# Why is Heisenberg picture used if Schrödinger picture is simpler?

In Heisenberg picture the operators evolve in time instead of states, which evolve in time in Schrödinger picture. But this also means that to solve the equations (at least numerically) one has to work with $N^2$ numbers instead of Schrödinger's $N$ (i.e. Hamiltonian matrix vs state vector).

So, why, despite being dramatically more inefficient for calculations, is Heisenberg picture still used?

• Note that the hamiltonian is still Hermitean. So, before you even start to consider degrees of freedom, you should consider that you're not really manipulating $N^{2}$ numbers. May 7, 2014 at 15:33
• @JerrySchirmer well, $\frac12N^2$. Still $\mathcal O(N^2)$. May 7, 2014 at 19:01
• well, consider that the hamiltonian always has a complete set of eigenvectors. How many independent components does it have when the hilbert space is expressed in this eigenbasis? May 7, 2014 at 19:15
• Well, if we always knew the eigenbasis, we'd have nothing to solve in many cases. May 7, 2014 at 19:27