# Eigenfunctions/bases/fourier series

1. Mar 25, 2005

### maple

I posted this in the general maths section but i think it'll fit better over here.

Anyways...

In quantum mechanics, a free particle in an infinite potential well has the wave function (ie. overlap <x/phi>). Its eigenfunctions take the form:

(2/a)^1/2 * sin(n*pi*x/a), n is ofcourse an integer.

My question is that do all eigenfunctions form a basis? And if so how can you represent an even function with eigenfunctions which are clearly odd- my understanding of the Fourier Series is that its equals a function by representing it as an infinite sum of both sin and cos terms, and if the function is even, the coefficients of the sin terms are zero.

Eg. why can I represent:

cos (pi*x/a) = Infinite Sum (A(subscript)n * sin(n*pi*x/a)).

Any assistance would be much appreciated

2. Mar 25, 2005

### seratend

You also have K*cos(n*pi*x/a) depending on the parity of n. (I suppose that You have a symmetric potential well (-a/2 to +a/2).

For all practical purposes yes for an obseervable. It is in fact included on the definition of an observable: it is a "gentle" hermitian operator where its eigen vectors form a basis of the Hilbert space. Howerver, the class of these operators with such a mathematical property is difficult to define/verify. Try googling spectral theory for more information.

Seratend.

3. Mar 25, 2005

### maple

Thanks Seratend,

although, the well goes from (0 to a): i don't think there are any cos terms that make up the enegy-eigenfns.

so if they form a basis, you can represent any function as linear combinations of these eigenfunctions. Fourier series seems to suggest that you cannot express even functions as linear combinations of odd ones. So cos(n*pi*x/a) is clearly even, so how can you represent it in terms of eigenfns which are sin terms.

also what do you mean by "gentle" Hermitian operator... an observable has to be hermitian because eigenvalues must be real... complex ones are unphysical?

4. Mar 25, 2005

### dextercioby

Yes,eigenvalues (more general,spectral values) need to be real (ergo the linear operators need to be self-adjoint,or hermitean at least),because these are numbers which can be experimentally obtained.You can't measure energy (par éxample) and get a complex (i.e.imaginary) value...

Daniel.

5. Mar 25, 2005

### nrqed

No

You are absolutely correct there.

No, you cannot do that.

Both the sine AND the cosine form the basis. You cannot express the cos as a linear combination of the sine, as you can not express the unit vector $${\vec i}$$ as a linear combination of the basis vectors $${\vec j}$$ and $${\vec k}$$.

So in general, you need both the sine and the cosine. In the case of an infinite square well from 0 to a, any physical wavefunction will vanish at x=0 and at x=a and therefore it does not contain any cosine "piece".

Regards,

Pat

6. Mar 25, 2005

### maple

Thanks Pat, that really cleared somethings up.

I solved the TISE again, and I still get that the eigenfunctions are simply K*sin(n*pi*x/a) where K is a normalised constant as before, applying the boundary condition that you mention alongside the condition that the gradient of the wavefunction must be continuos.

So the problem lies therein. Perhaps I'm solving it incorrectly?

7. Mar 25, 2005

### dextercioby

You can always check with a book.Almost all introductory texts on QM deal with the infinite potential (square) well.Hopefully i'm not telling you something new.

Messiah,Cohen-Tannnoudji,Griffiths,exercises in Flügge,Constantinescu & Magyari,etc.

Daniel.

8. Mar 25, 2005

### seratend

And if you shift your x axis by a/2 (you just change the origin of your problem), you obtain:

K*sin(n*pi*x-a/2/a)=K*sin(n*pi*x/a - n*pi/2)
= K*sin(n*pi*x/a ) if n is even
= K*cos(n*pi*x/a) if n is odd

What do you conclude now?

You need to understand what is a fourier series on a given interval (a,b) that is not symmetric respectively to 0 (mathematic problem). The quantum well is a good example to understand that.
In addition you must understand what is the hilbert space associated to the quantum well: the space of functions square integrable with given values on boundaries (in this case 0). Therefore it is normal that the eigenvectors that *form* a basis of this peculiar hilbert space follow these conditions and all the possible functions also. The functions on the square well are a subset of all the square integrable functions (addition of the boundary condition) on (a,b), they are not symmetric respectively to the origin (no signification) and they may be written as a sum of sin(n*pi*x/a).
If you develop a little you may see that you can construct odd or even functions respectively to a/2.

Seratend.

9. Mar 25, 2005

### maple

Aha! It all makes sense now; quite interesting stuff. Thanks!

10. Mar 25, 2005

### Hurkyl

Staff Emeritus
A basis of the solution space... not of all possible wavefunctions.

11. Mar 25, 2005

### SpaceTiger

Staff Emeritus
To follow up on what Hurkyl said, that's only within the solution space. If you can represent any function with them, then it means they're complete. It's not true that all eigenfunctions for all potentials are complete, but it's usually the case.