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How exactly is "normal-ordering an operator" defined?

+ 5 like - 0 dislike
64 views

(In this question, I'm only talking about the second-quantization version of normal ordering, not the CFT version.)

Most sources (e.g. Wikipedia) very quickly define normal-ordering as "reordering all the ladder operators so that all of the creation operators are to the left of all of the annihilation operators." This definition is extremely vague, and I want to make sure I understand the actual definition.

If I understand correctly, people use the phrase "normal-order an operator" to mean two inequivalent things. Sometimes they mean "use the (anti)commutation relations to rewrite the operator so that it is normal-ordered (without changing the operator itself)." Under this (unambiguous) definition, we have that the normal-ordered form of the operator $a a^\dagger$ is $a^\dagger a + 1$. We can use this definition to put any operator into canonical form (up to a sign, in the fermionic case. We can fix this sign ambiguity by specifying a canonical ordering of the single-site Hilbert spaces.)

But sometimes the verb "normal-order" is used in a different way, which can actually change the operator. I believe that this definition is the one usually represented by surrounding the operator with colons. If I understand correctly, this procedure is defined as "use the (anti)commutation relations $\left[ a_i, a_j \right]_\pm = \left[ a_i^\dagger, a_j^\dagger \right]_\pm = 0$ to move all the creation operators to the left of all the annihilation operators, while ignoring the $\left[ a_i, a_j^\dagger \right]_\pm = \delta_{ij}$ (anti)commutation relation and pretending that its RHS were zero."

This procedure obviously seems a bit arbitrary and unmotivated. Moreover, it doesn't seem entirely well-defined. It's fine for products of ladder operators, but the problem is that under this definition, normal-ordering does not distribute over addition:

$$ {:} a^\dagger a{:} \ =\ a^\dagger a\ =\ a a^\dagger - 1\ $$ but $${:} a a^\dagger{:} -1\ = a^\dagger a - 1.$$

It's therefore not clear how to define normal-ordering for a general operator, i.e. a general linear combination of products of ladder operators. And of course, whether or not an operator is a nontrivial sum of products of ladder operators depends on how you write it; we can equivalently write the same operator as $a^\dagger a$ (only one summand) or as $a a^\dagger - 1$ (multiple summands).

From this, I conclude that (under the second definition) "normal-ordering an operator" is actually an abuse of terminology; we can only meaningfully normal-order certain particular expressions for some operators. Is this correct? If not, how does one define the normal-ordering of a linear combination of products of ladder operators?

This post imported from StackExchange Physics at 2017-09-16 17:52 (UTC), posted by SE-user tparker
asked Jul 16 in Theoretical Physics by tparker (255 points) [ no revision ]
Related: physics.stackexchange.com/q/309410/2451 , physics.stackexchange.com/q/323801/2451 and links therein.

This post imported from StackExchange Physics at 2017-09-16 17:52 (UTC), posted by SE-user Qmechanic
@Qmechanic It seems that ACuriousMind's answer to the first question exactly answers my question - thanks!

This post imported from StackExchange Physics at 2017-09-16 17:52 (UTC), posted by SE-user tparker

3 Answers

+ 2 like - 0 dislike

The main point is that the normal ordering procedure $:~:$ does not take operators to operators, but symbols/functions to operators. This important point resolves various paradoxes created by abuse of language. For a full explanation, see e.g. this Phys.SE post and links therein.

This post imported from StackExchange Physics at 2017-09-16 17:52 (UTC), posted by SE-user Qmechanic
answered Jul 16 by Qmechanic (2,790 points) [ no revision ]
+ 2 like - 0 dislike

A good way to understand Wick algebras "of normal ordered operators" is as the Moyal deformation quantization of the algebra \(\mathcal{F}_{\mathrm{loc}}\) of local observables. Here a local observable is an expression in the field \(\Phi\) of the form \(\int_{\mathrm{spacetime}} b(x) \Phi(x)^{n} \partial_{\mu}\Phi(x) \mathrm{dvol}_\Sigma(x)\) or similar, where \(b\) is a bump function . This is a Poisson algebra under the canonical Poisson bracket of the free field theory (the Peierls bracket). Finding a formal deformation quantization means to find a non-commutative associative product \(\star\) on \(\mathcal{F}_{\mathrm{loc}}[ [ \hbar ] ]\)which coincides with the commutative product to 0th order in \(\hbar\) and is such that the coefficient of \(\hbar\) in the (anti-)commutator is the given Poisson bracket.

