Quantum Field Theory – Asymptotic States in Heisenberg and Schrödinger Pictures

hilbert-spacequantum-field-theorys-matrix-theoryscatteringvacuum

One can show that, in the interacting theory, the operators that create single-particle energy-momentum eigenstates from the vacuum are
\begin{align}
(a_p^{\pm\infty})^\dagger=\lim_{t\to\pm\infty}(a_p^t)^\dagger\tag*{(1)}
\end{align}
where
\begin{align}
(a_p^t)^\dagger=i\int{\rm d}^3x\,e^{-ipx}\overleftrightarrow{\partial_0}\phi(x)=\int{\rm d}^3x\,e^{-ipx}\left(\omega_{\boldsymbol{p}}\phi(x)-i\pi(x)\right).\tag*{(2)}
\end{align}
(see e.g. Srednicki QFT or this answer; I'm using superscript $t$ to emphasise that this operator does not undergo the usual unitary time evolution, but is instead redefined at each $t$). These are used to create the asymptotic states that go into the calculation of the $S$-matrix. For example, in $2\to n$ scattering, the "in" and "out" states would be
\begin{align}
|i\rangle&=|k_1k_2\rangle=(a_{k_1}^{-\infty})^\dagger (a_{k_2}^{-\infty})^\dagger|\Omega\rangle\tag*{(3)}\\
|f\rangle&=|p_1\cdots p_n\rangle=(a_{p_1}^{\infty})^\dagger\cdots (a_{p_n}^{\infty})^\dagger|\Omega\rangle,\tag*{(4)}
\end{align}
the inner product $\langle f|i\rangle$ of which is said to be the corresponding $S$-matrix element (Srednicki, Schwartz, et al.).

However, $\langle f|i\rangle$ is merely an inner product of products of single-particle states. How can this be right? Surely it's the Schrödinger picture states at asymptotic times that should be plane waves (or wave packets if you want to be more precise)? After all, what we want to know is the amplitude for an initial state of particles in the far past to evolve into some other state of particles in the far future (letting the limits be implicit):
\begin{align}
\langle f|S|i\rangle&=\langle p_1\cdots p_n|U(\infty,-\infty)|k_1\cdots k_m\rangle=\langle\Omega|a_{p_1}\cdots a_{p_n}U(\infty,-\infty)a_{k_1}^\dagger\cdots a_{k_m}^\dagger|\Omega\rangle\tag*{(5)}
\end{align}
But in going from the Heisenberg to the Schrödinger picture, the operators $(a_p^{\pm\infty})_S^\dagger=\lim_{t\to\pm\infty}U^\dagger(t_0,t)(a_p^t)^\dagger U(t_0,t)$ no longer create single-particle momentum eigenstates from the vacuum, so they cannot be used to construct the $S$-matrix elements $(5)$. In other words, I'm objecting to the claim that $(3)$ and $(4)$ represent the creation of particles at $t=\pm\infty$; they're applying the wrong operators!

So what gives?

Alternative construction

Just to illustrate my point further, here's an alternative construction that seems more reasonable to me (but also way too OP). Since the asymptotic operators $(a_p^{\pm\infty})^\dagger$ are time-independent anyway, we can just define operators in the Schrödinger picture by $A_p^\dagger=(a_p^\infty)^\dagger$ and create momentum eigenstates from the vacuum at asymptotic times (which still works because the vacuum is translation-invariant). In fact, it seems like this should work at whatever time we want, not just $t=\pm\infty$, then the $S$-matrix elements are
\begin{align}
\langle f|S|i\rangle&=\langle p_1\cdots p_n|U(t_f,t_i)|k_1\cdots k_m\rangle=\langle\Omega|A_{p_1}\cdots A_{p_n}U(t_f,t_i)A_{k_1}^\dagger\cdots A_{k_m}^\dagger|\Omega\rangle\tag*{(Schrödinger)}\\
&={_{\rm out}}\langle p_1\cdots p_n|k_1\cdots k_m\rangle_{\rm in}=\langle\Omega|A_{p_1}(t_f)\cdots A_{p_n}(t_f)A_{k_1}^\dagger(t_i)\cdots A_{k_m}^\dagger(t_i)|\Omega\rangle\tag*{(Heisenberg)}
\end{align}
where $A_{p}^\dagger(t)=U^\dagger(t,t_0)A_{p}^\dagger U(t,t_0)$ and I should emphasise that the Heisenberg states $|k_1\cdots k_m\rangle_{\rm in}$ and $|p_1\cdots p_n\rangle_{\rm out}$ are not products of momentum eigenstates.

