Why are fields in the interaction picture the same as free fields in the Heisenberg picture?

Question

The bottom part of Peskin & Schroeder Chap. 4.2 p. 83 has me puzzled.

What is the justification for the expression of $\phi(t_0, \mathbf{x})$? They claim that at any fixed time $t_0$, in an interacting theory or not, $\phi$ can be expanded in terms of ladder operators:

$$ \phi ( t_{0} ,\mathbf{x}) =\int \frac{d^{3} p}{( 2\pi )^{3}}\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left( a_{\mathbf{p}} e^{i\mathbf{p} \cdot \mathbf{x}} +a_{\mathbf{p}}^{\dagger } e^{-i\mathbf{p} \cdot \mathbf{x}}\right) $$

They just say "of course" but it's not obvious to me at all. $\phi$ should be a solution to the equation of motion, and since we've added interaction terms to the Lagrangian the equation of motion is no longer the Klein-Gordon equation, so $\phi$ should be different right?

In going from Eq. (4.14) to Eq. (4.15),

$$ \phi _{I}( t,\mathbf{x}) =e^{iH_{0}( t-t_{0})} \phi ( t_{0} ,\mathbf{x}) e^{-iH_{0}( t-t_{0})} =\int \frac{d^{3} p}{( 2\pi )^{3}}\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left( a_{\mathbf{p}} e^{-ip\cdot x} +a_{\mathbf{p}}^{\dagger } e^{ip\cdot x}\right)_{|x^{0} =t-t_{0}}, $$

I think they've used that $e^{iH_0t} a_{\mathbf{p}} e^{-iH_0t} =a_{\mathbf{p}} e^{-iE_{\mathbf{p}} t}$ from Eq. (2.46). But to derive Eq. (2.46) I think they used that the free Hamiltonian is

$$H_0=\int \frac{d^{3} p}{( 2\pi )^{3}} E_{\mathbf{p}} a_{\mathbf{p}}^{\dagger } a_{\mathbf{p}} \quad ,$$

which again was derived by solving the Klein-Gordon equation and plugging the expression for $\phi$ back into $H$. But since the equation of motion is something else now this no longer applies?

Answer 1

I largely agree with the sentiment of the answers above, expanding the fixed time boundary condition field in Fourier modes is always allowed, and the ladder operators have to obey the canonical commutation relations. I believe what is actually confusing you here, is the fact that you believe that these ladder operators are a priori assumed to annihilate some vacuum in some free theory. The short answer is, well, a priori, they don't need to. As a matter of fact their properties with the vacuum are never used in the rest of the perturbative expansion exposition by Peskin and Schroeder. Also, there seems to be some confusion about the equations of motion (EOM's) the fundamental field obeys. In particular, the field $\phi(t_0,x)$ does not obey any EOM's because it is given at a particular fixed moment of time. Hence it can be freely expanded into it's Fourier modes.

However, it just so happens that, regardless of the operator's $\phi(t_0,x)$ exact action on the physical Hilbert space, the interaction picture field $\phi_I$ as defined above obeys the Klein-Gordon equation of motion. The only assumption that need go into this is the requirement that the boundary field obeys canonical commutation relations. That allows an expansion of the interacting field in terms of Fourier modes, that in turn can be shown to be exactly the canonical ladder operators.

Let us show this. If $[\phi(t_0,x), \Pi(t_0, y)]=i\delta(x-y)$ and $$H_0=\int dx\frac{1}{2}(\Pi^2(t_0,x)+(\nabla\phi(t_0,x))^2+m^2\phi^2(t_0,x))$$

it is straightforward to show that

$$\dot{\phi}_I(t,x)=ie^{iH_0(t-t_0)}[H_0,\phi(t_0,x)]e^{-iH_0(t-t_0)}=\Pi_I(t,x)\\ \ddot{\phi}_I(t,x)=ie^{iH_0(t-t_0)}[H_0,\Pi(t_0,x)]e^{-iH_0(t-t_0)}=(-\nabla^2+m^2)\phi_I(t,x)$$

With this proven, there is no doubt in the world that in the canonical quantization scheme, this field will always be free, no matter how complicated the full Hamiltonian is. It is true that the representation the authors use for the fixed time boundary field is a bit ad hoc: why would it be a useful representation if you didn't know that the interacting field obeys the KG EOM and why would these ladder operators be related to the canonical ones? It may be relatively difficult to detect the connection. However, once you know this field's EOM, it becomes clear that the interacting field can be expressed in terms of such operators, and these operators can be shown to annihilate the perturbative vacuum.

