- #1

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instead of being the original S.E in terms of ψ*

or the equation in terms of ψ with the signs swapped

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- Thread starter QuantumDuality
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- #1

- 10

- 0

instead of being the original S.E in terms of ψ*

or the equation in terms of ψ with the signs swapped

- #2

Paul Colby

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- ##(a b)^\ast = a^\ast b^\ast##
- ##(a + b)^\ast = a^\ast + b^\ast##
- ##(i)^\ast = -i##
- ##(\frac{\partial \psi}{\partial t})^\ast = \frac{\partial \bar\psi}{\partial t}##

- #3

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I got it!. I always forget to use the product rule, Thanks

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