MATLAB Making a short test for self-adjointness

AI Thread Summary
The discussion centers on testing the self-adjointness of an operator in MATLAB using symbolic manipulation. The user attempts to validate the equation involving inner products of two functions, but receives feedback that the validity of the result for specific functions does not guarantee it holds for all functions. It's emphasized that if the difference calculated (O) is not zero, the operator cannot be considered self-adjoint. Additionally, there is a critique of the use of indefinite integrals instead of definite integrals for the inner product calculation. The user tests the operator on two functions over the interval from 0 to 2π, resulting in a non-zero value, which raises questions about the admissibility of the functions or the self-adjointness of the operator. They seek suggestions for a suitable pair of functions that would demonstrate self-adjointness effectively.
SemM
Gold Member
Messages
194
Reaction score
13
Hi, I made the following command in MATLAB to test for the following:

\begin{equation}
\langle T\psi,\phi\rangle=\langle \psi, T\phi\rangle
\end{equation}

%Self-adjoint test
syms x
a = (diff(exp(-x^2)))*exp(-2*x^2)
b = (diff(exp(-2*x^2)))*exp(-x^2)
W = int(a, x)
P = int(b, x)
O=W-P

and it is tested on two square integrable functions. Can someone validate that this is OK?
 
Physics news on Phys.org
No. Just because it is valid for a specific pair of functions doesn't mean that it will be true for all functions.
 
So one needs to add that if ##O \ne 0## then the operator is not self-adjoint.
 
SemM said:
So one needs to add that if ##O \ne 0## then the operator is not self-adjoint.
Yes, you can use it for counterproofs.

I am not very familiar with symbolic manipulations in Matlab, but the way you use int appears to be for indefinite integrals, not a definite integral as it should be for the inner product.
 
DrClaude said:
Yes, you can use it for counterproofs.

I am not very familiar with symbolic manipulations in Matlab, but the way you use int appears to be for indefinite integrals, not a definite integral as it should be for the inner product.

Sure. trying the following function on the differential operator (d/dx) on the two given functions in the space 0-2pi:

%Self-adjoint test
syms x
a = (diff(exp(-x^2)))*exp(-2*x^2)
b = (diff(exp(-2*x^2)))*exp(-x^2)
W = int(a, 0, 2*pi)
P = int(b, 0, 2*pi)
O = W-PGives:

1/3 - exp(-12*pi^2)/3

So these two functions are either inadmissible, or the operator is not self adjoint. Of course the latter is ridiculous, but in order to try, can you give me a pair of functions that should work?
 

Similar threads

Replies
1
Views
2K
Replies
4
Views
2K
Replies
10
Views
2K
Replies
9
Views
5K
Replies
4
Views
2K
Replies
59
Views
10K
Replies
12
Views
4K
Back
Top