===== Sept 21 (Mon)  SHO wrap up Algebraic approach and move on to Analytical =====
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====Schrodinger's Dog 13:58 9/19====
Does anyone know of a potential dependent on a differential operator? So far, all the potentials I seen in QM have non-differential operators (i.e. x, 1/x^2, etc..). Thanks!

====Andromeda 21:31 9/19====
I am still very confused by this which might be a very small thing but where exactly is the square root of(1/2hmω)factor comes from in page 42??? it seems inconsistent with eq. 2.45 and it didnt come up in the lecture yesterday. 

===Blackbox 22:41 9/19===
I think the factor in front of 2.47 is just multiplied in order to see the results better. As you know a± was defined in 2.47. After all mathmatical calculations have been done in page 43, finally the Hamiltonian can be written as 2.56. If you put 2.52 or 2.54 into 2.57 then you will get the same result as 2.45.
===Andromeda23:18 9/19===
so without hω factor in 2.47 we would get hω/2 instead of the 1/2 at the end...therefore they are just defining 2.47 like that to get a nicer factor at the end or is there more to it???

===Captain America 11:01 9/21===
I might be confused on the question being asked, but I think the hω becomes clearer when you follow the steps Griffiths takes down page 44.  Without hω making the final units after you operate into the units of energy, you wouldn't be able to get the final equation on the page, and the "(E-hω)" would be "(E-1)" but E-1 doesn't make sense because 1 doesn't have the same units as E.  

So it seems necessary to put the square root factor in there since you will be multiplying a± with itself, thus giving you 1/2hmω.  This will in turn lead to the correct units when finally using the operator to find the next step in the ladder.

====Hydra 14:30 9/20====
How does this evenly-spaced ladder model agree with the non-evenly-spaced levels we see in, say, hydrogen?
===Schrodinger's Dog 17:55 9/20===
Where does it say that the SHO potential uses a evenly-spaced ladder model? And for hydrogen, Griffths didn't use the the ladder operators, he used spherical coordinates and some techniques in solving coupled differential equation. This may be cause they are "non-evenly-spaced levels". If there is a book on using ladder method use on the columb potential with hydrogen I ll check it out.

====vinc0053 9/20 20:08====
How does Griffiths arrive at equations 2.65 (pg 48)?  I have looked over 2.57, 2.61, and 2.65 and cannot find the same conclusion.  Is he merely glossing over some aspects (as he often does) to cut to a nice proportionality?  I'm lost, and he uses 2.65 to simplify example 2.5.
===spillane 9/21 7:00===
It appears that if you 2.57=2.61 at EΨ then you can rearrange this expression to obtain 2.65. In one case subtracting 1/2 and in the other adding 1/2

====vinc0053 9/20 20:16====
I need a refresher on orthogonal relation.  What does it conceptually mean for two states to be orthogonal?  What useful consequences will this have in QM (other than Fourier's trick)?  I may be over-thinking this.

===Captain America 9/21 10:39===
I think the easiest conceptual way to picture the orthogonality relation is to consider the any bound particle in a potential well.  The infinite square well for example has specific energy states that are all orthogonal to each other, but what that really means is they have to be at certain energies to exist based off of the boundary conditions.  Since each stationary state in the infinite square well has to be able to be normalized, they have to go to zero at x= plus and minus infinity.  This allows only certain energy levels to exist in the well.

For 3-D objects (the electron cloud around a hydrogen atom for example), I still like to think of orthogonality in this way.  The wave function has to be of a certain energy level, thus giving us a certain number of energies that can exist, each successive state having one more node than the previous one.  It gives interesting probability densities in spherical harmonics, as seen on page 157.

Page 33 shows the proof of why each of the states in a infinite square well has to be orthogonal to each other state, if that helps to conceptualize it.

====chap0326 9/20 21:08====
Does anyone know what the formula would be for finding the expectation value of the momentum squared?
===Andromeda 22:13 9/20===
formula 1.36 in the book is the general function for finding an expectation value; so if you square the momentum operator and insert it in place of Q(x,p) in 1.36 you should have a formula for expectation value of momentum squared.
===Green Suit 9/21 20:45===
 Prof. Yuichi started this in class leaving us with: <math> <p^2>=\frac{-mh\omega}{2}\int\Psi_n^\ast [a_-^2 - (a_-a_+) - (a_+a_-)a_+^2] \Psi_n dx</math>. We also assertained that <math>a_-^2</math> and <math>a_+^2</math> both go to zero. From 2.65, <math> -(a_-a_+)=-(n+1)</math> and <math> -(a_+a_-)=-n</math>. These two are constants and get pulled out which now gives us: <math>-(2n+1)\frac{-mh\omega}{2}\int\Psi_n^\ast \Psi_n dx</math>. The two negatives cancle and solves the issue of negative momentum that Prof. Yuichi identified earlier.

===== time to move on =====

It's time to move on to the next Q_A: [[Q_A_0923]]