Undergrad Even/Odd solutions for particles in boxes

[MaestroBach]

TL;DR

I don't get the even/odd solutions

My book tells me that, for both infinite and finite particle in boxes, that:

Even solutions are for n = 2, 4, 6, etc and have a sin function form, while odd solutions are for n = 1, 3, 5 etc and have a cos function form.

I'm very confused though, because sin functions are odd and cosine functions are even. Do the even and odd terms just apply to the n values?

Also, as a side-note, do these solutions have n-1 nodes?

Thanks

 

[vanhees71]

It depends on how you localize your "box". If you choose the interval as $[0,a]$ you get much more simple
$$u_n(x)=\sin(n k x), \quad n \in \mathbb{N}=\{1,2,3,\ldots \}, \quad k=\frac{\pi}{a}.$$
"Even" and "Odd" refers to the symmetry of the problem under reflections at the midpoint $x_0=a/2$ of the interval, i.e., under the transformation $x \rightarrow a-x$. This is the parity operation $\hat{P}$. Indeed you get
$$\hat{P}u_n(x)=u_n(a-x)=(-1)^{n+1} u_n(x),$$
i.e., the odd-$n$ eigenfunctions have positive, the even-$n$ eigenfunctions negative parity.

If you make the interval $[-a/2,a/2]$, then you get cos and sin solutions, which directly are even and odd functions under reflections $x \rightarrow -x$. Of course these are the same solutions as with the other choice of the interval.

 

[MaestroBach]

It depends on how you localize your "box". If you choose the interval as $[0,a]$ you get much more simple
$$u_n(x)=\sin(n k x), \quad n \in \mathbb{N}=\{1,2,3,\ldots \}, \quad k=\frac{\pi}{a}.$$
"Even" and "Odd" refers to the symmetry of the problem under reflections at the midpoint $x_0=a/2$ of the interval, i.e., under the transformation $x \rightarrow a-x$. This is the parity operation $\hat{P}$. Indeed you get
$$\hat{P}u_n(x)=u_n(a-x)=(-1)^{n+1} u_n(x),$$
i.e., the odd-$n$ eigenfunctions have positive, the even-$n$ eigenfunctions negative parity.

If you make the interval $[-a/2,a/2]$, then you get cos and sin solutions, which directly are even and odd functions under reflections $x \rightarrow -x$. Of course these are the same solutions as with the other choice of the interval.

That makes a lot of sense, thanks.

Only thing I'm still confused about is that my book, which has the interval -a/2, a/2, associates the odd solutions with cosine, despite the fact that cosine is obviously even..

 

[PeroK]

That makes a lot of sense, thanks.

Only thing I'm still confused about is that my book, which has the interval -a/2, a/2, associates the odd solutions with cosine, despite the fact that cosine is obviously even..

The book must mean odd and even numbered.

 

[MaestroBach]

The book must mean odd and even numbered.

Well that's very confusing, thanks

 

[Vanadium 50]

he book must mean odd and even numbered.

So the odd solutions are even-numbered and the even solutions are odd-numbered.

That's really...odd.

 

[Nugatory]

That's really...odd.

No, it's really even, at least as long as

...the odd solutions are even-numbered and the even solutions are odd-numbered.

SCNR