A Derivation of QM limit of QFT in "QFT and the SM" by Schwartz

Hill
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How the time derivative moved into the braket?
In this derivation, a basis of one-particle states ##\langle x|=\langle \vec x,t|## is expressed with the field operator, $$\langle x|=\langle 0| \phi (\vec x, t)$$
"Then, a Schrodinger picture wavefunction is $$\psi (x)=\langle x| \psi \rangle$$
which satisfies $$i \partial _t \psi (x) = i \partial _t \langle 0| \phi (\vec x, t)|\psi \rangle = i \langle 0| \partial _t \phi (\vec x, t)| \psi \rangle$$

I need help to understand why the time derivative is applied to the field ##\phi## and not to the state vector ##\psi##.
 
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Hill said:
I need help to understand why the time derivative is applied to the field ##\phi## and not to the state vector ##\psi##.
Because the equation is written in the Heisenberg picture, where observables depend on time and state does not depend on time.
 
Demystifier said:
Because the equation is written in the Heisenberg picture, where observables depend on time and state does not depend on time.
Thank you.
 
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