Necessity of the field viewpoint

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In summary, the use of field operators in relativistic quantum mechanics is necessary due to several reasons. One reason is the lack of finite-dimensional unitary representations of the Poincaré group. Additionally, fields allow for the fulfillment of causality and locality, which cannot be achieved in the first-quantization formalism. Furthermore, in relativistic quantum mechanics, particles can be destroyed and created during collisions, which is more easily described using a quantum field theory. The use of bosonic commutation relations for field operators also ensures a positive definite Hamiltonian.
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simonjech
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I am learning QFT rn, so the question that I naturaly ask myself often is why do we have to use field operators in relativistic quantumn theory instead of operators with finite number of degrees of freedom which are used in non-relativistic quantumn mechanics?
One of the reasons that I came across is that there are no finite-dimensional unitary representations of the Poincaré group (Wigners theorem).
But I would like to now some other reasons why fields are necessary in relativistic quantumn mechanics. So could you please write me some arguments that you know?
 
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There are also no finite-dimensional unitary representations of the Galilei group.

The reason is that in relativity you have a more subtle "causality structure", which is most easily fulfilled by descriptions of the dynamics of physical entities in terms of local field theories. There's no way to fulfill locality and causality in the first-quantization formalism, while it's easy to establish using the microcausality constraint on local operators that represent local observables.

From an empirical point of view it's, because when particles collide at relativistic energies (i.e., when the energy exchange between the interacting particles gets in the order of magnitude of the masses of particles that are interacting with the colliding particles) there can always particles be destroyed and new ones created, and this is most naturally described with a quantum-field theoretical description.
 
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Thank you for your respond. Can you also please explain why it is impossible to fulfill causality without using fields?
 
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You can try the first-quantization formalism for, e.g., the Klein-Gordon field. Then you'll see that the time-evolution operator must be
$$\hat{U}(t)=\exp(-\mathrm{i} \sqrt{m^2 + \hat{\vec{p}}^2} t),$$
because energy should be positive definite, i.e., the negative-energy solutions must be abandoned, and the corresponding time evolution of the wave function with this ##\hat{U}(t)## does not fulfill the causality constraint, i.e., if you solve the KG equation with this time-evolution operator for the initial condition ##\Phi(t=0,\vec{x})=\delta^{(3)}(\vec{x})## you get for any ##t>0## a wave function ##\Phi(t,\vec{x})## which is non-zero at any ##\vec{x}##, i.e., you have faster-than-light signal-propagation through ##\Phi##, and this clearly violates relativistic causality.

The solution through QFT is that here you can write the free-field operator with both positive and negative frequency-modes without making the Hamiltonian unbound from below. That's achieved by using the Feynman-Stückelberg trick, i.e., writing an creation instead of an annihilation operator in front of the negative-frequency modes. For the Klein-Gordon field you must quantize everything as bosons to achieve all the good properties of the QFT, i.e., microcausality, ##[\hat{\Phi}(x),\hat{\Phi}(y)]=0## for ##(x-y)## spacelike and positive definiteness of the Hamiltonian:
$$\Phi(x)=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 p}{(2 \pi)^3 \sqrt{2 E_p}} [\hat{a}(\vec{p}) \exp(-\mathrm{i} x \cdot p) + \hat{b}^{\dagger}(\vec{p}) \exp(+\mathrm{i} x \cdot p]_{p^0=E_p}$$
with ##E_p=\sqrt{m^2+\vec{p}^2}##. The creation and annihilation operators have to fulfill the commutator relations
$$[hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{q})]=[\hat{b}(\vec{p}),\hat{b}^{\dagger}(\vec{q})]=(2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{q})$$
with all other combinations of operator pairs in the commutator giving 0.

This construction of the microcausal field operators shows that there are necessarily anti particles (annihilated by the ##\hat{b}## operators) with the same mass as the particles (annihilated by the ##\hat{a}## operators). At the same time, thanks to the use of bosonic commutation relations for the field and thsu the annihilation-creation operators, you find a positive semidefinite Hamiltonian,
$$\hat{H}=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 p}{(2 \pi)^3} E_p [\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}) +\hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p})].$$
For a very clear discussion on an introductory level, see

https://arxiv.org/abs/1110.5013
 
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What is tangentially related and really impressed me when I realized it is that QM IS often a QFT, except in 0 spatial and 1 temporal dimensions, except the fields are, for example, the position and momentum operators.
 
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vanhees71 said:
You can try the first-quantization formalism for, e.g., the Klein-Gordon field. Then you'll see that the time-evolution operator must be
$$\hat{U}(t)=\exp(-\mathrm{i} \sqrt{m^2 + \hat{\vec{p}}^2} t),$$
because energy should be positive definite, i.e., the negative-energy solutions must be abandoned, and the corresponding time evolution of the wave function with this ##\hat{U}(t)## does not fulfill the causality constraint, i.e., if you solve the KG equation with this time-evolution operator for the initial condition ##\Phi(t=0,\vec{x})=\delta^{(3)}(\vec{x})## you get for any ##t>0## a wave function ##\Phi(t,\vec{x})## which is non-zero at any ##\vec{x}##, i.e., you have faster-than-light signal-propagation through ##\Phi##, and this clearly violates relativistic causality.

The solution through QFT is that here you can write the free-field operator with both positive and negative frequency-modes without making the Hamiltonian unbound from below. That's achieved by using the Feynman-Stückelberg trick, i.e., writing an creation instead of an annihilation operator in front of the negative-frequency modes. For the Klein-Gordon field you must quantize everything as bosons to achieve all the good properties of the QFT, i.e., microcausality, ##[\hat{\Phi}(x),\hat{\Phi}(y)]=0## for ##(x-y)## spacelike and positive definiteness of the Hamiltonian:
$$\Phi(x)=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 p}{(2 \pi)^3 \sqrt{2 E_p}} [\hat{a}(\vec{p}) \exp(-\mathrm{i} x \cdot p) + \hat{b}^{\dagger}(\vec{p}) \exp(+\mathrm{i} x \cdot p]_{p^0=E_p}$$
with ##E_p=\sqrt{m^2+\vec{p}^2}##. The creation and annihilation operators have to fulfill the commutator relations
$$[hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{q})]=[\hat{b}(\vec{p}),\hat{b}^{\dagger}(\vec{q})]=(2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{q})$$
with all other combinations of operator pairs in the commutator giving 0.

This construction of the microcausal field operators shows that there are necessarily anti particles (annihilated by the ##\hat{b}## operators) with the same mass as the particles (annihilated by the ##\hat{a}## operators). At the same time, thanks to the use of bosonic commutation relations for the field and thsu the annihilation-creation operators, you find a positive semidefinite Hamiltonian,
$$\hat{H}=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 p}{(2 \pi)^3} E_p [\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}) +\hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p})].$$
For a very clear discussion on an introductory level, see

https://arxiv.org/abs/1110.5013
Thank you very much for the clarification.
 

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