A particle in the three-dimensional cubical box of Section 41.2 is in the ground state, where nX

A particle in the three-dimensional cubical box of Section...

Key Concepts

Integration Over Spatial Regions

Calculating the probability of finding a particle within a specific region involves integrating the probability density over that region. In cases where the wavefunction is separable, the integration in one spatial dimension can be performed independently, providing insights into how likely the particle is to be found within particular intervals along that axis.

Separation of Variables

Because the problem in a cubical box is separable, the overall three-dimensional wavefunction can be written as a product of three one-dimensional wavefunctions, each depending on a single spatial coordinate (x, y, or z). This technique simplifies the analysis of the system and the computation of probabilities over subregions.

Normalization

Normalization is the requirement that the total probability of finding the particle somewhere in the entire space must equal one. This is ensured by appropriately choosing the constant factors in the wavefunction, and it is fundamental for interpreting the squared wavefunction as a probability density.

Particle in a Cubical Box

This concept describes a quantum system where the particle is confined within a three?dimensional box with impenetrable (infinite potential) walls. The time-independent Schrödinger equation is solved in this domain, yielding quantized energy levels and discrete stationary states corresponding to standing wave solutions that satisfy the boundary conditions.

Probability Density

In quantum mechanics, the probability density is given by the square of the absolute value of the wavefunction. It quantifies the likelihood of finding the particle at a particular point in space; by integrating this density over a specific volume, one obtains the probability of locating the particle within that region.

Transcript

00:01 Okay, so in this problem we have a particle inside a three -dimensional box, cubicle box.

00:08 We know that a particle is in the ground state, which means nx, n y, n z equals 1, and therefore because of that, we also know the solution for the wave function of this particle.

00:30 Okay, so the solution for the wave function is basically, let me see, we have the sign, pi xl, sine, pi y, l, sign, pi, y, l, sign, pi zl.

00:58 All this with, okay, yeah, that's the solution.

01:05 All the multiplication of the three signs in here.

01:11 This is the wave function and the probability is what we must calculate here.

01:19 So in first item of this problem, we have to calculate the probability of finding the particle in a specific interval.

01:28 So how we calculate the probability.

01:30 So the probability p is described by the integral from minus infinity to infinity of.

01:38 The absolute value square of the wave function integrated in the axis.

01:49 Okay, so therefore we can say that the probability in first item, let's put here item a.

01:59 This is the first part of the problem.

02:02 It's going to be 2 divided by l cubic.

02:08 So this is the normalization constant.

02:10 That multiplies the integral from 0 to l half, l divided by 2, sine square pi x, l, this is an x, d x, times the integral from 0 to l, sine square pi y, l, d y, and the y, d y, and the integral from 0 to l of sine square phi z l d z so this is the integral we must do as we can see all three integrals are the same the only thing that changes is the interval the interval in x goes from zero to half of l and the integral of y and z goes from zero to l the entire size of the box so let's begin with the standard solution without specific in the intervals so we have psi square of pi xl d x and the integral here okay so this solution the standard solution is going to be first of all we use trigonometric relation and change the sign for 1 minus cosine of 2 pi xl divided by 2 d x and the solution for this integral here is going to be x divided by 2 minus sine 2 pi xl that multiplies l divided by 4 pi so this is the standard solution and we just need to put here the two intervals 0 to l and 0 to l divided by 2 in this way we can calculate the two solutions so without specifying yet the interval let's put here.

04:51 The first interval goes from 0 to y, which is also equals equals z.

05:04 That goes from l.

05:06 Okay, so this solution is going to be just l divided by 2.

05:12 And the second integral, we go from 0 to x to l divided by 2.

05:25 And this one is has a solution of l divided by 4.

05:32 Okay, knowing this, we just need to multiply all the solutions, and we have the probability.

05:38 So the probability is going to be 2 divided by l, cubic that multiplies l to the 4, and l divided by 2 square.

05:52 Multiplying this, we find out the probability is 0 .5...