# Particle in a box (1-D)

1. Mar 26, 2005

### Dathascome

Hi there, I'm having a bit of trouble with this problem. The book tells me that I have a particle in the gorund state of a box of length L. Then suddenly the box expands to twice it's size (symmetrically), leaving the wave function undisturbed. I supposed to show that the probability of finding the particle inm the ground state in the new box is (8/3pi)^2.
I'm not quite shure what to do. I know that the wave function is
S(x)= (2/L)^2 sin(n*pi*x/L) for n=even
=(2/L)^2 cos(n*pi*x/L) for n=odd
The first thing I thought was that the probabilty is just(sorry I'm new to the forum and don't know how to write integral sign here)
P=int from 0-2L of S*(x)S(x) dx
and I was just getting 1 but then realized, of course I'm just going to get on, I'm just finding the probability that the particle will be somewhere in the box which sure as hell should be 1 right?
So I'm not quite sure how to find the probabilty of finding something in the ground state (or any state for that matter).
Perhaps you could just steer me in the right direction without actually giving me the answer? Any help would be greatly appreciated.

2. Mar 26, 2005

### Berislav

Try solving the SE equation, it shouldn't be too mathematically demanding (1-D, no time). Then renormalize with the new lenght.

This might help:

$$P_1= \int_{0}^{2L} {\psi_1}^2 dx$$

Last edited: Mar 26, 2005
3. Mar 26, 2005

### Dathascome

Forgive my ingnorance, but I'm still a bit confused. If I do that and take (n=1 because it's the ground state)
int from 0-2L of (2/L)sin^2(pi*x/L)dx= int from 0-2L of 1/L(1-cos(2*pi*x/L)dx
=2
right?
Is that what I'm supposed to renormalize? I think I'm confused about the renormalization (if I did the first part right that is).
I'm also confused about the meaning of the equation I think. Didn't I solve the SE by finding that S(x)=(2/L)^1/2 sin(nx*pi/L) (forgetting the cosine for now)? Are you saying that instead of normalizing to get the (2/L)^1/2 I should use the fact that I'm in this bigger box and renormalize to that? And then using that renormalized S(x) find P(x)?
I think I'm also confused about the probability. Doesn't finding
P(x)= int of S*(x)S(X)dx
just give me the probability of finding the particle at some place in the box defined by my limits of integration? How does that correspond to them asking me to find the probabliity of finding the particle in the gorund state in the new box? What's confusing me perhaps is that they aren't asking for the prob in some interval but just the prob that the particle be in the ground state...I'm guessing they mean anywhere in the box then as long as it's in the ground state???
Sorry for all the confusion, again any and all help will be greatly appreciated.
Oh yeah, also, Do you kno how I canuse an integral sign or sqrt or any of those things and make them look proper as opposed to just typing them in the way I've been? Thanks again

Last edited: Mar 26, 2005
4. Mar 26, 2005

### Tom Mattson

Staff Emeritus
Yes, that's right. Your ground state will be different in this larger box.

No, that's the probability density for a particle in state S. If you are in the initial state $\psi_i(x)=\sqrt{\frac{2}{L_i}}sin(\frac{\pi x}{L})$ when the box suddenly expands, the ground state changes to $\psi_f(x)=\sqrt{\frac{2}{L_f}}sin(\frac{\pi x}{L})$, The probability of finding the particle in that state is the overlap integral of the two states:

$$P=\int_{all space}\psi_f^*(x)\psi_i(x)dx$$

5. Mar 26, 2005

### Tom Mattson

Staff Emeritus
This site is LaTeX-enabled. A tutorial can be found here:

https://www.physicsforums.com/showthread.php?t=8997

You can also click on any LaTeX images you see to get a pop up that contains the code used to generate it.

