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On page 83 of Peskin and Schroeder, they expand an interacting field $\phi(x)$ at a fixed time $t_0$ in terms of the ladder operators as $$\phi(t_0,\vec{x})=\int\frac{d^3p}{(2\pi)^3\sqrt{2E_{\vec{p}}}}\Big(a_{\vec{p}}e^{i\vec{p}\cdot\vec{x}}+a^\dagger_{\vec{p}}e^{-i\vec{p}\cdot\vec{x}}\Big),\tag{4.13b}$$ which is the Heisenberg picture field fixed time $t_0$ and coincides with the Schrodinger picture field.

I have three questions.

  1. First of all, why doesn't $t_0$ appear on the right side?

  2. Are these $a_p, a_p^\dagger$ same as those for the free field $\phi_{\rm free}$? From the discussion on page 88, it looks like $a_{\vec{p}}$ annihilates the free vacuum $|0\rangle$ (not the interacting vacuum $|\Omega\rangle$). Therefore, I think $a_p, a_p^\dagger$ are those for free theory.

  3. Also if they use the same $a_p, a_p^\dagger$ for $\phi$ as for $\phi_{\rm free}$, this $E_{\vec{p}}$ must be different from $E^{\rm free}_{\vec{p}}$. After all, something must distinguish $\phi$ and $\phi_{\rm free}$ if not the ladder operators.

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  • $\begingroup$ Q.1. is answered in physics.stackexchange.com/q/310353/84967 $\endgroup$ – AccidentalFourierTransform Feb 22 at 17:46
  • $\begingroup$ @AccidentalFourierTransform Thanks for the link. It does answer my question mostly. However, in that answer of yours, my hunch is that $\omega_{\vec k}$ is the eigenvalue of full $H$, has a complicated dispersion relation compared to $\omega_{\vec k}=ck$ of the free theory? Can you confirm if that is true? Thanks again. $\endgroup$ – mithusengupta123 Feb 25 at 7:57
  • $\begingroup$ All the expressions in that answer are true, regardless of the value of $\omega(\vec k)$ (as long as you use the same expression everywhere: in the measure, the definition of the conjugate momentum, and in the exponent). You could e.g. take it to be $\omega(\vec k)=k$, or any other expression you want. It is, after all, a definition, so you are free to define it to be whatever you like. $\endgroup$ – AccidentalFourierTransform Feb 27 at 0:27
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The whole expression is identical to the one for a free field. In terms of the interacting fields $\phi$ and $\pi$, one can define $a_\vec{p}$ and $a_\vec{p}^\dagger$ in terms of their spatial Fourier transforms (see here) at a given time $t_0$, and expand the fields as you have done.

The difference between the free and interacting fields is how they evolve in time. In the free theory, you can easily solve for the full time evolution $$ \phi_\mathrm{free}(t, \vec{x}) = \int\frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2E_\vec{p}}} \left(a_\vec{p} e^{-i (E_\vec{p} (t - t_0) - \vec{p} \cdot \vec{x})} + a_\vec{p}^\dagger e^{i (E_\vec{p} (t - t_0) - \vec{p} \cdot \vec{x})}\right) $$ because you have $\frac{d}{dt} a_\vec{p}(t) = i[H, a_\vec{p}] = -E_\vec{p} a_\vec{p}(t)$ (here $a_\vec{p}(t)$ is the Heisenberg picture operator and $a_\vec{p} = a_\vec{p}(t_0)$ is the Schrödinger picture operator). In the interacting theory, the time evolution is much more complicated, so you cannot write down $\phi(t, \vec{x})$ explicitly for $t \neq t_0$.

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  • $\begingroup$ Hi! Can you please address the 3rd point? Also, see my comment above :-) $\endgroup$ – mithusengupta123 Feb 25 at 8:12
  • $\begingroup$ The dispersion relation is still $E_\vec{p} = \sqrt{\vec{p}^2 + m_0^2}$ by definition, where $m_0$ is the parameter in the Lagrangian. The only difference is that the bare mass $m_0$ is no longer the same as the the real mass $m$ of the particle (the energy of the lowest excited state). This touches on renormalization; down that path, there be dragons. $\endgroup$ – Elias Riedel Gårding Feb 25 at 9:19
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I will make a few comments that will hopefully help.

1) The point is that we choose a reference time $t_0$ at which the Heisenberg and schrodinger picture fields coincide, called $\phi(x,t_0)$. This is unchanging in time, by definition of Schrodinger picture. The 'interaction' would be captured only in the time evolution of $a_p(t_0), a^\dagger_p(t_0)$, which will tell you about the Heisenberg picture field $\phi(x,t)$. Recall how the mode expansion for a free field $\phi(x,t)$ is carried out-you time-evolve the mode expansion for the schrodinger picture free field $\phi(x,t_0)$(which, by the way, has no time dependence-answers your first point), using the commutation relations between $H,a_p$ etc.

2) The 'interaction picture field' is simply a free field(confusingly, I admit) then; as it is simply time evolution with the free part of the hamiltonian. To convince yourself, you can indeed show that $\phi_I$ satisfies the FREE Klein Gordon equation. This justifies the mode expansion, and the claim that the $a_p$ are the same as that for free K-G field. To further convince yourself, calculate the hamiltonian and verify the $H,a$ commutation relations.

3) By the same token as above, all quantities in that mode expansion of $\phi_I(x,t)$ are free field quantities-because its a free field expansion. But the actual field-the heisenberg field $\phi(x,t)$-is obtained by time evolution with the full hamiltonian; and therefore contains information about the interaction.

4) Where does the interaction come in? Well, in

a) The Heisenberg picture fields; the correlation functions are what you get by sandwiching these HEISENBERG fields between the INTERACTING theory vaccuum.

b) The pole of the propagator $\frac{1}{p^2-m^2}$ is now at some $m$ which NOT the free field mass, i.e. if $L=\frac{1}{2}((\partial\phi)^2-m_0^2\phi^2)+V(\phi)$; then $m\neq m_0$. This will take us down the road of renormalization, bare mass vs renormalized mass etc.

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Firstly, I think one of the signs in the exponents must be negative. (I don't have my copy of the book with me, so I cannot check if there is a typo in the book, but suspect not.) So, if $\phi$ explicitly depended on $t$, then that would have appeared on the right-hand side as a factor of $\exp(\pm i\omega t)$. So by going to the Schroedinger picture, one effectively absorbs this factor into the ladder operators. Since one fixed $t=t_0$, which is a constant and not a variable, one need not show it explicitly on the right-hand side.

What the expression represents is a quantization of the scalar field. It is general practice to quantize the field as a free field, even when one wants to consider an interacting field theory. Afterward the interactions are introduced perturbations involving these free fields.

In this expression, $E_p$ is the temporal component of the momentum four-vector that is associated with the momentum three-vector over with the integral is evaluated. The association is given by the dispersion relation for the scalar field, which merely involves the rest-mass of the scalar field. Other than that, it is not affected by the scalar and also not by any interactions.

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  • $\begingroup$ Sorry. It was a typo by me. Corrected $\endgroup$ – mithusengupta123 Feb 21 at 14:35

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