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Quantum Physics
Creation/annihilation operators question
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[QUOTE="vanhees71, post: 6323968, member: 260864"] You stumbled over an issue that is usually not so carefully treated and then you run into trouble. That's the way QFT is usually taught and to some extent that's because there's no simple way to formulate it mathematically rigorously (at least not interacting field theories). The problem is that you build the Fock space for a particle in an infinite 3D space and thus your momentum components are observables with an entirely continuous spectrum. Indeed the momentum "eigenvalues" are just ##\vec{p} \in \mathbb{R}^3##. From ordinary quantum mechanics you already know that the momentum eigenfunctions are ##u_{\vec{p}}(\vec{x})=N \exp(\mathrm{i} \vec{p} \cdot \vec{x})## are not really "allowed functions", because they are not square integrable and you can only normalize them to ##\delta## distributions ##\label \vec{p}|\vec{p}' \rangle=\delta^{(3)}(\vec{p}-\vec{p}')##. Now all this is cured in some sense by putting the particles into a finite volume. You could try it with a "cavity", i.e., a potential cubic box with infinite walls. That's however not a clever choice when you want to study finally indeed scattering theory for particles in the entire infinite-volume space, because there you like to have well-defined momentum operators also for the finite-volume box (which you don't have with rigid boundary conditions at the boundaries of the box). That's why it's more suitable to consider a finite cube of length ##L## and impose periodic boundary conditions on the wave functions, $$\psi(\vec{x}+\vec{n} L)=\psi(\vec{x}) \quad \text{for all} \quad \vec{n} \in \mathbb{Z}^{3}.$$ Then it's easy to see that you still have a well-defined self-adjoint momentum operator (in the 1st-quantization formalism), ##\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}## and thus momentum eigensolutions ##u_{\vec{p}}=N \exp(\mathrm{i} \vec{p} \cdot \vec{x})##, but now due to the boundary conditions you have only a discrete set of momentum $$\vec{p} = \vec{n} \frac{2 \pi}{L}, \quad \vec{n} \in \mathbb{Z}.$$ Also the eigenmodes are normalizable and you have $$\langle \vec{p}|\vec{p}' \rangle=\int_{[0,L]^3} \mathrm{d} ^3 x u_{\vec{p}}^*(\vec{x}) u_{\vec{p}'}(\vec{x}) = \delta_{\vec{p},\vec{p}'},$$ where the ##\delta## here is a usual Kronecker ##\delta## which is simply 1 for ##\vec{p}=\vec{p}'## and 0 otherwise. Now you can build your Fock space without the troubles you recognized above. To get the "infinite-volume limit" you have to define useful physical quantities in such a way that the limit ##L \rightarrow \infty## makes sense, and then the sum over momentum eigenstates translates into an integral, i.e., in an operator sense $$\frac{1}{V} \sum_{\vec{p}} \rightarrow \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 p}{(2 \pi)^3}.$$ The first encounter with infinities is when looking at total energy and momentum, where in the relativistic case you get an infinity (even in the finite volume), because a naive expression for the total energy operator (the Hamiltonian of the free QFT) is $$\hat{H}=\sum_{\vec{p}} E(\vec{p}) [\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}) + \hat{b}(\vec{p}) \hat{b}^{\dagger}(\vec{p}).$$ Apply this to the vacuum state, you get from the 2nd term $$\hat{H} |\Omega \rangle=\sum_{\vec{p}} E(\vec{p}) |\Omega \rangle,$$ but since ##E(\vec{p})=\sqrt{m^2+\vec{p}^2}## the sum diverges. On the other hand the energy is only defined up to an arbitrary additive constant, and you can simply use the commutation relations of the creation and annihilation operators to write $$\hat{b}(\vec{p}) \hat{b}^{\dagger}(\vec{p}) = [\hat{b}(\vec{p}),\hat{b}^{\dagger}(\vec{p})] + \hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p})=\hat{1} +\hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p}) ,$$ which means you Hamiltonian gets $$\hat{H}=\hat{H}'+ \hat{1} \sum_{\vec{p}} E(\vec{p})$$ with $$\hat{H}'=:\hat{H}:=\sum_{\vec{p}} E(\vec{p}) [\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}) + \hat{b}^{\dagger} (\vec{p})].$$ Now ##\hat{H}'## has all the commutation properties with any operators as ##\hat{H}## but the infinite "vacuum energy" is subtracted and any proper Fock state (which are Fock states with a finite total number of particles!) has a definite finite energy. Now let's discuss the infinite-volume limit. First note that from the translation rule of the momentum sums to an integral it makes sense to rescale the annihilation operators $$\hat{a}^{(\infty)}(\vec{p}) \rightarrow \sqrt{V} \hat{a}(\vec{p}),$$ because then you can write $$\hat{H}'=V \sum_{\vec{p}} E(\vec{p}) [\hat{a}^{(\infty)\dagger}(\vec{p}) \hat{a}^{\infty}(\vec{p}) + \hat{b}^{(\infty)\dagger} (\vec{p}) \hat{b}^{\infty}(\vec{p})]$$ and now take the limit to the integral for ##L \rightarrow \infty##, which gives $$\hat{H}'=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} [\hat{a}^{(\infty)\dagger}(\vec{p}) \hat{a}^{\infty}(\vec{p}) + \hat{b}^{(\infty)\dagger} (\vec{p}) \hat{b}^{\infty}(\vec{p})].$$ For ##L \rightarrow \infty## now the commutator of the new annihilation and creation operator indeed goes to ##\infty## for ##\vec{p}=\vec{p}'## but stays 0 otherwise. The rescaling is such that you get $$[\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p}')]=(2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{p}').$$ where I skipped the ##(\infty)## supersecripts again, and the additional factor ##(2 \pi)^3## is put into make everything consistent with all momentum integrals always coming with the meaure ##\mathrm{d}^3 \vec{p}/(2 \pi)^3## which is the common convention among HEP physicists. [/QUOTE]
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