# Writing scalar quantum field as mode expansion form for interacting theory

We know that for Klein-Gordon Equation, quantum field can be written in the form $$\phi(\mathbf{x},t) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}[a_p e^{-ipx} + a^\dagger_p e^{ipx}]$$

It was stated somewhere that in interactive theory, I can write scalar field in the same form as this for a fixed time $t$ with creation and annihilation operator replaced as $a_p(t)$ and $a_p^\dagger (t)$, which satisfy the same algebra as the free theory.

The reason for this was because Hilbert space is the same at every time due to time-translation invariance, which I don't understand as a justification of this.

Can someone please elaborate on this point? Thank you.

## 1 Answer

Let's understand this statement in Hamiltonian formalism, where KG equation is equivalent to having the free scalar field hamiltonian and the Heisenberg equations of motion for the free fields.

Then $\phi(\vec{x},t) = \int \frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left( a_p e^{ipx} + a^\dagger_p e^{-ipx}\right)$ and the canonical conjugate $\pi(\vec{x},t) = \dot{\phi}(\vec{x},t)$, are the most general solution.

Now let's consider an interacting Hamiltonian $H = H_0 + \lambda V$, and DEFINE $$\Phi(\vec{x},t)\equiv e^{iHt}e^{-iH_0 t} \phi(\vec{x},t) e^{iH_0 t}e^{-iHt}$$ $$\Pi(\vec{x},t)\equiv e^{iHt}e^{-iH_0 t} \pi(\vec{x},t) e^{iH_0 t}e^{-iHt}$$ Then it is straight forward to show that $\Phi$ and $\Pi$ satisfy the canonical commutation relations, as well as the new interacting Heisenberg equations (notice that in the definition we use $H(\phi,\pi)$ and in the Heisenberg equations $H(\Phi,\Pi)$, we are allowed to do so because both are equal!). (Hint: to see that the new fields satisfy the full heisenberg equations notice that $\Phi(\vec{x},t) = e^{iHt}\phi(\vec{x},0)e^{-iHt}$) So in this sense they are the interacting fields, written in terms of the free fields.

Then we conclude that $$\Phi(\vec{x},t) = \int \frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left( \mathbb{a}(t)_p e^{ipx} + \mathbb{a}(t)^\dagger_p e^{-ipx}\right)$$ Where $$\mathbb{a}(t)_p\equiv e^{iHt}e^{-iH_0 t} a_p e^{iH_0 t}e^{-iHt}$$ and $\mathbb{a}_p(t)$ satisfies the required commutation relations as a consequence of its parents $\Phi$ and $\Pi$ doing so, or as can be verified directly using those of $a_p$.

Notice that this description is particularly useful for weakly coupled theories, since then $\mathbb{a}_p = a_p + \mathcal{O}(\lambda,a_p^2)$, then all our particle spectrum can be inferred from that of the free theory, unlike when this expansion is no longer valid, and the new creation operator can create states completely different in nature from what's contained in the free theory.

• Sorry, I don't think I follow how the new $\Phi$ and $\Pi$ satisfies the new Heisenberg equation, and what does this have to do with time-translation invariance of the Hilbert space? – Quantization Mar 5 '15 at 20:31
• Heisenberg equations after all just say that the Hamiltonian is the generator of time translations! $\dot{\phi} = i\left[ H,\phi\right] = \frac{\delta H}{\delta \phi}$ – Ali Moh Mar 5 '15 at 20:43
• I added a hint in the answer (in parentheses) so you can see how taking $\partial \Phi/\partial t$, in deed gives the commutator with the Full Hamiltonian. – Ali Moh Mar 5 '15 at 20:55
• in the above comment there is a typo, it should be $\frac{\delta H}{\delta \pi}$ – Ali Moh Mar 5 '15 at 21:07