Let us show a more streamlined way to do this: First, expand the field $\phi(t_0,x)$ in terms of it's Fourier modes

$$\phi(t_0,x)=\int\frac{d^3 p }{\sqrt{2E_p}}\left(f_p e^{i\mathbf{p x}}+f^{\dagger}_pe^{-i \mathbf {px}}\right)$$

We know by standard classical KG theory that the solution to

$$\left(\frac{\partial}{\partial t^2}-\nabla^2+m^2\right)\phi_I(t,x)=0~~~,~~~ \phi_I(t_0,x)=\int \frac{d^3p}{(2\pi)^3}G(p)e^{i\mathbf{px}}\Rightarrow\\\phi_I(t,x)=\int \frac{d^3p}{(2\pi)^3}G(p)e^{i\mathbf{px}-iE_p (t-t_0)} $$

This carries over operatorially, as Peskin and Schroeder note, to

$$\phi_I(t,x)=\int \frac{d^3p}{\sqrt{2E_p}}(f_pe^{i\mathbf{px}-iE_p (t-t_0)}+f^\dagger_pe^{-i\mathbf{px}+iE_p (t-t_0)})$$

Now it is a simple matter to find $\Pi(t_0,x)=\dot{\phi}(t_0,x)$:

$$\Pi(t_0,x)=-i\int d^3 p \sqrt{\frac{E_p}{2}}(f_pe^{i\mathbf{px}}-f^\dagger_pe^{-i\mathbf{px}})$$

Finally, now we can show that

$$f_q=\sqrt{\frac{2}{E_q}}\int\frac{d^3 x}{(2\pi)^3}(E_q\phi(t_0,x)+i\Pi(t_0,x))\\ f^\dagger_q=\sqrt{\frac{2}{E_q}}\int\frac{d^3 x}{(2\pi)^3}(E_q\phi(t_0,x)-i\Pi(t_0,x))$$

Now we cancompute the commutation relations between these operators and express $H_0$ in terms thereof, which finally allows us to show that when $H_0|0\rangle=0\Rightarrow f_q|0\rangle=0 $, showing that these operators are indeed free-field ladder operators in a non ad-hoc manner.

Exercise: Show that if $\phi(t_0,x)$ obeys canonical quantization relations, then the Heisenberg field $\phi(t,x)=e^{iH(t-t_0)}\phi(t,0)e^{-iH(t-t_0)}$, with $\mathcal H= \mathcal{H}_0+\mathcal{V}[\phi(t_0,x)]$ obeys the interacting EOM

$$\left(\frac{\partial}{\partial t^2}-\nabla^2\right)\phi(t,x)+\frac{\partial \mathcal {V}}{\partial \phi}=0$$

Answer 2

1. why the field in the Schrodinger picture is always unchanged when interactions are added (if that is the case)

The field in Schrodinger picture is simply a quantum operator with a quantum-classical correspondence with an observable (the value of the scalar field in the given space point). It is actually hard to imagine what it actually means that "operator changes".
Suppose you study the quantum mechanics of a particle in harmonic oscillator but you decide to add a quartic term $V' = \epsilon x^4$ to the potential so that the Hamiltonian now reads $H = \frac {\hat p^2} {2m} + \frac 1 2 m \omega^2 \hat x^2 + V'(\hat x)$. If, for a given energy state, you calculate, say, mean value of the $\hat {x}^2$ operator, your result changes if you consider this extra interaction $V'$. But it does not mean that the operator itself has changed - what changed is its mean value $\langle n|\hat x^2|n \rangle$ for a certain state $|n \rangle $. But the operator is still an operator of the square of the positional coordinate.
Analogously, $\phi(x)$, or more precisely, $\hat \phi(\vec x)$, is an operator of the value of the scalar field $\phi$ in the point $\vec x$, regardless of what is the full Hamiltonian of your theory.

EDITED: If your question is "why can I still rewrite $\phi$ in terms of $a$'s and $a^\dagger$'s", then notice that these ladder operators create a non-interacting excitation (particle).
Again, let me use the analogy with the harmonic oscillator in QM: here the ladder operators can be expressed $\hat a^{(\dagger)} = \sqrt{m\omega/2\hbar} \left(\hat x \pm i\hat p/(m\omega)\right)$ or inversely $\hat x = \sqrt{\hbar /(2m\omega)} (\hat a^\dagger + \hat a)$. In coordinate representation, $\hat a^\dagger =\sqrt{\hbar /(2m\omega)}\left( x - \frac{\hbar}{m\omega} \frac{\mathrm d}{\mathrm d x}\right)$. If you add a quartic term $\lambda x^4$ to the potential, the energy levels change but you can still use the above expressions, even though $a^\dagger$ no longer excites the true energy eigenstates to the next ones. Analogously, you can still use the prescription for $\hat \phi(x)$ although $a_p^\dagger$'s create energy+momentum-eigenstates of the simpler (non-interacting) theory.

2. why the field in the interaction picture is always the same as the free field in the Heisenberg picture (if that is the case)

Because it is defined so. Again, since we're dealing with quantum field theory, what you call the field is in fact the operator of the field. In the interaction picture, all operators evolve under the free evolution operator $e^{-iH_0t}$ by definition. The mismatch between $e^{-iH_0t}$ and $e^{-iHt}$ is then saved by evolving the state accordingly etc. etc. See "Dirac picture" in any reasonable book of QM basics or even wikipedia if you just need to shortly recall this stuff.