00:04
Schrodinger's equation for the quantum harmonic oscillator is going to be minus h bar squared over 2m d squared d x squared so that's our momentum operator times si plus u which u is equal to one -half m omega squared x squared times si and then this is going to be equal to e si and if this is equal to e -sci, if we find functions that satisfy this, these are energy - eigenfunctions with that corresponding energy.
00:44
So we're told to check solutions of the form, si0 equals a knot, e to the minus alpha squared x squared over 2.
01:01
So this is a gaussian solution and alpha squared is equal to m omega over hbar.
01:10
So this is going to look something like this, these solutions.
01:17
So we check them by plugging into here and here, sine not, and then seeing if we spit out some constant energy.
01:32
So first we need to find the derivatives of sin not.
01:36
So d by d x, si not, we just use the properties of the gaussian and or the exponential function and the chain rule.
01:48
We get that it's minus alpha squared x times si not.
01:58
So all that is is just the gaussian or an exponential function returns itself times the derivative of its argument, which is this right here.
02:09
So now we take the second derivative of si not, and taking the second derivative, we get that it's equal to alpha to the fourth times x squared minus alpha squared times si not.
02:39
So if we plug this in to our equation, so we're plugging this in right here, we get minus h plus.
02:52
Bar squared over 2m times alpha to the fourth x squared minus alpha squared times si -0 plus 1 half m omega squared x squared psi zero however this right here is coefficient on the x squared if we compare to this then we can see that one -half one -half m omega squared is equal to the same thing as alpha to the fourth times h bar squared over to m so noticing that we can rewrite this equation can get rid of this and just write this as plus h bar squared over 2m alpha to the fourth well then this term right here cancels with this term right here.
04:19
And what we're left with is that minus alpha squared over sine not, sorry, positive alpha squared over side not because of this negative sign right there, is equal to, now equating it with the energy of the schroding equation from right here.
04:41
So we can conclude that e0, so the emerson, of whatever state this is, sine knot, is equal to m squared omega squared over hbar squared.
05:01
Sorry, get rid of those squares.
05:03
Alpha is already squared.
05:04
So it's just equal to this quantity right here.
05:08
So we get that m0 is equal to m omega over hbar.
05:15
And this is what we wanted to confirm.
05:19
So we have this.
05:21
But now we're asked to do one more thing.
05:29
We are also told to find the normalization condition for our wave function.
05:33
The normalization condition says that the integral from negative infinity to infinity of psi -not star, which is a complex conjugate of si -not, times psi -0, dx, will be equal to 1...
