- #1
JD_PM
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- Homework Statement
- Problem 3.6 Introduction to QM by Griffiths; second edition (I have rewritten it, as my issue is finding the eigenvalues and eigenvalues only)
Consider the following operator
$$\hat Q = \frac{d^2}{d \phi^2}$$
Find its eigenfunctions and eigenvalues.
Hint: Note we are using functions ##f(\phi)## on the finite interval ##0 \leq \phi \leq 2 \pi##
- Relevant Equations
- ##\frac{d^2}{d \phi^2} f(\phi) = q f(\phi)##
##f(\phi +2\pi ) = f(\phi)##
The eigenvalue equation is
$$\frac{d^2}{d \phi^2} f(\phi) = q f(\phi)$$
This is a second order linear homogeneous differential equation. The second order polynomial associated to it is
$$\lambda ^2 - q = 0 \rightarrow \lambda = \pm \sqrt{q}$$
As both roots are real and distinct, the solution to the differential equation is
$$f(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$$
Once here I have two issues:
a) My solution does not match the provided one: ##f_{\pm} (\phi) = C e^{\pm \phi \sqrt{q}}## (maybe they are equivalent and I do not see it XD)
b) Then the solution says 'The periodicity condition ##f(\phi +2\pi ) = f(\phi)## implies that ##\sqrt{q} 2\pi = 2in \pi## but I do not see why.
Thanks
$$\frac{d^2}{d \phi^2} f(\phi) = q f(\phi)$$
This is a second order linear homogeneous differential equation. The second order polynomial associated to it is
$$\lambda ^2 - q = 0 \rightarrow \lambda = \pm \sqrt{q}$$
As both roots are real and distinct, the solution to the differential equation is
$$f(\phi) = A e^{\phi \sqrt{q}} + B e^{-\phi \sqrt{q}}$$
Once here I have two issues:
a) My solution does not match the provided one: ##f_{\pm} (\phi) = C e^{\pm \phi \sqrt{q}}## (maybe they are equivalent and I do not see it XD)
b) Then the solution says 'The periodicity condition ##f(\phi +2\pi ) = f(\phi)## implies that ##\sqrt{q} 2\pi = 2in \pi## but I do not see why.
Thanks