Harmonic oscillator's energy levels

[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?