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glS
glS
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In first quantization  ,the the operator $J$ assumes the form $J=\sum_{i}j(x_i)$.  In Fock space  , it is instead written as $J=\int dx \psi^\dagger(x)j(x)\psi(x)$,where where $\psi^\dagger,\psi$$\psi^\dagger, \psi$ are the field operators.Is j(x)

Is $j(x)$ an operator in fockFock space or simply a function of x?If

If $j(x)={\nabla_x}^2$$j(x)=\nabla^2_x$,then then does it commute with $\psi^\dagger,\psi$$\psi^\dagger$ and $\psi$?

In first quantization  ,the operator $J=\sum_{i}j(x_i)$.In Fock space  $J=\int dx \psi^\dagger(x)j(x)\psi(x)$,where $\psi^\dagger,\psi$ are the field operators.Is j(x) an operator in fock space or simply a function of x?If $j(x)={\nabla_x}^2$,then does it commute with $\psi^\dagger,\psi$ ?

In first quantization, the operator $J$ assumes the form $J=\sum_{i}j(x_i)$.  In Fock space, it is instead written as $J=\int dx \psi^\dagger(x)j(x)\psi(x)$, where $\psi^\dagger, \psi$ are the field operators.

Is $j(x)$ an operator in Fock space or simply a function of x?

If $j(x)=\nabla^2_x$, then does it commute with $\psi^\dagger$ and $\psi$?

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Annie
Annie
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Representation of operators in Fock Space and Second Quantizationspace

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Annie
Annie
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Fock Space and Second Quantization

In first quantization ,the operator $J=\sum_{i}j(x_i)$.In Fock space $J=\int dx \psi^\dagger(x)j(x)\psi(x)$,where $\psi^\dagger,\psi$ are the field operators.Is j(x) an operator in fock space or simply a function of x?If $j(x)={\nabla_x}^2$,then does it commute with $\psi^\dagger,\psi$ ?