Is positron creation operator times electron creation operator equal to the ground state? This is part of a larger problem, but the important part is that at one point I have:
$$
bb^\dagger+bd+d^\dagger b^\dagger + d^\dagger d + b^\dagger b +db + b^\dagger d^\dagger+d d^\dagger
$$
where $b^\dagger$ and $b$ are the creation and annihilation operators for electrons, while $d^\dagger$ and $d$ are the same for positrons.
The first step is to recognize the anti-commutators of the operators
$$
\{b,b^\dagger\}=\{d,d^\dagger\}=(2\pi)^3\delta^3(k-k')\delta_{rs}
$$
$$
\{b,b\}=\{b^\dagger,b^\dagger\}=\{d,d\}=\{d^\dagger,d^\dagger\}=0
$$
But that leaves me with:
$$
2(2\pi)^3\delta^3(k-k')\delta_{rs}+\{b,d\}+\{b^\dagger,d^\dagger\}
$$
it is this second pair of anti-communators which I have absolutely no idea how to tackle, I've checked the lecture notes but they never consider this pairing. 
I know that they must be $0$ because if I just ignore them I get the result I'm trying to prove, but I can't justify at all why they should be $0$
 A: The proof that the anticommutator of different annihilation operators and its complex conjugate can be shown as follows (demonstrated in Srednicki's notation, i.e. in "west coast Minkowski metric", in particular $e^{ipx} = e^{i(\mathbf{p}\mathbf{x} - \omega_p t)}$  , furthermore it requires some relations of spinor algebra which are not shown here):
We start off with the general development of the fermion field operator : 
$$\Psi(x) = \sum_{s=\pm}\int \frac{d^3k}{(2\pi)^3 2\omega} [b_s(\mathbf{k})u_s(\mathbf{k})e^{ikx} + d_s^\dagger(\mathbf{k})v_s(\mathbf{k})e^{-ikx}]$$
and "project out" the coefficients. But it will only shown for $b_s(\mathbf{k})$, because for $d_s(\mathbf{k})$ it is analogous. For doing so we multiply by $e^{-ipx}$ and integrate over $d^3x$:
$$\int d^3x e^{-ipx}\Psi(x) = \sum_{s=\pm}\int\frac{d^3k}{(2\pi)^3 2\omega}[b_s(\mathbf{k})u_s(\mathbf{k})(2\pi)^3\delta^3(\mathbf{k}-\mathbf{p})e^{i(-\omega_k+\omega_p)t} + d_s^\dagger(\mathbf{k})v_s(\mathbf{k})(2\pi)^3\delta^3(-\mathbf{k}-\mathbf{p})e^{i(\omega_k+\omega_p)t}]$$
In order to get the operator $b_s$ we multiply by $\overline{u}_s(\mathbf{p})\gamma^0$, use $\overline{u}_s(\mathbf{p})\gamma^0 u_{s'}(\mathbf{p})=2\omega_p \delta_{ss'}$ and $\overline{u}_s(\mathbf{p})\gamma^0 v_{s'}(-\mathbf{p})=0$. We obtain the desired opertor:
$$b_s(\mathbf{p}) = \int d^3x e^{-ipx}\overline{u}_s(\mathbf{p})\gamma^0 \Psi(x)$$
The computation for the expression for $d^\dagger_s$ is analogous, essential difference is to multiply $\Psi(x)$ at the beginning with $e^{ipx}$. The result is:
$$d^\dagger_s(\mathbf{p}) = \int d^3x e^{ipx} \overline{v}_s(\mathbf{p})\gamma^0 \Psi(x)$$
In order to evaluate the desired the anticommutator we have to take the hermitian conjugate: 
$$d_s(\mathbf{p}) = \int d^3x e^{-ipx} \overline{\Psi(x)}\gamma^0 v_s(\mathbf{p}) $$
The very last step is the  evaluation of the anticommutator: 
$$\{b_s(\mathbf{p}),d_{s'}(\mathbf{p'})\} = \int d^3x d^3y e^{-ipx-ip'y} \overline{u}_s(\mathbf{p})\gamma^0\{\Psi(x),\overline{\Psi(y)}\}\gamma^0 v_{s'}(\mathbf{p'})=\int d^3x e^{-i(p+p')x}\overline{u}_s(\mathbf{p})\gamma^0\gamma^0\gamma^0 v_s'(\mathbf{p'})= (2\pi)^3 \delta^3(\mathbf{p}+\mathbf{p'}) \overline{u}_s(\mathbf{p})\gamma^0 v_{s'}(\mathbf{p'})=\overline{u}_s(\mathbf{p})\gamma^0 v_{s'}(-\mathbf{p})=0$$.
The second anticommutator is just the hermitian conjugate of the first one, so if the first is zero, the second is zero too. I hope that no typo slipped in.
