Question:
Recall the definition of the parity operator $ \hat{I} $ from lecture 5. What is $ \hat{I}(xe^{-\frac{x^2}{2l^2}}) $?
Answer:
This one is really simple, the operator does nothing but just substitute $ x $ by $ -x $, so we have the answer is $ (-x)e^{-\frac{(-x)^2}{2l^2}} = -xe^{-\frac{x^2}{2l^2}} $
Recall the definition of the parity operator $ \hat{I} $ from lecture 5. What is $ \hat{I}(xe^{-\frac{x^2}{2l^2}}) $?
Answer:
This one is really simple, the operator does nothing but just substitute $ x $ by $ -x $, so we have the answer is $ (-x)e^{-\frac{(-x)^2}{2l^2}} = -xe^{-\frac{x^2}{2l^2}} $
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