Harmonic potential, quantum mechanics

[fluidistic]

Homework Statement

Consider a particle of mass m in a unidimensional harmonic potential of the form $$V(x)=\frac{kx^2}{2}$$.
1)Show that a function of the form $$\Psi (x)=Ce^{-\alpha x^2}$$ is an eigenfunction of the time-independent Schrödinger's equation. Calculate the values of E, C and $$\alpha$$.
This solution corresponds to the minimum energy of the unidimensional harmonic oscillator.
2)Verify the uncertainty principle for a particle that have this state of energy.
3)From the uncertainty principle, estimate the the ground state energy for a unidimensional harmonic oscillator and compare it with the value obtained in 1).

Homework Equations

$$-\frac{\hbar ^2}{2m} \frac{d ^2 \Psi}{dx^2} + V(x)\Psi = E \Psi$$.

The Attempt at a Solution

What I've done is calculate the first 2 derivatives of the given $$\Psi$$, with respect to x.
It gave me $$\Psi ' (x)=2x \alpha C e^{-\alpha x^2}$$ and $$\Psi '' (x)=-2\alpha C e^{-\alpha x^2} (1+2x^2 \alpha)$$.
Then replacing $$\Psi ''(x)$$ and $$\Psi (x)$$ into Schrödinger's equation, I reached $$-\frac{\hbar ^2}{2m} (1+2x^2 \alpha)+E-\frac{kx^2}{2}=0$$ if I assume $$C\neq 0$$ which I think is fair.
Now I'm totally stuck. I've isolated E from the equation, which gave me $$E=\frac{kx^2}{2}+\frac{\hbar ^2}{m}(1+2 x^2 \alpha)$$ and I can recognize the potential energy part and what would be the kinetic energy?
The minimum kinetic energy would be when x=0, in this case the potential energy is also minimum (worth 0J), and it gives the ground state energy $$\frac{\hbar ^2}{m}$$. But I'm afraid, I've no idea what $$\alpha$$ and $$C$$ might be.
Maybe I can get C from a normalization? That is, by setting $$C^2\int _{-\infty}^{\infty} |e^{-\alpha x^2}|^2 dx=1$$. But it seems I have 2 unknowns and 1 equation...

I would like to know if I'm doing things right and ideas on how to proceed a bit further. Thanks a lot.

 

[fzero]

Your equation

$$-\frac{\hbar ^2}{2m} (1+2x^2 \alpha)+E-\frac{kx^2}{2}=0$$

isn't quite right, since you left out a numerical factor from $$\Psi''$$. Assuming that you can recheck your algebra, what you'll find is some expression

$$E = M + N x^2$$

where M,N are numerical factors given in terms of $$\hbar, m, \alpha$$, etc. For E to be an eigenvalue it must be independent of x, so if you can solve N =0, then E will be an eigenvalue. This will determine $$\alpha$$. Then as you correctly suspect, C can be determined from normalizing $$\Psi$$.

 

[fluidistic]

Thank you fzero for your help.
Ok, I've done a small error in calculating $$\Psi '' (x)$$. Now it gives me $$\Psi '' (x)=2C \alpha e^{- \alpha x^2} (2x^2 \alpha -1)$$.
Replacing into Schrödinger's equation, I get $$2Ce^{-\alpha x^2} \left [ \alpha (2x^2-1) +\frac{m}{\hbar} \left ( E-\frac{kx^2}{2} \right ) \right ]=0$$. I assume I can divide by $$2Ce^{-\alpha x^2}$$ and I'm still left with a dependence of x...
Now I don't see what I'm doing wrong.
Edit: nevermind, I guess it's right.
Edit 2: I get $$\alpha =\sqrt {\frac{mk}{4 \hbar}}$$. I'll work on C now.

 

[gabbagabbahey]

It gave me $$\Psi ' (x)=2x \alpha C e^{-\alpha x^2}$$ and $$\Psi '' (x)=-2\alpha C e^{-\alpha x^2} (1+2x^2 \alpha)$$.

Be careful with those negative signs

 

[fluidistic]

Be careful with those negative signs

Yeah... I thought I could do the algebra in my head, until someone pointed me out I had made an error. I then redone the algebra with much more care, writing down all the steps on a sheet of paper and I think I got it right now.

Hmm, in order to get C, I have that $$C^2 e^{-\sqrt{\frac{mk}{\hbar}}} \int _{-\infty}^{\infty} e^{x^2}dx=1$$.
Just looking at the integrand, is it a famous integrand that is known not have a closed form primitive? If so, what do I do?!

 

[fzero]

Yeah... I thought I could do the algebra in my head, until someone pointed me out I had made an error. I then redone the algebra with much more care, writing down all the steps on a sheet of paper and I think I got it right now.

