Question:
Prove that the only eigenvalues of the parity operator, $ \hat{I} $, are 1 and −1.
Solution:
Recall that $ \hat{I}(f(x)) = f(-x) $, and let $ \lambda $ be its eigenvalue, so we have:
$ \begin{eqnarray*} f(x) &=& \hat{I}(f(-x)) \\ &=& \lambda f(-x) \\ &=& \lambda (\hat{I}(f(x))) \\ &=& \lambda (\lambda f(x)) \\ 1 &=& \lambda^2 \end{eqnarray*} $
So the only eigenvalues are 1 and -1.
Prove that the only eigenvalues of the parity operator, $ \hat{I} $, are 1 and −1.
Solution:
Recall that $ \hat{I}(f(x)) = f(-x) $, and let $ \lambda $ be its eigenvalue, so we have:
$ \begin{eqnarray*} f(x) &=& \hat{I}(f(-x)) \\ &=& \lambda f(-x) \\ &=& \lambda (\hat{I}(f(x))) \\ &=& \lambda (\lambda f(x)) \\ 1 &=& \lambda^2 \end{eqnarray*} $
So the only eigenvalues are 1 and -1.
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