Best Answer

It took me a while, but I think I've finally figured it out.

The problem was that I was writing down all these momentum eigenstates without paying attention to which momentum operators they are actually eigenstates of!

That is, the operators $(a_p^{\pm\infty})^\dagger$ don't just "create momentum eigenstates", they specifically create eigenstates of the asymptotic momentum operators $\hat{p}(\pm\infty)=U^\dagger(\pm\infty,t_0)\,\hat{p}\,U(\pm\infty,t_0)$. Likewise, the $a_p^\dagger$ create eigenstates of the Schrödinger-picture operator $\hat{p}$. Not only does this immediately invalidate my alternative construction, it explains why this whole business works in the first place.

In the Schrödinger picture, the state is evolved between asymptotic times, which makes transparent the fact that the initial and final states are particles hanging around at those asymptotic times. But it was unclear to me how fixed momentum eigenstates in the Heisenberg picture could also be interpreted as particles at asymptotic times, because I was implicitly thinking of them as being the same eigenstates of the same operator $\hat{p}$. But they're eigenstates of $\hat{p}(\pm\infty)$, and that's the point: the states in the Heisenberg picture are fixed and the operators evolve such that the predictions come out the same. In the Schrödinger picture, the initial state is a collection of eigenstates of the (fixed) momentum operator that evolves into a complicated mess during the interaction. In the Heisenberg picture the (fixed) state is a collection of momentum eigenstates of the initial momentum operator, but looks like a complicated mess to the evolved momentum operator. Ultimately, the inner product $\langle f|i\rangle$ is between collections of eigenstates of different operators, making it very non-trivial indeed.

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[Physics] Scattering, Perturbation and asymptotic states in LSZ reduction formula

The first question we have to ask is: what is a one particle state in an interacting theory? It is reasonable to require that they are states that are both momentum eigenstates and energy eigenstates. (In fact, as the Hamiltonian and the momentum operator commute, these are not two different conditions.) Weinberg, in his famous textbook, says that particle states are those which transform under an irreducible representation of the Poincare group, but we need not fuss around with the Poincare group here.

All we will say is that, in the interacting theory, there are some single particle states, labelled by

$$|\lambda k \rangle$$

where $k$ is the four-momentum, and $\lambda$ is whatever other labels we need for our particles. (In this answer I will be working with just a real scalar field, but even in the spin-0 case there can still be extra data that distinguishes our particles in an interacting theory.)

Now, we know know we have a set of momentum and energy eigenstates $|\lambda k\rangle$ that represent the stable particles of our theory. We can now "smudge out" these definite momentum states into wave packets, using a Gaussian window function $f_W$ that has some momentum uncertainty $\kappa$. We will denote these smudged out, approximate energy and momentum eigenstates with a subscript $W$ for "window."

$$|\lambda k\rangle_W \equiv \int d^3{\mathbf{k}'} f_W(\mathbf{k} - \mathbf{k}') |\lambda k\rangle$$

We will come back to these.

Now, the free vacuum $|0\rangle$ of $\hat H_0$ and the true vacuum $|\Omega\rangle$ of $\hat H = \hat H_0 + \hat H_{\rm int}$ are very different states. Particles in the interacting theory must indeed be defined to be formed from the action of the "creation operator" on the true vacuum, as long as we properly define what we mean by the "creation operator" in the interacting theory.

To create an annihilate particles, we will use the Klein Gordon inner product. (We suppress $\hbar$ and $c$.)

$$(\psi_1, \psi_2)_{KG} \equiv i \int d^3 x (\psi^*_1 \partial_t \psi_2 - \partial_t \psi^*_1 \psi_2)$$

The motivation for defining this is that in the FREE theory, the Klein Gordon inner product gives us an inner product between single particle states. If we have two single particle states (in the free theory) $|\Psi_1\rangle$ and $|\Psi_2 \rangle$, we have

$$\langle \Psi_1 | \Psi_2 \rangle = ( \psi_1, \psi_2 )_{KG}$$

where we used the "single particle wave functions" of the states defined by

$$\psi_i(x) \equiv \langle 0| \hat \phi(x) |\Psi_i\rangle$$

The niceties of free field theory come from the simple algebra of the creation and annihilation operators, combined with the fact that the annihilation operator annihilates the vacuum. We will try to recreate those relationships using the Klein Gordon inner product. However, to do this, we will need to use widely separated wave packets.

