# Reducing unitary evolution operator of a two-spin system to the evolution operator of one of the spins

Consider a system of two spins $s_1$ and $s_2$, each of which can be in one of two states, represented by 0 or 1. A basis for the Hilbert space of this system would be {|0,0>,|0,1>,|1,0> and |1,1>}, where the first index of each basis vector is the value of $s_1$ and the second is the value of $s_2$.

Let $U(T,0)$ be the operator that evolves the wavefunction of the system at time $t=0$ to time $t=T$.

Let us construct a new operator $X(T,0)$ that evolves the wavefunction of $s_1$ at $t=0$ to $t=T$. Adopting a sum over possibilities approach, would this operator be: $<j|X|i> = \sum_{k} \sum_{l} <jk|U(T,0)|il>$ , where $|i>$represents the state where $s_1 = i$ ?

Would this operator be unitary? Why?

• if your spins do not interact it would be possible to evolve them separately if not then not. To construct a time evolution operator you would have to say more about your Hamiltonian, or at least the type of Hamiltonian i guess. Commented Feb 10, 2015 at 10:42

The time evolution of the two spins can be separated if they are independent, i.e. if they don't interact. Under this assumption the time operator splits in the tensor product $$U_1\otimes U_2=(U_1\otimes I)(I\otimes U_2)$$ and therefore it is clear how to define the time evolution for the single spin: for the $j$th particle one simply needs to take the unitary operator $U_j$.