Harmonic oscillator's energy levels

[vela]

Again, it's impossible to see where you're going wrong when all you do is post your incorrect results. This is especially relevant in your case because, to put it bluntly, your math skills are very poor. You seem to be making basic mistakes that you shouldn't be making if you're taking this course.

Show your calculation of the normalization constant N. Note it should absolutely nothing to do with the expansion in terms of eigenfunctions.

 

[Rorshach]

ok, normalization constant is equal to $N=\frac{2\sqrt{2}\sqrt[4]{\frac({m\omega}{\hbar})^7}}{\sqrt{15}\sqrt[4]{\pi}}$. Then I have the wave function with substituted values $\psi(\xi)=H(\xi)exp(-\frac{\xi^2}{2})$, where $H(\xi)$ are hermite polynomials, so in my case it should look like $\psi(\xi)=H_0(\xi)exp(-\frac{\xi^2}{2})+H_1exp(-\frac{\xi^2}{2})+H_2exp(-\frac{\xi^2}{2})+H_3exp(-\frac{\xi^2}{2})$ where $H_n(\xi)= \sum_{n=0}^\infty a_n\xi^n$. Where in all this is the normalization constant? Is it somewhere in $a_n$? Also, in the previous posts You told me about $C_n$, for which the formula is $C_n=\frac{1}{\sqrt{2^n n!}}(\sqrt[4]{\frac{m\omega}{\pi\hbar}}$ which also I don't understand where it came from. The only difference I see is the lower index of $\psi(\xi)$ function: $\psi(\xi)=H_n exp(-\frac{\xi^2}{2})$ and $\psi_n (\xi)=C_n H_n(\xi)exp(-\frac{\xi^2}{2})$. How to understand this?

 

[vela]

No, that's wrong. Use $\psi$ only for the given wave function. Use $\phi_n$ for the eigenfunctions.

Write out explicitly what the first four normalized eigenfunctions $\phi_0(\xi)$, $\phi_1(\xi)$, $\phi_2(\xi)$, and $\phi_3(\xi)$ are. Figure out what those are first.

 

[Rorshach]

ok, so it should go like this:
$\phi_0(\xi)=\sqrt[4]{(\frac{m\omega}{\pi\hbar}}exp(-\frac{\xi^2}{2})$
$\phi_1(\xi)=\frac{1}{\sqrt{2}}\sqrt[4]{(\frac{m\omega}{\pi\hbar}}2\xi exp(-\frac{\xi^2}{2})$
$\phi_2(\xi)=\frac{1}{2\sqrt{2}}\sqrt[4]{(\frac{m\omega}{\pi\hbar}}(4\xi^2-2)exp(-\frac{\xi^2}{2})$
$\phi_3(\xi)=\frac{1}{4\sqrt{3}}\sqrt[4]{(\frac{m\omega}{\pi\hbar}}(8\xi^3-12\xi)exp(-\frac{\xi^2}{2})$

 

[vela]

Now show your calculations for $N$.

 

[Rorshach]

I didn't calculate the integral myself, I used Wolframalpha to do this, but the result came out as $\frac{\sqrt{15}\sqrt[4]{\pi}}{2\sqrt{2}\sqrt[4]({\frac{m\omega}{\hbar})^7}}$

 

[vela]

What integral?

 

[Rorshach]

the integral from minus infinity to plus infinity of squared wavefunction (wavefunction times its conjugate)

 

[vela]

Will you please provide a complete explanation of what you're doing rather just than bits and pieces?

 

[Rorshach]

Ok, I normalized the given wave function $\psi(x)=Nx^3exp(-\frac{m\omega x^2}{2\hbar})$ with the integral $\int_{-\infty}^ {+\infty} N^2 x^6 exp(-\frac{m\omega x^2}{\hbar})\,dx$. Thanks to that I obtained $N$. Then I substituted the values $\xi=x\alpha$, $\lambda=\frac{2E}{\hbar\omega}$ and $\alpha=\sqrt{\frac{m\omega}{\hbar}}$. Thanks to this I obtained the formula
$\psi(\xi)=H(\xi)exp(-\frac{\xi^2}{2})$.

 

[vela]

Substituted into what? What's $H(\xi)$ supposed to be, and why are you writing it like that? Where does $\lambda = 2E/\hbar\omega$ come into this?

