Normal ordering

If I understood correctly there are two terms called normal ordering:

1. $:c c^\dagger: = c^\dagger c \hspace{.5cm}$so shifting all creation operators to the left and all annihilation operators to the right.
2. $:c^\dagger c:=c^\dagger c- \langle c^\dagger c \rangle\hspace{.5cm}$ so subtracting the contraction

Both of these are called normal ordering (I think) and use the same notation. But what is the connection?

If I got something wrong, please correct and explain!

• Normal ordering is the ordering of a product of operators such that acting on the vacuum, the product gives zero. – PPR Aug 13 '14 at 13:16
• @PPR Are you sure that condition uniquely characterizes normal ordering? – joshphysics Aug 13 '14 at 15:28
• The two definitions are equivalent for most purposes, but this is not true for any CFT. Have a look at Polchinski's String theory book I, pages 36 and 60 for a brief discussion. Polchinski also uses different symbols for the two definitions. – Heterotic Aug 13 '14 at 18:30

As far as my experience goes (theoretical nuclear physics), they are pretty much equivalent. To make this clearer, I'll try to give an intuitive example for bosons:

$$\langle cc{}^\dagger\rangle = cc{}^\dagger-:cc{}^\dagger:$$

which becomes

$$\langle cc{}^\dagger\rangle = cc{}^\dagger-c{}^\dagger c = \left[c,c{}^\dagger\right]$$

so in this case $\langle cc{}^\dagger\rangle$ is just the commutator of $c$ and $c{}^\dagger$. So now lets look at what you wrote above.

1. $:cc{}^\dagger: = c{}^\dagger c$
2. $:cc{}^\dagger: = cc{}^\dagger - \langle cc{}^\dagger\rangle = cc{}^\dagger - \left[c,c{}^\dagger\right] = c{}^\dagger c$

You can show the same for fermions using the anticommutator relationships. Basically, that which we call a contraction, (in the case of creation and annihilation operators) just comes from the commutators of the terms we commute to get the normal ordering. You can check for cases with more operators and you will get analogous situations. (You can also check the examples on the Wikipedia page on Wick's theorem.)