- #1
stunner5000pt
- 1,461
- 2
Homework Statement
Consider the following operators
a) Reflection: [tex]\hat{I}\Psi(x)=\Psi(-x)[/tex] , [tex] x\in(-\infty,\infty)[/tex]
b) Translation: [tex] \hat{T_{a}}\Psi(x)=\Psi(x+a)[/tex], [tex] x\in(-\infty,\infty)[/tex]
c) [tex] \hat{M_{c}}: \hat{M_{c}}\Psi(x) = \sqrt{c}\Psi(x) [/tex]
d) [tex] \hat{c}\Psi(x)= (\Psi(x))^* [/tex]
e) [tex] i\frac{\partial}{\partial x}, x\in(-\infty,\infty)[/tex]
f) [tex] i\frac{\partial}{\partial r}, r\in[0,\infty)[/tex] where r is the radial variable
Are the operators linear? Find operators which are complex conjugate, transposed, Hermitian conjugate, and inverse with respect to the given operators
2. The attempt at a solution
Ok so the book (Pauli's) says If F is a linear operator than corresponding to every eigenfunction [itex]u_{n} [/itex] there is an expansion
[tex] (Fu_{n}) ~\sum_{k} u_{k} F_{kn} [/tex]
with [tex] F_{kn} = \int u_{k}^* ((Fu_{n}) dq[/tex]
For the reflection operator then would writing something like this be sufficient?
[tex] \hat{I}\Psi(x)=\sum_{n}u_{n} \Psi_{n}(x) I_{kn} = \Psi(-x)[/tex]
i alwyas thought that to prove the linearity of an operator we would have to do something like this
For two function [itex] \Psi_{1} [/itex] and [itex] \Psi_{2} [/itex]
[tex]\Hat{I} (\Psi_{1} + \Psi_{2})(x) = \Psi_{1}(-x) +\Psi_{2}(-x) [/tex] which means it is linear
is there something else i nthe first method that i have not pointed out??
i am actually not sure how to find the inverse of an operator... it doesn't seem to be given in the textbook for my class (Griffiths)
This is for an assignment but its pointless doing it for the assignment... i need to find the explicit formula (i think that one has to do the an integral with integrating by parts and so on) Just point out the book in which i need to look that's all i ask
thanks