From here on out, everything will be in the interacting theory.

For a given function $\psi$, we define the creation and annihilation operators that "create" the state corresponding to that wave function as follows. $$ \hat a^\dagger_i (t) \equiv -\big( \psi^*_i(t, \cdot), \hat \phi(t, \cdot) \big)_{KG} $$ $$\hat a_i(t) = \big( \psi_i(t, \cdot), \hat \phi(t, \cdot) \big)_{KG}$$

(In the free theory, this creation operator literally would create the single particle state with the single particle wave function $\psi_1$.)

(Something I must mention about these operators is their time evolution. It is a point of notational confusion that $\hat a^\dagger_{1}(t)$ depends explicitly on a time $t$, given that we usually have defined time dependence such that $e^{i \hat H t} \hat{O}(t') e^{-i \hat H t} = \hat {\mathcal{O}}(t'+t)$. This is not the case here.)

Now, sadly, in the interacting theory, the annihilation operator defined above will not annihilate the vacuum. However, we can recover something close:

$$\langle \Omega| \hat a_1(t) |\Omega\rangle = i \int d^3{x}\langle \Omega| \big( \psi_1^*(t, \vec x) \partial_t \hat \phi (t, \vec x) - \partial_t \psi_1^*(t, \vec x) \hat \phi(t, \vec x) \big) |\Omega\rangle$$ $$= i \int d^3{x} \big( \psi_1^*(t, \vec x) \partial_t \langle \Omega| \hat \phi(t, \vec x) |\Omega\rangle - \partial_t \psi^*_1(t, \vec x) \langle \Omega| \hat \phi(t, \vec x) |\Omega\rangle \big)$$ $$= i \langle \Omega| \hat \phi(t, \vec x) |\Omega\rangle \int d^3{x} (- \partial_t \psi_1^*(t, \vec x))$$

The fact that $\partial_t \langle \Omega| \hat \phi(t, \vec x) |\Omega\rangle = 0$ follows directly from the fact that the vacuum state has zero energy, so $e^{- i \hat H t} |\Omega\rangle = |\Omega\rangle$. Now as we want $\langle \Omega| \hat a_1(t) |\Omega\rangle = 0$ for any $\psi_1$, we can see that this is achieved if and only if $\langle \Omega| \hat \phi(x) |\Omega\rangle = \langle \Omega| \hat \phi(0) |\Omega\rangle = 0$. We will assume this is the case.

In the free theory, $\langle 0| \hat a_1(t) \hat a_2^\dagger(t) |0\rangle = \langle \Psi_1 | \Psi_2\rangle = (\psi_1, \psi_2)_{KG}$. In an interacting theory, for any $\hat a_1$ and state $|\Psi_2\rangle$ (not just a single particle state) we have

$$\langle \Omega| \hat a_1(t) |\Psi_2\rangle = \langle \Omega| \big( \psi_1(t, \cdot) , \hat \phi(t, \cdot) \big)_{KG}|\Psi_2\rangle$$ $$= \big( \psi_1(t, \cdot), \langle \Omega| \hat \phi(t, \cdot) |\Psi_2\rangle\big)_{KG} $$ $$= \big( \psi_1(t, \cdot), \psi_2(t, \cdot) \big)_{KG}$$

$$\langle\Psi_2| \hat a_1(t) |\Omega\rangle = \big( \psi_1(t, \cdot) , \psi_2^*(t, \cdot) \big)_{KG}$$

Remember our single particle states? We're now going consider the "single particle wave function" of those states. Namely, they have to be plane waves.

$$\langle \Omega| \hat \phi(x) |\lambda k\rangle = C_\lambda e^{-ikx}$$

where $C_\lambda$ is a constant that depends on $\lambda$.

