# A particle in an infinite well

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Homework Statement:
Relevant Equations:

(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}##

Last edited:
Delta2

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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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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.

docnet
docnet
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

Last edited:
docnet
This still does not make much sense to me, because ##sin(\pi n)=0## means ##P=0##

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The hint was to look for a trig identity for ##\sin^3\theta##.

docnet
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The hint was to look for a trig identity for ##\sin^3\theta##.

Hint: $$\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}~\Rightarrow~\sin^3\theta=\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^3=\dots$$

docnet
docnet
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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That looks right to me.

docnet
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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: $$\psi(x) = \frac{3}{\sqrt {10}} \psi_1(x) - \frac{1}{\sqrt {10}} \psi_3(x)$$

docnet
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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.

docnet
docnet
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

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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)##

docnet
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)##