Lysario - by Panagiotis Chatzichrisafis

Lysario - by Panagiotis Chatzichrisafis » Quantum Mechanics https://lysario.de "ούτω γάρ ειδέναι το σύνθετον υπολαμβάνομεν, όταν ειδώμεν εκ τίνων και πόσων εστίν ..." Thu, 15 Apr 2021 17:28:24 +0000 de-DE hourly 1 Liboff: “Introductory Quantum Mechanics”, 2nd edition p.86 exercise 3.16 https://lysario.de/solved_problems/liboff-introductory-quantum-mechanics-2nd-edition-p86-exercise-316 https://lysario.de/solved_problems/liboff-introductory-quantum-mechanics-2nd-edition-p86-exercise-316#comments Fri, 02 Jan 2009 16:55:15 +0000 Panagiotis https://lysario.de/?p=6 [1, p. 86 ex. 3.16] asks to prove that if the eigenfunctions and eigenvalues of an operator $\widehat{A}$ are $\{\phi_{n}\}$ and $\{a_{n}\}$ , respectively ( $\ensuremath{\widehat{A}}\phi_{n}=a_{n}\phi_{n}$ ) then the eigenfunctions of a function $f(x)$ having an expansion of the form:

$f(x)=\sum\limits^{\infty}_{l=0}b_{l}x^{l}$

(1)

will also be $\phi_n$ with corresponding eigenvalues $f(a_n)$ , $n=0,1...$ . That is $f(\widehat{A})\phi_n=f(a_n)\phi_n$ .
Solution:To prove this feature, one must have the distributive property of operators in mind which can be stated as: $\widehat{A}^{l}=\widehat{A}^{l-1}\cdot \widehat{A}.$ Iterative application of the operator $\widehat{A}^{l}$ to the eigenfunction $\phi_n$ gives the following result:

$\widehat{A}^{l}\phi_n=\widehat{A}^{l-1}\widehat{A}\phi_n = a_n\widehat{A}^{l-1}\phi_n =...= a^{k}\widehat{A}^{l-k}\phi_n=...=a^{l}_{n}\phi_n.$

(2)

The function $f(\widehat{A})$ can be written with the use of the given expansion as: $f(\widehat{A})=\sum^{\infty}_{l=0}b_{l}\widehat{A}^{l}$ Applying the right hand side to an eigenfunction $\phi_{n}$ results in:

$f(\widehat{A})\phi_n$	$=$	$\sum\limits^{\infty}_{l=0}b_{l}\widehat{A}^{l}\phi_n$
	$=$	$\sum\limits^{\infty}_{l=0}b_{l}a^{l}_{n}\phi_n$
	$=$	$f(a_n)\phi_n$	(3)

The last result (3) means that the eigenfunctions of an operator $f(\widehat{A})$ are the same as for the operator $\widehat{A}$ but the eigenvalues are given by $f(a_n)$ where $a_n$ are the eigenvalues of the operator $\widehat{A}$ ; Q.E.D.. ]]> https://lysario.de/solved_problems/liboff-introductory-quantum-mechanics-2nd-edition-p86-exercise-316/feed 0