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I am still unclear what the hamilitonian of a quantized field is, but what I do know is the hamilitonian of the boson field is defined as \begin{align} H_{\text{boson}} &=& \frac{1}{2}\sum_{\mathbf{k},\mu=-1,1} \hbar \omega_{\mathbf{k},\mu} \Big({a^\dagger}^{(\mu)}(\mathbf{k})\,a^{(\mu)}(\mathbf{k}) + a^{(\mu)}(\mathbf{k})\,{a^\dagger}^{(\mu)}(\mathbf{k})\Big) \\ &=& \sum_{\mathbf{k},\mu} \hbar \omega_{\mathbf{k},\mu} \Big({a^\dagger}^{(\mu)}(\mathbf{k})a^{(\mu)}(\mathbf{k}) + \frac{1}{2}\Big) \\ &=& \sum_{\mathbf{k},\mu} \hbar \omega_{\mathbf{k},\mu} \Big( N_{\mathbf{k},\mu} + \frac{1}{2}\Big) , \end{align} however something that appears scarcely mentioned in the literature is the Hamiltonian of the fermion field. Interpolating from the boson field Hamilton, my guess is the Hamiltonian of the fermion field would be \begin{align} H_{\text{fermion}} &=& \frac{1}{2}\sum_{\mathbf{k},\mu=-1,1} \hbar \omega_{\mathbf{k},\mu} \Big({b^\dagger}^{(\mu)}(\mathbf{k})\,b^{(\mu)}(\mathbf{k}) + b^{(\mu)}(\mathbf{k})\,{b^\dagger}^{(\mu)}(\mathbf{k})\Big) \\ &=& \frac{1}{2}\sum_{\mathbf{k},\mu=-1,1} \hbar \omega_{\mathbf{k},\mu} \Big({b^\dagger}^{(\mu)}(\mathbf{k})\,b^{(\mu)}(\mathbf{k}) - {b^\dagger}^{(\mu)}(\mathbf{k})\,b^{(\mu)}(\mathbf{k})+1\Big) \\ &=& \sum_{\mathbf{k},\mu} \hbar \omega_{\mathbf{k},\mu} \Big( 0_{\mathbf{k},\mu} + \frac{1}{2}\Big) . \end{align}

I found $H_{\text{fermion}}$ to be very different from $H_{\text{boson}}$, because $H_{\text{boson}}$ depends on the number of bosons $N_{\mu,\boldsymbol{k}}$, however $H_{\text{fermion}}$ is independent of the number of fermions in each of its states, but rather seams to equal the vacuum energy only. Is $H_{\text{fermion}}$ real and what is its physical interpretation?

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    $\begingroup$ I believe your Hamiltonian density is wrong, it should be of the form: $$ \mathcal{H}_{\text{fermion}} \propto \sum\limits_{s=1}^2 \left[a_s^\dagger(\mathbf{p}) a_s(\mathbf{p}) + b_s^\dagger(\mathbf{p}) b_s(\mathbf{p}) \right] $$ where I have already normal ordered it. $\endgroup$ – Hunter Jul 26 '14 at 1:46
  • $\begingroup$ As for the interpretation, we can use the Hamiltonian and momentum operator to show that $a^\dagger_s(\mathbf{p})$ and $b^\dagger_s(\mathbf{p})$ create particle particles with energy $E_{\mathbf{p}}$ and momentum $\mathbf{p}$. This is very similar to the Klein-Gordon field, but now these particles are fermions and anti-fermions. $\endgroup$ – Hunter Jul 26 '14 at 1:53
  • $\begingroup$ @Hunter That is very strange. Thank you for pointing that out. I just discovered that the boson field uses the same operators $a$ and $a^\dagger$, however fermion field uses two different operators $a$ and $a^\dagger$ and $b$ and $b^\dagger$ in its field. Why the big difference? $\endgroup$ – linuxfreebird Jul 26 '14 at 9:47
  • $\begingroup$ The use of two different operators is necessary by the existence of quasi-particles. One has to do this for bosons as well. However, if you consider real bosons, they are their own anti-particles and hence in the end you get a formula for one kind of operators alone. $\endgroup$ – Lorenz Mayer May 11 at 5:23

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