Hmm, in order to get C, I have that $$C^2 e^{-\sqrt{\frac{mk}{\hbar}}} \int _{-\infty}^{\infty} e^{x^2}dx=1$$.
Just looking at the integrand, is it a famous integrand that is known not have a closed form primitive? If so, what do I do?!

You still have a sign problem, because

$$\int _{-\infty}^{\infty} e^{x^2}dx$$

does not converge. The Gaussian integral

$$\int _{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}.$$

does converge.

Edit: Oh

$$C^2 e^{-\sqrt{\frac{mk}{\hbar}}} \int _{-\infty}^{\infty} e^{x^2}dx=1$$

seems to be the result of some dodgy manipulations of the exponential.

$$e^{-\sqrt{\frac{mk}{\hbar}}x^2} \neq e^{-\sqrt{\frac{mk}{\hbar}}} e^{x^2}.$$

You want to make a change of variable

$$u^2 = \sqrt{\frac{mk}{\hbar}}x^2$$

in order to do the integral.

 

[fluidistic]

You still have a sign problem, because

$$\int _{-\infty}^{\infty} e^{x^2}dx$$

does not converge. The Gaussian integral

$$\int _{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}.$$

does converge.

Oh right. Even worse than that, I'm making other errors. I have $$C^2 \int _{-\infty}^{\infty} |e^{-\sqrt{\frac{mk}{4\hbar}}x^2}|^2 dx=1$$.
I'll try to finish this out.

 

[gabbagabbahey]

Yeah... I thought I could do the algebra in my head, until someone pointed me out I had made an error. I then redone the algebra with much more care, writing down all the steps on a sheet of paper and I think I got it right now.

Hmm, in order to get C, I have that $$C^2 e^{-\sqrt{\frac{mk}{\hbar}}} \int _{-\infty}^{\infty} e^{x^2}dx=1$$.
Just looking at the integrand, is it a famous integrand that is known not have a closed form primitive? If so, what do I do?!

You mean $$|C|^2 \int _{-\infty}^{\infty} e^{-\sqrt{\frac{mk}{\hbar^2}}x^2}dx=1$$, right?

Remember, $e^{ab}\neq e^ae^b$, and double check your value of $\alpha$ again.

As for evaluating the integral, start with a substitution $\overline{x}=\sqrt{\alpha}x$, then either look up the result of the integral $$\int_{-\infty}^{\infty}e^{-\overline{x}^2}d\overline{x}$$, or calculate it yourself using the following trick:

$$\int_{-\infty}^{\infty}e^{-\overline{x}^2}d\overline{x}=\int_{-\infty}^{\infty}e^{-\overline{y}^2}d\overline{y}$$

So...

$$\begin{aligned}\left(\int_{-\infty}^{\infty}e^{-\overline{x}^2}d\overline{x}\right)^2 &= \left(\int_{-\infty}^{\infty}e^{-\overline{x}^2}d\overline{x}\right)\left(\int_{-\infty}^{\infty}e^{-\overline{y}^2}d\overline{y}\right) \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\overline{x}^2-\overline{y}^2}d\overline{x} d\overline{y} \\ &= \int_{0}^{\infty}\int_0^{2\pi}e^{-\overline{r}^2}\overline{r}d\overline{r}d\overline{\theta}\end{aligned}$$

 

[fzero]

Oh right. Even worse than that, I'm making other errors. I have $$C^2 \int _{-\infty}^{\infty} |e^{-\sqrt{\frac{mk}{4\hbar}}x^2}|^2 dx=1$$.
I'll try to finish this out.

I edited my last post to point out an error you seem to have made. Just posting so that you don't miss it as it will save you some time.

 

[fluidistic]

Ok thanks guys. I can't believe I made this error with the exponential, I wasn't even tired! Thanks for the tips to solve the integral, I will try them.
I triple checked my value for $$\alpha$$ and I still get the same error.
Following post #2's notation, I think my value for N is wrong. Because it seems to lead to my value for $$\alpha$$. My N is worth $$\frac{k}{2}-\frac{2\alpha ^2 \hbar}{m}$$. And as said, I solved for N=0 in order to get $$\alpha$$.
I really don't see my error for $$\alpha$$... how did you get $$\alpha =\sqrt {\frac{mk}{\hbar}}$$?
Oh now I see another error I made. Ok I redo everything from scratch.
Edit: done. I get $$\alpha = \frac{\sqrt {mk}}{2\hbar}$$. Is this possible?

 

[vela]

Edit: done. I get $$\alpha = \frac{\sqrt {mk}}{2\hbar}$$. Is this possible?

Yeah, that's right.