Scalar Yukawa theory derivation - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-11-18T23:43:09Z https://physics.stackexchange.com/feeds/question/182381 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/182381 2 Scalar Yukawa theory derivation SuperCiocia https://physics.stackexchange.com/users/37677 2015-05-06T10:12:24Z 2015-05-08T02:49:15Z <p>I am using Tong's notes for QFT, and on page 59 there is a derivation for the scattering amplitude of $\psi\psi \rightarrow \psi\psi$ in Scalar Yukawa theory.</p> <p>It goes from here:</p> <p>$$\langle p_1',p_2'|:\psi(x_1)^\dagger\psi(x_1)\psi(x_2)^\dagger\psi(x_2):|p_1,p_2\rangle$$</p> <p>to here:</p> <p>$$\langle p_1',p_2'|\psi(x_1)^\dagger\psi(x_2)^\dagger|0\rangle\langle0|\psi(x_1)\psi(x_2)|p_1,p_2\rangle.$$ </p> <p>I honestly have no idea what the step is there. </p> <p>Shoulnd't the identity be $\int |i\rangle \langle i|$ and not just the vacuum? How can he move the fields around?</p> <p>The question had already been asked <a href="https://physics.stackexchange.com/questions/48157/scattering-processes-in-scalar-yukawa-theory">here</a> but no answer was given, hence my new question.</p> https://physics.stackexchange.com/questions/182381/-/182736#182736 2 Answer by physicus for Scalar Yukawa theory derivation physicus https://physics.stackexchange.com/users/41637 2015-05-08T02:07:47Z 2015-05-08T02:07:47Z <p>I am not aware of an elegant answer to this question, but writing out the expansion of the fields $\psi$ in creation- and annihilation operators gives an answer. $\psi$ is a complex scalar with $$\psi(x)=\int\frac{d^3 k}{(2\pi)^3\sqrt{2E_{\vec{k}}}}\left(b_{\vec{k}}e^{-ikx}+c_{\vec{k}}^\dagger e^{ikx}\right)$$ The initial and the final state only contain the $b$-type particle, not its anti-particle. This is not apparent from your question, but in Tong's lecture notes you can see that \begin{align} |p_1,p_2\rangle &amp;= \sqrt{2E_{\vec{p}_1}}\sqrt{2E_{\vec{p}_2}}b_{\vec{p}_1}^\dagger b_{\vec{p}_2}^\dagger|0\rangle\\ |p_1',p_2'\rangle &amp;= \sqrt{2E_{\vec{p}_1'}}\sqrt{2E_{\vec{p}_2'}}b_{\vec{p}_1'}^\dagger b_{\vec{p}_2'}^\dagger|0\rangle \end{align} Since the normal ordering moves all annihilation operators to the right and the creation operators to the left and the $c_{\vec{k}}$, $c_{\vec{k}}^\dagger$ operators commute with $b_{\vec{p}}^\dagger$, all terms in the product $$\langle p_1',p_2'|:\psi(x_1)^\dagger\psi(x_1)\psi(x_2)^\dagger\psi(x_2):|p_1,p_2\rangle$$ containing either $c_{\vec{k}}$ or $c_{\vec{k}}^\dagger$ will vanish. That leaves us with an expression of the following form, leaving away all normalization factors and exponentials etc, just focusing on the ladder operator structure: $$\int d^3k_1d^3k_2d^3k_3d^3k_4 \langle 0| b_{\vec{p}_1'} b_{\vec{p}_2'} b_{\vec{k}_1}^\dagger b_{\vec{k}_2}^\dagger b_{\vec{k}_3} b_{\vec{k}_4} b_{\vec{p}_1}^\dagger b_{\vec{p}_2}^\dagger|0\rangle$$ The crucial point is, that this expression is equal to one where we insert a $|0\rangle\langle 0|$ into the middel in the following way: $$\int d^3k_1d^3k_2d^3k_3d^3k_4 \langle 0| b_{\vec{p}_1'} b_{\vec{p}_2'} b_{\vec{k}_1}^\dagger b_{\vec{k}_2}^\dagger |0\rangle\langle 0| b_{\vec{k}_3} b_{\vec{k}_4} b_{\vec{p}_1}^\dagger b_{\vec{p}_2}^\dagger|0\rangle$$ A quick way to see this is by looking at the annihilation operator $b_{\vec{p}_2'}$, the second from the left in the original expression. One can use the commutator twice to move it to the right. On the way, there are two $\delta$-functions arising from the commutator, but these equally arise in the expression with the inserted $|0\rangle\langle 0|$. The remaining term, in which the annihilation operator has been moved to the right by two positions is then $$\langle 0| b_{\vec{p}_1'} b_{\vec{k}_1}^\dagger b_{\vec{k}_2}^\dagger b_{\vec{p}_2'} b_{\vec{k}_3} b_{\vec{k}_4} b_{\vec{p}_1}^\dagger b_{\vec{p}_2}^\dagger|0\rangle$$ which vanishes, because there are three annihilation operators acting after two creation operators on the vacuum.</p> <p>Of course the argument can be made more explicit by calculating the whole expression more carefully. It is much more easy to see that $$\langle p_1',p_2'|\psi(x_1)^\dagger\psi(x_2)^\dagger|0\rangle\langle0|\psi(x_1)\psi(x_2)|p_1,p_2\rangle$$ will lead to the same expression we have found for starting with the normal ordered expression.</p> <p>The question remains, why David Tong has chosen to write the expression in this form. Maybe he considered the step to be obvious. </p>