Where is the particle most likely to be found?

1. Feb 26, 2013

Levi Tate

1. The problem statement, all variables and given/known data

Where is the particle most likely to be found in the first excited state (n=2) ? It is in a rigid box of length a

2. Relevant equations

wave function for a particle in a rigid box ψ(x)=(2/a)^1/2*sin(n[pi]x/a)

3. The attempt at a solution

The answer is a/4 and 3a/4 . I know I need to maximize the function by setting it equal to zero, but I haven't really done this since calc 1 and I kind of forgot.

If somebody could just guide me as to how to maximize this function,

Much appreciated.

2. Feb 26, 2013

haruspex

I think you want to maximise the probability function, which is real, by differentiation etc. The wave function is complex, so does not have a maximum in the same sense.

3. Feb 26, 2013

Levi Tate

The function I gave IS the normalized wavefunction for the particle in the rigid box mate. This is a 300 level intro to quantum mechanics class, it's not full out grad school type stuff yet. I just need to know how to maximize this function to get those values. It's an example in the book, Taylor 'Modern Physics' 2e. He just skips the step where he shows the maximization and I am having trouble getting those results. This comes down to a single variable calculus problem, which I am unfortunately struggling with.

4. Feb 26, 2013

tms

What can you say about any differentiable function at an extremum?

5. Feb 26, 2013

Levi Tate

Yeah it equals zero it's basic calc 1, i am having problems with the computation, not any of the ideas.

6. Feb 26, 2013

Dick

What did you get for the derivative?

7. Feb 26, 2013

tms

The function doesn't equal zero, its derivative does. (That may be what you meant, but it isn't what you said, and it's best to be absolutely clear.)

8. Feb 26, 2013

Levi Tate

It takes too long to write out the derivate and everything. This is a purely math issue I realized that I can resolve in a much shorter time with my teacher tomorrow.

Thanks a lot though mates, much appreciated. I'll save space and time to use when I actually have a question that isn't just a math issue, of which there will be several.

Thanks again though.

9. Feb 26, 2013

tms

Actually, you can just about get it by inspection. Where does the derivative of a sine equal zero?

10. Feb 26, 2013

Staff: Mentor

The wave function doesn't give you the probability directly. What do you have to do to it to get the probability distribution?

And you shouldn't have use the "set the derivative to zero" thing to find the maximum in this case. You should be able to figure it out by using the basic properties of trig functions.

11. Feb 26, 2013

haruspex

12. Feb 26, 2013

Levi Tate

Okay now I'm getting really confused about the physics.

Where I'm at in my studies here is that we're looking at a particle in a rigid box and the wavefunction for that, which was what I gave.

We are solving 2d and 3d problems currently but I'm trying to catch up in optics so I haven't looked at them yet.

Yeah you're exactly right, it's the maxima of the probability density that give the most probable value for the particle (x).

See in the example they just show a graph with two maxima at a/4 and 3a/4 for the probability density..

So what I was doing, sorry for the confusion, I was setting the square of the wavefunction equal to zero and messing around with it..

That's where it just turns into a math problem. Now about these trig functions, I'm not following you, if I plug in pi/2 for x, I get something like 2pi^2/a^3.

I don't know, it just shows a graph of the function from zero to a with two maxima and says that it is easy to see the maxima occur at these places. I just don't see it.

It's some really interesting stuff, we're just getting into the 3d Schrodinger (sp) wave equation which is probably the coolest thing i've ever seen.

13. Feb 26, 2013

tms

You want the derivative to be zero, not the function itself.
Don't plug in for $x$, but for the argument to the appropriate trig function, and then solve for $x$.

14. Feb 26, 2013

Staff: Mentor

$\pi/2$ looks like an angle (radians), not a position (meters or whatever).

15. Feb 26, 2013

haruspex

To be clear, the derivative of the square of the modulus of the wavefunction. That will be zero wherever either the modulus is zero or its derivative is. Then, of course, you have to figure out which are maxima and which minima.

16. Feb 26, 2013

Levi Tate

I just want to say that i am actually not as dumb as i sound here. I did my Grandmothers taxes all day. I am going to try what tms said here, note jtbell, affectionately call him jtballer, and solve this thing tomorrow.

I actually have a question I am really confused about, solving the time independent differential equation, there is this strange cavat that seems at odds with what i learned about linear homogeneous first order differential equations, and i think it might be at the energy cannot be zero or less, that there is a physical thing rather than just mathematics in the solution, but i've checked my ODEs text and it doesn't make sense to me. My teacher told me 'make sure the solution works'. I think he works on computers a lot and doesn't bother with trying to solve any differential equations.

17. Feb 26, 2013

Levi Tate

^oh I will post it soon, i think it's a good question. That wasn't the question statement up there, I just was casually talking about it.

18. Feb 26, 2013

Levi Tate

How can it be that a cosine function and a sine function will have the same maxima and minima here in this physics problem? I'm not disputing you, i just don't understand it.

Also, why do you have the respective title of science advisor? Do they give that title to experimental or theoretical physicists? I want that title some day. I like these forums very much. Someday I am going to come back and answer all the questions that I asked, you know, give back to the community.

19. Feb 27, 2013

Staff: Mentor

You've got the square of the wave function which is (ignoring the normalization constant) $\sin^2 (n\pi x / a)$. Think of it as $sin^2 \theta$ where $\theta = n\pi x / a$. What's the maximum possible value of $\sin^2 \theta$, and what values of $\theta$ give you that maximum? This is something you should be able to read off a graph of $\sin^2 \theta$.

"Science Advisor" is a title that the Mentors (moderators) here give to posters that we think give reliably good answers and advice. "Homework Helper" is similar, but it's for people who hang out mainly in the homework forums.

Last edited: Feb 27, 2013
20. Feb 27, 2013

Levi Tate

Cool thanks for the info mate.

See I just did this problem here, I didn't ignore the 2/a, I just divided it out. The problem says for the first excited state, so n=2, so I ended up with θ=Arcsin(2[pi]x/a), (n=2 for the first energy state)

I dunno I just am not getting this problem, because arcsin(sin^2x)≠x I don't think.

I am just going to ask my teacher this one. I don't think he solves ODEs and uses a computer, so I have a better question, it's pretty straightforward.

Thanks for the help.