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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 $\Psi^\dagger$.

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  • $\begingroup$ 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 Jun 8 '18 at 14:27
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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.

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