1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – pindakaas Feb 10 '15 at 10:42
1
$\begingroup$

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 $U_j$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.