6. Mar 27, 2005

### Dathascome

I'm still doing something wrong, and am not getting (8/3*pi) as the probability of finding the particle in the groundstate in the new expanded box. First off I renormalized my wave function in the expanded box and came out with $\psi_f(x)=\sqrt{\frac{1}{L}}sin(\frac{\pi x}{L})$ . But then doing this integral $$P=\int_{all space}\psi_f^*(x)\psi_i(x)dx$$ gives me $$P=\int_{all space}\sqrt{\frac{1}{L}}\sqrt{\frac{2}{L}}sin^2(\frac{\pi x}{L})dx$$
(not really over all space though, just from 0-2L because outside of this my wave function is 0)
and I know that the sin^2(pi*x/L) part goes to 1/2(1-cos2x)
and I get nothing that even looks like (8/3*pi)
Is there something really stupid that I'm doing wrong and missing here??? This is making me crazy
There's probably something really silly that I'm missing? Any suggestions?

7. Mar 27, 2005

### Berislav

Are you sure $\psi_f$ is correctly renormalized?

8. Mar 28, 2005

### Dathascome

I'm not sure if you're asking me or telling me. Eiter way, I'm pretty sure that it is correctly renormalized. I did it explicitly but I think it also makes sense that the renormalization constant would be $$\sqrt{\frac{1}{L}}$$ Seeing as the normalization constant for the not-expanded box was$$\sqrt{\frac{2}{L}}$$. Can anyone see what I'm doing wrong?

9. Mar 28, 2005

### Davorak

$$A_n = \int_{all space} u_{n}^{*}\psi(x)dx$$
$$P(n) = |A_n|^2$$
Where $u_n$ is the state you want to find the probability of tranistion to, in you case it is $u_1$ the ground state, $\psi(x)$ is the wave function before expansion.

From the fact that you have been using sine functions as the wave function it seems like you are not placing you box symmetrically on the axis.
$-L\ {to}\ L$ expands to $-2L\ {to}\ 2L$.
Since the problem requires that the box expand symmetrically it is best to place the box symmetrically on the axis.

Last edited: Mar 28, 2005
10. Mar 29, 2005

### Dathascome

You're right, but isn't that sort of a pendantic point in this case? Whether it's a sin or cos (in this specific example) won't change anything right? When I integrate the sin^2 or cos^2 will go to 1/2(1-+cos(2x)) (+or - for cos or sin)
but when integrating which ever on I chose would drop out. So I'm still doing something wrong and not getting anything that even looks like (8/3*pi)^2,
I don't even see where I would get a (1/Pi)^2 from. It's gotta be something stupid, what am I missing

Also the box is of size L so doing it symmetrically would be from -L/2 to L/2, and then in the expanded from -L to L.

Concerning this
I don't understand this. Why do I take the overlap of the two states, and how do I know which to take the complex conjugate of?

11. Mar 29, 2005

### Davorak

Yes you can use sin or cosine it is just easier to treat symmetric problems symmetrically. If the original well is from 0 to 2L(same length as my last example) the expanded well will then be from –L to 3L.

Or if we cut the length of the well in half:
0 to L expands to –L/2 to 3/2 L

If you express the wave function for the ground state of the expand well with a sine term then you will have to include a phase factor(translation on x axis.)

This will lead to a messy integral which can be avoid by putting the well symmetrical on the axis.

I pointed it out because in your integral is wrong:
$$P=\int_{all space}\sqrt{\frac{1}{L}}\sqrt{\frac{2}{L}}sin^2(\frac{\pi x}{L})dx$$
You gave the sine wave from each wave function(both initial and final) the same wave length $2 {L}$. The wave function for the expanded ground state will have a larger wavelength.

I think when Tom posted
I think he accidentally left off the subscripts i and f in side the sine function for L.

I do not know now if Tom Mattson meant P in his equation was supposed to be probability. If it was it is wrong.
quote was edited to included tex formatting. From:
http://www.astro.cf.ac.uk/undergrad/module/PX4114/aqm/node8.html [Broken]
also Gasiorowicz.

The quantum levels for any quantum system are considered to form an orthonormal basis for a vector space. Much like $\hat{i}\ \hat{j}\ \hat{k}$ form a basis for a three dimensional space.

