# Stuekelberg interpretetion of anti-particles

1. Aug 6, 2013

### cabrera

Dear Forum,

I have discovered by accident the method developed by Feynman and Stuekelberg to explain the origin of antiparticles ( I have the little book "Elementary particles and the laws of physics).

As you might be all aware by using basic perturbation theory and assuming that negative energies are not possible because as Feynman quoted "we could dump particles into this negative energy and run the world with the extra energy" it is found that particles traveling faster that the speed of light contribute to the amplitude. The perturbation expansion calculates the amplitude of a particle to go state from θ to state θ by means of a perturbation by a potential U1 at t1 (change state to p) and second perturbation at t2 by potential U2 that brings the particle back to state θ.

I am having problem to understand the following part of the argument. Then it is assumed that such particles correspond to a different relativistic framework by which the order of two events: t1 particle scattered by potential U1 to state p from state θ and event t2 from state p back to state θ happen in reverse order. Under the new framework we will see that t2 correspond to the creation of a particle - antiparticle and t1 annihilation of the original particle.

First question is: what is the physical origin of the potentials U1 and U2? Do they create and destroy particles?

Second question :Could we claim that antiparticles are just particles traveling back in time because of simple relativistic consideration?

Third question: What is the place of Stuekelberg theory in modern particle physics?

2. Aug 7, 2013

### vanhees71

The Feynman-Stückelberg approach can be made much more understandable by simply using the quantum-field theoretical formalism of creation and annihilation operators (for free particles of course!) than by these handwaving arguments of particles running backwards in time.

Let's take complex scalar fields, i.e., Klein-Gordon particles as an example. The Lagrangian reads
$$\mathcal{L}=\partial_{\mu} \phi^* \partial^{\mu} \phi-m^2 \phi^* \phi.$$
The conjugate field momenta are
$$\Pi=\frac{\partial \mathcal{L}}{\partial \dot{\phi}}=\dot{\phi}^*, \quad \Pi^*=\frac{\partial \mathcal{L}}{\partial \dot{\phi}^*}=\dot{\phi}.$$
Thus when quantizing the field to describe bosons (as it turns out that's the only way it works, which is a special case of the spin-statistics theorem), we have the commutation relations
$$[\hat{\phi}(t,\vec{x}),\hat{\phi}(t,\vec{y})]=[\dot{\hat{\phi}}(t,\vec{x}),\dot{\hat{\phi}}(t,\vec{y})]=0,$$
$$[\hat{\phi}(t,\vec{x}),\dot{\hat{\phi}}^{\dagger}(t,\vec{y})=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y})$$
and so on.

The equations of motion (in the Heisenberg picture) read
$$(\Box+m^2) \hat{\phi}(x)=0.$$
Fourier transforming to momentum space leads to the mode functions,
$$u_{\vec{p}}^{(\pm)}(x)=\frac{1}{\sqrt{(2 \pi)^3 2 \omega(\vec{p})}} \exp(\mp \mathrm{i} \omega(\vec{p}) t + \mathrm{i} \vec{p} \cdot \vec{x})$$
with $\omega(\vec{p})=+\sqrt{\vec{p}^2+m^2}$. The general solution thus reads
$$\hat{\phi}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} [\hat{a}_+(\vec{p}) u_{\vec{p}}^{(+)}(x)+\hat{a}_-(\vec{p}) u_{\vec{p}}^{(-)}(x)].$$
Now, the negative-frequency solutions have a time dependence as a annihilation term for holes in non-relativistic many-body physics, written in terms of "2nd quantization", i.e., non-relativistic QFT. To make the whole expression manifestly covariant we thus reinterpret the term with negative frequencies as a creation contribution to the field operator of particles of another kind running in opposite direction, i.e., we set
$$\hat{a}_-(\vec{p})=\hat{b}^{\dagger}(-\vec{p}).$$
Substituting then $$\vec{p} \rightarrow -\vec{p}$$ in this contribution to the field operator and also setting
$$\hat{a}_+(\vec{p})=\hat{a}(\vec{p}),$$
we get
$$\hat{\phi}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} [\hat{a}(\vec{p}) u_{\vec{p}}^{(+)}(x)+\hat{b}^{\dagger}(\vec{p}) [u_{\vec{p}}^{(+)}(x)]^*].$$
It turns out that the operator of total energy, i.e., the Hamiltonian, is plagued by operator-ordering problems. One has to subtract a c-number valued diverging expression, which leads to the normal-ordering prescription. Thanks to the bosonic commutator relations we started with, this leads to a positive definite energy. The commutator relations for the annihilation and creation operators read
$$[\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p}')]=[\hat{b}(\vec{p}),\hat{b}^{\dagger}(\vec{p}')]=\delta^{(3)}(\vec{p}-\vec{p}'),$$
with all other possible combinations vanishing, and thus the normal-ordered Hamiltonian reads
$$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \omega(\vec{p}) [\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p})+\hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p})],$$
which is a positive semidefinite operator in Fock space, whose occupation-number basis is built out of the vacuum state of lowest energy by applying successively creation operators.

3. Aug 7, 2013

### andrien

You can see it when people draw Feynman diagram and in the way they draw an antiparticle on this diagram as compared to particle.

4. Aug 7, 2013

### vanhees71

The direction of arrows in Feynman diagrams is by definition in the direction of the current with a relative sign between particles and anti particles. Thus an "incoming" ("outgoing") external anti-particle line has the meaning of a final (initial) anti-particle state in the transition amplitude symbolized by the Feynman diagram.