0
$\begingroup$

enter image description here

Here $\phi_I$ is just the free Klein-Gordon field. So, this field is decomposed of two components shown above. Now let $N$ be the normal ordering operator. Then, I think that $N(\phi_I^+(x)\phi_I^-(y))=N(\phi_I^-(y)\phi_I^+(x))=\phi_I^-(y)\phi_I^+(x)$. Is this true? If so, then $N([\phi_I^+(x), \phi_I^-(y)])=0$ must hold. Is this right?

!!!Addtional question!!!

enter image description here

This is page 89 of Peskin and Schroeder's QFT book. First, from the definition (4.35) and my original question, $N(\text{contraction of}\; \phi(x)\; \text{and} \; \phi(y))$ must be zero. However, from (4.36) and the definition of the Feynman propagator (which seems to contain no operators $a_\textbf{p}$, $N(\text{contraction of}\; \phi(x)\; \text{and} \; \phi(y))=N(D_F(x-y))=D_F(x-y)$. So I am extremely confused... Also, (4.37) equation seems to be wrong. The right side seems to be just equal to $N(\phi(x)\phi(y))$ and it does not match with the left side. For the last, what exactly is the definition of the contraction of non-adjacent fields? For example what exactly is $\phi_1\phi_2\phi_3\phi_4$ with $\phi_1$ and $\phi_3$ contracted? I mean the thing inside $N$ of the left side of the last line of the picture I posted.

$\endgroup$
0
$\begingroup$

Normal-ordering of the identity operator is zero, i.e. $$ N(I) = 0. $$ The idea of normal-ordering is introduced so that a normal-ordered operator has vanishing vacuum expectation value $$ \langle 0| N({\cal O}) |0 \rangle = 0 \qquad \qquad (1) $$ The way this is done is follows. Let ${\cal O}(x)$ be an operator with $\langle 0| {\cal O}(x) |0 \rangle = a\neq0$.

Then, define the normal ordering of the operator by $$ N({\cal O}(x)) = {\cal O}(x) - a I $$ Then, by construction (1) is true. The following are now exercises left to the reader -

  1. Convince yourself that this definition is in fact the same as the one defined in Peskin and Schroeder, which defines the operation in terms of creation and annihilation operators. It asks us to move all annihilation operators to the right and all creation operators to the left. That is the same as the procedure above.

  2. By the above definition $N(I) = 0$.

$\endgroup$
  • $\begingroup$ Yes I understand this construction. Then, is the equation (4.37) in my picture true? What exactly is the left side of equation (4.37)? $\endgroup$ – Keith Mar 23 '18 at 20:45
  • $\begingroup$ (4.37) should read $T(\phi(x) \phi(y)) = N(\phi(x) \phi(y) ) + contraction ( \phi (x) \phi(y) )$. $\endgroup$ – Prahar Mar 23 '18 at 20:50
  • $\begingroup$ So is it a typo? Then what becomes (4.38) and (4.39)? The book seems to be clearly wrong then... $\endgroup$ – Keith Mar 23 '18 at 20:52
  • $\begingroup$ (4.38) has the normal-ordering sign on all operators on the RHS, but not the identity operator. Strictly speaking, you are right of course, the book is wrong. However, I rather think of it as abuse of notation. Its more cumbersome to write out the exact result as is and so PS abuse notation and write it in a short but loose way. $\endgroup$ – Prahar Mar 23 '18 at 20:55
  • $\begingroup$ Where is an identity operator in the RHS of (4.38)? As a double major in math, I just cannot tolerate anything that is not in the most exact form....Is this attitude an impediment to studying physics? $\endgroup$ – Keith Mar 23 '18 at 21:07

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.