Is this correct? (eigenfunctions)

  • Thread starter Thread starter hoch449
  • Start date Start date
  • Tags Tags
    Eigenfunctions
hoch449
Messages
13
Reaction score
0

Homework Statement



a) Show that the functions f=sin(ax) and g=cos(ax) are eigenfunctions of the operator \hat{A}=\frac{d^2}{dx^2}.

b) What are their corresponding eigenvalues?

c)For what values of a are these two eigenfunctions orthogonal?

d) For a=\frac{1}{3} construct a linear operator of f and g which is orthogonal to f

The Attempt at a Solution



a) \hat{A}f=\frac{d^2}{dx^2}sin(ax)=-a^2sin(ax)
\hat{A}g=\frac{d^2}{dx^2}cos(ax)=-a^2cos(ax)

b) the eigenvalues are -a^2

c)orthogonality condition is: \int f^*gdx=0

so to satisfy the above condition a would have to be \pm\frac{n\pi}{2} \pm\n\pi where n=\pm1,\pm2,\pm3...

d) I have no clue how to do this one.. Any help?
 
Physics news on Phys.org
hoch449 said:

Homework Statement



a) Show that the functions f=sin(ax) and g=cos(ax) are eigenfunctions of the operator \hat{A}=\frac{d^2}{dx^2}.

b) What are their corresponding eigenvalues?

c)For what values of a are these two eigenfunctions orthogonal?

d) For a=\frac{1}{3} construct a linear operator of f and g which is orthogonal to f

The Attempt at a Solution



a) \hat{A}f=\frac{d^2}{dx^2}sin(ax)=-a^2sin(ax)

\hat{A}g=\frac{d^2}{dx^2}cos(ax)=-a^2cos(ax)
Okay.

b) the eigenvalues are -a^2
Okay.

c)orthogonality condition is: \int f^*gdx=0
Integrated over what interval?

so to satisfy the above condition a would have to be \pm\frac{n\pi}{2} \pm\n\pi where n=\pm1,\pm2,\pm3...[/quote]
How did you get that?

d) I have no clue how to do this one.. Any help?
What does it mean for a linear functional be be orthogonal to a (function). For that matter why are they asking for a linear operator of f and g? Why not just a linear operator on the set of function f and g belong to?
 
For part c) the integral is performed under all space.

When you write out the integral, the integrand is [sin(\frac{x}{3})]^*cos(\frac{x}{3}) with a constant on the outside of the integral. Now for this to be 0. X would have to be those values that I wrote in my previous post. Does this seem right?

and for part d) I believe what they are asking is to make a new function (the linear combination of {f and g}) that will then be orthogonal to f. Using the above orthogonality relation. I am not to sure how to construct the linear combination though..
 
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