Particle in a Box: Max Probability at x=xj

[shamone]

Homework Statement

Show that the probability of finding a particle p=[tex]\Psi^2[/tex] is at max at x=xj
where xj=(2j+1/2n)a where j=0,1,2,3...n-1 and a is width of well

Homework Equations

p(x) = (2/a)sin^2((n*pi*x)/a)

The Attempt at a Solution

Im so confused I don't know where to start, please help

 

[sgd37]

well how do you find the maxima of any function

 

[shamone]

But getting the maxima wouldn't figure out this question as it asking about it in general and not really a specific point and anyway I have no numbers to work with

 

[diazona]

But getting the maxima wouldn't figure out this question as it asking about it in general and not really a specific point and anyway I have no numbers to work with

Yes it will. The question asks about the points at which the probability of finding the particle is a maximum. You have a function which gives you the probability. Now, do you know how to find the points at which that function has its maxima?

 

[shamone]

So you are saying I should differentiate the function? but what would I do with the constants like a and n like what values would i plug in and how does that relate to xj?

 

[diazona]

You don't need to do anything with the constants. They're constant, so when it comes to taking a derivative, they just act like ordinary numbers.

Go ahead and differentiate the function and see what you get. Then what's the next step you need to take to find the maxima?

 

[shamone]

Its been a while since iv done proper differentiating so not sure if this is right
(4*n*pi/a^2)sin(n*pi/a)x*cos(n*pi/a)x

To ensure its a maxima wouldn't i have to dfferentiate again but i don't know what values to plug in??

 

[diazona]

That's almost right. Somehow your x's wound up outside the arguments to the sine and cosine functions, but they should be inside [e.g. it should be sin(n*pi*x/a)].

You're getting a little ahead of yourself with the second derivative. First you need to find the points at which the function has a local extremum. How would you do that?

After you do that, you will need to check and see which of those points are maxima and which are minima. Taking the derivative again is (one step of) one way you can do that, but it's not the only way.

 

[shamone]

I guess I would put that equation equal to zero and solve for x to find the local max and local min? But I have so many unknowns and its such a complex equation I am not sure if I could do this.
If I did have these values the usual way to find the max and min is to differentiate again then plug the values in and whichever is negative is the max.

If I did manage to do all this I don't see how it would still solve the initial problem of showing the probability of finding the particle is max at xj??

 

[diazona]

I guess I would put that equation equal to zero and solve for x to find the local max and local min? But I have so many unknowns and its such a complex equation I am not sure if I could do this.

Yes, you can do it. Just try it ;)

If I did have these values the usual way to find the max and min is to differentiate again then plug the values in and whichever is negative is the max.

First find the extrema (i.e. set the equation equal to zero and solve), then worry about figuring out which is which.

If I did manage to do all this I don't see how it would still solve the initial problem of showing the probability of finding the particle is max at xj??

You will be finding a general expression (a formula) for the positions at which the probability is a maximum. Once you do that, you should be able to compare it to the formula given for xj and see that they match.

 

[shamone]

perfect I know what to do now thanks!