# Infinite Square Well with polynomial wave function

• Zack K
In summary: Wait a minute. Did you cover the time-dependent Schrodinger equation yet?In summary, the conversation discussed a particle in a one-dimensional infinite square well potential and determining the constant A to normalize the wave function. It also touched upon the validity of the wave function and the evaluation of the integral for quantum number n. The conversation then moved on to the measurement of the first three energy levels and calculating the energy expectation and probability for each. The main points to take away are the relationship between a wave function and the Schrodinger equation, the concept of energy eigenstates, and the use of superposition to find the probabilities for different energy levels.
Zack K
Homework Statement
A particle is in a one-dimensional infinite square well potential of width ##a## such that:
##V(x)=0 ## for ## 0\leq x\leq a## and ##V(x)=\infty## otherwise. At ##t=0##, the particle has wave function ##\psi(x,0)=Ax(a-x)## for ##0\leq x\leq a##. Determine the constant ##A## to normalize the wave function.
Relevant Equations
##\int_{-\infty}^\infty \psi*_n\psi_n dx=1##
Some questions:
Why is this even a valid wave function? I thought that a wave function had to approach zero as x goes to +/- infinity in all of space. Unless all of space just means the bounds of the square well.

Since we have no complex components. I am guessing that the ##\psi *=\psi##.

If this is true, how do I evaluate the integral for quantum number n? I first chose n=1 then evaluated the integral and managed to normalize it. But if I evaluate for ##\psi_n##, I get a nasty function which did not normalize. is it wrong that ##\psi_n=A_nxn(a-xn)?##

Zack K said:
Homework Statement:: A particle is in a one-dimensional infinite square well potential of width ##a## such that:
##V(x)=0 ## for ## 0\leq x\leq a## and ##V(x)=\infty## otherwise. At ##t=0##, the particle has wave function ##\psi(x,0)=Ax(a-x)## for ##0\leq x\leq a##. Determine the constant ##A## to normalize the wave function.
Relevant Equations:: ##\int_{-\infty}^\infty \psi*_n\psi_n dx=1##

Some questions:
Why is this even a valid wave function? I thought that a wave function had to approach zero as x goes to +/- infinity in all of space. Unless all of space just means the bounds of the square well.
The problem should have stated that the wavefunction is ##\psi(x,0)=Ax(a-x)## for ##0\leq x\leq a## and ##\psi(x,0)=0## for ##x## not in the region ##0\leq x\leq a##.

Since we have no complex components. I am guessing that the ##\psi *=\psi##.
The constant ##A## could be a complex number. However, it is probably intended that you assume ##A## is real and positive. Otherwise, there is not a unique answer for ##A##. For real ##A##, ##\psi## will be real.

If this is true, how do I evaluate the integral for quantum number n? I first chose n=1 then evaluated the integral and managed to normalize it. But if I evaluate for ##\psi_n##, I get a nasty function which did not normalize. is it wrong that ##\psi_n=A_nxn(a-xn)?##
The infinite well will have an infinite number of energy eigenstates (or "stationary states") that are labeled with the quantum number ##n##. However, a particle in the well does not have to be in one of these energy eigenstates. At time ##t = 0## you are given the state ##\psi(x, 0)## for the particle. This is not one of the energy eigenstates. So, there is no label ##n## for this state. Your state will be normalized if ##\int_{-\infty}^\infty \psi^*\psi dx=1##

TSny said:
This is not one of the energy eigenstates.
So for the wave function to be an eigenfunction, does it have to satisfy the Schrodinger equation?

TSny said:
Your state will be normalized if ∫∞−∞ψ∗ψdx=1
I see. For a later part of the question, it wants me to measure its first three energy levels. So would I now assume the general solution (sinusoidal) of the Schrodinger equation?

Zack K said:
So for the wave function to be an eigenfunction, does it have to satisfy the Schrodinger equation?
Yes. More specifically, for a wave function to be an energy eigenfunction, it must satisfy the time-independent Schrodinger equation. The wave function that you are given does not satisfy the time-independent Schrodinger equation. However, it can be considered as the wavefunction at ##t = 0## of a solution to the time-dependent Schrodinger equation.
For a later part of the question, it wants me to measure its first three energy levels. So would I now assume the general solution (sinusoidal) of the Schrodinger equation?
I'm not quite sure what is meant by "measure its first three energy levels". The wavefunction that you are given (at ##t = 0##) can be represented as a superposition of energy eigenstates:

##\psi(x, 0) = \sum_{n = 1}^\infty a_n \psi_n(x)##

where the ##\psi_n## are energy eigenstates and the ##a_n## are constants that you can calculate. If you were to measure the energy of the particle at time ##t = 0##, then ##a_n## is related to the probability that you would find the energy level ##E_n## corresponding to the energy eigenstate ##\psi_n##. Hopefully, you've covered all of this. So, maybe they want you to find the first three nonzero ##a_n##'s.

