# Evolution of a 'state' in the Heisenberg picture

Suppose that we have a Hamiltonian, $\hat{H}$, and an operator $\hat{A}$ which satisfies the Heisenberg equation$^{[a]}$ $$i \frac{d}{dt} \hat{A} = [\hat{A},\hat{H}].$$

Can we create a 'state' by acting on, for example, the vacuum with the operator $\hat{A}$? $$\lvert \psi \rangle = \hat{A}(t) \lvert 0 \rangle$$

If we investigate the time evolution of this state, then $$i \frac{d}{dt} \lvert \psi \rangle = i \frac{d}{dt} \hat{A}(t) \lvert 0 \rangle = [\hat{A},\hat{H}] \ \lvert 0 \rangle.$$ If the state $\lvert 0 \rangle$ is such that it is annihilated by the Hamiltonian, then $$[\hat{A},\hat{H}] \ \lvert 0 \rangle = - \hat{H} \hat{A} \lvert 0 \rangle,$$ so $$i \frac{d}{dt} \lvert \psi \rangle = - \hat{H} \lvert \psi \rangle,$$ which has the opposite sign from what I would have expected. Is there a problem with constructing/manipulating 'states' this way in the Heisenberg picture? Does the reversed sign from the ordinary Schrodinger equation make sense in this case?

$[a]$: Where we write $\hat{H}$ we mean what would normally be written $\hat{H}/\hbar$. In other words, $\hat{H}$ here has dimensions of frequency.

• Assume $H$ is time-independent for simplicity, then $A(t)=e^{iHt}Ae^{-iHt}$ where $A\equiv A(0)$. So $A(t)|0\rangle = e^{iHt}A|0\rangle$ assuming $H|0\rangle=0$. It is exactly the state $A|0\rangle$ evolved back in time, and you got the right sign from your calculation. But there is no reason to expect some random state you construct to satisfy the Schrodinger equation. – Meng Cheng Aug 9 '15 at 23:56
• You are mixing up the Schrodinger and Heisenberg pictures of quantum mechanics. The first equation is true only in the Heisenberg picture where states are independent of time. The Schrodinger equation however holds in Schrodinger picture where states evolve in time according to the equation. If you are careful with the picture you are working in, everything is consistent. – Prahar Aug 10 '15 at 5:44