Before expansion the system in is in the ground state and can be described as a single vector in fact it is a member the quantum system’s basis.

The expansion of the well also changes the systems basis. The wave function does not immediately change, but it also can not be represented by a single basis vector like before either.

Instead it is represented by a linear combination of the new basis vectors:
$$A_1|1>+A_2 |2>+A_3 |3>...$$
This is a superposition of quantum states to produce the wave function that before expansion could be explained by one quantum state.

If you take a measurement you can not realistically find the particle in many quantum states and therefore many different energies at the same time. So when you measure the system it takes on the properties of only one state. This is sometimes described as the wave function collapsing.

$$P(n) = |A_n|^2$$
The above equation describes the probability of finding the wave function in a particular state.

Edit:

Correct me if I am wrong but since you will eventually take the absolute value of A_n then any phase fact introduced by the complex conjugate should be eliminated therefore you should be able to place the complex conjugate on the final rather then initial.

Last edited by a moderator: May 2, 2017
12. Mar 29, 2005

### Dathascome

You're confusing me a bit, when you mention the thing that Tom Mattson said about not having the i and f on the L's inside the sine, I think you are wrong about this. One of the things the problem says is that the wave function remains undisturbed when the box is expanded. So shouldn't it just be L's without any subscript (or maybe just an i I guess to indicate that it is the initial length)?
Also doesn't what I just mentioned negate this part of what you say
Because the wave function is undisturbed I thought the only thin that would change in the expanded case would be the normalization constant no?

Also thanks for explaining the thing about the superposition of states, that makes sense now

Last edited: Mar 29, 2005
13. Mar 29, 2005

### Tom Mattson

Staff Emeritus
OK, it was late when I posted and I made a couple of mistakes. Let me fix them now.

I am assuming that the potential is zero on $-L/2 \leq x \leq L/2$ and infinite elsewhere. That means that the ground state before expansion is:

$$\psi_{0i}(x)=\sqrt{\frac{2}{L}}cos(\frac{\pi x}{2L})$$

After the expansion the ground state of the particle is

$$\psi_{0f}(x)=\sqrt{\frac{1}{L}}cos(\frac{\pi x}{L})$$

The amplitude (not the probability) for the transition from $\psi_{0i}$ to $\psi_{0f}$ is:

$$A=\int_{allspace}\psi_{0f}^*(x)\psi_{0i}(x)dx$$

The probability is $P=|A|^2$.

You take the overlap of the two states because you are projecting one onto the other. Remember that this overlap integral is the inner product in our function space, much like the dot product is the inner product in $\mathbb{R}^n$. You're really finding the projection of one vector onto another in an abstract vector space.

It doesn't matter, because you're taking the norm of the amplitude to get the probability. That was the part that I had left out last time.

Last edited: Mar 29, 2005
14. Mar 29, 2005

### Davorak

If you look at what Tom wrote again you will see that Tom used subscripts in the Ls in the normalization factor but not in sine function.(i = initial, f= final) It makes no sense if it applies to the normalization factor and not the sine function.

The final wave function is the ground state of the expand well will not have the same L as the wave function for the initial well.

If a well expands as 0 to L -> 0 to 2L then The ground state wave function is not the same. In fact the ground state(n=1) of the initial well is the n=2 state of the expand well. Plot or draw it out you will see.

The final wave function is not the wave function right after you expand the well, rather it is the wave function after you measure the system. Measuring the system will collapse the wave function to one of the quantum states of the new well.

The wave function is the same immediately before and after expansion(undisturbed like you said). However the basis does change.(initial quantum well basis and the expanded quantum well basis) The basis of both systems are similar because they are both quantum wells but that does not mean they are the same.

Edit:
Nice post Tom. Hope you are able to get more sleep in the future. I know I do :grumpy: .

Never mind Tom posted his correction. Looks good to me
Don’t forget though that in Tom's Example the initial wave function is only in -L/2 to L/2 and zero everywhere else. So the integral has the bounds of -L/2 to L/2.

Last edited: Mar 29, 2005
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