We now want to see what our states $\hat a^\dagger_1|\Omega\rangle$ have to do with these true particle wave packets $| \lambda k \rangle_W$. To do this, we will see what the inner product of these two states are. Just from our simple algebra above, for an annihilation operator $\hat a_{\lambda_1 k_1} = (\psi_{k_1}, \hat \phi)_{KG}$ where $k_1^2 = m_{\lambda_1}^2$, we have

\begin{equation*} \begin{split} \langle \Omega | \hat a_{\lambda_1 k_1} (t) |\lambda_2 k_2\rangle_W = \big( \psi_{ k_1}(t, \cdot), \langle \Omega |\hat \phi(t, \cdot)|\lambda_2 k_2\rangle_W \big)_{KG} = C_{\lambda_2}\big(\psi_{ k_1}(t, \cdot), \psi_{ k_2 }(t, \cdot) \big)_{KG}\\ {}_W \langle \lambda_2 k_2 | \hat a_{\lambda_1 k_1}(t) |\Omega\rangle = \big( \psi_{ k_1}(t, \cdot), {}_W \langle \lambda_2 k_2 |\hat \phi(t, \cdot) |\Omega\rangle \big)_{KG} = C_{\lambda_2}\big(\psi_{ k_1}(t, \cdot), \psi^*_{ k_2 }(t, \cdot) \big)_{KG}. \end{split} \end{equation*} We desire for the top expression to be $\propto \delta_{\lambda_1 \lambda_2} \delta^3(\mathbf{k}_1 - \mathbf{k}_2)$ and for the bottom expression to be 0. If this were the case, then the only single particle state $\hat a^\dagger_{ k_1}(t) |\Omega\rangle$ would overlap with would be $|\lambda_1 k_1\rangle$, and $\hat a_{k_1}(t)$ could still functionally "annihilate" the vacuum, even though we need to keep ${}_W \langle \lambda k |$ on the left. Defining $\omega_{\lambda k} \equiv (m^2_{\lambda} + \mathbf{k}^2)^\frac{1}{2}$, we have

\begin{equation*} \begin{split} \big(\psi_{ k_1}(t, \cdot), \psi_{ k_2 }(t, \cdot) \big)_{KG} = (2 \pi)^3 \int d^3{\mathbf{k}} f_W(\mathbf{k}_1 - \mathbf{k}) f_W(\mathbf{k}_2 - \mathbf{k}) (\omega_{\lambda_1 k} + \omega_{\lambda_2 k})e^{it(\omega_{\lambda_1 k} - \omega_{\lambda_2 k})} \\ \big(\psi_{ k_1}(t, \cdot), \psi_{ k_2 }^*(t, \cdot) \big)_{KG} = (2 \pi)^3 \int d^3{\mathbf{k}} f_W(\mathbf{k}_1 - \mathbf{k}) f_W(\mathbf{k}_2 + \mathbf{k}) (\omega_{\lambda_1 k} - \omega_{\lambda_2 k})e^{it(\omega_{\lambda_1 k} + \omega_{\lambda_2 k})}. \\ \end{split} \end{equation*}

The top expression is not $\propto \delta_{\lambda_1 \lambda_2} \delta^3_W(\mathbf{k}_1 - \mathbf{k}_2)$ and the bottom expression is not $0$. However, if we take $\kappa \ll |\mathbf{k}_1 - \mathbf{k}_2|$ and also take $t \to \pm \infty$, they are! This hinges on our assumption that $m_{\lambda_1} \neq m_{\lambda_2}$ if $\lambda_1 \neq \lambda_2$. The $e^{it (\ldots)}$ term will oscillate wildly in both integrals if $\lambda_1 \neq \lambda_2$, causing them to be 0. In the top integral, this oscillation does not occur when $\lambda_1 = \lambda_2$. Furthermore, the top integral will be negligible unless $\mathbf{k}_1 = \mathbf{k}_2$. Taking the $f_W(\mathbf{k}) \to \delta^3(\mathbf{k})$ and $t \to \pm \infty$ limit, we can now write

\begin{equation*} \begin{split} \langle \lambda_2 k_2 | \hat a_{\lambda_1 k_1}^\dagger (\pm \infty) |\Omega\rangle = C_{\lambda_2} (2 \pi)^3 2 \omega_{\lambda_2 k_2} \delta_{\lambda_1 \lambda_2} \delta^3(\mathbf{k}_1 - \mathbf{k}_2) \\ \langle \lambda_2 k_2 | \hat a_{\lambda_1 k_1} (\pm \infty) |\Omega\rangle = 0. \end{split} \end{equation*} These properties are even more important than I let on. This is because the states $| \lambda k \rangle$ are so generally defined: they are just momentum eigenstates with all the extra necessary data stuffed into $\lambda$. As they diagonalize the momentum operator, they form a basis of our entire state space! Therefore, we can immediately see from the first equation that

