A particle in an infinite well

In summary, the given function ##\psi(x)## can be expressed as a linear combination of the energy eigenfunctions ##\psi_n(x)##, with coefficients ##C_1=\frac{3}{\sqrt{10}}## and ##C_3=\frac{-1}{\sqrt{10}}##. The two coefficients were found by applying the trig identity ##sin^3x=\frac{3}{4}sinx-\frac{1}{4}sinx## and normalizing the resulting coefficients to satisfy the probability constraint. The energy values that can be measured are ##E_1=\frac{ħ^2}{2m}\frac{(\pi )^2}{a^2}## and
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Homework Statement
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Relevant Equations
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Screen Shot 2021-02-09 at 10.16.55 PM.png


(a) I guess I should find ##C_n## by normalizing ##\psi_n##.
$$∫_{∞}^∞|C_nψn(x)|^2 dx=C_n^2 \frac{2}{a}∫_0^a sin^2(\frac{πnx}{a})dx=1$$
$$C_n^2 \frac{2}{a}[\frac{a}{2}−\frac{a}{4πn}sin(\frac{2πna}{a})]=1⇒C_n=1$$

(b) $$Hψ_n(x)=\frac{-ħ^2}{2m}\frac{\partial^2}{\partial x^2}\sqrt{\frac{2}{a}}sin(\frac{πnx}{a})$$
$$Hψ_n(x)=\frac{-ħ^2}{2m}\frac{(\pi n)^2}{a^2}\sqrt{\frac{2}{a}}sin(\frac{πnx}{a})$$

##\frac{-ħ^2}{2m}\frac{(\pi n)^2}{a^2}## gives the ##n##th energy value with probability ##\frac{1}{n^2}##
 
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  • #2
You have completely misunderstood the question. You are asked to express the given function ##\psi(x)## as a linear combination of the energy eigenfunctions ##\psi_n(x)##.

What you did in part b) makes no sense.
 
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  • #3
I am afraid @PeroK is correct, what you do in part a) is to calculate the normalized constants of the already normalized eigenfunctions, that's why afterall you calculate ##C_n=1## for all ##n##.
But the question doesn't ask you to do that anyway, he asks you to find the ##C_n## such that $$\psi(x)=\sum C_n\psi_n(x)$$, a task similar to calculating the Fourier coefficients in a Fourier series if that rings a bell to you.
 
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  • #4
PeroK said:
You have completely misunderstood the question. You are asked to express the given function ##\psi(x)## as a linear combination of the energy eigenfunctions ##\psi_n(x)##.

What you did in part b) makes no sense.
I agree with you that did not make much sense. Let's try this again.

$$\psi=\sum_n^\infty C_n \sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})$$

The ##nth## coefficient is given by the projection of ##\psi_n## onto ##\psi##.

$$C_n=<\psi_n|\psi>=\int^a_0\psi_n\psi dx=\int^a_0 \sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a}) \frac{4}{\sqrt{5a}}sin^3(\frac{\pi x}{a})dx$$
By Mathematica
$$C_n=\frac{24\sqrt{10}}{5}\frac{sin(\pi n)}{\pi(n^4-10n^2+9)}$$

The hamiltonian ##H## operates on ##\psi_n## to give the energy by ##H\psi_n=E_n\psi_n##
$$Hψ_n(x)=\frac{ħ^2}{2m}\frac{(\pi n)^2}{a^2}\sqrt{\frac{2}{a}}sin(\frac{πnx}{a})⇒E_n=\frac{ħ^2}{2m}\frac{(\pi n)^2}{a^2}$$

The probability of measuring the nth energy eigenstate is given by
$$|Cn|^2=(\frac{24\sqrt{10}}{5}\frac{sin(\pi n)}{\pi(n^4-10n^2+9)})^2$$

edited for grammar
 
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  • #5
This still does not make much sense to me, because ##sin(\pi n)=0## means ##P=0##
 
  • #6
The hint was to look for a trig identity for ##\sin^3\theta##.
 
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  • #7
PeroK said:
The hint was to look for a trig identity for ##\sin^3\theta##.
Or derive your own identity.

Hint: $$\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}~\Rightarrow~\sin^3\theta=\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^3=\dots$$
 
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  • #8
Final attempt: The term ##sin^3x## reduces with the trig identity ##sin^3x=\frac{3}{4}sinx-\frac{1}{4}sinx## to give $$\psi=\frac{3}{4}\psi_1-\frac{1}{4}\psi_3$$
Which tells us ##\psi## is a linear superposition of states ##\psi_1## and ##\psi_3##. The two coefficients are ##C_1=\frac{3}{4}## and ##C_3=\frac{-1}{4}##.

The hamiltonian ##H## operates on ##\psi_n## to give the energy by ##H\psi_n=E_n\psi_n##
$$Hψ_n(x)=\frac{ħ^2}{2m}\frac{(\pi n)^2}{a^2}\sqrt{\frac{2}{a}}sin(\frac{πnx}{a})⇒E_n=\frac{ħ^2}{2m}\frac{(\pi n)^2}{a^2}$$
So the values of energy that are measured are
$$E_1=\frac{ħ^2}{2m}\frac{(\pi )^2}{a^2}$$
$$E_3=\frac{ħ^2}{2m}\frac{(3\pi)^2}{a^2}$$
The probability of measuring the nth energy eigenstate is given by
$$|C_1|^2\times N=\frac{9}{16}\times \frac{16}{10}= \frac{9}{10}$$
$$|C_3|^2\times N=\frac{1}{16}\times \frac{16}{10}=\frac{1}{10}$$
where ##N## is the normalization factor ##\frac{16}{10}##.

edited for grammar
 
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  • #9
That looks right to me.
 
