The discussion clarifies that the equation ψ*∇^2 ψ = ∇ψ*⋅∇ψ is not generally true. The correct relationship involves a volume integral where the function vanishes on the boundary, leading to the equation ∫_V ψ^* ∇^2 ψ dV = - ∫_V (∇ψ^*)⋅(∇ψ) dV. This relationship highlights the importance of recognizing that the derivative operator is anti-hermitian, contrary to the assumption that it could be treated as hermitian.
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No. It is definitely not generally true.DuckAmuck said:Is the following true?
ψ*∇^2 ψ = ∇ψ*⋅∇ψ
It seems like it should be since you can change the direction of operators.
You can do that for hermitian operators. But the derivative is not a hermitian operator. It is an anti-hermitian operator, due to the minus sign explained in the previous post.DuckAmuck said:It seems like it should be since you can change the direction of operators.