# Is $:A: \; \;= A - \left<0\right|A\left|0\right>$ a correct definition of normal ordering?

My course notes say that normal ordering is defined as

$$:A: \;\; = A - \left< 0\right| A \left| 0\right>.\tag{1}$$

This works for $A = aa^\dagger$ and all already normal ordered expressions.

When $A = a a^\dagger a$, though, or anything that is not normal ordered but has at least one annihilation operator furthest right, the second term is immediately $0$ and the expression returned is simply $A$, which is not normal ordered in this case.

\begin{align*} :aa^\dagger a: \; \; &= a a^\dagger a - \left<0\right| a a^\dagger a \left| 0 \right> \\ &= a a^\dagger a - 0 \\ &= a a^\dagger a \end{align*}\tag{2}

$A = a a^\dagger a^\dagger$ also doesn't work.

Have I misunderstood something, or are my notes incorrect?

• I think $A$ is only allowed to be a series of creation or annihilation operator 'symbols', which in the second term are supposed to be commutated with each other until they annihilate by their usual rules on the vacuum. I can't find this exact definition anywhere except in my notes, though, and my notes don't elaborate on the interpretation. – perilousGourd Jul 18 '18 at 3:45