Calculating a Quantum Nomalization Constant

[Seda]

Homework Statement

If I have a wave function that = Ce^{((-mwx2)/(2*hbar))}

(Its the wave function of the ground state of a simple harmonic oscilator)How do I calculate C?

Homework Equations

Quantum Normalization condition I think is all i need.

The Attempt at a Solution

C² is pulled out of the integral because its a constant

Leaving me with the integral from - infinity to infinity of e^-mwx2/hbar

How do I integrate that? Is there an easier way to solve for C?

Everythings 1D

 

[Niles]

Here's the proof of the integral:

http://quantummechanics.ucsd.edu/ph130a/130_notes/node87.html

 

[Seda]

Ok, now I know that

C= sqrt[1/(sqrt((pi*hbar)/(m*w)))]However, I don't know w.

Odd...

 

[Seda]

Ok, now I need to solve for <x²>

Which means I obviously end up with:

C² times the integral from - infinity to inifinty of x²*e^-ax2

where a = mw/hbar

I can seem to find a solution to this integral in my handbook. How do you intgrate that?

 

[Niles]

https://www.physicsforums.com/library.php?do=view_item&itemid=139

It's a good trick - learn to use it, because it is not the last time you will have to integrate a Gaussian when doing quantum mechanics.

 

[DeShark]

https://www.physicsforums.com/library.php?do=view_item&itemid=139

It's a good trick - learn to use it, because it is not the last time you will have to integrate a Gaussian when doing quantum mechanics.

I can't get that link to work... It says I don't have permission.. Any ideas?

 

[Niles]

<br /> <br /> \begin{align*}<br /> \int_{ - \infty }^\infty {x^2 \exp \left( { - \lambda x^2 } \right)} dx = - \int_{ - \infty }^\infty {\frac{\partial }{{\partial \lambda }}\exp \left( { - \lambda x^2 } \right)} dx = - \frac{\partial}{{\partial\lambda }}\int_{ - \infty }^\infty {\exp \left( { - \lambda x^2 } \right)} dx = - \frac{d}{{d\lambda }}\sqrt {\frac{\pi }{\lambda }}<br /> \end{align*}<br /> <br />

- this is what the link contains. It is called differentiation under integration.

Of course partial integration can be used too, but the above method simplifies things greatly.