# Time factor in time independent Schrödinger Equation

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?

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

You are mistakenly assuming that $|\sum_k a_k|^2 = \sum_k |a_k|^2$.