$$\hat a^\dagger_{\lambda k}(\pm \infty) |\Omega\rangle = -C_\lambda \big( e^{ikx}, \hat \phi( x)\big)_{KG} |\Omega\rangle \vert_{t = \pm \infty} = | \lambda k \rangle$$ where we have chosen the normalization $\langle\lambda k | \lambda' k' \rangle = C_{\lambda}^* C_{\lambda'} (2 \pi)^3 (2 \omega_{\lambda k}) \delta_{\lambda \lambda'} \delta^3(\mathbf{k} - \mathbf{k}')$. From the second equation, we can immediately see that

$$\langle\Psi | \hat a_{\lambda k}(\pm \infty) |\Omega\rangle = 0 \hspace{0.15 cm} \text{ for all } \langle\Psi | \hspace{0.5 cm} \Longrightarrow \hspace{0.5 cm} \hat a_{\lambda k}(\pm \infty) |\Omega\rangle = 0.$$ Apparently our asymptotic creation and annihilation operators behave almost exactly like our good old creation and annihilation operators from the free theory!

There's another important property I must mention, which is that two creation/annihilation operators that have different $\lambda k$ data will commute. This is a direct consequence of the fact that our creation/annihilation operators are spacial integrals weighted by wave packets that are spatially separated at large times. (For operators with the same $k$ but different $\lambda$, as $m_\lambda$ is different the wave packets will propagate at different speeds and will still succeed to separate.) Note that spatial separation is a property of wave packets but not of plane waves. This is another place where it is necessary to view plane waves as a limit of wave packets in order to properly understand your theory. In fact, the operators will not commute unless they are defined with this limiting procedure.

We are finally ready to define our incoming and outgoing multi-particle states. As our asymptotic creation operators only change the ground state in localized spatial regions and each spatial excitation is justifiably called a "particle state" we can say that acting with a few of them on the ground state will create a perfectly good multi-particle state. We will now define our incoming (created at $t = -\infty$) and outgoing (created at $t = + \infty$) multi-particle asymptotic states.

$$ |\lambda_1 k_1, \ldots, \lambda_n k_n\rangle_{\rm in} \equiv \hat a^\dagger_{\lambda_1 k_1}(-\infty) \ldots \hat a^\dagger_{\lambda_n k_n}(-\infty) |\Omega\rangle \\ |\lambda_1 k_1, \ldots, \lambda_n k_n\rangle_{\rm out} \equiv \hat a^\dagger_{\lambda_1 k_1}(+\infty) \ldots \hat a^\dagger_{\lambda_n k_n}(+\infty) |\Omega\rangle$$

The four-momenta $k_i$ will have masses $k_i^2 = m_{\lambda_i}^2$ and no $|\lambda_i k_i\rangle$ is allowed to equal another. Some people prefer to rescale $\hat \phi$ in order to hide those $C_\lambda$ prefactors but I will not. The nature of these prefactors will be explored much later. It is important to note that the total momenta of these states are approximately the sum of all $\mathbf{k}_i$, and the energy is approximately the sum of all $\omega_{\lambda_i k_i}$. This lends more credence to the notion that these are "multi-particle" states.

Now that we have successfully defined our incoming and outgoing asymptotic multi particle states and derived some important properties of our newly constructed asymptotic creation and annihilation operators, we have completed the framework necessary to derive the LSZ reduction formula. Using the properties defined here, you should be able to justifiably go through the steps as outlined in Srednicki.

To answer your doubt 2: In order to get our states to have the right properties, we needed these to be wave-packets that are widely separated in the distant past and future. Therefore, these states are only approximately momentum and energy eigenstates (although you can get as close as you want). As they're not perfect energy eigenstates, some time evolution will occur. You particles will start far apart, come together, interact, then (different) particles will leave.

TLDR: If you define creation and annihilation operators properly, using the Klein Gordon inner product with widely separated wave packets in the far past/future, you will get your actual particle states when acting with these operators on the true vacuum $|\Omega\rangle$.

[Physics] S-matrix and time-evolution operator

Review of the picture transform

So the "picture transform" is a quantum coordinate transform which basically starts from the usual facts of quantum mechanics. These are roughly: systems are described by vectors in a Hilbert space; everything you can observe is modeled by a Hermitian operator on that Hilbert space, with eigenvalues being the actual values you observe; the only prediction of QM in any particular case is an average value of some observable $A$ in some state $\psi$ given by $\langle A \rangle_\psi = \langle \psi|\hat A |\psi\rangle$; some observable corresponding to total energy $\hat H$ also governs the time evolution of the state vectors in the Hilbert space as $i\hbar \partial_t|\psi\rangle = \hat H |\psi\rangle$.

