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Nixom
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In the free case,we decomposite the free Hamiltonian into the creation and annihilation operators, i just wonder why this ad hoc method can not be used to the interaction theory?
Nixom said:OK,if I originally take the a+ and a- as the Fourier coefficients, can they be interpreted as creation and annihilation operators definitely, or it is just depend on some specific form in Hamiltonion, e.g. the quadratic form?
Nixom said:OK,if I originally take the a+ and a- as the Fourier coefficients, can they be interpreted as creation and annihilation operators definitely, or it is just depend on some specific form in Hamiltonion, e.g. the quadratic form?
If I expand [tex]x\propto(a+a^\dagger)[/tex] in the Hamiltonian [tex]H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2+\lambda x^4[/tex], I can't get the results such as [tex][H,a^\dagger]=\omega a^\dagger[/tex] or [tex][H,a]=-\omega a[/tex], so how can I interpret them as creation and annihilation operators?You can (and we do!) expand fields out in terms of "creation" and "annihilation" operators.
The commutation relations of these operators derived from the algebraic relations of the field(coordinate) and the conjugate momentum, so dose it have some relationship with the symplectic constructure of the Hamiltonian?The interpretation as creation and annihilation operators works simply b/c of their (purely algebraic) commutation relations
How do we interpret the in and out state? I can't understand that since they are the eigenstate of the full Hamiltonian, why they are used to describe the "free" particles?They are called in and out operators and they create "free" particles in the interacting theory that are eigenstates of the full hamiltonian. Look in Weinberg vol. 1
I think it should be interaction Hamiltonians instead of Lagrangians, since fields in Lagrangians aren't field operators and we can only write field operators in terms of creation and annihilation operators. In addition, to remember the difference between creation/annihilation operators in in/out field operators (Heisenberg picture) and them in field operators of interaction picture is important.K^2 said:Interaction Lagrangians in QFT are written in terms of creation and annihilation operators for the fields involved in the interaction. So yeah, that's exactly what we do.
No, creation and annihilation operators are typically used in quantum field theory to describe the interactions between particles, but they may not be applicable in other types of interactions, such as classical interactions.
In the interaction case, creation and annihilation operators are used to describe the creation and annihilation of particles during an interaction. They act on the quantum states of the particles involved in the interaction, and can be used to calculate the probabilities of different outcomes.
Yes, creation and annihilation operators can be used simultaneously in an interaction to describe the creation and annihilation of different types of particles. However, they must be used correctly and in the appropriate order to ensure accurate calculations.
One limitation is that creation and annihilation operators may not accurately describe interactions that involve more than two particles. In these cases, higher-order operators may be needed to fully describe the interaction.
The use of creation and annihilation operators in interactions helps to maintain the conservation of certain properties, such as energy and momentum. The operators must be used in a way that conserves these properties in order for the calculations to be valid.