49 a particle is trapped inside an infinite one dimensional...

4.9 A particle is trapped inside an infinite one-dimensional square well of width a in the first excited state (n = 2). (a) You make a measurement to locate the particle. At what positions are you most likely to find the particle? At what positions are you least likely to find it? (b) Calculate $\langle p^2 \rangle$ for this particle.

Transcript

00:01 Okay, so in this problem, we have a particle confined in a box, and we know this particle is in the first excited state.

00:11 And we need to find the average position and the most likely position that this particle can be found.

00:22 So let's begin with the wave function of this particle.

00:28 So we know that since this particle is in the first.

00:32 Excited state the wave function of this particle is going to be equal 2 divided by l the square root of 2 divided by l the sign of 2 pi xl therefore if you want to find in part a the average position what we want is just the expectation value of the position and this is described by the integral from 0 to l, which is the length of the box, of the psi to star, which is the complex conjugated function of x that multiplies the position that multiplies the psi to x d x.

01:32 Okay, so if we substitute the wave function in this integral, what we have this, let's put in here, we have the integral is going to be equal 2 divided by l of the integral from 0 to l of x, the sign square of 2 pi x divided by l, dx.

02:05 Therefore, since we have in here a sine square, we can use a trigonometric, which states that the sine square of x is going to be equal 1 minus the cosine of 2x divided by 2 therefore this integral as simplify to just 2 divided by l of the integral from 0 to l of x divided by 2 minus x cosine of 2 pi sorry not 2 4 pi x divided by l d x okay so as we can see we have two terms in here so let's solve let's separate this as the first term and this one as the second term.

03:21 So let's solve the first integral.

03:23 The first integral here is a power integral, so it's pretty straightforward.

03:31 We have the integral from 0 to l of x divided by 2 d x.

03:38 This is just x squared divided by 4 from 0 to l, which gives us l 2 divided by 4.

03:51 So that's the first solution.

03:55 Now let's look to the second integral.

03:57 The second integral is put in blue.

04:02 The second integral is integration by parts...