# How can I show that Second Quantization Hamiltonian is Hermitian?

How can I show that the non relativistic second quantization hamiltonian $$\hat{H}=\lmoustache d^3x \hat{\psi}^\dagger_\alpha(x) T(x)\hat{\psi}^\dagger(x)+\frac{1}{2}\lmoustache d^3x\lmoustache d^3x' \hat{\psi}^\dagger_\alpha(x) \hat{\psi}_\beta(x')V(x,x')\hat{\psi}_\beta(x')\hat{\psi}^\dagger_\alpha(x)\bigr)$$ is hermitian?

And how can I show, for both bosons and fermions, that: $$[\hat{N},\hat{H}]=0,\\ [\hat{P},\hat{H}]=0$$ where $$\hat{N}=\lmoustache d^3x \hat{\psi}^\dagger_\alpha(x) \hat{\psi}^\dagger(x)_\alpha\\ \hat{P}=\lmoustache d^3x \hat{\psi}^\dagger_\alpha(x) p(x)\hat{\psi}^\dagger(x)_\alpha$$ are respectively the total number and momentum operetor.

The following commutation relations are given: for bosons:

$$[\hat{\psi}_\alpha(x),\hat{\psi}^\dagger_\beta(x')]=\delta(x-x')\delta_{\alpha\beta}$$ for fermions:

$$\{\hat{\psi}_\alpha(x),\hat{\psi}^\dagger_\beta(x')\}=\delta(x-x')\delta_{\alpha\beta}$$ I would really appreciate your help,please let me know.

• What commutation relations are you given? – probably_someone Mar 23 '18 at 3:50