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The Dirac field has the expansion $$\Psi(x)=\int\frac{d^3p}{\sqrt{(2\pi)^32E_p}}\sum\limits_{s=1,2}\Big(b_s(p)u^s(p)e^{-ip\cdot x}+d^\dagger_s(p)v^s(p)e^{+ip\cdot x}\Big)$$ where $b_s$ and $d_s$ are the annihilation operators for the particle and antiparticle respectively with momentum $p$ and spin projection $s$. For a scalar field, such an expansion can be rigorously derived. But I have not seen a derivation of this expansion for $\Psi$; it's written down as if it is very obvious.

Peskin and Schroeder has a derivation but it goes back and forth between Schrodinger and Heisenberg picture while I would like to stick to Heisenberg picture.

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    $\begingroup$ "written down" where? please remember to always cite your sources! $\endgroup$ – AccidentalFourierTransform Dec 5 '19 at 14:39
  • $\begingroup$ Sorry about that. See eq. 4.44 of L. H. Ryder's Quantum Field Theory: books.google.co.in/… $\endgroup$ – mithusengupta123 Dec 5 '19 at 14:45
  • $\begingroup$ Though they use a different normalization, this too is equally discomforting $\endgroup$ – mithusengupta123 Dec 5 '19 at 14:48
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Each component of $\Psi$ satisfies the Klein-Gordon equation, and so we can write (cf. this PSE post) $$ \Psi_\alpha(x)=\int\frac{d^3p}{\sqrt{(2\pi)^32E_p}}\Big(a_\alpha(p)e^{-ip\cdot x}+b^\dagger_\alpha(p)e^{+ip\cdot x}\Big) $$ for some operators $a_\alpha,b_\alpha$. If we now require $\Psi$ to satisfy the Dirac equation, we get the algebraic conditions $$ (\not p+m)a(p)=(\not p-m)b(p)=0 $$

We solve these as follows. Let $u_s(p)\in \mathbb C^4$ with $s=1,2$ be the two linearly linearly independent solutions to $(\not p+m)u(p)=0$, and let $v_s(p)\in \mathbb C^4$ with $s=1,2$ be the two linearly independent solutions to $(\not p-m)v(p)=0$ (there are two and only two solutions because the matrices $\not p\pm m$ have rank 2, as is easily checked). As $u_s,v_s$ are four linearly independent vectors, they are a basis of $\mathbb C^4$, which means we can expand any other vector as linear combinations of them. Thus, we can write $$ a(p)=\sum_{s=1,2}b_s(p)u_s(p),\qquad b(p)=\sum_{s=1,2} d_s^\dagger v_s(p) $$ for some scalar operators $b_s,d_s$. Finally, plugging this back into our previous expression, we get $$ \Psi_\alpha(x)=\int\frac{d^3p}{\sqrt{(2\pi)^32E_p}}\sum_{s=1,2}\Big(b_s(p)u_s(p)e^{-ip\cdot x}+d_s^\dagger v_s(p) e^{+ip\cdot x}\Big) $$ as required.

For more details see Srednicki §37.

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  • $\begingroup$ Thank you so much. This answer is too good. $\endgroup$ – mithusengupta123 Dec 7 '19 at 18:43
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Modes decomposition comes from the solution of motion equation. You should start from Dirac equation, $$[i(\gamma\cdot\partial)-m]\psi=0$$ and consider the following ansatz for $\psi$, $$\psi=\sum_s\int_{\bf p}\frac{1}{\sqrt{2E_{\bf p}}}\left(b_su_s(p)e^{-ip\cdot x}+d_s^{\dagger}v_s(p)e^{+ip\cdot x}\right).$$ Then you can rewrite your equation in terms of 2$\times$2 block-matrices and consider that $\psi=(\phi,\chi)^T$ (bispinor with two spinors component). Dirac equation gives you system of two equations for $\phi$ and $\chi$. It is convenient to start from $\phi$ and then find $\chi$. Negative frequency solution, $v_s$, can be obtained by charge-congutation matrix $\mathcal{C}=-i\gamma^2\gamma^0$ (it depends on representation of $\gamma$-matrices).

Modes decompoisition arises as you can create/annihilate particle with momentum ${\bf p}$ in different point in space-time. Integration over ${\bf p}$ means that momentum of particle can be arbitrary. Generally, with external fields, spinors $u_s$ and $v_s$ may have more complicated structure.

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  • $\begingroup$ How do you motivate that ansatz? $\endgroup$ – mithusengupta123 Dec 5 '19 at 14:43
  • $\begingroup$ @mithusengupta123 , I said: you can create/annihilate a particle with a momentum at a point in space-time $\endgroup$ – Artem Alexandrov Dec 5 '19 at 15:34

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