# Is electric field operator in Schrödinger picture time-dependent?

We know that in the Schrödinger picture, operators are time-independent if they do not have explicit time-dependence.

So do electric field and vector potential field operators have time dependence in Schrödinger's picture?

I ask this because, in canonical quantization of EM theory, this point is never addressed, after dealing with particular issues related to gauge fields, it is quantized exactly as Klein Gordon fields. I suppose they assume that in Schrödinger's picture, those operators do not have any explicit time dependence.

However, the confusing part is, in general, the classical electric field and vector potential have explicit time dependence; am I wrong? So why in Schrödinger's picture, do these operators not have explicit time dependence?

• Well, to start with, do you understand how the operator $\hat{x}$ for a single particle can be time-independent, even though $x(t)$ has explicit time dependence in Newton's laws? Because the answer is really exactly the same. Jun 23 '20 at 21:01

The classical electric field is the expectation value of the electric field operator applied to a particular quantum state of the electromagnetic field $$|\psi\rangle$$: $$\langle \mathbf{E}\rangle=\langle \psi|\mathbf{E}|\psi\rangle$$

If the state evolves in time, then the expectation value of the electric field operator can also evolve in time.

For example, the QED vacuum state $$|0\rangle$$ has a time-independent expectation value. The electric field fluctuates about this expectation value due to the uncertainty principle, so the instantaneous value of the electric field at a point fluctuates from measurement to measurement, but the average over many measurements (i.e. the classical electric field) will be zero, at any time.

• so in Schrodinger picture electric field operator is time-independent? Jun 23 '20 at 16:05
• @physshyp Yes, if you examine the definition of the quantized electric field operator (see e.g. en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field), you'll see it consists only of constants and sums over polarization vectors and the creation and annihilation operators (which are time-independent in the Schrodinger picture). Jun 23 '20 at 16:15

@probably_someone gave you the answer, but your comment suggests you don't see how constant operators in the Schrodinger picture can give you time-dependent, indeed, oscillating, expectation values. Free quantum fields are just an infinity of elegantly packaged dumb oscillators.

For simplicity, non-dimensionalize to $$\hbar=1$$, and pick up just one momentum mode of the Heisenberg time-dependent field; that is, pick up one plain canonical oscillator mode, of frequency $$\omega=1$$, fixed by further nondimensionalization, with conventional, non-QFT normalization, $$H= a^\dagger a +1/2 , ~~~~~[a,a^\dagger]=1, ~~~ a\equiv a_S,\\ a_H(t)= e^{iHt}a e^{-itH}= e^{-it} a_H(0) = e^{-it} a= e^{-it} a_S.$$

As you learned in your elementary QM course, the Heisenberg states are time-independent, $$|\psi_{H~~0}\rangle = |0\rangle, ~~~|\psi_{H~~1}\rangle = a^\dagger |0\rangle, ~~~\leadsto \\ H |\psi_{H~~0}\rangle = 1/2 |\psi_{H~~0}\rangle, ~~~H |\psi_{H~~1}\rangle = 3/2 |\psi_{H~~1}\rangle,...$$ but the Schrödinger states are time-dependent, by virtue of Schroedinger's equation, $$|\psi_{S~~0}(t)\rangle =e^{-it/2} |0\rangle, ~~~~~|\psi_{S~~1}(t)\rangle = e^{-i3t/2}a^\dagger |0\rangle =e^{-i3t/2}|\psi_{H~~1}\rangle ,...$$ (In coordinate space, they are the customary recondite Hermite functions... You may well appreciate why all this superfluous huffing and puffing left Heisenberg cold.)

Now consider the simplest flip-flop state, $$|\psi_{H}\rangle\equiv {1\over \sqrt{2}} ( |\psi_{H~~0}\rangle + |\psi_{H~~1}\rangle), \leadsto \\ |\psi_{S}(t)\rangle = {e^{-it/2}\over \sqrt{2}} ( |\psi_{H~~0}\rangle + e^{-it}|\psi_{H~~1}\rangle ),...$$ and the identical time-dependent expectation value in either picture, $$\langle \psi_H | a_H(t) |\psi_H \rangle = \langle \psi_H | e^{-it} a |\psi_H \rangle= e^{-it}/2 , \\ \langle\psi_{S}(t) | a |\psi_{S}(t)\rangle = e^{-it}/2 ~!$$ Adding the hermitean conjugate, this yields a cosine oscillating expectation value. So, in the Schrödinger picture, the operator assayed is constant, but, naturally, the expectation value needs be the same, oscillating, otherwise the two pictures would fail equivalence!

But you see how the time-independent Fock space is superior to Schroedinger time-dependent states, which is why you'd virtually never use the Schrödinger picture. In real life, latching on interactions to the free hamiltonian, you'd use a hybrid interaction picture, but that's a whole different story...