1
$\begingroup$

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?

| cite | improve this question | | | | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Oh! I see. In the back of my mind I was still using the commutators and not anti-commutators between the operators. Thank you. $\endgroup$ – fogof mylife Mar 22 at 3:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.