# I Probability density and kinetic

1. Mar 14, 2016

### Incand

energy.
Consider a particle in a box of the form
$V(x)= \begin{cases}0 \; \; \; -L < x < 0\\ V_0 \; \; \; 0<x<L\\ \infty \; \; \; \text{ elsewhere}.\end{cases}$
One can show that the probability density
$P(x) = \Psi^* \Psi$ is greater in the region of lower kinetic energy (that is at higher potential energy).

What's the physical explanation for this? My notes say something along the lines of "lower velocity give higher probability density" which seems very vague.
I'm guessing this is also related to how the Boltzmann distribution, where the probability decreases with the energy.

Last edited: Mar 14, 2016
2. Mar 14, 2016

### vanhees71

Is this a homework problem? Then post it in the homework forum. Please post a complete problem description. Otherwise we have to guess, what the problem is!

Otherwise, I guess they ask to find the energy eigenvalues and eigenfunctions and calculate the position-probability density distribution for these eigenstates. That's done by solving the time-independent Schrödinger equation.

3. Mar 14, 2016

### Staff: Mentor

A lower kinetic energy means a lower speed, and a slower-moving particle takes longer to leave a given region... So intuitively you would expect the particle to spend more time in the regions of low kinetic energy just because once there it won't be in any hurry to leave.

It's worth taking a moment to calculate the classical trajectory of a classical particle bouncing back and forth between the walls of the box. In one cycle, how much time does the particle spend in the region between -L and 0, and how much time does the particle spend in the region between 0 and L? The ratio between the two determines the probability of finding the particle In either region at any randomly selected moment.

Last edited: Mar 14, 2016
4. Mar 14, 2016

### Incand

It's not a homework problem, I just gave an example to describe what I mean. I'm able to show the above result (that would have been a homework problem!), what I was wondering was the physical motivation behind it.

That makes sense, thanks! And the example nicely illustrates this as well:
If we assume the speed in the first half is $v$ and in the second half $c_1 v$ with $c_1 > 1$ then a bounce from the "midpoint to back to the midpoint" takes $T_1 =2L/v$ and $T_2 = 2L/(c_1v)$ respectively.
The probability of finding the particle in each place is then
$P_1 =\frac{T_1}{T_1+T_2}= \frac{1}{1+1/c_1}$and
$P_ 2= \frac{T_2}{T_1+T_2} = \frac{1}{1+c_1}$
so for $c_1 >1$ we have $P_2 < P_1$ so the particle clearly spends more time in region $1$.

Edit: fixed a mistake, think it's correct now

Last edited: Mar 14, 2016