Does the superposition principle affect the space of quantum states? I am confused about the set of quantum states.
I have seen it written that in classical physics, the set of all states is a simplex. (I think this refers to the probability simplex.)
In quantum physics, the set of all states is not a simplex. Does this have anything to do with the superposition principle? If we didn't have the superposition principle in quantum mechanics, would the set of states be a simplex, like in the classical case?
Thanks for your help!
 A: I think that yes, but it's a bit intricate.
The problem here is that different notions are mixed up. Usually, what people mean by "quantum superposition" is the fact that you can have a pure state that is partly in one state and partly in another. Now, as Timo points out, pure states in classical physics are just points in phase space. No simplex there, the set of states is just phase space.
However, we can also consider mixtures of state (ensembles). For convenience, let's take a finite index set (i.e. there is a finite number of possible classical pure states). Then the states correspond to probability distributions on the index set and these are just positive normalized vectors. In this sense, the state space forms a simplex. The corresponding quantum states are normalized density operators and these do NOT form a simplex. Now, a "superposition" could just be the fact that a convex combination of states is still a state, which holds both for classical and for quantum mixtures. Therefore, this "superposition" principle states that the set of states should be convex. Note that the extremal points of the set of states are exactly the pure states (both in the classical and the quantum case).
The normal superposition of pure states works differently here - adding the density matrices of two different pure states gives a mixed state and you can't (by definition) obtain a pure state by adding some states. A superposed pure state is simply a different sate and as such a different extremal point of the set of states. 
And here is the connection to simplices: For simplicity, let's take the smallest possible system, a qubit in the quantum case and a bit in the classical case. The classical system has two pure states, 0 and 1. Let's represent them as vectors in $\mathbb{R}^2$ with a $1$ either in the first or the second component and a zero otherwise. The mixed states are then any positive vectors in $\mathbb{R}^2$ whose components sum up to one - that's a line segment and hence a simplex. Now to the quantum case. Here, the pure states are $|0\rangle$ or $|1\rangle$ or any superposition of this. The pure states are given by $\alpha |0\rangle + \beta |1\rangle$ where $\alpha$ is a real and $\beta$ is a complex number such that $\alpha^2+|\beta|^2=1$ (note that a global phase is irrelevant). All these pure states are extremal points of the set of mixed states. But they are not finite in number, hence the state space cannot be a simplex (it's actually the Bloch sphere).
A: In classical physics, the state of a system is represented by points in configuration space or phase space. However in quantum mechanics, a state is represented by a vector and the state vectors are assumed to come from a vector space (or more precisely Hilbert space). From linear algebra you should know that we can add up two vectors from a vector space and the resultant vector is still inside the vector space. Similarly, the superposition of two quantum states from a Hilbert space is again another quantum state in the same Hilbert space.
So basically, there involves two of the seven assumptions of quantum theory. (1)a quantum state is represented by a vector from a Hilbert space (2)the superposition principle
