I'm taking an introductory course in QFT. During quantization of the Dirac field, my textbook gives a lot of information on how annihilation and creation operators act on vacuum, but nothing about how they act on non-vacuum states. I need these to compute $$ \int \frac{\mathrm d^3 p}{(2\pi)^3} \sum_s \left( {a^s_ {{\vec{p}}}}^\dagger a^s_ {{\vec{p}}} - {b^s_ {{\vec{p}}}}^\dagger b^s_ {{\vec{p}}} \right) |\vec{k},s \rangle, $$ where ${a^s_ {{\vec{p}}}}^\dagger, {b^s_ {{\vec{p}}}}^\dagger $ are the creation operator for fermions and anti-fermions respectively and $ {a^s_ {{\vec{p}}}},{b^s_ {{\vec{p}}}}$ are the annihilation operators of fermions and anti-fermions repectively. I have searched google, but I couldn't find anything after about 1 hour of searching.

Are you able to tell me how ${a^s_ {{\vec{p}}}}^\dagger, {b^s_ {{\vec{p}}}}^\dagger, {a^s_ {{\vec{p}}}},{b^s_ {{\vec{p}}}}$ act on non-vacuum states?


If you need to compute $$ \int \frac{d^3 p}{(2\pi)^3} \sum_s ( {a^s_ {{\vec{p}}}}^\dagger a^s_ {{\vec{p}}} - {b^s_ {{\vec{p}}}}^\dagger b^s_ {{\vec{p}}} ) |\vec{k},r \rangle, $$ you will need ${a^s_ {{\vec{p}}}}^\dagger a^s_ {{\vec{p}}}|\vec{k},r \rangle$ and ${b^s_ {{\vec{p}}}}^\dagger b^s_ {{\vec{p}}} |\vec{k},r \rangle$. Since you are dealing with Dirac fields, you get these using the anti-commutation relations (with the proper normalization factors - and I don't know which convention you are using): $$ \{{a^s_ {{\vec{p}}}},{a^r_ {{\vec{q}}}}^\dagger\}=\delta_{sr}\delta(\vec{p}-\vec{q}),\\ \{{b^s_ {{\vec{p}}}},{b^r_ {{\vec{q}}}}^\dagger\}=\delta_{sr}\delta(\vec{p}-\vec{q}),\\ \{{a^s_ {{\vec{p}}}},{b^r_ {{\vec{q}}}}^\dagger\}=\{{b^s_ {{\vec{p}}}},{a^r_ {{\vec{q}}}}^\dagger\}=0.\\ $$ and knowing that ${a^s_ {{\vec{p}}}}|0\rangle={b^s_ {{\vec{p}}}}|0\rangle=0$.

It follows the answer with the same procedure @flippiefanus used.


The basic procedure is as follows: $$ a_r(\mathbf{k}_1) |\mathbf{k}_2,s\rangle = a_r(\mathbf{k}_1) a_s^{\dagger}(\mathbf{k}_2) |0\rangle = \{a_r(\mathbf{k}_1), a_s^{\dagger}(\mathbf{k}_2) \}|0\rangle = |0\rangle (2\pi)^2\omega_1 \delta(\mathbf{k}_1-\mathbf{k}_2) \delta_{rs} , $$ where $|\mathbf{k}_2,s\rangle$ is assumed to be a fermion state. For an anti-fermion state one would use the $b$-operators, instead. The reason why one can express this in terms of the anti-commutator is because $ a_r(\mathbf{k}_1) |0\rangle = 0$. The detail of the final expression depends on the particular anti-commutation relation that you use. Here I've used a Lorentz convariant version.

  • $\begingroup$ Thanks, are you also able to explain how $b_r(\vec{k}_1)$, $b_r^\dagger(\vec{k}_1)$ and $a_r^\dagger(\vec{k}_1)$ acts on $ | \vec{k}_2,s\rangle$ or just state the result? Thanks $\endgroup$ – Mikkel Rev Oct 27 '16 at 11:55
  • $\begingroup$ And add $b_r^\dagger(\vec{k}) | 0 \rangle$ for completeness? :) $\endgroup$ – Mikkel Rev Oct 27 '16 at 12:13
  • $\begingroup$ Perhaps you can add in your question the definitions for $a_s(\mathbf{k})$, $b_s(\mathbf{k})$, etc. $\endgroup$ – flippiefanus Oct 27 '16 at 13:00
  • $\begingroup$ Thank you for your response. I added in the definitions as requested. $\endgroup$ – Mikkel Rev Oct 30 '16 at 13:49
  • $\begingroup$ I offer bounty +50 for the answer now $\endgroup$ – Mikkel Rev Oct 30 '16 at 14:00

All you need is the (anti-)commutation relations and the definitions of the states in terms of creation operators acting on vacuum state.

e.g. a state $|\psi\rangle$ of two particles: $$ c_k|\psi\rangle =c_k\left(\sum_{i<j}\psi_{ij}|i,j\rangle\right)= \sum_{i<j}\psi_{ij}c_k c_i^{\dagger}c_j^{\dagger}|0\rangle $$

Then commutes $c_k$ with $c_i^{\dagger}$ and $c_j^{\dagger}$ until hit the vacuum state and annihilate it. $$ \sum_{i<j}\psi_{ij}\left(\left[ c_k ,\, c_i^{\dagger}\right]_+ - c_i^{\dagger}c_k\right) c_j^{\dagger}|0\rangle=\sum_{i<j}\psi_{ij}\left(\left[ c_k ,\, c_i^{\dagger}\right]_+c_j^{\dagger} - c_i^{\dagger} \left[ c_k ,\, c_j^{\dagger}\right]_+ \right) |0\rangle = \sum_{i<j}\psi_{ij}\left(\left[ c_k ,\, c_i^{\dagger}\right]_+|j\rangle - \left[ c_k ,\, c_j^{\dagger}\right]_+ |i\rangle \right) $$


