# Spectrum of Dirac Hamiltonian

The Dirac Hamiltonian is given by, \begin{aligned} H &=\sum_s\int \frac{d^{3} p}{(2 \pi)^{3}} E_{\vec{p}}\left[b_{\vec{p}}^{s \dagger} b_{\vec{p}}^{s}+c_{\vec{p}}^{s \dagger} c_{\vec{p}}^{s}\right]. \end{aligned} Now the Hamiltonian satisfies nice commutation relations with creation and annihilation operators as below and thus the spectrum of the Hamiltonian can be built from these, \begin{aligned} &\left[H, b_{\vec{p}}^{r}\right]=-E_{\vec{p}} b_{\vec{p}}^{r} \quad \text { and } \quad\left[H, b_{\vec{p}}^{r \dagger}\right]=E_{\vec{p}} b_{\vec{p}}^{r \uparrow}\\ &\left[H, c_{\vec{p}}^{r}\right]=-E_{\vec{p}} c_{\vec{p}}^{r} \quad \text { and } \quad\left[H, c_{\vec{p}}^{r \dagger}\right]=E_{\vec{p}} c_{\vec{p}}^{r \uparrow} \end{aligned} I am unable to derive these relations. Let me show you by trying the first one. \begin{align} [H,b^r_{\vec{q}}]&=\sum_s\int \frac{d^{3} p}{(2 \pi)^{3}} E_{\vec{p}}\left[b_{\vec{p}}^{s \dagger} b_{\vec{p}}^{s}+c_{\vec{p}}^{s \dagger} c_{\vec{p}}^{s}\,\,,\,\,b^r_{\vec{q}}\right]\\ &=\sum_s\int \frac{d^{3} p}{(2 \pi)^{3}} E_{\vec{p}}\left[b_{\vec{p}}^{s \dagger}\,\,,\,\,b^r_{\vec{q}}\right] b_{\vec{p}}^{s} \end{align} Now we know that $$\left\{b_{\vec{p}}^{s \dagger}\,\,,\,\,b^r_{\vec{q}}\right\}=(2\pi)^3\delta^{(0)}(\vec{p}-\vec{q})\delta^{rs}$$ Thus, $$\left[b_{\vec{p}}^{s \dagger}\,\,,\,\,b^r_{\vec{q}}\right]=(2\pi)^3\delta^{(0)}(\vec{p}-\vec{q})\delta^{rs}-2b^r_{\vec{q}}b_{\vec{p}}^{s \dagger}$$ Using this relation I get, \begin{align} [H,b^r_{\vec{q}}]&=\sum_s\int \frac{d^{3} p}{(2 \pi)^{3}} E_{\vec{p}}\left[b_{\vec{p}}^{s \dagger}\,\,,\,\,b^r_{\vec{q}}\right] b_{\vec{p}}^{s}\\ &=\sum_s\int \frac{d^{3} p}{(2 \pi)^{3}} E_{\vec{p}}\left[(2\pi)^3\delta^{(0)}(\vec{p}-\vec{q})\delta^{rs}-2b^r_{\vec{q}}b_{\vec{p}}^{s \dagger}\right]b_{\vec{p}}^{s} \end{align} Now I don't know how to get the commutation relation I quoted above. Can someone show where I got wrong and please show me the derivation of at least one of the commutators?

• $[AB,C]=A\{B,C\}-\{A,C\}B$, perhaps? Mar 21 '20 at 20:43

In $$\left[b_{\vec{p}}^{s \dagger} b_{\vec{p}}^{s}+c_{\vec{p}}^{s \dagger} c_{\vec{p}}^{s}\,\,,\,\,b^r_{\vec{q}}\right] = \left[b_{\vec{p}}^{s \dagger}\,\,,\,\,b^r_{\vec{q}}\right] b_{\vec{p}}^{s}$$, you assumed that $$b^r_{\vec{q}}$$ commutes with all the other operators. However, it anti-commutes (and therefore does not commute).
Using the relation $$[AB,C] = A\{B,C\} - \{A,C\}B$$, you will get the right result.