How Do You Normalize the Wave Function for a Particle in an Infinite Well?

[docnet]

Homework Statement

please see below

Relevant Equations

please see below

(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 $\frac{1}{n^2}$

 

[PeroK]

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.

 

[Delta2]

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]

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

 

[docnet]

This still does not make much sense to me, because $sin(\pi n)=0$ means $P=0$

 

[PeroK]

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

 

[kuruman]

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

Or derive your own identity.

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

 

[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

 

[kuruman]

That looks right to me.

 

[PeroK]

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

 

[kuruman]

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]

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

 

[PeroK]

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 $\psi(x) = Nf(x)$

 

[kuruman]

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.

 

[docnet]

Thank you for the help. I ♥ you all