# Put the eigenvalue function in self-adjoint form

• GGGGc
GGGGc
Homework Statement
The original equation is -(1-x^(2))y’’+xy’=ky. How to put it in self-adjoint form?
Also, if let x=cos(theta) how to put that form in d^(2)y/dx^(2)=-ky form?
Relevant Equations
-(1-x^(2))y’’+xy’=ky
Here’s my work:
The integrating factor I find is (x^(2)-1)^1/2. The self adjoint form I find is
-d/dx (((1-x^(2))^(3/2))*dy/dx))=k(x^(2)-1)^(1/2).
Am I right?

The left hand side is of the form $$-u^2y'' - uu'y' = -u(uy')'$$ for some $u$.

GGGGc said:
Homework Statement: The original equation is -(1-x^(2))y’’+xy’=ky. How to put it in self-adjoint form?
Also, if let x=cos(theta) how to put that form in d^(2)y/dx^(2)=-ky form?
Relevant Equations: -(1-x^(2))y’’+xy’=ky

Here’s my work:
The integrating factor I find is (x^(2)-1)^1/2. The self adjoint form I find is
-d/dx (((1-x^(2))^(3/2))*dy/dx))=k(x^(2)-1)^(1/2).
Am I right?
Please use ## ## tags and code to render Latex

## What does it mean to put the eigenvalue function in self-adjoint form?

Putting the eigenvalue function in self-adjoint form means expressing the operator in a way that it equals its own adjoint. This ensures that all eigenvalues are real and the eigenfunctions form a complete basis, which is particularly useful in quantum mechanics and other fields of physics and engineering.

## Why is it important to have an operator in self-adjoint form?

Having an operator in self-adjoint form is important because it guarantees that the eigenvalues are real, which is essential for physical observables in quantum mechanics. It also ensures that the eigenfunctions are orthogonal and form a complete basis, making it easier to solve differential equations and perform other mathematical operations.

## How can you determine if an operator is self-adjoint?

An operator is self-adjoint if it equals its own adjoint. Mathematically, an operator $$A$$ is self-adjoint if $$A = A^\dagger$$, where $$A^\dagger$$ is the adjoint of $$A$$. This means that for all functions $$f$$ and $$g$$ in the domain of $$A$$, the inner product $$\langle Af, g \rangle$$ equals $$\langle f, A g \rangle$$.

## What are the steps to convert an operator to its self-adjoint form?

To convert an operator to its self-adjoint form, you typically need to:1. Identify the domain where the operator is defined.2. Ensure that the operator is symmetric, meaning $$\langle Af, g \rangle = \langle f, Ag \rangle$$.3. Extend the domain if necessary so that the operator becomes self-adjoint.4. Verify that the extended operator satisfies $$A = A^\dagger$$.

No, not all operators can be made self-adjoint. Some operators are inherently non-self-adjoint and cannot be extended to a self-adjoint operator. However, many important operators in physics and engineering can either be made self-adjoint or approximated by self-adjoint operators within a certain domain.

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