6
$\begingroup$

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?

$\endgroup$
2
+50
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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. $$ \sum_{i<j}\psi_{ij}\left(\left[ c_k ,\, c_i^{\dagger}\right]_+ - c_i^{\dagger}c_k\right) c_j^{\dagger}|0\rangle=\sum_{i<j}\psi_{ij}\left(\left[ c_k ,\, c_i^{\dagger}\right]_+c_j^{\dagger} - c_i^{\dagger} \left[ c_k ,\, c_j^{\dagger}\right]_+ \right) |0\rangle = \sum_{i<j}\psi_{ij}\left(\left[ c_k ,\, c_i^{\dagger}\right]_+|j\rangle - \left[ c_k ,\, c_j^{\dagger}\right]_+ |i\rangle \right) $$

$\endgroup$
1
$\begingroup$

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

$\endgroup$

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.