Zack K and PeroK
TSny said:
I'm not quite sure what is meant by "measure its first three energy levels". The wavefunction that you are given (at t=0t=0t = 0) can be represented as a superposition of energy eigenstates:

ψ(x,0)=∑∞n=1anψn(x)ψ(x,0)=∑n=1∞anψn(x)\psi(x, 0) = \sum_{n = 1}^\infty a_n \psi_n(x)

where the ψnψn\psi_n are energy eigenstates and anana_n are constants that you can calculate. If you were to measure the energy of the particle at time t=0t=0t = 0, then anana_n is related to the probability that you would find the energy level EnEnE_n corresponding to the energy eigenstate ψnψn\psi_n. Hopefully, you've covered all of this. So, maybe they want you to find the first three nonzero anana_n's.
I worded that poorly. This is getting into another part of the question. The question wants me to find the energy expectation for ##E_n##. I know that ##\langle E\rangle=\int_{\infty}^\infty \psi*\hat H\psi dx##. So would ##\langle E_n\rangle=\int_{\infty}^\infty \psi*\hat H\psi dx## where:
TSny said:
##\psi(x, 0) = \sum_{n = 1}^\infty a_n \psi_n(x) ##
I seem to be getting confused with how to implement the general solution and the linear combinations of each particular solution.

Hmm. "find the energy expectation for ##E_n##" seems a little odd to me. Can you write out this part of the problem statement word for word?

hutchphd
"Calculate the probability that the measurement of the energy would yield the value ##E_n##".

If it helps,the energy level question I was referring to is "in particular, calculate the numerical values of the probability to measure ##E_1,E_2## or ##E_3##"

Ok. How is the probability to measure ##E_1## related to the value of ##a_1##? How would you calculate ##a_1##?

 ##\langle E_1\rangle=\int_{0}^a A_1\psi^*_1\hat HA_1\psi_1 dx## (I think?).

Zack K said:
 ##\langle E_1\rangle=\int_{0}^a A_1\psi^*_1\hat HA_1\psi_1 dx## (I think?).
##E_1## has a fixed value; that of the first energy level. You want to find the probability of getting ##E_1## as an energy measurement.

PS can you simplify that integral on the RHS?

PeroK said:
##E_1## has a fixed value; that of the first energy level. You want to find the probability of getting ##E_1## as an energy measurement.

PS can you simplify that integral on the RHS?
I'm not sure how to find the probability of being in an energy state. Are we using the real valued eigenfunction for n=1 then finding its probability distribution?

Zack K said:
I'm not sure how to find the probability of being in an energy state. Are we using the real valued eigenfunction for n=1 then finding its probability distribution?
An eigenfunction (or any wavefunction) is a probability distribution. Or, at least, the square of its modulus is. But that's a probability distribution for particle position.

You seem more than a little bit confused by these concepts. Did you understand what was said in post #4?

PeroK said:
An eigenfunction (or any wavefunction) is a probability distribution. Or, at least, the square of its modulus is. But that's a probability distribution for particle position.

You seem more than a little bit confused by these concepts. Did you understand what was said in post #4?
It was said that :
 ##\psi(x, 0) = \sum_{n = 1}^\infty a_n \psi_n(x)##
So expanding the first three n terms: ##Ax(a−x)= B\psi_1+C\psi_2+D\psi_3##. Where I am assuming ##\psi_n=\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})##.

TSny said:
##a_n## is related to the probability that you would find the energy level ##E_n##
This is the relation that I do not know. How is ##E_n=\frac{n^2 \pi^2\hbar}{2ma^2}## related to the coefficients B,C and D?

Zack K said:
It was said that :
 ##\psi(x, 0) = \sum_{n = 1}^\infty a_n \psi_n(x)##
So expanding the first three n terms: ##Ax(a−x)= B\psi_1+C\psi_2+D\psi_3##. Where I am assuming ##\psi_n=\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})##.

Well, we must have:

##Ax(a−x)= B\psi_1+C\psi_2+D\psi_3 + E\psi_4 +F\psi_5 \dots##

Do you know how to find ##B, C, D \dots## using integration and the orthonormality of eigenfunctions? Have you heard of "Fourier's trick"?