 

[Rorshach]

After the substitution the Schrodinger equations becomes $\frac{d^2 \psi(\xi)}{d \xi^2}+(\lambda-\xi^2)\psi(\xi)=0$ for the region $|\xi|\rightarrow\infty$. For finite value of E the quantity $\lambda$ becomes negligible with respect to $\xi^2$ in the limit $|\xi|\rightarrow\infty$, so equation reduces to $(\frac{d^2}{d\xi^2})\psi(\xi)=0$. It is verified for large $|\xi|$ that equation is satisfied by functions $\psi(\xi)=\xi^p exp(\pm\frac{\xi^2}{2})$, where p is some finite value. The wave function has to bounded everywhere, so only physically acceptable solution must contain the minus sign in the exponent, so I am looking for the equation of form $\psi(\xi)=H(\xi)exp(-\frac{\xi^2}{2})$.

 

[Rorshach]

and according to textbook now I should do this $\frac{d^2 H}{d\xi^2}-2\xi\frac{dH}{d\xi}+(\lambda-1)H=0$

 

[vela]

You're going off on a tangent. Reread post 16. Use the eigenfunctions you wrote down in post 39.

 

[Rorshach]

It came to solving following equation:
$\frac{2\sqrt{2}(\frac{m\omega}{\hbar})^\frac{7}{4}}{\sqrt{15}\sqrt[4]{\pi}} x^3 exp(-\frac{m\omega x^2}{2\hbar})=exp(-\frac{m\omega x^2}{2\hbar})\sqrt[4]{\frac{m\omega}{\hbar\pi}}[(a_0+a_1\frac{1}{\sqrt{2}}\sqrt{\frac{m\omega}{\hbar}}x+a_2 \frac{1}{2\sqrt{2}}(\frac{m\omega}{\hbar}x^2-2)+a_3\frac{1}{4\sqrt{3}}(8x^3 \frac{m\omega}{\hbar}\sqrt{\frac{m\omega}{\hbar}}-12\sqrt{\frac{m\omega}{\hbar}}x))]$. I hope I didn't make any typos.

 

[vela]

Good! You still have a few typos, and it would have been easier to rewrite the lefthand side in terms of $\xi$ to get
$$\sqrt{\frac{8}{15}} \sqrt[4]{\frac{m\omega}{\hbar\pi}} \, \xi^3 e^{-\xi^2/2} = \sqrt[4]{\frac{m\omega}{\hbar\pi}} \, e^{-\xi^2/2} \left[a_0 + \frac{1}{\sqrt{2}} a_1 (2\xi) + \frac{1}{\sqrt{8}} a_2 (4\xi^2-2) + \frac{1}{\sqrt{48}} a_3 (8\xi^3-12\xi)\right]$$ or, equivalently,
$$\sqrt{\frac{8}{15}} \, \xi^3 = a_0 + \frac{1}{\sqrt{2}} a_1 (2\xi) + \frac{1}{\sqrt{8}} a_2 (4\xi^2-2) + \frac{1}{\sqrt{48}} a_3 (8\xi^3-12\xi).$$ Can you take it from there?

 

[Rorshach]

Yes, thank You:D I finally calculated the coefficients $a_1=\sqrt{\frac{3}{5}}$ and $a_3=\sqrt{\frac{2}{5}}$. I hope that his time they are correct. Now I need to calculate the energies corresponding to those eigen functions by solving schrodinger equation, and coefficients squared are their probabilities, right?

 

[vela]

Those look reasonable. The squares of the coefficients are indeed the probabilities; however, while it's a good exercise to go through at least once, you should not have to solve the Schrodinger equation for this problem. It's probably already done in your textbook. You should be familiar with the basic results.

 

[Rorshach]

of course You were right, I forgot about the most basic formula for energy levels:P Now there is last part of the problem-calculate the probability of finding the particle in the classically forbidden region- I thought about using this formula: $P_n=2 \int_|x_n|^{+\infty} |\psi(\xi)|^2 \,d\xi.$

 

[Rorshach]

about the second part of the problem: I tried to calculate the probability of finding a particle of energy $e_3=\frac{3}{2}\hbar\omega$ using a formula $P(E=E_1)=\frac{2}{\sqrt{\pi}2^1 1!}\int_\sqrt{3}^{+\infty} exp(-\xi^2){H_1}^2(\xi)\,dx$, but the result comes out as 0,06696. Where am I making a mistake?

 

[Rorshach]

Could someone please tell me if the formula I chose is right?