# Bloch sphere for 2 and more qubits

if I represent my qubit in the Bloch sphere as

$|ψ⟩=a|u⟩+b|v⟩$

where

$|ψ⟩=\cos(θ_2)|u⟩+e^{iϕ}\sin(θ_2)|v⟩$,

I can see where my qubit is inside the sphere.

Now imagine I have 2 qubits or more and I want to check whether they lie on the xy-plane.

If I have my state for 2 qubits as, for example

$|ψ⟩=[0.5;0.5i;0.5i;-0.5]$,

how can I check their position on the Bloch sphere?

Thank you very much in advance.

The best way to think about the Bloch sphere is as a Bloch ball. In particular instead of considering only pure states we should consider mixed states ($\rho$) as well. Any mixed state can be completely specified by the expectation values of a complete set of mutually orthogonal observables, where orthogonality of observables A and B is given by $Tr(A B)=0$. So for example the state of a qubit can be complete specified by the three Pauli matrices (X,Y,Z). The expectation values of these three observables corresponds precisely with the three orthogonal axes of the space where the Bloch ball lives, and the shape of the Bloch ball is defined by the constraint that the density matrix $\rho$ has a non-negative spectrum, where $\rho$ can be given by $$\rho = \frac{\mathbb{1}}{Tr(\mathbb{1})}+\alpha_x X +\alpha_y Y + \alpha_z Z.$$ So the vector $\vec{\alpha}$ corresponds with a point in the three dimensional space of expectation values, and the set of points $\vec{\alpha}$ corresponding to non-negative $\rho$ defines the Bloch ball.
This way of thinking allows us to define the appropriate analogue of the Bloch ball for two qubits, or for any finite dimensional quantum system we wish to consider. For two qubits we can specify a complete set of mutually orthogonal observables given by the tensor products of Pauli matrices (XI, YI, ZI, IX, IY, IZ, XX, XY etc.), the dimension of the space is then 15 instead of 3. The shape of our "Bloch ball" will be given by the set of 15 dimensional points $\vec{\alpha}$ such that the density matrix $$\rho=\frac{\mathbb{1}}{Tr(\mathbb{1})}+\alpha_{IX} IX +\alpha_{IY} IY + \alpha_{IZ} IZ+ ...$$ has a non-negative spectrum. It turns out that this shape is no longer a ball! A good name for it is a spectrahedron. Understanding the structure of this shape and its higher dimensional cousins is a great way to spend your time.
However, you can consider the one qubit reduced density matrices of the two qubit state, and ask where inside the Bloch Ball they live, in particular, if you have a 1 qubit reduced density matrix $\rho_1$ you can specify its position on the xy-plane by the point $\{Tr(X \rho_1), Tr(Y \rho_1)\}$.