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I'm on the very beginning of learning quantum mechanics. When we solve the time independent Schrödinger Equation as far as I understand we will get the general solution:

$$\Psi(r,t)=\sum c_n\cdot \psi(r)\cdot \exp(-iE_nt/\hbar)$$

But I have learnt that $\Psi$ has no physical meaning, and that we have to use $\mid \Psi \mid^2$ for a physical interpretation. Describing the probability of finding a particle at a given neighbourhood. We know that $(\exp(-iE_nt/\hbar))^2=1$

Does that mean that the factor $\exp(-iE_nt/\hbar)$ is totally useless, and that for any system we may change our solution to $\Psi(r,t)=\sum c_n\cdot \psi(r)\cdot "1"$ and still have an equivalent physical system?

I clearly see that this is a mathematical sin, but would it be okay from a purely physical perspective?

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  • $\begingroup$ $A^2\neq |A|^2$. And if you change that factor to 1, $\Psi$ is no longer a solution of the Schrödinger equation. $\endgroup$ – Demosthene Jan 23 '17 at 9:41
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You are mistakenly assuming that $|\sum_k a_k|^2 = \sum_k |a_k|^2$.

Make sure you understand why this relation does not hold (find a counterexample), and you should be able to apply it to your problem.

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It's not quite correct that the wavefunction lacks physical meaning. If you left out the imaginary rotation term, you couldn't get interference effects. It'd be kinda okay if your system was in an energy eigenstate, as those do have time symmetry when you look at the probability density with respect to time, but the same would not be true for a superposition of two energy eigenstates, which should more or less oscillate between the two.

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