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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?

Qmechanic
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userManyNumbers
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  • 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. – userManyNumbers Jul 18 '18 at 03:45

1 Answers1

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OP is right:

  1. Eq. (1) is strictly speaking meaningless since the argument for the normal ordering on the lhs. should be a function/symbol, not an operator. See my Phys.SE answer here for a necessary condition.

  2. Normal ordering takes a symbol/function into an operator with creation (annihilation) operators to the left (right), respectively. It does not take an operator into an operator, cf. e.g. my Phys.SE answer here.

Qmechanic
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