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In chapter 4 of Srednicki's QFT book (introducing the spin-statistics theorem for spin-0 particles), he introduces nonhermitian field operators (just taking one as an example):
$$\varphi^+(\mathbf{x},0) = \int \tilde{dk}\text{ }e^{i \mathbf{k}\cdot\mathbf{x}}a(\mathbf{k})$$
and time-evolves them in the Heisenberg picture with the Hamiltonian:
$$H_0 = \int \tilde{dk} \text{ }\omega\text{ } a^{\dagger}(\mathbf{k})a(\mathbf{k})$$
to get:
$$\varphi^+(\mathbf{x},t) = e^{iH_0t}\varphi^+(\mathbf{x},0)e^{-iH_0t} = \int \tilde{dk}\text{ }e^{i kx}a(\mathbf{k})$$
where ##kx = \mathbf{k}\cdot\mathbf{x}-\omega t##. I'm trying to derive this last line. So far, my approach has been to expand the ##e^{iH_0t}## operators in a Taylor series and calculate each term of the ##\varphi^+## directly. Based on my progress (I've only done through first order in ##H_0##), I figure I'll eventually get:
$$e^{iH_0t}\varphi^+(\mathbf{x},0)e^{-iH_0t} = \int \tilde{dk}\text{ }a(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}}(1-i\omega t-\frac{1}{2}\omega^2 t^2 + \cdots)$$
Two questions:
1) Second order gives the term ##a(\mathbf{k}')a^{\dagger}(\mathbf{k})a(\mathbf{k})a^{\dagger}(\mathbf{k})a(\mathbf{k})+a^{\dagger}(\mathbf{k})a(\mathbf{k})a(\mathbf{k}')a^{\dagger}(\mathbf{k})a(\mathbf{k})+a^{\dagger}(\mathbf{k})a(\mathbf{k})a^{\dagger}(\mathbf{k})a(\mathbf{k})a(\mathbf{k}')##, and third order will give me permutations of terms with 7 creation/annihilation operators. Et cetera for higher-order terms. Is there a more clever way to do this that will make it clear that these terms all collapse to powers of ##\omega##?
2) In carrying out the evaluation of the first-order term, part of my integrand ended up being proportional to ##a(\mathbf{k})a(\mathbf{k}')-a(\mathbf{k}')a(\mathbf{k})##. I only get ##i\omega t## if I take this term to be zero; i.e., if the creation operators obey commutation relations. But that's what the rest of the chapter is trying to prove in the first place! Is there a way to time evolve this operator that doesn't involve assuming commutation (instead of anticommutation) at the outset?
$$\varphi^+(\mathbf{x},0) = \int \tilde{dk}\text{ }e^{i \mathbf{k}\cdot\mathbf{x}}a(\mathbf{k})$$
and time-evolves them in the Heisenberg picture with the Hamiltonian:
$$H_0 = \int \tilde{dk} \text{ }\omega\text{ } a^{\dagger}(\mathbf{k})a(\mathbf{k})$$
to get:
$$\varphi^+(\mathbf{x},t) = e^{iH_0t}\varphi^+(\mathbf{x},0)e^{-iH_0t} = \int \tilde{dk}\text{ }e^{i kx}a(\mathbf{k})$$
where ##kx = \mathbf{k}\cdot\mathbf{x}-\omega t##. I'm trying to derive this last line. So far, my approach has been to expand the ##e^{iH_0t}## operators in a Taylor series and calculate each term of the ##\varphi^+## directly. Based on my progress (I've only done through first order in ##H_0##), I figure I'll eventually get:
$$e^{iH_0t}\varphi^+(\mathbf{x},0)e^{-iH_0t} = \int \tilde{dk}\text{ }a(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}}(1-i\omega t-\frac{1}{2}\omega^2 t^2 + \cdots)$$
Two questions:
1) Second order gives the term ##a(\mathbf{k}')a^{\dagger}(\mathbf{k})a(\mathbf{k})a^{\dagger}(\mathbf{k})a(\mathbf{k})+a^{\dagger}(\mathbf{k})a(\mathbf{k})a(\mathbf{k}')a^{\dagger}(\mathbf{k})a(\mathbf{k})+a^{\dagger}(\mathbf{k})a(\mathbf{k})a^{\dagger}(\mathbf{k})a(\mathbf{k})a(\mathbf{k}')##, and third order will give me permutations of terms with 7 creation/annihilation operators. Et cetera for higher-order terms. Is there a more clever way to do this that will make it clear that these terms all collapse to powers of ##\omega##?
2) In carrying out the evaluation of the first-order term, part of my integrand ended up being proportional to ##a(\mathbf{k})a(\mathbf{k}')-a(\mathbf{k}')a(\mathbf{k})##. I only get ##i\omega t## if I take this term to be zero; i.e., if the creation operators obey commutation relations. But that's what the rest of the chapter is trying to prove in the first place! Is there a way to time evolve this operator that doesn't involve assuming commutation (instead of anticommutation) at the outset?