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About Normal Ordering

  1. Mar 8, 2014 #1
    In normal ordering method the anhillation operators are put in the left and creation operators on the right. What happens when we try to normal order the product of two anhillation operators or two creation operators (as in the case of complex scalar fields). What are we doing there actually..??

    :[itex]a^\dagger[/itex][itex] b^\dagger[/itex]: = ??
    Last edited: Mar 8, 2014
  2. jcsd
  3. Mar 9, 2014 #2
    If you have two annihilation operator or two creation operation sandwiched between vacuum states,you will not be able to get the vacuum after annihilating two times vacuum(annihilation operator acting on vacuum will give automatically zero) or creating something out of vacuum twice.You have to create something and then annihilate it to get vacuum again.
    In general,you have to decompose the field into sums of some fields like,
    $$ψ(x)=ψ^+(x)+ψ^-(x)$$,where one part will be responsible for creation and other part for annihilation.
  4. Mar 9, 2014 #3
    In quantizing the complex field, there will be creation of particle and antiparticle -terms. They will come as a product in hamiltonian. After the normal ordering process I cant see such terms involving in the hamiltonian. Thats why the question.

    Expecting the answer...!
  5. Mar 9, 2014 #4
    when you are writing the product of fields,you have to decompose the fields into creation and annihilation part as I written above,and then you can get a normal ordering.After doing it you use the commutation relations between fields and try to get a contribution for energy.You have to define some number operators like $$N_a=a^{\dagger}a$$
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