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$\newcommand{\ket}[1]{\left\vert#1\right\rangle}$In Quantum mechanics, we have the interaction picture. When the Hamiltonian is in the form of $H=H_{0}+V$, we can transform the evolution equation into $\frac{d}{dt}\ket{\phi}_{I}=V_{I}\ket{\phi}_{I}$, where $\ket{\phi}_{I}=e^{iH_{0}t}\ket{\phi}$, and $V_{I}=e^{iH_{0}t}Ve^{-iH_{0}t}$.

In QFT, we apply the interaction picture as well. But for most of the text I have seen, take $V=\frac{g}{4}\phi^{4}$ as an example, when calculating the Dyson series or Feynman amplitudes, the interaction Hamiltonian is still $\frac{g}{4}\phi^{4}$, not $e^{iH_{0}t}Ve^{-iH_{0}t}$. Why is that the case?

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2 Answers 2

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This confused me as well. Remember that the $\phi$ in $\frac{g}{4}\phi^4$ that we used in the Interaction picture has time dependence. More precisely, the $\phi$ that we use is the free field solution, i.e. it is the solution to $\frac{d \phi}{dt}=-i[\phi, H_{0}]$, which is the same as the value of $e^{iH_0t}\phi e^{-iH_0t}$ (You can differentiate this to get the commutator equation)

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    $\begingroup$ Exactly, $$e^{iH_0t}{\phi^4(\mathbf{x})} e^{-iH_0t} = {\phi^4(\mathbf{x}, t)}.$$ $\endgroup$
    – AfterShave
    Oct 25, 2022 at 13:47
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In an attempt to make the previous answers clearer consider the following example using the Klein-Gordon density $$ \mathcal{L}_0=\frac{1}{2}\partial_\mu\phi(x)\partial^\mu\phi(x)-\frac{1}{2}m^2\phi^2(x)+\Omega_{0}, \tag{1}\label{1} $$ where the metric signature is $(+,-,-,-)$ and $x\equiv (t,\boldsymbol{x})$. The constant $\Omega_{0}$ is there to regulate the zero-point divergence of the theory.

The equations of motion for \eqref{1} read $$ (\square+m^2)\phi(x)=0, \label{2}\tag{2} $$ and the solutions after canonical quantization can take the following form $$ \phi(x)=\int\frac{\mathrm{d}^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\left[a_{\boldsymbol{k}}e^{-i\kappa\cdot x}+a^{\dagger}_{\boldsymbol{k}}e^{i\kappa\cdot x}\right], \tag{3}\label{3} $$ where $\omega_{\boldsymbol{k}}=\sqrt{\boldsymbol{k}^2+m^2}$ and $\kappa=(\omega_{\boldsymbol{k}},\boldsymbol{k})$. Now, the Legendre transformation of $\mathcal{L}_0$ will result in the Hamiltionian desnity $\mathcal{H}_{0}$ and from there one can evaluate the Hamiltonian $H_0$ from $$ H_0=\int\mathrm{d}^3x\mathcal{H}_{0}=\int\frac{\mathrm{d}^3k}{(2\pi)^3}\left|\omega_{\boldsymbol{k}}\right|\left(a^{\dagger}_{\boldsymbol{k}}a_{\boldsymbol{k}}+\frac{1}{2}(2\pi)^3\delta^{(3)}(0)\right)-\Omega_{0}\int\mathrm{d}^3x. \tag{4}\label{4} $$ Please note that in the process of deriving the above it is assumed that $H_{0}$ will act on an arbitrary state meaning that terms $\sim a^{\dagger}_{\boldsymbol{k}}a^{\dagger}_{\boldsymbol{k}}$ and $\sim a_{\boldsymbol{k}}a_{\boldsymbol{k}}$ will have no contribution. Also note that $H_0$ is now time-independent meaning that this operator is in the Schrödinger picture. Now fix $\Omega_0$ in such a way that $H_0$ becomes $$ H_0=\int\frac{\mathrm{d}^3k}{(2\pi)^3}\left|\omega_{\boldsymbol{k}}\right|a^{\dagger}_{\boldsymbol{k}}a_{\boldsymbol{k}}. \tag{5}\label{5} $$ From there use the appropriate version of Baker–Campbell–Hausdorff formula to show that $$ e^{iH_0t}a_{\boldsymbol{k}}e^{-iH_0t}=a_{\boldsymbol{k}}e^{-i\omega_{\boldsymbol{k}}t} \quad \text{and} \quad e^{iH_0t}a^{\dagger}_{\boldsymbol{k}}e^{-iH_0t}=a^{\dagger}_{\boldsymbol{k}}e^{i\omega_{\boldsymbol{k}}t}. \tag{6}\label{6} $$ Combine \eqref{3} and \eqref{6} to get $$ \phi(x)=e^{iH_0t}\underbrace{\int\frac{\mathrm{d}^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\left[a_{\boldsymbol{k}}e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+a^{\dagger}_{\boldsymbol{k}}e^{-i\boldsymbol{k}\cdot \boldsymbol{x}}\right]}_{\phi(\boldsymbol{x})}e^{-iH_0t}=e^{iH_0t}\phi(\boldsymbol{x})e^{-iH_0t}. \tag{7}\label{7} $$ Let's take a step back and stare at \eqref{7} for a while. We know from quantum mechanics that operators that depend explicitly on time are in either the Heisenberg picture or the interaction picture, hence the field $\phi(x)$ is in one of them. In order to determine which one, consider the interaction picture first. For the theory to be in the interaction picture the Hamiltonian (which is in the Schrödinger picture) should have two parts say $H=H_{1,S}+H_{2,S}$ and the rule for jumping from the Schrödinger picture to the interaction picture for an arbitrary operator $A$ should be something like $$ A_{I}(x)=e^{iH_{1,S}t}A_{S}(\boldsymbol{x})e^{-iH_{1,S}t}, \tag{8}\label{8}$$ meaning that we only evolve the operator with only a specific part of the Hamiltionian and not the whole complete $H$.

