# Help graphing wave functions and probability densities.

1. Oct 26, 2014

### Ascendant78

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Well, I felt like the first part wasn't too bad and graphed the potential like so:

However, to be honest, I'm not even sure if I did that right, as I wasn't sure whether he wanted it as a 2-D or 3-D graph (and I can't even figure out how to graph 2-D in Mathematica). I have a group of 7 of us from class that work together and none of us have ever done anything like this, so we are at a loss? We don't know how to go about plugging the multiple variables (wave function) into Mathematica to get what we want and the instructor gave us no assistance at all. The equations from pg 224 are completely different than the ones given on the homework and none of us know why. We are just completely lost and desperately need help.

2. Oct 26, 2014

### Staff: Mentor

I don't think you are expect to do a 3D plot (on which you can't see anything). Here, $k$ turns out to be simply a scaling constant. You can plot everything in terms of $x/\sqrt{k}$.

If you mean plotting many curves on the same plot, simply use {} to specify the list of plots, for example:
Code (Text):

Plot[{x^2, x^3},{x,-10,10}]

Completely different? They are the same. It is time for you to get accustomed to varying (arbitrary) choices of notation. For instance, can you figure out the relationship between $b$ and $\alpha$?

3. Oct 26, 2014

### Ascendant78

Ok, I'm at a loss as to how 1/2kx^2 would be converted to 1/2x/(sqrt(k)), unless I misunderstood that explanation? Also, I still don't have any idea how to plot in 2-d in Mathematica, nor does anyone else from our group. Is there maybe some free software out there I could do this in that is more user-friendly?

What I meant as far as plotting the wave function, I meant how to plot it with both a variable k and m in the function, as I'm assuming from the homework, he is expecting us to expand b into those components for each of the 3 states? Also, as far as plugging in a value for h, we are not sure whether we should be using the Js value or eVs value of h?

From our best understanding, we think we use the harmonic oscillator function on the quiz and plug in each hermite value for the 3 states. However, we are not sure whether or not we should substitute the "b" value in the exponent with the given value of b=m/k...etc., or just leave b as-is in the function?

Also, yes, we did see what you were saying as far as comparing a to b in the functions. We were just so lost on the graphing concept that we really didn't take the time to try and manipulate the values like that. Thank you for clarifying though.

4. Oct 27, 2014

### Ascendant78

Ok, so now we are using Graphmatica and things got a lot easier. However, we are still unsure what to do with the potential k for each value of n? The book doesn't seem to show the relation of how k changes with respect to each energy state? Can someone help out please?

5. Oct 27, 2014

### haruspex

Perhaps DrClaude meant x√k.

6. Oct 27, 2014

### Staff: Mentor

Yes, indeed.

But I must apologize since I missed the last line of the assignment, which explicitly tells you you can use $y=bx$ to remove the dependency. If you look, this is exactly what is done in the plot you took from a book: the bottom axis is $\sqrt{\alpha} x$. Do that, and the dependence on $k$ will go away.

As I indicated above, plotting in Mathematica is not very complicated. You can also try plotting in Excel. As for free software, you can try gnuplot (although I wouldn't call it more user-friendly than Mathematica) or Grace. The advantage of Mathematica is that you don't need to calculate discrete numerical values for the functions: just give the equation and Mathematica will plot it.

You want to scale the energy the same way $x$ is scaled. If you take $y=bx$ and use it in the equation for $V(x)$ to get a function of $y$ instead, you can then find what you need to multiply $V$ by to get something dimensionless.