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How can I show that the non relativistic second quantization hamiltonian \begin{equation} \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) \end{equation} is hermitian?

And how can I show, for both bosons and fermions, that: \begin{equation} [\hat{N},\hat{H}]=0,\\ [\hat{P},\hat{H}]=0 \end{equation} where \begin{equation} \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 \end{equation} are respectively the total number and momentum operetor.

The following commutation relations are given: for bosons:

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

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

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

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