From this small discussion it should be obvious that \eqref{7} indicates that $\phi(x)$ is in the Heisenberg picture since we are evolving with the respect to the complete Hamiltonian $H_0$. We may now define $\phi_{H_0}(x):=\phi(x)$ where the subscript $H_0$ indicates the Heisenberg picture with respect to $H_0$ and $\phi_{S}(\boldsymbol{x}):=\phi(\boldsymbol{x})$ where $S$ indicates the Schrödinger picture and rewrite \eqref{7} as $$ \phi_{H_0}(x)=e^{iH_0t}\phi_{S}(\boldsymbol{x})e^{-iH_0t}. \tag{9}\label{9} $$ We can slightly generalize the above by evaluating \eqref{3} at $t=t_0$ to get $$ \phi_{H_0}(t_0,\boldsymbol{x})=\int\frac{\mathrm{d}^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\left[a_{\boldsymbol{k}}e^{-i\omega_{\boldsymbol{k}}t_0 + i\boldsymbol{k}\cdot\boldsymbol{x}}+a^{\dagger}_{\boldsymbol{k}}e^{i\omega_{\boldsymbol{k}}t_0 - i\boldsymbol{k}\cdot\boldsymbol{x}}\right], \tag{10}\label{10} $$ and since $a^{\dagger}_{\boldsymbol{k}},a_{\boldsymbol{k}}$ are arbitrary and $e^{i\omega_{\boldsymbol{k}}t_0}$ is a constant (this is important), it can be absorbed by the creation and annihilation operators yielding $$ \phi_{H_0}(t_0,\boldsymbol{x})=\int\frac{\mathrm{d}^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\left[a'_{\boldsymbol{k}}e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+a'^{\dagger}_{\boldsymbol{k}}e^{- i\boldsymbol{k}\cdot\boldsymbol{x}}\right]=\phi_{S}(t_0,\boldsymbol{x}). \tag{11}\label{11} $$ Combining \eqref{9} and \eqref{11} gives the following general formula for $t\ne t_0$ $$ \phi_{H_0}(x)=e^{iH_0(t-t_0)}\phi_{S}(t_0,\boldsymbol{x})e^{-iH_0(t-t_0)}. \tag{12}\label{12} $$

We may now proceed with the interactions. Since we have already solved \eqref{1} exactly, it would make more sense to introduce interactions in such way that we can somehow take full advantage of the solutions of the free theory. From there we consider the following interaction as an example $$ \mathcal{L}_{\text{int}}=-\frac{\lambda}{4!}\phi^4_{H_0}(x) \quad \rightarrow \quad \mathcal{H}_{\text{int}}=\frac{\lambda}{4!}\phi^4_{H_0}(x) \quad \rightarrow \quad H_{\text{int}}=\int\mathrm{d}^3x\frac{\lambda}{4!}\phi^4_{H_0}(x), \tag{13}\label{13} $$ so the complete Hamiltonian now reads $H=H_0+H_{\text{int}}$. The Heisenberg picture for the complete theory will now be given by $$ \phi_{H}(x)=e^{iH(t-t_0)}\phi_{S}(t_0,\boldsymbol{x})e^{-iH(t-t_0)}, \tag{14}\label{14} $$ where you can safely assume that $\phi_{S}(t_0,\boldsymbol{x})$ will have the same form as in \eqref{11}. Now as $\lambda\rightarrow 0$ \eqref{14} becomes $$ \phi_{H}(x)\Big|_{\lambda\rightarrow 0}=e^{iH_0(t-t_0)}\phi_{S}(t_0,\boldsymbol{x})e^{-iH_0(t-t_0)}=\phi_{H_0}(x). \tag{15}\label{15} $$ Comparing \eqref{14} and \eqref{15} we understand that $\phi_{H_0}(x)$ (which is in the Heisenberg picture with respect to the free theory) is identified with the interaction picture field with respect to the interacting theory or $$ \phi_{H_0}(x)=\phi_{I}(x), \tag{16}\label{16} $$ where $I$ indicates the interaction picture.

In conclusion, we do not explicitly transform the fields into the interaction picture because they are already in the interaction picture!

I hope this resolves the confusion.

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