Where is the particle most likely to be found?

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In summary, the conversation discusses a problem in quantum mechanics involving a particle in a rigid box and the corresponding wave function. The main question is how to maximize the probability density function to determine the most probable position of the particle. The conversation also touches on the use of trigonometric functions and solving differential equations. The speaker expresses confusion and seeks clarification on certain concepts.
  • #1
Levi Tate
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Homework Statement



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


Homework Equations



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

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.
 
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  • #2
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
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
What can you say about any differentiable function at an extremum?
 
  • #5
Yeah it equals zero it's basic calc 1, i am having problems with the computation, not any of the ideas.
 
  • #6
Levi Tate said:
Yeah it equals zero it's basic calc 1, i am having problems with the computation, not any of the ideas.

What did you get for the derivative?
 
  • #7
Levi Tate said:
Yeah it equals zero it's basic calc 1, i am having problems with the computation, not any of the ideas.
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
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
Actually, you can just about get it by inspection. Where does the derivative of a sine equal zero?
 
  • #10
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.
 
  • #12
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
Levi Tate said:
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.
You want the derivative to be zero, not the function itself.
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.
Don't plug in for [itex]x[/itex], but for the argument to the appropriate trig function, and then solve for [itex]x[/itex].
 
  • #14
Levi Tate said:
Now about these trig functions, I'm not following you, if I plug in pi/2 for x,

##\pi/2## looks like an angle (radians), not a position (meters or whatever).
 
  • #15
tms said:
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.
You want the derivative to be zero, not the function itself.
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
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
^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
haruspex said:
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.

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
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:
  • #20
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 don't know 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.
 
  • #21
And definitely, the graph is right next to the problem showing the maxima. Another thing is I just don't get that if a is a constant where the infinite square well ends how I know that, ah! I am letting this one go. It's not even a question that my teacher asked or anything, either is the other one really. Which is irrelevant, but this isn't my homework, this is just me wondering about physics, or really in this case math.
 
  • #22
I think you're going at it sort of backwards. Try the approach I gave in my previous post, which is actually the first of two steps.

First step: figure out which values of ##\theta## give you the maximum value of ##\sin^2 \theta##. This is where a graph of ##\sin^2 \theta## versus ##\theta## (not ##\psi## versus x!) comes in handy.

Second step: use ##\theta = n \pi x / a## to find the corresponding values of x.
 
  • #23
Levi Tate said:
How can it be that a cosine function and a sine function will have the same maxima and minima here in this physics problem?
That isn't what I said. In hindsight, my post may have been confusing because I was taking an unnecessarily general approach. Suppose f = f(x), and you want the extrema of f2(x). Standard calculus says to find extrema of a function you look for where the derivative of the function is zero, so that will be d f2(x) /dx = 0. But of course that reduces to f(x) f'(x) = 0, which will be true wherever either f(x) = 0 or f'(x) = 0. And since f2(x) ≥ 0, you know straight away that f(x) = 0 will give the minima (and only the minima), while f'(x) = 0 will give all the maxima (but maybe repeats some of the minima).
As jtbell points out, you don't actually need any calculus in the present case. You just need to know what arguments maximise the sine function.

Wrt "Science Advisor", please remember it only indicates a general reliability, not infallibility. I don't think anyone has made Science Pope yet.
 
  • #24
Your user name, everytime i read it, i read 'harpsex'. I'm like, he has a bit of a funny name, oh.. ! Then i realize i don't know how to read.

We actually took the Schrodinger (time independent) to three dimensions today and while it was extremely beautiful, I do not know what we were doing. I'm trying to catch up with Optics but hopefully you fine fellows might be able to provide some assistance, to be honest, in 3d at some point I didn't even know what we were looking for, or doing, or trying to solve anymore.
 
  • #25
haruspex said:
I don't think anyone has made Science Pope yet.
We'll know when that happens because white smoke will pour from our computers.
 
  • #26
They should just have the retiring Pope be science Pope, it will be easier because he's already been a Pope.
 

1. Where is the particle most likely to be found in an atom?

The particle is most likely to be found in the electron cloud, which surrounds the nucleus of an atom.

2. Can the precise location of a particle be determined?

No, according to the Heisenberg uncertainty principle, it is impossible to know both the exact position and momentum of a particle at the same time.

3. How does the location of a particle affect its behavior?

The location of a particle can affect its behavior, as it can determine the strength of its interactions with other particles and the probability of it being affected by external forces.

4. Is it possible for a particle to exist in multiple locations at once?

According to quantum mechanics, particles can exist in a state of superposition, meaning they can be in multiple locations simultaneously until observed or measured.

5. Can a particle be found outside of its expected location?

Yes, particles can exhibit wave-like behavior and have a probability of being found in unexpected locations, known as quantum tunneling.

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