Interaction Picture Ladder Operators vs Free Field Operators

Question

Consider the case of an interacting scalar field theory with bare mass $m_0$. After having derived the LSZ reduction formula, all that is left is to compute the time ordered products of the Heisenberg picture full interacting field operators $\phi (t, \vec{x})$ acting on the full interacting field vacuum. $$\left< \Omega \right|T(\phi(x)\phi(y))\left| \Omega \right>$$

We would like to relate the quantities involved in this expression to free field operators and the vacuum of the free theory.

Consider the interaction picture operators $$\Phi_I(t,\vec{x}) = e^{-iH_0(t-t_0)}\phi(t_0,\vec{x})e^{iH_0(t-t_0)}$$ $$\Pi_I(t, \vec{x}) = e^{-iH_0(t-t_0)}\pi(t_0,\vec{x})e^{iH_0(t-t_0)}$$ where $\phi(t_0, \vec{x})$ and $\pi(t_0,\vec{x})$ are the Heisenberg picture field and conjugate momentum operators respectively of the full interacting theory at some time $t_0$. We then show that $\Phi_I(t,\vec{x})$ satisfies the Klein-Gordon equation $$(\square +m_0^2)\Phi_I(t,\vec{x})=0$$ We thus justify that $\Phi_I(t,\vec{x})$ admits the form of a mode expansion $$\Phi_I(t,\vec{x})=\int \frac{d^3p}{(2\pi)^3}(a_{p,I}(t)e^{-ip.x}+a_{p,I}^{\dagger}(t)e^{ip.x})$$ for some operator $a_{p,I}(t)$.

Using the equal time commutation relation on the Heisenberg fields $[\phi(t_0, \vec{x}), \pi(t_0,\vec{y})] = i\delta^{(3)}(\vec{x}-\vec{y})$ and the fact that $\partial_t \Phi_I(t,\vec{x}) = \Pi_I(t, \vec{x})$ we deduce that $[a_{p,I}(t), a_{p',I}^{\dagger}(t)] = i \delta^{(3)}(\vec{p}-\vec{p'})$. We have thus shown that the interaction picture fully interacting field and conjugate momentum operators mimic the exact same algebraic relations as those of the free field and conjugate momentum operators in the Heisenberg picture. Needless to say, the fact that the two sets of operators satisfy the same algebraic relations among themselves does not imply that their action on the state kets $\left | \psi \right> \in H$ of the "Hilbert space" of wavefunctionals are identical. There is some highly nontrivial relation between the ladder operators $a_{p,I}(t)$ that appear in the mode expansion above and the $a_p$ operators that appear in the mode expansion of the free theory. The states annihilated by the $a_I$ operators need not coincide with those annihilated by $a_p$ in general.

Why does the author in this reference https://www.thphys.uni-heidelberg.de/courses/weigand/QFT1-12-13.pdf page 54 eq 2.79 conclude that the operator $a_{p,I}(t)$ annihilates the vacuum state of the free theory $\left |0 \right>$? Why should the state $\left |0 \right>$ annihilated by $H_0$ be also annihilated by $a_{p,I}$? $H_0$ is expressed in terms of $a_p$ of the free theory, not $a_{p,I}$.

I am aware of similar questions being asked before but they just go on about justifying the mode expansion above, no reference to the vacuum state misunderstanding presented here.

Answer 1

I think your reference makes the following two assumptions (stated in the two sentences above (2.79)):

Assumption 1: A non-zero state $|\psi\rangle$ exists such that $H_0 |\psi\rangle = a_I(\vec{p}) |\psi\rangle = 0$ for all $\vec{p}$.

Assumption 2: $0$ is an eigenvalue of multiplicity 1 of the free field Hamiltonian $H_0$, i.e. the free field vacuum vector $|0\rangle$ is (up to a non-zero complex number) the only non-zero vector that is annihilated by $H_0$.

Taking these two assumptions into account, the conclusion is $|\psi\rangle = c|0\rangle$, $c\in\mathbb{C}\backslash\{0\}$, and $a_I(\vec{p})|0\rangle = 0$.

Ad Assumption 1: The free Hamiltonian $H_0\geq 0$ is bounded from below. Let $|\psi\rangle$ be a state of lowest energy (it is implicitly assumed that such a state exists), i.e. $H_0|\psi\rangle = 0$. Moreoever, we have the commutator identity $[H_0,a_I(\vec{p})] = -E_p a_I(\vec{p})$, $E_p = \sqrt{|\vec{p}|^2+m^2}$ (eq. (2.78)). Thus, $$H_0 a_I(\vec{p})|\psi\rangle = [H_0,a_I(\vec{p})]|\psi\rangle + a_I(\vec{p}) H_0|\psi\rangle = -E_p a_I(\vec{p})|\psi\rangle.$$ This implies that $a_I(\vec{p})|\psi\rangle$ is a (generalised) eigenstate of $H_0$ with eigenvalue $-E_p < 0$, contradicting $H_0 \geq 0$ unless $a_I(\vec{p})|\psi\rangle = 0$.

Answer 2

Consider the Hilbert space for the full interacting theory Hamiltonian, $H = H_0 + H_{int}$. $H_0$ has the same dependence on the field and the conjugate momentum operators as the free Klein-Gordon theory, and $H_{int}$ is an additional part that leads to interactions. In an abuse of notation, we use the same symbol $H_0$ here as we do for the Hamiltonian of the free theory. But they are not the same, because this is an interacting theory, and $H_0$ is not the Hamiltonian for this theory. For example, the fields (based on which $H_0$ is defined) in this theory are not the same as the free fields, and the vacuum is not the same as the free vacuum.

As stated in the Weigand reference on p. 53, we establish a correspondence between the field operators and ladder operators in the Interaction Picture for this full interacting theory Hamiltonian, with the field operators and ladder operators for the free theory. This is done by noting that the Interaction Picture field satisfies the free Klein-Gordon equation with mass $m_0$, the Fourier transform relations between the Interaction Picture field and the Interaction Picture ladder operators take the same form as for the free theory, and the commutation relations for the Interaction Picture field and the Interaction Picture ladder operators take the same form as for the free theory. Furthermore, $H_0$ in the Interaction Picture remains $H_0$, namely: $(H_0)_I = H_0$ (as stated in the sentence after (2.78) in Weigand).

This means that if we restrict ourselves to the Interaction Picture $H_0$ and the Interaction Picture fields and the Interaction Picture ladder operators, those operators have the same relations as the relations between their counterparts in the free theory.

Therefore, there exists a unique state in the Hilbert space for this interacting theory that is annihilated by both $H_0$ and $a_I(\vec{p})$. We call this unique state the "free vacuum" and denote it as $|0\rangle$ in the current context, where it is understood that it is not the same as the "true free vacuum" of the free Klein-Gordon theory (because that "true free vacuum" is presumably in a different Hilbert space). To avoid confusion, in the remainder of this answer we denote this as the "Interaction Picture free vacuum".

Then we express the interacting vacuum $|\Omega\rangle$ in terms of the Interaction Picture free vacuum, and the Heisenberg Picture fields in terms of the Interaction Picture fields. This lets us arrive at expressions that look like $\langle 0 | T \phi_I(x) \phi_I(y) | 0 \rangle$. And now we can use the fact that $a_I(\vec{p})$ (the Interaction Picture annihilation operator) annihilates the Interaction Picture free vacuum to use Wick's theorem to simplify this to just the fully contracted terms.