Eigenfunctions and their Eigenvalues

g782k936
Messages
3
Reaction score
0
If I have two eigenfunctions of an operator with the same eigenvalue how do I construct linear combinations of my eigenfunctions so that they are orhtogonal?

My eigenfunctions are: f=e^(x) and g=e^(-x)

and the operator is (d)^2/(dx)^2
 
Physics news on Phys.org
Orthogonal wrt what?

You need a scalar product.

Daniel.
 
I want to have linearly independent combinations of f and g that are orthognal on the interval from (-1,1) I'm guesing that they need to be wrt f and g.
 
g782k936 said:
I want to have linearly independent combinations of f and g that are orthognal on the interval from (-1,1) I'm guesing that they need to be wrt f and g.
No, that was not the question. "Orthogonal" means that the inner product is 0 so whether or not two vectors are orthogonal depends on the inner product used.

The most common inner product for real valued functions on an interval (a, b) is \int_a^b f(x)g(x)dx.

Since, if two eigenvectors correspond to the same eigenvalue, any linear combination is also an eigenvector corresponding to that eigenvalue, a simple "orthogonal projection" will work.

If u and v are two vectors in an inner product space, then the "projection of v onto u" is given by
\frac{<u,v>}{<u,u>}\vec{u}
The "orthogonal projection" is v minus that:
\vec{v}- \frac{<u,v>}{<u,u>}\vec{u}

Calculate that with u= ex, v= e-x, and inner product <u,v>= \int_{-1}^1 u(x)v(x)dx.
 
O.k.

I think that worked. I had been trying the integral in a slightly different way using f + g instead of fg.

Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top