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 $\Psi^\dagger$.

  • $\begingroup$ 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. $\endgroup$
    – Jon
    Commented Jun 8, 2018 at 14:27

2 Answers 2


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.


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