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Why should an equation (TDSE) in which first time derivative is related to second space derivative have a solution that contains $i$?The wave function is supposed to be complex, but I am unable to understand why it can also be assumed to be complex directly from my previous statement as stated in Quantum Physics by Eisberg & Resnick. Can anyone help me develop an intuition?

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marked as duplicate by John Rennie, Qmechanic Oct 21 '13 at 0:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Possible duplicates: physics.stackexchange.com/q/8062/2451 and links therein. –  Qmechanic Oct 20 '13 at 12:47
    
Why the downvotes? –  Ben Crowell Oct 20 '13 at 20:03

1 Answer 1

A first answer that cames to my mind is "why shouldn't it?".

More seriously (if your question is serious). First, you should notice that the original version of TDSE is $$i\hbar\dot\psi = H\psi$$ so you actually already have an $i$ there. However, notice that this is not required to have complex solutions. For example, the equation (second order in space) $$ \frac{d^2f}{dx^2} = -af $$ has absolutely real coefficients, but also has complex solutions, for example of the kind $$ Ae^{ix}+Be^{-ix}$$ You might be interested in knowing that those can also be written in real form as a linear combination of sine and cosine.

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