# Particle in a box that suddenly expands

• LCSphysicist
In summary: Yes, but the system is in the ground state of the original well. Of length ##L##. That is the state that needs to be expressed in the new basis.It's only because the problem is very simple that you get the same numerical answer the wrong way. Imagine the system was originally in a superposition of energy eigenstates. Let's say:$$\sqrt{\frac{1}{3}}\psi_1(x) + \sqrt{\frac 2 3}\psi_2(x)$$How would you approach the problem... or would you still do it the wrong way?I would use the same procedure as you have mentioned. But then,
LCSphysicist
Homework Statement
.
Relevant Equations
.
A particle is in the ground state of a box of length L. Suddenly the box
expands (symmetrically) to twice its size, leaving the wave function undisturbed. Show that
the probability of finding the particle in the ground state of the new box is ##(8/З\pi)^2.##
My approach was: $$\psi_{2g} = \sqrt{1/L}sin(\frac{n \pi x}{2L}) = \sum c_n \sqrt {\frac{2}{L}} sin(\frac{n \pi x}{L})$$

Summarizing what i have done after that (Fourier series and sum of infinite series), we get the result realizing that, what we want matematically is ##P## $$P = \frac{(c_1)^2}{\sum_{n=1}^{n=\infty} (c_n)^2}$$

The main thing that bothers me is the necessity to calculate the infinite series. It is possible, indeed, even if we use wolframalpha https://www.wolframalpha.com/input?i=\sum+of+(n^2/(1-4n^2)^2)+from+1+to+infinity
BUT, i could not use it on a test, neither would have time to do that. So even so i have got the right answer, i want to know if there is a better approach to this problem.

Thank you

Delta2
You have a particle in the original ground state ##\psi_1##. The system is changed so that the eigenfunctions are ##\varphi_n##, say. You know that you can express ##\psi_1## in terms of these new eigenfunctions:
$$\psi_1(x) = \sum_{n = 1}^{\infty}c_n\varphi_n(x)$$You can then use the usual technique to extract the coefficient ##c_1##, using the orthonormality of the set ##\varphi_n##:
$$\langle \varphi_1(x)|\psi_1(x) \rangle = \sum_{n = 1}^{\infty} \langle \varphi_1(x)|c_n\varphi_n(x)\rangle = c_1$$And, you should find that ##|c_1|^2## is the probability you are looking for.

LCSphysicist
PeroK said:
You have a particle in the original ground state ##\psi_1##. The system is changed so that the eigenfunctions are ##\varphi_n##, say. You know that you can express ##\psi_1## in terms of these new eigenfunctions:
$$\psi_1(x) = \sum_{n = 1}^{\infty}c_n\varphi_n(x)$$You can then use the usual technique to extract the coefficient ##c_1##, using the orthonormality of the set ##\varphi_n##:
$$\langle \varphi_1(x)|\psi_1(x) \rangle = \sum_{n = 1}^{\infty} \langle \varphi_1(x)|c_n\varphi_n(x)\rangle = c_1$$And, you should find that ##|c_1|^2## is the probability you are looking for.
##|c1|^2## is not exactly the probability i suppose, because if you do the sum of the square of the coefficients (as i have done, and even posted the link above), it is different than 1, and so is not normalized.
As i said, this is the main problem, and the necessity to evaluate such a sum makes me wonder if there is not another approach

Expanding on that: Assuming that the original state is properly normalized, you would have
$$\sum_n |c_n|^2 =1$$
by default. No need to compute it.

LCSphysicist
Orodruin said:
Expanding on that: Assuming that the original state is properly normalized, you would have
$$\sum_n |c_n|^2 =1$$
by default. No need to compute it.
Strange... Let me see if i have done something wrong.

LCSphysicist said:
Strange... Let me see if i have done something wrong.
If you have calculated every coefficient ##c_n##, and the sum of modulus squared is not ##1##, then you must have done something wrong. Note that:
$$|\psi_1(x)|^2 = \sum_{n = 1}^{\infty}|c_n|^2$$

LCSphysicist
LCSphysicist said:
Homework Statement:: .
Relevant Equations:: .

My approach was: $$\psi_{2g} = \sqrt{1/L}sin(\frac{n \pi x}{2L}) = \sum c_n \sqrt {\frac{2}{L}} sin(\frac{n \pi x}{L})$$
Isn't this the wrong way round? That looks like the new ground state expressed in the original eigenbasis.

