Orthogonality of eigenfunctions with continuous eigenvalues

In summary, the conversation involves using knowledge of the orthogonality conditions for eigenfunctions with discrete eigenvalues to determine the orthonormal set for eigenfunctions with continuous eigenvalues. The definition of completeness is then used to show that the square of a(k) is equal to 1. The conversation also includes a discussion on integrating the right hand side with respect to k, and clarifying the correct definition of psi(x) in terms of a(k) and phi(k,x).
  • #1
jazznaz
23
0

Homework Statement



With knowledge of the orthogonality conditions for eigenfunctions with discrete eigenvalues, determine the orthonormal set for eigenfunctions with continuous eigenvalues. Use the definition of completeness to show that | a(k) |^2 = 1.

2. The attempt at a solution

The first step is:

http://img55.imageshack.us/img55/5229/81115215vg1.jpg
http://g.imageshack.us/img55/81115215vg1.jpg/1/

Since the integral is only equal to 0 when k' = k. (The same condition as the kronecker delta.)

Next:

http://img352.imageshack.us/img352/9519/94213081to0.jpg
http://g.imageshack.us/img352/94213081to0.jpg/1/

After here I get a bit lost, even though I think this is almost the solution. My work differs from the course notes and a QM book I've looked through. They both have an integration with respect to k for the RHS, not x.

Any pointers would be greatly appreciated.
 
Last edited by a moderator:
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  • #2
jazznaz said:

Homework Statement



With knowledge of the orthogonality conditions for eigenfunctions with discrete eigenvalues, determine the orthonormal set for eigenfunctions with continuous eigenvalues. Use the definition of completeness to show that | a(k) |^2 = 1.

2. The attempt at a solution

The first step is:

http://img55.imageshack.us/img55/5229/81115215vg1.jpg
http://g.imageshack.us/img55/81115215vg1.jpg/1/

Since the integral is only equal to 0 when k' = k. (The same condition as the kronecker delta.)

no. the right hand side is always equal to zero except when k'=k

Next:

http://img352.imageshack.us/img352/9519/94213081to0.jpg
http://g.imageshack.us/img352/94213081to0.jpg/1/

After here I get a bit lost, even though I think this is almost the solution. My work differs from the course notes and a QM book I've looked through. They both have an integration with respect to k for the RHS, not x.

Any pointers would be greatly appreciated.

The very first line is incorrect. The RHS should not be integrated over x but rather over k.
 
Last edited by a moderator:
  • #3
Should I integrate the RHS w.r.t. k, multiply both sides by the conjugate of Phi and then integrate both sides over x?
 
  • #4
jazznaz said:
Should I integrate the RHS w.r.t. k, multiply both sides by the conjugate of Phi and then integrate both sides over x?

First... get the definition of [itex]\psi(x)[/itex] in terms of a(k) and [itex]\phi_k(x)[/itex] correct.

Then square psi(x) (which is in terms of a(k) and phi(k,x) now) and integrate over x using the "first step" you mention in the OP.

Use the fact that psi(x) is normalized to see that a(k) is is normalized.
 

Related to Orthogonality of eigenfunctions with continuous eigenvalues

What is the concept of "orthogonality" in the context of eigenfunctions with continuous eigenvalues?

In mathematics, orthogonality refers to the property of two mathematical objects being perpendicular or at right angles to each other. In the context of eigenfunctions with continuous eigenvalues, orthogonality refers to the property of two eigenfunctions having an inner product of zero when integrated over a specific interval. This means that the two eigenfunctions are perpendicular in a certain sense and do not overlap.

Why is the orthogonality of eigenfunctions with continuous eigenvalues important in mathematics?

The orthogonality of eigenfunctions with continuous eigenvalues is important because it allows us to separate complex systems into simpler, independent components. This is particularly useful in solving differential equations and analyzing physical systems. It also allows us to express any function as a linear combination of orthogonal eigenfunctions, which can greatly simplify calculations and analyses.

How is the orthogonality of eigenfunctions with continuous eigenvalues mathematically expressed?

The orthogonality of eigenfunctions with continuous eigenvalues is mathematically expressed through an inner product, which is a generalization of the dot product for vector spaces. For two eigenfunctions, f(x) and g(x), the inner product is defined as ∫f(x)g(x)dx over a specific interval. If this integral evaluates to zero, then the two eigenfunctions are considered orthogonal.

Can eigenfunctions with continuous eigenvalues be both orthogonal and non-orthogonal?

No, eigenfunctions with continuous eigenvalues cannot be both orthogonal and non-orthogonal. This is because orthogonality is a binary property; two functions are either orthogonal or they are not. If the inner product of two eigenfunctions is zero, then they are orthogonal, and if it is non-zero, then they are non-orthogonal.

What are some real-world applications of the orthogonality of eigenfunctions with continuous eigenvalues?

The orthogonality of eigenfunctions with continuous eigenvalues has numerous applications in physics, engineering, and other scientific fields. Some examples include analyzing the behavior of a vibrating string or a vibrating membrane, solving the Schrödinger equation in quantum mechanics, and decomposing signals in signal processing. It also has applications in image and sound processing, as well as in statistics and data analysis.

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