Harmonic oscillator's energy levels

[Rorshach]

Homework Statement

Okay, this one confuses me a bit:

A particle is in a one-dimensional harmonic oscillator. At time t = 0 is given by its wave function

ψ(x)=Nx³exp(-mωx²/2hbarred)

a) At this point you measure the particle's energy. What measurement values ​​are available? Also determine the corresponding probabilities!
b) After the power supply that gave outcome E = 3hbarredω/2 measure the particle's position immediately. What is the probability of finding the particle in the classically forbidden region? (The classically forbidden region is defined by the condition that V (x)> = E_total)

Homework Equations

E_n=(n+1/2)hbarredω

The Attempt at a Solution

I don't really have an idea how to come up with energy levels that are available, or how to calculate the probabilities of those levels.

 

[BruceW]

"one-dimensional harmonic oscillator" should imply the form of the Hamiltonian. You can almost guess what it should look like. If it is not in your notes, then you could google it if you are having trouble.

 

[Rorshach]

I got notice on violation of rules of the forum by posting a problem that I did not try to solve on my own- while it is not entirely true, I understand it could be perceived as such. I am sincerely sorry, won't happen again. I tried to come with idea how to find out how many possible energy levels are available for my function, but only thing that comes to my mind is this basic formula for energy E=(n+1/2)hbarredω. Hamiltonian for my function is of course
h=-hbarred²/2m(d/dx)+1/2mω²x² , but how to connect it to my search for available energy levels?

 

[fluidistic]

I don't really understand why would the Hamiltonian help to solve for a).
At first glance the psi function they give you is a supperposition of only 2 eigenstates (can you find them?) that have a definite energy. I guess that these energies are the one you could possibly measure.

As for the energy level formula $E_n =\hbar \omega (n+1/2)$, it comes from solving the Schrödinger's equation and assuming that the solution behaves as $f(x)v(x)$ where v(x) is a decaying exponential that solves the Schrödinger's equation when x tends to infinity and the function f(x) are the Hermite polynomials that you can determine when solving the original Schrödinger's equation using a power series "guess function" for f(x). You get that the solution doesn't diverge if and only if the power series is truncated for some n. This condition gives precisely that $E_n =\hbar \omega (n+1/2)$.

 

[Rorshach]

I don't know how to find those two eigenstates, how can You tell from first glance that there will be two levels?

 

[fluidistic]

I don't know how to find those two eigenstates, how can You tell from first glance that there will be two levels?

Hint: Look at the first few Hermite polynomials and at the psi(x) they give you.

 

[Rorshach]

So it is connected to power to which x is taken right? if x was to the power of 10 there would be 6 levels?

 

[fluidistic]

So it is connected to power to which x is taken right? if x was to the power of 10 there would be 6 levels?

If you only had x to the power of 10, I think so. But if you had an x^10 as well as say an x^9 in your psi(x), then no, not only necessarily 6 levels.
The idea is that the $\psi _n (x)$ form a basis for any function $\Psi (x,t)$ for a fixed t. In your case t is fixed at $t=0$.
So that you can write your $\Psi (x,0)$ as a linear combination of the $\psi _n(x)$. One writes $\Psi (x,0)=\Psi(x)=\sum _{n=0}^\infty c_n \psi _n (x)$.
We're lucky enough that the from the $\Psi (x)$ they give you, it's not hard to determine the coefficients $c_n$ and the $\psi _n (x)$ involved.
I'll also tell you a "trick". The associated probabilities with each energy levels are simply the values of $|c_n|^2$. But in your exercise, I think you're meant to find them by integration (I've never done it, so I'll have to do it myself too!). Though it's good to know 2 methods so you can compare the results afterwards.

 

[Rorshach]

okay, but how do I determine the c_n and the ψ_n(x)? It sounds similar to distributing the function into the Fourier series, but I have never had a clue how to perform Fourier series distribution on a function... if I am right and this is the thing to do, could you show me on example how to perform Fourier series distribution?

 

[fluidistic]

okay, but how do I determine the c_n and the ψ_n(x)? It sounds similar to distributing the function into the Fourier series, but I have never had a clue how to perform Fourier series distribution on a function... if I am right and this is the thing to do, could you show me on example how to perform Fourier series distribution?

I'm not sure it's a "Fourier expansion" but the idea is pretty similar. We must exploit the orthogonality property of the eigenstates; or the fact that they form a basis for any psi(x).
We have $\Psi(x)=\sum _{n=0}^\infty c_n \psi _n (x)$. Multiply both sides by $\psi ^* _m (x)$ and integrate from - infinity to positive infinity. After a bit of algebra you'll reach that $c_m = \int _{-\infty}^\infty \psi ^* _ m (x) \Psi (x) dx$.