PeroK said:
Well, we must have:

##Ax(a−x)= B\psi_1+C\psi_2+D\psi_3 + E\psi_4 +F\psi_5 \dots##

Do you know how to find ##B, C, D \dots## using integration and the orthonormality of eigenfunctions? Have you heard of "Fourier's trick"?
I remember that two functions are orthogonal if ##\int_a^b f(x)g(x)dx=0##.
So then do the eigenfunctions form a basis of funtions?

Zack K said:
I remember that two functions are orthogonal if ##\int_a^b f(x)g(x)dx=0##.
So then do the eigenfunctions form a basis of funtions?
Energy eigenfunctions do indeed form an orthonormal basis. Can you work with that?

PeroK said:
Energy eigenfunctions do indeed form an orthonormal basis. Can you work with that?
I guess I could get the coefficients by using ##\int_0^a \psi(x,0)\psi_ndx=0## because the functions would be orthogonal.

If this is true, still how would I relate the coefficients to the energy and energy probability?

Zack K said:
I guess I could get the coefficients by using ##\int_0^a \psi(x,0)\psi_ndx=0## because the functions would be orthogonal.

If this is true, still how would I relate the coefficients to the energy and energy probability?

What you want to do is integrate both sides of the following equation with ##\psi_n##.

##Ax(a−x)= B\psi_1+C\psi_2+D\psi_3 + E\psi_4 +F\psi_5 \dots##

PeroK said:
What you want to do is integrate both sides of the following equation with ##\psi_n##.

##Ax(a−x)= B\psi_1+C\psi_2+D\psi_3 + E\psi_4 +F\psi_5 \dots##
I see. So all the terms on the right except for what is ##\psi_n## will go to zero? So then $$\int_0^aAx(a-x)\psi_ndx=\int_0^aB^2_n\psi^2_ndx$$

Zack K said:
I see. So all the terms on the right except for what is ##\psi_n## will go to zero? So then $$\int_0^aAx(a-x)\psi_ndx=\int_0^aB^2_n\psi^2_ndx$$
That's half way there. Can you simplify the RHS using what you know about ##\psi_n##?

PS How did you get ##B_n^2## instead of ##B_n##?

PeroK said:
PS How did you get B2nBn2B_n^2 instead of BnBnB_n?
That was a typo.
Evaluating the right hand side: ##\int_0^aB_n\sqrt{\frac{2}{a}}^2sin^2(\frac{n\pi x}{a})dx = B_n\frac{(2\pi n - sin(2\pi n))}{2\pi n}##. If calculated right, will let me solve for ##B_n##

Zack K said:
That was a typo.
Evaluating the right hand side: ##\int_0^aB_n\sqrt{\frac{2}{a}}^2sin^2(\frac{n\pi x}{a})dx = B_n\frac{(2\pi n - sin(2\pi n))}{2\pi n}##. If calculated right, will let me solve for ##B_n##
You should know that ##\int |\psi_n(x)|^2dx = 1##, from the normality of ##\psi_n##.

The difficult bit is the other integral with ##\psi(x, 0)##.

I'm going offline now.

PeroK said:
You should know that ##\int |\psi_n(x)|^2dx = 1##, from the normality of ##\psi_n##.

The difficult bit is the other integral with ##\psi(x, 0)##.

I'm going offline now.
Thank you, I've learned quite a bit from this thread. I think I can finish from here on.

## 1. What is the Infinite Square Well with polynomial wave function?

The Infinite Square Well with polynomial wave function is a concept in quantum mechanics that describes a particle confined to a square well potential with infinitely high walls. The wave function of the particle is described by a polynomial equation.

## 2. What is the significance of the polynomial wave function in the Infinite Square Well?

The polynomial wave function is significant because it represents the probability amplitude of the particle in the Infinite Square Well. It allows us to calculate the probability of finding the particle at a certain position within the well.

## 3. How is the polynomial wave function derived in the Infinite Square Well?

The polynomial wave function is derived by solving the Schrödinger equation for the Infinite Square Well potential. This involves finding the eigenvalues and eigenfunctions of the Hamiltonian operator for the system.

## 4. What are the limitations of using a polynomial wave function in the Infinite Square Well?

Using a polynomial wave function in the Infinite Square Well assumes that the particle has a constant energy throughout the well, which is not always the case in real systems. Additionally, the polynomial wave function may not accurately describe the behavior of particles in more complex potentials.

## 5. How does the polynomial degree of the wave function affect the behavior of the particle in the Infinite Square Well?

The polynomial degree of the wave function determines the number of nodes and the energy levels of the particle in the Infinite Square Well. A higher degree polynomial will have more nodes and a higher energy, while a lower degree polynomial will have fewer nodes and a lower energy.

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