# Particle in Spherical Well : Sudden Approximation

• Xyius
This is what I did. But the way you are suggesting is by showing the summation is 1 for all values of n except n=2. Then the difference is 1/2 and that is the probability for that transition. Correct?Yes. But make sure you use L'Hopital's rule correctly; it is not as simple as applying it to the numerator and denominator separately. You have to show that the ratio of the derivatives of the numerator and denominator both tend to zero in the limit n -> 2.Also, be careful with your notation. You wrote ##\frac{1}{\sqrt{2}}## but you should have written ##\sqrt{2}##.
Xyius

## Homework Statement

In a spherical well in which..

$$V= \begin{cases} 0,\text{for }0 \le r < R \\ ∞, \text{for } r > R \end{cases}$$

the s-wave eigenstates are

$$\phi_n(r)=\frac{A}{r}\sin\left( \frac{n\pi r}{R} \right)$$

where A is a normalization constant. If a particle is in the ground state and R suddenly
increases to ##R'=R/\epsilon##, where ##0 \le \epsilon \le 1##

(i) Compute and plot the probability of ﬁnding the particle in the new ground state as a
function of ##\epsilon##.
(ii) Compute and plot the probability of ﬁnding the particle in a new state with quantum
number n as a function of ##\epsilon## and n (this will be a 3D plot). What are the most important
features you see in your plot?
(iii) Sum all probabilities in (ii) for = 1=2 to show that they add to 1.
Note: The relations

$$\sum_{n=0}^{∞}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$

$$\sum_{n=0}^{∞}\frac{1}{(2n+3)^2(2n-1)^2}=\frac{\pi^2}{64}$$

may be useful here.

## Homework Equations

Obviously the ones given in the problem.

Also, this problem deals with the sudden approximation. Which says that when the system is changed very rapidly, the state remains unchanged initially and then is allowed to evolve.

## The Attempt at a Solution

My solution is in the attached images. I know typing it out would have been ideal but it would have been too much and I could give more of my work this way. I scanned them to make sure they were high quality images.

I don't know where my thought process went wrong, but what I am getting for part (iii) is obviously wrong since it diverges at n=2.

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Generally, your work looks good. But,

(1) Make sure you distinguish between probability amplitude and probability.

(2) When R is switched to R/ε, the normalization constant for the new eigenstates will depend on ε.

(3) It is not true that ##\frac{\sin(n\pi/2)}{n^2-4} = \frac{(-1)^{2n+1}}{n^2-4}##. For example, let n = 4. Also, n = 2 will need special treatment.

1 person
Oh I forgot to replace n with 2n+1 in the denominator. After doing this, I get the second form of the second equation listed in the problem, except, It needs to be squared. In other words, I get.. (We will call it P)

$$P=-\frac{8}{\pi}\frac{(-1)^{2n+1}}{(2n-1)(2n+3)}$$

So in order to get the summation, I need to first square this, THEN sum over n. I cannot determine if this makes sense or not. That means I am summing up the ##P^2## values, not the ##P## values.

Your expression for P is off by an overall numerical factor because you have not normalized the wave functions correctly for the case where the radius of the well is ##R/ \epsilon##.

When you correct for this, you will have expressions for all of the transition amplitudes from the initial state to final states with odd quantum number ##\psi_{2n+1}##. You found that the transition amplitudes to states with even quantum number are zero. But there is an exception: the transition amplitude to ##\psi_2## is not zero.

1 person
I believe I got it. Re-writing the normalization constant, I got a factor of 1/2. Meaning the summation I showed in my previous post is equal to 1/2. I then used L'Hopitals rule for when n=2 and got that term to be equal to 1/2. So 1/2+1/2=1.

Hopefully, this is it. Thank you very much for your help!

Good.

Another option for handling the n = 2 case is to set up the transition amplitude integral specifically for n = 2 and ##\epsilon = 1/2## and show the integral is ##\frac{1}{\sqrt{2}}##. So the probability for this transition is 1/2.

## 1. What is a particle in a spherical well?

A particle in a spherical well is a theoretical model used in quantum mechanics to describe the behavior of a particle confined to a spherical potential. The particle is assumed to be in a three-dimensional spherical box with infinitely high walls.

## 2. What is the sudden approximation in relation to a particle in a spherical well?

The sudden approximation is a simplifying assumption used in the analysis of a particle in a spherical well. It assumes that the particle is initially in an eigenstate of the potential, and that the potential is suddenly changed to a new value. This simplifies the calculations and allows for the study of the time evolution of the particle's wavefunction.

## 3. How does the sudden approximation affect the energy levels of a particle in a spherical well?

The sudden approximation does not affect the energy levels of a particle in a spherical well. The energy levels are determined by the size and shape of the potential well, and the sudden approximation only simplifies the calculations for the time evolution of the particle's wavefunction.

## 4. What are the limitations of the sudden approximation for a particle in a spherical well?

The sudden approximation is a simplifying assumption and therefore has limitations. It assumes that the potential change is instantaneous and that the particle is in an eigenstate of the potential. This may not accurately reflect the physical reality and can lead to errors in the calculations.

## 5. How is the sudden approximation used in practical applications?

The sudden approximation is used in practical applications to study the time evolution of a particle in a spherical well. It allows for the calculation of transition probabilities and can be used to understand the behavior of particles in various quantum systems, such as atoms and molecules.

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