Result: The only thing you'll really need for this calculation is the definition of one-(anti)-particle states (given below) and the application of annihilation operators on those, given by

$$a_{\vec p_1}^{s_1} |\vec p_2, s_2;0,0\rangle =\delta_{s_1, s_2} \delta^3\left(\vec p_1 - \vec p_2\right) |0\rangle,\\b_{\vec q_1}^{r_1} |0,0;\vec q_2, r_2\rangle=\delta_{r_1, r_2} \delta^3\left(\vec q_1 - \vec q_2\right) |0\rangle.\\\\$$

Derivation: You were asking for the action of creation and annihilation operators on one-particle states, given by $$|\vec p, s; \vec 0, 0\rangle = a_{\vec{p}}^{s\dagger}|0\rangle\\ |0,0;\vec p, s\rangle = b_{\vec{p}}^{s\dagger}|0\rangle.$$

It makes sense to also define the following two-particle states, which are only non-zero if again all ${\vec p_i, s_i}$ and ${\vec q_j, s_j}$ are respectively distinct. $$|\vec p, s; \vec q, r\rangle = \frac{1}{2}\left(a_{\vec{p}}^{s\dagger}b_{\vec{q}}^{r\dagger}-b_{\vec{q}}^{r\dagger}a_{\vec{p}}^{s\dagger}\right)|0\rangle\\ |\vec p_1, s_1, \vec p_2, s_2;\vec 0,0\rangle = \frac{1}{2}\left(a_{\vec{p}_1}^{s_1\dagger}a_{\vec{p}_2}^{s_2\dagger}-a_{\vec{p}_2}^{s_2\dagger}a_{\vec{p}_1}^{s_1\dagger}\right)|0\rangle\\|\vec 0,0;\vec q_1, r_1, \vec q_2, r_2\rangle = \frac{1}{2}\left(b_{\vec{q}_1}^{r_1\dagger}b_{\vec{q}_2}^{r_2\dagger}-b_{\vec{q}_2}^{r_2\dagger}b_{\vec{q}_1}^{r_1\dagger}\right)|0\rangle$$ where we just decided to use a (anti)symmetrical definition - it is clear that using the appropriate anticommutation-relations, all of those states can be written without the difference of two terms.

Now, to find the action of those operators we are going to use the mentioned anticommutation relations $$\{a_{\vec p}^s, a_{\vec q}^r\}=0 \qquad \{a_{\vec p}^{s\dagger}, a_{\vec q}^{r\dagger}\}=0\\ \{a_{\vec p}^s, a_{\vec q}^{r\dagger}\}=\delta^{rs} \delta^3(\vec p - \vec q)$$ and similar for the $b$-operators. Also, every $b$ anticommutes with every $a$.

Note, that the above states are adequately normalized, provided the vacuum $|0\rangle$ is: $$\langle \vec p, s; \vec 0, 0|\vec q, r; 0, 0\rangle = \langle 0| a_{\vec{q}}^{r}a_{\vec{p}}^{s\dagger}|0\rangle\\ = \langle 0|\{a_{\vec{q}}^{r},a_{\vec{p}}^{s\dagger}\}|0\rangle\\=\delta^{rs} \delta^3(\vec p-\vec q)$$ From the fact that all b's and a's anticommute we can immediately derive $$b_{\vec p}^s |\vec q, r;0,0\rangle = 0, \\a_{\vec p}^s |0,0;\vec q, r\rangle = 0.$$ Also, because the creation operators anticommute with themselves, we have $$\left(a_{\vec p}^{s\dagger}\right)^2 = 0 =\left(b_{\vec p}^{s\dagger}\right)^2$$ so that $$a_{\vec p}^{s\dagger} |\vec p, s; 0, 0\rangle = 0 = b_{\vec p}^{s\dagger} |0,0;\vec p, s\rangle.$$ Of course, if we act with creation operators with different momenta and/or spins on the one-particle states, we are going to create the above two-particle (and particle-antiparticle states). We can combine this with the last formula in the following way: $$a_{\vec p_1}^{s_1\dagger} |\vec p_2, s_2;0,0\rangle = (1-\delta_{s_1, s_2}\delta_{\vec p_1, \vec p_2})|\vec p_1, s_1, \vec p_2, s_2; 0,0\rangle\\ b_{\vec p_1}^{s_1\dagger} |0,0;\vec p_2, s_2\rangle = (1-\delta_{s_1, s_2}\delta_{\vec p_1, \vec p_2})|0,0;\vec p_1, s_1, \vec p_2, s_2\rangle\\ a_{\vec p}^{s\dagger} |0,0;\vec q, r\rangle = |\vec p, s; \vec q, r\rangle\\ b_{\vec q}^{r\dagger} |p, s;0,0\rangle = -|\vec p, s; \vec q, r\rangle $$

Now, the really interesting$^{1}$ thing happens, if we annihilate a particle from the one-particle state (or an anti-particle from the one-anti-particle state).

$$a_{\vec p_1}^{s_1} |\vec p_2, s_2;0,0\rangle = a_{\vec p_1}^{s_1}a_{\vec p_2}^{s_2\dagger}|0\rangle \\=\{a_{\vec p_1}^{s_1}, a_{\vec p_2}^{s_2\dagger}\}|0\rangle \\=\delta_{s_1, s_2} \delta^3\left(\vec p_1 - \vec p_2\right) |0\rangle$$ and analogously $$b_{\vec q_1}^{r_1} |0,0;\vec q_2, r_2\rangle=\delta_{r_1, r_2} \delta^3\left(\vec q_1 - \vec q_2\right) |0\rangle$$


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