I have a few questions about how in and out states are defined in QFT. I was reading Schwartz's book but I found some confusing stuff about some of the things he does. I'm going to work with a real scalar field for simplicity.

Schwartz derives creation and annihilation operators simply as the expansion of the field $\phi(x)$ in a plane wave basis:

$\phi(x) = \int \frac{d^3 \mathbf{k}}{\left(2 \pi \right)^3 \sqrt{2 \omega_k}} a_k(t) e^{-ikx} + a_k^{\dagger}(t) e^{ikx}$.

This I can understand: at any specific time the full Heisenberg evolution with the hamiltonian $H=H_0+V$ can be expanded in such a basis with time dependent creation and destruction operators.

He then proceeds claiming that the final state $|f>$ can be obtained by acting on the vacuum state with creation operators at time $t=+ \infty$, and similarly for the initial state $|i>$:

$|i> = \sqrt{2 \omega_1 ... 2 \omega_k} a_{p_1}^{\dagger}(-\infty) ... a_{p_k}^{\dagger}(-\infty) |\Omega>$

$|f> = \sqrt{2 \omega_{k+1} ... 2 \omega_n} a_{p_{k+1}}^{\dagger}(+\infty) ... a_{p_n}^{\dagger}(+\infty) |\Omega>$.

So far so good.

I lose track when he claims: $ <f|S|i> = \sqrt{2 \omega_1 ... 2 \omega_n} <\Omega|a_{p_{k+1}}(+\infty) ... a_{p_n}(+\infty) a_{p_{1}}^{\dagger}(-\infty) ... a_{p_k}^{\dagger}(-\infty) |\Omega>$.

Where did the time evolution operator $S$ go? It seems important to me because the initial state state $|i>$ needs to propagate before I take the bra with the final state.

My attempt at a solution:

I read a lot about in and out states even here on SE and I came up with my own way to derive this formula putting together all I could find. I'm not sure whether what I did even makes sense, so I'm in your hands here.

Following what many do I decided to define annihilation operators in a different way than a Fourier expansion:

$a_k (t) \equiv i \int d^3 \mathbf{x} \; W \left( \frac{e^{ikx}}{\sqrt{2 \omega_k}}, \phi(x) \right)$, where $W(f(\mathbf{x}, t), g(\mathbf{x}, t)) = fg'-f'g$, where the prime means derivative over time. The creation operator is simply its hermitian conjugate.

This time dependent creation operator reduces to the usual time independent operator in the case of a free field. Furthermore it is easy to show that in fact $\phi(x) = \int \frac{d^3 \mathbf{k}}{\left(2 \pi \right)^3 \sqrt{2 \omega}} a_k(t) e^{-ikx} + a_k^{\dagger}(t) e^{ikx}$. Also, with this new form we can simply derive an equation of motion for the operators $a$ and $a^{\dagger}$:

$\frac{d}{dt} a_k (t) = \frac{\partial}{\partial t} i \int d^3 \mathbf{x} \; W \left( \frac{e^{ikx}}{\sqrt{2 \omega_k}}, \phi(x) \right) + i \left[ H_0 + V, a_k (t) \right] = - \omega_k a_k (t) + \omega_k a_k (t) + i \left[ V, a_k (t) \right]$, so:

$\frac{d}{dt} a_k (t) = i \left[ V, a_k (t) \right]$

Which I think is a quite nice result: annihilation operators in Heisenberg representation evolve as if it was interaction picture in a sense.

In Schroedinger representation an out state would simply be: $|f>_S = \sqrt{2 \omega_{k+1} ... 2 \omega_n} a_{p_{k+1}}^{\dagger}(-\infty) ... a_{p_n}^{\dagger}(-\infty) |\Omega>$, now:

$ S^{\dagger} (+\infty, -\infty) |f>_S = \sqrt{2 \omega_{k+1} ... 2 \omega_n} S^{\dagger} (+\infty, -\infty) a_{p_{k+1}}^{\dagger}(-\infty) ... a_{p_n}^{\dagger}(-\infty) |\Omega> = \sqrt{2 \omega_{k+1} ... 2 \omega_n} S^{\dagger} (+\infty, -\infty) a_{p_{k+1}}^{\dagger}(-\infty) S (+\infty, -\infty) S^{\dagger} (+\infty, -\infty) ... S^{\dagger} (+\infty, -\infty) a_{p_n}^{\dagger}(-\infty) S (+\infty, -\infty) S^{\dagger} (+\infty, -\infty) |\Omega> = \sqrt{2 \omega_{k+1} ... 2 \omega_n} a_{p_{k+1}}^{\dagger}(+\infty) ... a_{p_n}^{\dagger}(+\infty) S^{\dagger} (+\infty, -\infty) |\Omega>$.

I am left with an extra factor $S^{\dagger} (+\infty, -\infty)$ acting on the ground state, but that should not be of any concern because the ground state is stable, so $S^{\dagger} (+\infty, -\infty) |\Omega> = \lambda |\Omega> $, and I can simply disregard the extra $\lambda$ factor. I couldn't come up with a better way to eliminate this extra factor, so I hope maybe one of you could tell me if there is one.

Is this correct?

  • $\begingroup$ See physics.stackexchange.com/questions/626768/… $\endgroup$
    – Jan Bos
    May 29, 2021 at 13:07
  • $\begingroup$ @JanBos So basically since the vectors are in Heisenberg representation and the in and out states are in the same Hilbert space the S-matrix is just the identity? $\endgroup$ May 29, 2021 at 13:33
  • $\begingroup$ What the answer in the linked question hints to is that S matrix is defined as $<f|S|i> (heis) = <f, t = -\infty | i, t = \infty> (schroed) $ hence the expression you found in the book $\endgroup$
    – Jan Bos
    May 29, 2021 at 14:00
  • $\begingroup$ @JanBos Aren't the $a(+\infty)$ Heisenberg operators? If so why do you say the bra $< \Omega |a(+\infty)...a(+\infty) $ is a Schrodinger bra, which would justify the absence of the S operator? Also, I do not understand why the S operator is there in Heisenberg picture but not in Schrodinger picture. Shouldn't it be the other way around? Kets evolve in S picture so I expect a time evolution. In H picture they don't so I can simply construct an amplitude without any evolution in between. $\endgroup$ May 29, 2021 at 14:29
  • $\begingroup$ You can convert <f,t=−∞|i,t=∞> (schroed) to the Heisenberg picture and this operator S arises as a functional of V. Note the Heisenberg states are equal to the Schroedinger states at a certain reference time. It can be expanded into a power series in V leading to Feynman diagrams. $\endgroup$
    – Jan Bos
    May 30, 2021 at 1:10


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