> Can you do QM with just the real valued probability distribution?
You can't. The key fact is that other observables, such as momentum, depend on the complex part of the wave function.
You can't. The key fact is that other observables, such as momentum, depend on the complex part of the wave function.
If you were to append some fictitious spin system to whatever quantum state, you could put all the real amplitude on the spin down state and all the imaginary amplitude on the spin up state.
Perhaps "the simulation" is on a quantum computer and some of the qubits are not directly encoding things we can touch.
QM has very specific rules when it comes to how two random variables are dependent. That specific rule comes up when you have quantum interference and this is best described by the conjugate square of two complex functions A^2 + B^2 + A*B + B*A. This is just a special case of the union probability: P(A) + P(B) - P(AB).
It's the interaction between variables that is key.
More precisely a quantum observable is a map (a function) that takes in a quantum state and outputs a probability distribution, representing the probabilities of the various outcomes you could get if you measure that observable on that state. The equivalent statement is also true of classical observables and classical states.
Under classical rules it turns out that if you have many observables acting on the same system you can come up with a joint observable, that maps a state to a joint probability distribution for all the observables. For incompatible quantum observables this is emphatically not the case. Given two quantum observables there is generally not a joint observable representing simultaneous measurement of them.