 

[Rorshach]

I'm sorry, but I still don't know how to get ψ_n. It seems to me that without it I cannot do anything.

 

[fluidistic]

I'm sorry, but I still don't know how to get ψ_n. It seems to me that without it I cannot do anything.

You get them when you solve the -normalized- Schrödinger's equation ($\psi '' + (2\varepsilon - y^2)\psi =0$ where $y=x(m \omega / \hbar )^{1/2}$ and $\varepsilon = E/(\hbar \omega )$) and assume that the solutions you're looking for are of the form $\psi (y)=v(y)\exp (-y^2/2)$. And in order to calculate v(y) you propose a power series of the form $v(y)=\sum _{k=0}^\infty C_k y^k$.
In order for $\psi (y)$ to converge you'll get that the power series for v(y) must be truncated at some value for k=n.
You'll also get the recurrence relation $C_{k+2}=C_k \frac{2k+1-2\varepsilon }{(k+2)(k+1)}$. So for any value of n, you get that v(y) is a Hermite polynomial.
And thus that the $\psi (y)$ turns out to be of the form $\psi _n (y)= H_ n (y)e^{-y^2/2}$.
After replacing the variables back and normalizing, one reaches that $\psi _n (x)=\left ( \frac{m\omega }{\pi \hbar 2^{2n}n!^2} \right ) ^{1/4} \exp \left ( - \frac{m\omega x^2}{2 \hbar} \right ) H_n \left [ \left ( \frac{m\omega }{\hbar} \right ) ^{1/2 }x \right ]$.

So yes, in your exercise I don't know how you could solve it without knowing the eigenfunctions $\psi _n (x)$. Either you look at them in a book/Internet, either you find them yourself (which is a pain but maybe necessary).

 

[Rorshach]

Ok, but how do I know into which polynomials I am supposed to distribute my wave function? I only know that the hermite polynomial of the third power of x is equal to 8x³-12x. Am I supposed to look for some coefficients that will convert that polynomial into my wave function?

 

[vela]

What are the first few eigenfunctions for the one-dimensional harmonic oscillator?

 

[Rorshach]

Those are the functions that follow the formula ψ_n=C_nexp(\frac{-\xi^2}{2})H_n(\xi), right?

 

[vela]

Right, assuming you have the correct expressions for the normalization constant and $\xi$. I'm going to use $\phi_n$ to denote the eigenfunctions, so you don't confuse it with the given wavefunction $\psi$.

You want to find the appropriate coefficients $a_0, a_1, a_2, \dots$ such that
$$\psi(x) = a_0 \phi_0(x) + a_1 \phi_1(x) + a_2 \phi_2(x) + a_3\phi_3(x) + \cdots$$ Substitute in for $\psi$, $\phi_0$, etc., and see if you can figure out how to solve for the coefficients. You should easily be able to see why $a_n=0$ for $n\ge 4$.

 

[Rorshach]

a_n=0, because the highest power in the formula is 3, right? I got to the point where I am supposed to match the coefficients, but I am somewhat confused by hermitian H_n(\xi)=\sum_{n=0}^\infty c_n\xi^n, namely: I know that hermitian with the power of \xi being equal to 3 is H_3(\xi)=8\xi^3-12\xi am I supposed to fix the constants for the elements of the hermitian sum so it would match the formula for the hermitian of third power exp(-\frac{1}{2}\xi^2)C_3(8\xi^3-12\xi), or I am supposed to fix the constants so it would only leave \xi^3?

 

[vela]

I don't understand what you're asking. Write it out explicitly in math.

 

[Rorshach]

ok, so my wavefunction is \psi(x)=Nx^3exp(-\frac{m\omega x^2}{2\hbar}). Now we substitute certain values: \omega=\sqrt{\frac{k}{m}}, \alpha=\sqrt{\frac{m\omega}{\hbar}}, \lambda=\frac{2E}{\hbar\omega}, \xi=x\alpha. Then we can write the formula for the eigenfunctions: \psi(\xi)=C_nH(\xi)exp(-\frac{\xi^2}{2}). We can write that H_n(\xi)\sum_{n=0}^\infty\ c_n\xi^n, right? Now I know that hermit polynomial H(\xi)for \xi^3 is equal to 8\xi^3-12\xi. And now I don't know what coefficients I am supposed to be looking for: the ones that will give me just \xi^3, or the ones that will give me 8\xi^3-12\xi.

 

[vela]

No. Reread post #16.