Delta2 and LCSphysicist
Ok, the point is. On evaluating cn, it comes with an extra ##\sqrt{2}##. $$c_n =2 \int_{0}^{L} \frac{sin(n \pi x /L) sin (n \pi x /2L)}{\sqrt{2}} = -\frac{8\left(-1\right)^n n}{\pi \left(2n+1\right)\left(2n-1\right) \sqrt{2}}$$
$$\implies$$ $$\sum c_n^2 = \frac{64}{2 \pi^2} (\sum (n^2/(1-4n^2)^2) = \frac{\pi ^2}{64}) = 1/2$$

$$c_1^2 = \frac{64}{2*9*\pi^2} \implies P = c_1^2 / \sum c_n^2 = (\frac{8}{3 \pi })^2$$

Delta2
PeroK said:
Isn't this the wrong way round? That looks like the new ground state expressed in the original eigenbasis.
Yes, i have expressed the new ground state in function of our original eigenbasis, that is right. I thought it would be fair to think this way

LCSphysicist said:
Yes, i have expressed the new ground state in function of our original eigenbasis, that is right. I thought it would be fair to think this way
That's wrong. The system is not in the new ground state. It's in the original ground state, which is a superposition of eigenstates for the new, larger system.

malawi_glenn and Delta2
PeroK said:
That's wrong. The system is not in the new ground state. It's in the original ground state, which is a superposition of eigenstates for the new, larger system.
Although, you get the same numerical answer for both problems. If you are in the ground state of the larger well and it suddenly contracts, then we have:
$$\varphi_1(x) = \sum_{n = 1}^{\infty} a_n \psi_n(x)$$where$$a_1 = \langle \psi_1(x)|\varphi_1(x) \rangle = c_1^*$$And we have the same transition probability in each case, as ##|a_1|^2 = |c_1|^2##.

Delta2
PeroK said:
Although, you get the same numerical answer for both problems. If you are in the ground state of the larger well and it suddenly contracts, then we have:
$$\varphi_1(x) = \sum_{n = 1}^{\infty} a_n \psi_n(x)$$where$$a_1 = \langle \psi_1(x)|\varphi_1(x) \rangle = c_1^*$$And we have the same transition probability in each case, as ##|a_1|^2 = |c_1|^2##.
I see your point, and i agree with the physics behind it, i think my problem here is with the math, can't i express any state in any basis i want? And isn't ##| \langle a | b \rangle |^2 = |\langle b | a \rangle| ^2 ##? I have done the other way round, couldn't it be still right? WHy there is this "anomaly" with the normalizations?

LCSphysicist said:
I see your point, and i agree with the physics behind it, i think my problem here is with the math, can't i express any state in any basis i want? And isn't ##| \langle a | b \rangle |^2 = |\langle b | a \rangle| ^2 ##? I have done the other way round, couldn't it be still right?
Yes, but the system is in the ground state of the original well. Of length ##L##. That is the state that needs to be expressed in the new basis.

It's only because the problem is very simple that you get the same numerical answer the wrong way. Imagine the system was originally in a superposition of energy eigenstates. Let's say:
$$\sqrt{\frac{1}{3}}\psi_1(x) + \sqrt{\frac 2 3}\psi_2(x)$$How would you approach the problem now?

LCSphysicist
PeroK said:
Yes, but the system is in the ground state of the original well. Of length ##L##. That is the state that needs to be expressed in the new basis.

It's only because the problem is very simple that you get the same numerical answer the wrong way. Imagine the system was originally in a superposition of energy eigenstates. Let's say:
$$\sqrt{\frac{1}{3}}\psi_1(x) + \sqrt{\frac 2 3}\psi_2(x)$$How would you approach the problem now?
That's enough. I got it with the example, thank you!

LCSphysicist said:
On evaluating cn, it comes with an extra 2.
Then you are not using an inner product where your basis is orthonormal. Or, alternatively, you are not using an orthonormal basis for your inner product.

Delta2
I think there is a problem with the interpretation of "expands symmetrically". Judging from the initial eigenfunction, we have a well with walls at ##x=0## and ##x=L##. OP wants to write the ground state of this well, ##\psi_1(x)=\sqrt{\frac{1}{L}}\sin\left(\frac{\pi~x}{L}\right)##, as a linear combination of the eigenstates of a well with walls at ##x=0## and ##x=2L.## How is that symmetrically expanded if the expansion is entirely in the ##+x## direction?

I would use the ground state eigenfunction of a well with walls at ##x=\pm~\frac{L}{2}## and expand that in terms of the eigenfunctions of a well with walls at ##x=\pm~L.##

Delta2, LCSphysicist and Orodruin

## 1. What is a "Particle in a box that suddenly expands"?

A "Particle in a box that suddenly expands" is a theoretical concept in quantum mechanics where a particle is confined to a specific region (the "box") and then the size of the box suddenly increases, allowing the particle to move freely within the larger space.

## 2. Why is this concept important in science?

This concept is important because it helps us understand how particles behave in confined spaces and how they respond to sudden changes in their environment. It also has practical applications in fields such as nanotechnology and materials science.

## 3. How does the sudden expansion of the box affect the particle?

The sudden expansion of the box causes the particle to gain more energy and momentum, allowing it to move freely within the expanded space. This can lead to changes in the particle's behavior and properties.

## 4. Can this concept be observed in real life?

While it is not possible to observe a literal "Particle in a box that suddenly expands" in real life, the concept has been demonstrated in experiments with atoms and other particles confined in a potential well that is suddenly changed.

## 5. What are the practical applications of this concept?

This concept has practical applications in various fields, including nanotechnology, materials science, and quantum computing. It can also help us understand the behavior of particles in extreme conditions, such as in the early universe or in black holes.

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