- #1
Hill
- 735
- 576
- TL;DR
- 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##.
"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##.