Question

So I am working on a solve the infinite square well problem and I don't understand something.

I know I should know this but I am drawing a blank. I don't understand how:

*****The question is in the comments made by me I used the equation writer (SORRY)****

I assume there is a rule that is missing but which one?

----I commented below so you can see what it is supposed to look like with the equation writer.-----

Answer 1

\[B _{1}\cos (\frac{ \pi n }{ a } x)\]
and
\[B _{2}\sin (\frac{ \pi n }{ a } x)\]
equal\[B\sin (\frac{ \pi n }{ a } x + \frac{ \pi n }{ 2 })\]

Answer 2

Generally speaking it doesn't... where did this come up? You should have gotten sine and cosine as your possible solutions, but then your boundary conditions would exclude cosine.

Answer 3

This is where I got it from.

You will find the statement where it says it is convenient to write both cases using a single formula.

Answer 4

Oh, I see. The logic goes as follows:
\[ B \sin\left( \frac{n\pi}{a} x + \frac{n\pi}{a} \right) = B\sin\left(\frac{n\pi}{a} x\right)\cos\left(\frac{n\pi}{a} \right) + B\cos\left(\frac{n\pi}{a}x\right)\sin\left(\frac{n\pi}{a}\right) \]

Answer 5

Oops, sorry -- that should be cos(n pi/2) and sin(n pi/2) in the first and second terms respectively

Answer 6

Anyway, if n is even, cos(n pi /2) = +1 or -1 and sin(n pi/2) = 0, so you get
\[ \pm B \sin\left(\frac{n\pi}{a} x\right) \]
if n is odd, the opposite is true, yielding
\[ \pm B \cos\left(\frac{n\pi}{a} x \right) \]

Answer 7

So that single expression accounts for both cases.

Answer 8

Ok I thank you. But I have to say that leads to a (what some could say) messy integral in order to normalize. Is there a simpler way to solve SE without using "the single expression that accounts for both cases"?
Like you said earlier that cos becomes excluded once you apply the boundary conditions.

Answer 9

Sorry I fine finding the energy for a finite well a bit simpler since you don't have to solve the SE.

Answer 10

No, not the way you've done it. Typically the infinite square well has the boundary walls at 0 and a, which allows only the sine functions. If you have boundaries at -a/2 and a/2, the problem becomes a little more complicated and your expression is in terms of sine and cosine, depending on the state you're looking at (n=1,2,3...)

Answer 11

So problems that are symmetric about the origin will normally have a solution that is like the one I'm doing now?

Answer 12

The integral is really quite easy to solve, and I'll show you why. If I have
\[ \int_{-a/2}^{a/2} \sin^2(n\pi x/a + n\pi / 2) dx\]
I can rewrite that as
\[ \int_{-a/2}^{a/2} \frac{1}{2}- \frac{1}{2}\cos(2n\pi x/ a + n\pi) dx\]
Because of a simple trig identity. But notice that the second term will be zero for every value of n. Therefore, you're left with only
\[ \int_{-a/2}^{a/2} \frac{1}{2} dx = \frac{a}{2} \]

Answer 13

Yes. The reason one might do the problem like this, although its notationally more complicated and cumbersome, is that it introduces you to the idea of parity, and the connection between symmetry in the potential and symmetry in the solutions. In this solution, all odd n solutions are odd, or antisymmetric about the origin, while all even n solutions are even, or symmetric, about the origin.

Answer 14

I see it now. Thank you, I think I was psyching my self out for no reason. So really it is just a matter of figuring out how to simplify the two separate equations. Which will almost always have some variant of that solution from the looks of it.

Answer 15

I'm not sure what you mean by that last part.

Answer 16

NVM that last part thanks.