Normal order of operator product

  • Thread starter paweld
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  • #1
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I'm not sure if the normal orderdering of operator product is a well defined procedure.
I assume that normal order of a sum of operators is a sum of normal orders. It follows
from it that for bosonic creation and anihilation operators holds:
[tex]
a^\dagger a= ~: [a,a^\dagger ]:~ =~ :[a^\dagger,a]+1:~ = ~ :[a^\dagger,a]: +:1:=
a^\dagger a + :1:
[/tex]
So we have [tex] :1: =0[/tex]. Is it OK?
 

Answers and Replies

  • #2
A. Neumaier
Science Advisor
Insights Author
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3,955
I'm not sure if the normal orderdering of operator product is a well defined procedure.
I assume that normal order of a sum of operators is a sum of normal orders. It follows
from it that for bosonic creation and anihilation operators holds:
[tex]
a^\dagger a= ~: [a,a^\dagger ]:~ =~ :[a^\dagger,a]+1:~ = ~ :[a^\dagger,a]: +:1:=
a^\dagger a + :1:
[/tex]
So we have [tex] :1: =0[/tex]. Is it OK?

No. First of all, your commutators don't make sense.
But let me assume that you intended to say
a^*a = :aa^*: = :a^*a+1: = :a^*a:+:1: = a^*a+:1:, so that :1:=0.
This looks on the surface correct. Nevertheless, :1:=1.


If A is a formal linear combination of products of numbers, creation
operators and annihilation operators, the normal ordering :A: of A is
the operator defined by moving in each product term the creation
operators to the left of the annihilation operators.

In particular,
:AB: = :BA: = 0 for any A,B,
since AB and BA contain the same creation and annihilation operators
with the same multiplicity, only in different order. Therefore, we
also have :[A,B]: = :AB:-:BA: = 0.

This may look a bit strange since :[a,a^*]:=0 and :1:=1 although [a,a^*]=1. But nothing is wrong here, since [a,a^*] and 1, although equal as operators, are different as formal expressions.

Conclusion:

Normal ordering is applied only to_expressions_ involving operators,
not to operators themselves. The normal ordering of an operator is not
well-defined unless you specify a particular form of writing it.
 
  • #3
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Thank you a lot.
So normal ordering is well defined only when the form of expression is given.
For example :H: does not make sens if we know only that:
[tex]
H = ( \pi^2 + (\nabla \phi)^2 + m^2 \phi^2)/2
[/tex]
and [tex] \pi[/tex], [tex] \phi[/tex] can be expressed in terms of creation and
anihilation opperators.
 
  • #4
cgk
Science Advisor
524
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That is right. Normal ordering applied to expressions, not to operators. And unless the concrete expressions are given (or if there is a standard convention of assigning them), normal ordering is not a well-defined concept.
 
  • #5
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The same is true with time ordered product, because T(A B) \neq T(A C D)
even if B=C D.
 
  • #6
A. Neumaier
Science Advisor
Insights Author
8,059
3,955
Thank you a lot.
So normal ordering is well defined only when the form of expression is given.
For example :H: does not make sens if we know only that:
[tex]
H = ( \pi^2 + (\nabla \phi)^2 + m^2 \phi^2)/2
[/tex]
and [tex] \pi[/tex], [tex] \phi[/tex] can be expressed in terms of creation and
anihilation opperators.

Well, one can do all the algebra that doesn't depend on specific equations relating the symbols. Thus if you write pi and phi as linear combinations of c/a operators and expand and normal order, you get correct results as long as you use neither commutation relations nor field equations.
 

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