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  • #10
docnet said:
Final attempt: The term ##sin^3x## reduces with the trig identity ##sin^3x=\frac{3}{4}sinx-\frac{1}{4}sinx## to give $$\psi=\frac{3}{4}\psi_1-\frac{1}{4}\psi_3$$
Which tells us ##\psi## is a linear superposition of states ##\psi_1## and ##\psi_3##. The two coefficients are ##C_1=\frac{3}{4}## and ##C_3=\frac{-1}{4}##.
This is not quite right. You should have: $$\psi(x) = \frac{3}{\sqrt {10}} \psi_1(x) - \frac{1}{\sqrt {10}} \psi_3(x)$$
 
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  • #11
PeroK said:
This is not quite right. You should have: $$\psi(x) = \frac{3}{\sqrt {10}} \psi_1(x) - \frac{1}{\sqrt {10}} \psi_3(x)$$
It's not right if one assumes that ##\psi(x)## is supposed to be normalized. OP introduced the normalization factor later and the probabilities came out right in the end.
 
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  • #12
The term ##sin^3x## reduces with the trig identity ##sin^3x=\frac{3}{4}sinx-\frac{1}{4}sinx## to give $$\psi=\frac{3}{4}\psi_1-\frac{1}{4}\psi_3$$
Which tells us ##\psi## is a linear superposition of states ##\psi_1## and ##\psi_3##. The two coefficients ##C_1=\frac{3}{4}## and ##C_3=\frac{-1}{4}## need to be normalized.

$$(\frac{3N}{4})^2+(\frac{N}{4})^2=1⇒N=\frac{4}{\sqrt{10}}$$

So the two coefficients are ##C_1=\frac{3}{4}\times N= \frac{3}{\sqrt{10}}## and ##C_3=\frac{-1}{4}\times N= \frac{-1}{\sqrt{10}}##

The hamiltonian ##H## operates on ##\psi_n## to give the energy by ##H\psi_n=E_n\psi_n##
$$Hψ_n(x)=\frac{ħ^2}{2m}\frac{(\pi n)^2}{a^2}\sqrt{\frac{2}{a}}sin(\frac{πnx}{a})⇒E_n=\frac{ħ^2}{2m}\frac{(\pi n)^2}{a^2}$$
So the values of energy that are measured are
$$E_1=\frac{ħ^2}{2m}\frac{(\pi )^2}{a^2}$$
$$E_3=\frac{ħ^2}{2m}\frac{(3\pi)^2}{a^2}$$
The probability of measuring the nth energy eigenstate is given by
$$|C_1|^2= \frac{9}{10}$$
$$|C_3|^2=\frac{1}{10}$$

edited for typo
 
  • #13
kuruman said:
It's not right if one assumes that ##\psi(x)## is supposed to be normalized. OP introduced the normalization factor later and the probabilities came out right in the end.
The functions ##\psi(x)## and ##\psi_n(x)## given in the question are normalised. If we introduce a non-normalised function, it needs a different name. E.g. $$f = \frac 3 4 \psi_1 - \frac 1 4 \psi_3$$ Where ##\psi(x) = Nf(x)##
 
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  • #14
PeroK said:
The functions ##\psi(x)## and ##\psi_n(x)## given in the question are normalised. If we introduce a non-normalised function, it needs a different name. E.g. $$f = \frac 3 4 \psi_1 - \frac 1 4 \psi_3$$ Where ##\psi(x) = Nf(x)##
I stand corrected.
 
  • #15
Thank you for the help. I ♥ you all
 
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Related to A particle in an infinite well

1. What is a particle in an infinite well?

A particle in an infinite well is a theoretical model used in quantum mechanics to study the behavior of a particle confined within a potential well that extends infinitely in all directions.

2. What is the significance of studying a particle in an infinite well?

Studying a particle in an infinite well helps us understand the behavior of quantum particles and their energy levels in confined spaces, which has important applications in fields such as nanotechnology and solid state physics.

3. How does the energy of a particle in an infinite well change?

The energy of a particle in an infinite well is quantized, meaning it can only take on certain discrete values. As the size of the well changes, the energy levels also change accordingly.

4. What is the wave function of a particle in an infinite well?

The wave function of a particle in an infinite well is a mathematical representation of the probability of finding the particle at a certain position within the well. It describes the behavior of the particle in terms of its position and momentum.

5. How does the behavior of a particle in an infinite well differ from that of a particle in a finite well?

A particle in an infinite well has a continuous energy spectrum and can exist in any energy state, while a particle in a finite well has a discrete energy spectrum and can only exist in certain energy states. Additionally, the wave function for a particle in an infinite well is a sine or cosine function, while the wave function for a particle in a finite well is a combination of sine and cosine functions.

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