# Can we derive free field expansion formula for the spin-1/2 Dirac field?

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.

• "written down" where? please remember to always cite your sources! – AccidentalFourierTransform Dec 5 '19 at 14:39
• Sorry about that. See eq. 4.44 of L. H. Ryder's Quantum Field Theory: books.google.co.in/… – mithusengupta123 Dec 5 '19 at 14:45
• Though they use a different normalization, this too is equally discomforting – mithusengupta123 Dec 5 '19 at 14:48

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.

• Thank you so much. This answer is too good. – mithusengupta123 Dec 7 '19 at 18:43

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.

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