QFT question about using momentum raising and lowering operators

[arnshch]

How did you find PF?: Google

I know how to express Hamiltonian for scalar field written in field operators through the raising and lowering momentum operators, but I can't figure out how to do the same for the number of particles written in field operators: the 1/2E coefficient within the corresponding integral, doesn't go away in the latter expression, as it does in the former one, and I cannot figure out how to deal with it. Any advise?

 

[arnshch]

Sorry, I actually want to do the opposite: from the number of particles written through momentum raising and lowering operators to the same number expressed in field operators.

 

[HomogenousCow]

Start with $$\phi(\vec {x})=\int \frac{d^{3}k}{\sqrt{2 \omega_k}} a_k ~e^{i \vec{k} \cdot \vec{x}}+ a_k^\dagger~e^{-i \vec{k} \cdot \vec{x}}$$ $$\pi(\vec {x})=-i\int d^{3}k \sqrt{\frac{\omega_k}{2}} a_k ~e^{i \vec{k} \cdot \vec{x}}- a_k^\dagger~e^{-i \vec{k} \cdot \vec{x}}$$ and invert these, should be straightforward from there.

 

[vanhees71]

I'm not sure about what the OP's question is. I'd recommend to start with a finite volume with periodic boundary conditions on the fields (operators) to get rid of all kinds of problems with $\delta$ distributions.

Then it depends on how you normalize your annihilation-creation operators in the mode decomposition, which factors enter into the "number operators". If you want to get the simple relation $\hat{N}(\vec{k},\sigma)=\hat{a}^{\dagger}(\vec{x}) \hat{a}(\vec{x})$ you have to normalize by the (anti-)commutator relations
$$[\hat{a}(\vec{k}),\hat{a}^{\dagger}(\vec{k}')]_{\pm}=\delta_{\vec{k},\vec{k'}},$$
where here in the finite volume limit we have Kronecker-$\delta$'s.

Then for the non-interacting particles you get
$$\hat{H}=\sum_{\vec{k}} \vec{k} \hat{N}(\vec{k}), \hat{\vec{P}}=\sum_{\vec{k}} \vec{k} \hat{N}(\vec{k})$$
etc.