 

[Rorshach]

So it is supposed to be something like \psi(x)=c_0H_0(\xi)exp(-\xi^2/2)+c_1H_1(\xi)exp(-\xi^2/2)+c_3H_3(\xi)exp(-\xi^2/2)?

 

[vela]

Yes, but you don't want to neglect the normalization constants, $C_n$.
$$\psi(\xi) = a_0 [C_0 H_0(\xi) e^{-\xi^2/2}] + a_1 [C_1 H_1(\xi) e^{-\xi^2/2}] + a_2 [C_2 H_2(\xi) e^{-\xi^2/2}] + a_3 [C_3 H_3(\xi) e^{-\xi^2/2}]$$ Write out the lefthand side explicitly as well.

 

[Rorshach]

ok, but how do I know how to estimate a_n coefficient, if powers of \xi repeat? How do I know if it should be zero or some different value?

 

[Rorshach]

I came up with coefficients a_1=\frac{3\sqrt{3}}{2},a_3=\frac{\sqrt{3}}{2}, other coefficients should be equal to zero. Is it correct?

 

[vela]

Show your work.

 

[Rorshach]

\psi(\xi)=a_0[c_0 exp(-\frac{\xi^2}{2})]+a_1[c_1 2\xi exp(-\frac{\xi^2}{2})]+a_2[c_2 (4\xi^2-2)exp(-\frac{\xi^2}{2})]+a_3[c_3(8\xi^3-12\xi)exp(-\frac{\xi^2}{2})]
\psi(\xi)=a_0[\sqrt[4]{\frac{m\omega}{\pi\hbar}}exp(-\frac{\xi^2}{2})]+a_1[\frac{1}{2}2\xi\sqrt[4]{\frac{m\omega}{\pi\hbar}}exp(-\frac{\xi^2}{2})]+a_2[\frac{1}{2\sqrt{2}}\sqrt[4]{\frac{m\omega}{\pi\hbar}}(4\xi^2-2)exp(-\frac{\xi^2}{2})]+a_3[\frac{1}{4\sqrt{3}}\sqrt[4]{\frac{m\omega}{\pi\hbar}}(8\xi^3-12\xi)exp(-\frac{\xi^2}{2})]
I figured that since there are only supposed to be $\xi^3$, then in front of the equation with $\xi^3$ I should put something that would change $\frac{1}{4\sqrt{3}}\sqrt[4]{\frac{m\omega}{\pi\hbar}}(8\xi^3-12\xi)exp(-\frac{\xi^2}{2})]$ into just $\xi^3$, so $a_3=\frac{\sqrt{3}}{2}$, and $a_1=\frac{3\sqrt{3}}{2}$. Where am I making mistake?

 

[vela]

You're getting closer. How do you know the coefficient of $\xi^3$ is supposed to be $\sqrt[4]{\frac{m\omega}{\pi\hbar}}$ without knowing specifically what $\psi(\xi)$ equals? Did you ever work out what the normalization constant $N$ is equal to?

 

[Rorshach]

Actually I did, but completely forgot about it:P thank You:) It should be equal $N^2=\frac{2m^2\omega^2}{exp(m\omega x^2)(m\omega x^2-1)}$. Now I should substitute it to the equation on the left side right?

 

[vela]

That's not right. $N$ is a constant. It doesn't depend on $x$.

 

[Rorshach]

ok, it should be indefinite integral, so $N^2=\frac{8(-\hbar m\omega)^\frac{7}{2}}{15\sqrt{\pi}}$, right?
now it should be ok to substitute it to the left side of the equation

 

[vela]

That's close, but you seem to have made a few algebra errors. You should get
$$N^2 = \frac{8}{15\sqrt{\pi}}\alpha^7$$ where $\alpha$ is as you defined it above.

 

[Rorshach]

I tried again, but I came to conclusion that $N$ should stand in front of one expression, in this case
$Na\frac{1}{4\sqrt{3}}\sqrt[4]{\frac{m\omega}{\pi\hbar}}(8\xi^3-12\xi)exp(-\frac{\xi^2}{2})$, and the coefficient came out as $N\frac{4\sqrt{3}}{8c_3}$. Is it correct?

 

[vela]

I have no idea what that means.

 

[Rorshach]

I got lost. I don't know where to include the $N$ factor in the formula for the $\psi(\xi)$. I know that $c_n=\frac{1}{\sqrt{2^n n!}}\sqrt[4]{\frac{m\omega}{\pi\hbar}}$,but I don't know how to mix the constant $N$ into this- should it just stand in front of one of the expressions, or should it truncate- I have no idea now.

 

[Rorshach]

I am at the point where I know that my function is supposed to be a sum of eigenfunctions, but the constants are something I don't know how to calculate...

 

[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.