# Why do we use $\psi$ instead of a straightforward probability?

What is the advantage/purpose of using $\psi$ for wavefunctions and getting the probability with $|\psi|^2$ as opposed to just defining and using the probability function?

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We have got time-evolution equation (Schrodinger equation) for $\psi$ and not for $|\psi|^2$ –  user35952 Feb 26 '14 at 6:33
One motivation is the double slit interference. It is (i) the superposition of wavefunction; (ii) $|\psi|^2$ to get the diffraction pattern. If we use $\rho:=|\psi|^2$ directly without superposition of wavefunction, superposition of probability is incorrect, and it will be awkward to think about square root of $\rho$ and possibly with phase factor.. –  user26143 Feb 26 '14 at 8:46
This thread should answer the question. –  NikolajK Feb 26 '14 at 13:03
See @NikolajK's thread, which is indeed wonderful. But in my experience this question is often asked with the mistaken belief that quantum mechanics and complex $\psi$ are "harder" to understand than probability. IMO this is absolutely the wrong way around, see here! –  WetSavannaAnimal aka Rod Vance Feb 26 '14 at 22:36

One purpose of using $\Psi$ rather than just the probability density is to match observation. Dealing only with probability density isn't sufficient.

Imagine you can send particles through two adjacent slits toward a detector screen. You'll find an interesting pattern that looks like an interference phenomenon is happening. This is called the double slit experiment if you want to take a look. There's no way to explain the pattern that appears based only on the probability densities corresponding to each slit alone. In other words, if slit #1 alone produces a density $\rho_1(x)$ on the screen, and slit #2 alone produces a density $\rho_2(x)$ on the screen, there is no way to combine $\rho_1$ and $\rho_2$ to get the actual density $\rho$ that's observed.

Using the probability amplitude $\Psi$ allows us to predict the actual pattern.

I should mention that, historically, this was not the motivation for introducing $\Psi$, but it's a convenient way to view the need after-the-fact. I believe $\Psi$ first appeared in Schrodinger's equation as a way to get the allowed energies of a bound system. This is an equation in $\Psi$, not $\rho$. It wasn't until after the concept of $\Psi$ that the connection $\rho=|\Psi|^2$ was made. Nonetheless, as user36952 pointed out, we have an equation describing how $\Psi$ develops in time, not how $\rho$ develops.

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