A particle is described by the normalized wave function ψ(x, y, z) = Axe^-^a^x^2e^-^β^y^2e^-^y^z

A particle is described by the normalized wave function ψ(x,...

A particle is described by the normalized wave function $\psi$$(x, y, z)$ = $Axe$${^-}{^a}{^x}^2$$e$${^-}{^\beta}$${^y}^2$$e$${^-}{^y}^z$$^2$, where $A$, $\alpha$,$\beta$, and $\gamma$ are all real, positive constants. The probability that the particle will be found in the infinitesimal volume $dx$ $dy$ $dz$ centered at the point $(x_0$, $y_0$, $z_0$) is $\mid$$\psi$$(x_0$, $y_0$, $z_0$)$\mid$$^2$ $dx$ $dy$ $dz$. (a) At what value of $x_0$ is the particle most likely to be found? (b) Are there values of $x_0$ for which the probability of the particle being found is zero? If so,at what $x_0$$?$

Transcript

00:01 In this exercise you have a wave function, psi of x, y, and z that is given in a cartesian coordinate as a times x, times z to the minus alpha x squared, times e to the minus beta, y squared, times e to the minus gamma z squared, where a, alpha, beta, and gamma are positive and real.

00:30 And in question a, our goal is to find what is the x position, that we're going to call x0, where it's most probable to find the particle.

00:45 So notice that the probability density, p of x, y, and z, is equal to the square of the magnitude of the wave function.

01:01 And the square of the wave function is a squared, x squared, e to the minus 2, alpha, x squared, e to the minus beta, 2 beta, i'm sorry, y squared, e to the minus 2, gamma z squared.

01:21 Notice that this is equal to x squared, e to the minus 2, gamma, x squared times the function, f of y, and and z.

01:33 Okay? and since we want to find the most probable value of x, then we want the probability to be maximized.

01:43 So in order to find the maximum, i'm going to differentiate the probability with respect to x and search for the point where the derivative of p with respect to x is equal to 0.

01:58 Notice that the pdx is equal to according to the multiplication rule, 2x times z to the minus 2 alpha x squared f of y z plus x squared times e to the minus 2 alpha x squared times the derivative of what's inside the exponential by the chain rule.

02:27 So we have minus 4 alpha x times f of y z.

02:35 And we want this to be zero.

02:38 Okay, so i'm just going to factor out to, i'm going to factor out to x, e to the minus 2 alpha x squared, f of y z times 1 minus 2 alpha x squared...

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