# Deriving anti-commutation relation between creation/annihilation operators for Dirac fermions

Starting from Dirac fields:

$$\Psi(x) = \dfrac{1}{(2\pi)^{3/2}} \int \dfrac{d^3k}{\sqrt{2\omega_k}}\sum_r\left[ c_r(k)u_r(k)e^{-ikx}+d^\dagger_r(k)v_r(k)e^{-ikx} \right]_{k_0=\omega_k}$$

$$\Psi^\dagger(x) = \dfrac{1}{(2\pi)^{3/2}} \int \dfrac{d^3k}{\sqrt{2\omega_k}}\sum_r\left[d_r(k)v^\dagger_r(k)e^{-ikx} + c^\dagger_r(k)u^\dagger_r(k)e^{ikx}\right]_{k_0=\omega_k}$$

where $\omega_k = \sqrt{\vec{k}^2+m^2}$.

The canonical quatization condition reads:

$$\begin{cases} \{\Psi_\alpha(x), \Psi^\dagger_\beta(y)\}_t = \delta^{(3)}(\vec{x}-\vec{y})\ \ \delta_{\alpha\beta}\\ \{\Psi_\alpha(x), \Psi_\beta(y)\}_t = 0\\ \{\Psi^\dagger_\alpha(x), \Psi^\dagger_\beta(y)\}_t = 0\\ \end{cases}$$

In order to derive the quantization condition for the creation/annihilation operators I have to rewrite $c,c^\dagger,d,d^\dagger$ in terms of $\Psi$ and $\Psi^\dagger$.

For instance in order to derive the canonical quantization condition between $c,c^\dagger$ I can rewrite them as:

$$c_r(k) = \dfrac{1}{\sqrt{2\pi}^{3}} \int \dfrac{d^3x}{\sqrt{2\omega_k}} u_r^\dagger(k) \Psi(x) e^{ikx}$$

$$c^\dagger_s(p) = \dfrac{1}{\sqrt{2\pi}^{3}} \int \dfrac{d^3y}{\sqrt{2\omega_p}} \Psi^\dagger(y) u_s(p) e^{-ipy}$$

and then explicitly calculate the anti-commutator:

$$\begin{split} \{c_r(k), c^\dagger_s(p)\}_t &= \dfrac{1}{(2\pi)^3} \int \dfrac{d^3xd^3y}{\sqrt{2\omega_k 2\omega_p}} \left[ u_r^\dagger(k) \Psi(x)\Psi^\dagger(y) u_s(p) + \Psi^\dagger(y) u_s(p)u_r^\dagger(k) \Psi(x) \right]e^{i(kx-py)}\\ &= \dfrac{1}{(2\pi)^3} \int \dfrac{d^3xd^3y}{\sqrt{2\omega_k 2\omega_p}} \left[ u_r^\dagger(k) \{\Psi(x), \Psi^\dagger(y)\} u_s(p)\right]e^{i(kx-py)}\\ &= \cdots \end{split}$$

But here I miss something: I don't understand why I can swap $u_s(p)$ and $u_r^\dagger(k)$ in the second term in order to recover the anti-commutator between $\Psi$ and $\Psi^\dagger$.

• In all the sacred textbooks is the other way round. You postulate the anticommutation relations for $c_r$ and $d_r$ and then, you are a simple step away from deriving the anticommutation relations for the fields. – Jon Jun 8 '18 at 14:27

## 2 Answers

You should proceed in contraction of the spinor indices and recall that e.g a pair $u^{\dagger}(k,r) \Psi(x) \equiv u^{\dagger}_{\alpha}(k,r) \Psi_{\alpha}(x)$, with $u^{\dagger}_{\alpha}(k,r)$ labelling the $\alpha^{\text{th}}$ component of the Dirac spinor $u^{\dagger}(k,r)$. Therefore,$$\left\{c(k,r),c^{\dagger}(p,s)\right\} \sim \int d^3 x d^3 y \left(u^{\dagger}_{\alpha}(k,r) \Psi_{\alpha}(x)\Psi^{\dagger}_{\beta}(y) u_{\beta}(p,s) + \Psi^{\dagger}_{\beta}(y) u_{\beta}(p,s)u^{\dagger}_{\alpha}(k,r) \Psi_{\alpha}(x)\right) \\= \int d^3 x d^3 y \left(u^{\dagger}_{\alpha}(k,r) \left\{\Psi_{\alpha}(x),\Psi^{\dagger}_{\beta}(y)\right\} u_{\beta}(p,s)\right) = \dots =$$

Finish the exercise using your canonical quantisation field anticommutators together with orthogonality/completeness relations amongst the $u$'s.

Here $$\Psi(x)$$ is a column matrix having 4 components and $$\Psi^\dagger(x)$$ is a row matrix with 4 row elements (of course these elements are functions of $$x$$).

And when you are taking the anti-commutation, you are choosing one component(or element) $$\Psi_\alpha(x)$$ from (4$$\times$$1) column matrix $$\Psi(x)$$ and similarly you should choose one component $$\Psi^\dagger_\beta(x)$$ from (1$$\times$$4) row matrix $$\Psi^\dagger(x)$$. \begin{align} \Psi_\alpha(x) = \dfrac{1}{(2\pi)^{3/2}} \int \dfrac{d^3k}{\sqrt{2\omega_k}}\sum_{r=1,2}\left[ c_r(k)u_{r,\alpha}(k)e^{-ikx}+d^\dagger_r(k)v_{r,\alpha}(k)e^{-ikx} \right]_{k_0=\omega_k}\\ \Psi^\dagger_\beta(x) = \dfrac{1}{(2\pi)^{3/2}} \int \dfrac{d^3k}{\sqrt{2\omega_k}}\sum_{r=1,2}\left[d_r(k)v^\dagger_{r,\beta}(k)e^{-ikx} + c^\dagger_r(k)u^\dagger_{r,\beta}(k)e^{ikx}\right]_{k_0=\omega_k} \end{align} So, now you can see there $$u$$ and $$v$$ has two indices r and $$\alpha$$(or $$\beta$$), here $$r$$ can take values 1 and 2 while $$\alpha$$(or $$\beta$$) can take values 1,2,3,4.

Or simply $$u_{r,\alpha}$$ (or $$v_{r,\alpha}$$) is the $$\alpha$$-th component(or $$\alpha$$-th matrix element) of the (4$$\times$$1) column matrix $$u_r$$ (or $$v_{r}$$).

And similarly, $$u_{r,\alpha}^\dagger$$ (or $$v_{r,\alpha}^\dagger$$) is the $$\alpha$$-th component (or $$\alpha$$-th matrix element) of the (1$$\times$$4) row matrix $$u_r^\dagger$$ (or $$v_{r}^\dagger$$).

Thus $$u_{r,\alpha}$$, $$v_{r,\alpha}$$, $$u_{r,\alpha}^\dagger$$, $$v_{r,\alpha}^\dagger$$ all these are just numbers or the matrix elements, not the matrices.

So, you proceed the way you were going and easily swap $$u_{s,\alpha}(p)$$ and $$u_{r,\alpha}(k)$$ (in your notation) as they are just the numbers or the components(or elements) of the corresponding matrices.