The picture transform says: hey, that above equation is solved by $|\psi\rangle = U(t) |\psi_0\rangle$ for some unitary $U$ given by $\hat H$ as $$i \hbar \partial_t U = \hat H ~U.$$But suppose the Hamiltonian were instead $\hat h$ for some arbitrary Hermitian $\hat h.$ Then we would find $|\psi\rangle = u(t) |\psi_0\rangle$ for some other unitary $u$. Now insert $u u^\dagger = 1$ into the above to find, $$\langle A\rangle_\psi = \langle \psi_0| U^\dagger~u u^\dagger ~\hat A ~u u^\dagger ~U |\psi_0\rangle.$$Defining $A' = u^\dagger \hat A u$ and $|\psi'\rangle = u^\dagger U|\psi_0\rangle$ must therefore preserve all predictions of quantum mechanics, with the only cost being that now we must consider $$i\hbar \partial_t A' = u^\dagger (\hat A \hat h - \hat h \hat A) u=A'h' - h' A',$$ while for our Hamiltonian we find,$$ i\hbar \partial_t|\psi'\rangle = u^\dagger (- \hat h + \hat H) U |\psi_0\rangle = (H' - h')|\psi'\rangle.$$(Both formulas straightforwardly map to the right-hand-sides after strategically inserting $uu^\dagger = 1$ terms.)

Just like you can mentally understand $U$ as being some sort of time-ordered continuous product and write down a helpful mnemonic,$$U(t) = \mathcal T \exp\left(\frac1{i\hbar} \int_{t_0}^t d\tau ~ \hat H(\tau)\right),$$ if you define $\eta' = H' - h'$ you will find a helpful mnemonic that the new unitary evolution operator looks like,$$U'(t) = u^\dagger U = \mathcal T \exp\left(\frac1{i\hbar} \int_{t_0}^t d\tau ~ \eta'(\tau)\right).$$I claim that this last expression is exactly what was intended in the expression you have above, with $H_I(t)$ taking the place of $\eta'.$

Why the S-matrix tells us everything

The S-matrix is just a matrix expansion of $U'.$ (Equivalently, $U$, when taking the picture transform as the identity transform, the so-called "Schrödinger picture.") But suppose we have a basis $|n\rangle$ for the Hilbert space, $1 = \sum_n |n\rangle\langle n|.$ Then our above expression for the expectation value can be calculated as:$$\langle A\rangle_\psi = \sum_{mnpq} \langle \psi_0|m\rangle\langle m|{U'}^\dagger|n\rangle\langle n|A'|p\rangle\langle p|U'|q\rangle\langle q|\psi_0\rangle. = (\psi_0)_m^*~U^\dagger_{mn}~A_{np} ~U_{pq}~(\psi_0)_q.$$Therefore $U_{pq} = \langle p | U' | q\rangle$ really does in some sense contain all of the mystery of the time-evolution. And in your case it seems like you're probably dealing with, say, plane waves coming in from infinity getting scattered out to other plane waves travelling out to infinity: this is a great set of states to get a very simple picture, "here is what my Hamiltonian does."

Mixing these together in the interaction picture.

Usually when we're doing this we have some Hamiltonian which we can solve analytically, in your case it seems to be something like $\hat h = \hbar \omega_p a^\dagger_p a_p,$ where $p$ indexes a particle in a definite momentum-state.

When you do this, the presciption above basically just puts phase factors $e^{\pm~i\omega_p t}$ on all of your creation/annihilation operators, and on your definite-momentum states.

If that's right then your question becomes a non-issue: if you paid very careful attention to the $\propto$ sign, and tried to convert it into an $=$ sign, you would have to insert phase factors all over the place in order to keep this thing sane. I think Peskin and Schroeder are probably implicitly saying "hey, you've got this big complicated phase factor out front which does technically matter, but we are explicitly not considering it as important for the physics that we are about to discuss."

I don't have the book but that's where my gut would point me. They are proportional except for some sort of awkward phase factor which would make the equation too long to fit cleanly on one line, so it got absorbed into a $\propto$ symbol.