we find that its eigenfunctions are sinusoidal functions, in particular, the ground state is a sinusoidal function for which only half a cycle fits inside the well.

let $L$ be the well's length, the boundary conditions for the wavefunction are:

$\psi(0) = \psi(L) =0$

And Schrödinger's equation is:  

$- \frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} = E\psi$

The solution to this equation are sines and cosines, but the boundary conditions only allow for sine waves. As we pointed out, the ground state is the sine wave with the largest wavelength possible (that is, with the smallest energy).

$\psi_0(x)=\sqrt[]{\frac{2}{L} }\, \sin(\frac{\pi x}{L} )$

here the leading constant is just there to normalise the wavefunction.

Now, if we know the wavefunction, we can know what the probability density function is, it is:

$f_X(x) = |\psi|^2$

So in our case:

$f_X(x) = \frac{2}{L} \sin^2(\frac{\pi x}{L})$

And to find the probability of finding the particle in a strip at the centre of the well of width 2% of L we only have to integrate:

$P(X \in [0.49 L, 0.51L ])= \int\limits^{0.49L}_{0.51L } {\frac{2}{L} \sin^2(\frac{\pi x}{L})} \, dx$

If we do a substitution:

$x = u \, L$

We get the integral:

$\int\limits^{0.49}_{0.51 } 2\, \sin^2(\pi u)} \, du$

This integral can be computed analytically, and it's numerical value is .0399868, that is, almost a 4% probability.