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DOTDO

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S. Weinberg says in his book, "The Quantum Theory of Fields Volume I", that

Since electrons carry a charge, we would not like to mix annihilation and creation operators, so we might try to write the field as $$\psi(x)=\sum_{k}u_k (x)e^{-i\omega_k t}a_k$$

where ##u_k (x)e^{-i\omega_k t}## are a complete set of orthonormal plane-wave solutions of the Dirac equation with ##k## labelling the 3-momentum, spin, and sign of the energy.

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At the first time, I thought it's because ##\sum_{k}## already involves the negative ##\omega_k## so that ##b_k## and ##b^\dagger _k## in ##\psi(x)=\sum_{k}u_k (x)e^{-i\omega_k t}b_k+u_k (x)e^{i\omega_k t}b^\dagger _k## can be merged into ##a_k##.

But he says it's because of the electron's charge and explains no more. Can someone explain, please?

Thank you.

Since electrons carry a charge, we would not like to mix annihilation and creation operators, so we might try to write the field as $$\psi(x)=\sum_{k}u_k (x)e^{-i\omega_k t}a_k$$

where ##u_k (x)e^{-i\omega_k t}## are a complete set of orthonormal plane-wave solutions of the Dirac equation with ##k## labelling the 3-momentum, spin, and sign of the energy.

------

At the first time, I thought it's because ##\sum_{k}## already involves the negative ##\omega_k## so that ##b_k## and ##b^\dagger _k## in ##\psi(x)=\sum_{k}u_k (x)e^{-i\omega_k t}b_k+u_k (x)e^{i\omega_k t}b^\dagger _k## can be merged into ##a_k##.

But he says it's because of the electron's charge and explains no more. Can someone explain, please?

Thank you.

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