In good situations (such as for fields described by a normally hyperbolic linear differential operator on a globally hyperbolic spacetime) the deformed algebra \(\left( \mathcal{F}_{\mathrm{loc}}[ [\hbar ] ], \star \right)\)may be represented by suitable operators on a Hilbert space. Then the composite "quantization map"

\(\mathcal{F}_{\mathrm{loc}} \longrightarrow \mathcal{F}_{\mathrm{loc}}[ [ \hbar ] ] \overset{\rho}{\longrightarrow} End(\mathcal{H})\)

from "classical" local observables to the operators representing their formal deformation quantization is the "normal ordering map", and under this identification the star-product is given by the usual recipe.

Filling in the details of this story requires a little microlocal analysis of the causal propagator. It is spelled out in detail here:

nLab: Wick algebras

This observation that normal ordering and Wick algebras is a special case of formal deformation qunatuization applied to the covariant phase space of free fields is originally due to

  • J. Dito,
    "Star-product approach to quantum field theory: The free scalar field"
    Letters in Mathematical Physics, 20(2):125–134, 1990 (spire)

and was then further amplified in

  • Michael Dütsch, Klaus Fredenhagen,
    "Perturbative algebraic field theory, and deformation quantization",
    in Roberto Longo (ed.),
    "Mathematical Physics in Mathematics and Physics, Quantum and Operator Algebraic Aspects", volume 30 of Fields Institute Communications, pages 151–160. American Mathematical Society, 2001 (arXiv:hep-th/0101079)

Later it was understood that in fact all of perturbative quantum field theory (i.e. perturbations beyond the free field theory) is an example of formal deformation quantization:

  • Giovanni Collini,
    "Fedosov Quantization and Perturbative Quantum Field Theory"
    (arXiv:1603.09626)
answered Sep 18 by Urs Schreiber (5,735 points) [ revision history ]
+ 1 like - 0 dislike

In quantum mechanics with finitely many degrees of freedom, normal ordering always means applying the commutation rules to move creators to the left of annihilators, resulting in a (up to ordering creators or annihilators among themselves) unique final expression equal to the original one in the operator sense.

In quantum field theory (i.e., quantum mechanics with infinitely many degrees of freedom) proceeding this way usually is impossible as it leads to ill-defined coefficients. Therefore, in quantum field theory, normal ordering always means permuting creators to the left of annihilators without any consideration of commutation rules (i.e., on the classical expression). However the sign is changed when permuting two fermionic operators. In this way otherwise ill-defined expressions make sense at least as quadratic forms. In spacetime dimension $>2$, further renormalization is needed to turn the expressions into true operators.

The resulting operator is uniquely defined as a quadratic form, though the expression is again unique only up to ordering creators or annihilators among themselves.


This post imported from StackExchange Physics at 2017-09-16 17:52 (UTC), posted by SE-user Arnold Neumaier

answered Aug 10 by Arnold Neumaier (12,355 points) [ revision history ]
edited Sep 17 by Arnold Neumaier
I don't think that normal-ordered expressions are unique, because the order of the creation operators and the order of the annihilation operators are separately arbitrary, right?

This post imported from StackExchange Physics at 2017-09-16 17:52 (UTC), posted by SE-user tparker
@tparker - yes, but these (anti)commute, so the operator defined is independent of the remaining ordering ambiguity. Any precedence order will make it fully unique.

This post imported from StackExchange Physics at 2017-09-16 17:52 (UTC), posted by SE-user Arnold Neumaier

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