Quantum field theory

1. Sep 19, 2015

shereen1

Hi all
I am studying quantum field theory and i want to just to check something. We have said that the problem with klein gordon equation for real field is that is predict positive and negative energies in addition to the negative probability density. For the complex klein gordon field we have discove the antiparticle so we can said that the problem of negative energy is solved for the complex field. So why do they say that the problem of negative energy was first solved by dirac for spin 1/2 particles.
Thank you

2. Sep 19, 2015

my2cts

3. Sep 19, 2015

vanhees71

The reason that Dirac discovered antiparticles in context of his theory for spin-1/2 particles is historical. It's an example for the fact that flawed concepts can lead to great ideas, if thought over by a genius like Dirac. Even better, Dirac's work lead to the insight that a single-particle interpretation of relativistic wave functions are a flawed concept in the first place, and that relativistic quantum theory for interacting particles is necessarily a many-body theory. Nowadays that's evident, because we produce and destroy particles in relativistic particle accelerators everyday, but that was not the case in the late 1920ies, when Dirac did this work.

Indeed, equations of motion for relativistic fields should be quite symmetric in space and time derivatives such that you can write them in a covariant way. The most simple case is the scalar field, and you want a scalar partial differential equation for it. So with the four-gradient $\partial_{\mu}$ you can just build the scalar operator of 2nd order $\Box=\partial_{\mu} \partial^{\mu}=\partial_t^2-\vec{\nabla}^2$. For free particles you want $p^2=E^2-\vec{p}^2=m^2$, and using the usual Einstein-de Broglie-heuristics to guess wave equations, you end up with the Klein-Gordon equation
$$(\Box+m^2) \phi=0.$$
Now this is troublesome, because for a general wave packet solving this free equation you need indeed the solutions of this dispersion relation $p^0=\pm E_{\vec{p}}=\pm \sqrt{m^2+\vec{p}^2}$, and this would mean that the particles described by such an equation would have both positive and negative energies, and thus the energy spectrum wouldn't be bounded from below and thus there would be no stable ground state.

In addition there was no way to find a continuity equation for a positive definite probability measure. There's a conserved current for the complex KG field,
$$j^{\mu}=\mathrm{i} [\phi^* \partial_t \phi -(\partial_t \phi^*) \phi],$$
but
$$j^0=\mathrm{i} [\phi^* \dot{\phi}-\dot{\phi*} \phi],$$
is not positive definite and thus cannot be taken as a probability density in the sense of a single-particle interpretation of the wave function as with the nonrelativistic Schrödinger equation, where the corresponding "charge density" of the conserved current is simply $|\psi|^2$, which is positive semidefinite and can be interpreted as probability density for the position of the particle.

Of course, this was due to the 2nd-order nature of the equation. Dirac's idea was inspired by this success of the non-relativistic Schrödinger equation which is in first order of the time derivative. So he looked for a first-order equation and found his famous equation with the Dirac matrices and all that. So he thought, he's done and has solved everything, because here indeed the conserved current is $j^{\mu} =\overline{\Psi} \gamma^{\mu} \Psi$ with the "charge density" $j^0=\Psi^{\dagger} \Psi$ positive semidefinite. So at the first glance you could interpret this as a probability density for position. Still there were negative-energy states, but now Dirac immediately figured out that his wave equation describes particles with spin 1/2, and the electron was a good candidate for that. So he dealt with fermions, and thus his idea was to consider as the ground state of the system a state, where all negative-energy states are occupied, the socalled Dirac sea. Then the excitations are either particles with positive energy (electrons) or holes in the Dirac sea. Since holes are the absence of any property it means that a hole is a state with positive energy, because it's due to the excitation of a particle (electron) occupying a state with negative energy to positive energy. So at the end you have an electron and a hole in a Dirac sea, but the hole know can be interpreted as a particle with positive energy (the absence of negative energy is equivalent to the presence of positive energy). Now it's also absence of negative charge (electrons are negatively charged by convention), i.e., what Dirac had found out was that he could make sense of his model when assuming that automatically you also describe particles with the same properties as an electron but with positive charge. After some wrong idea to interpret these positively charged particles as protons and a corresponding hint by Oppenheimer, who showed that the holes must be interpreted as particles with the same mass as an electron, the antiparticles of electrons, dubbed positron, had to be predicted. They were, of course, found later, and this was a big breakthrough.

Although you can build a consistent relativistic quantum theory in terms of this hole theory (in fact Dirac's hole theory with the electromagnetic interaction is equivalent to modern QED), it's also quite evident that the very concept of a wave function is flawed, because if you interpret the wave function in the single-particle sense you are forced automatically to reinterpret it in terms of a many-body theory with non-conserved particle numbers, and as we know nowadays, this situation is much better attacked in terms of a quantum field theory (it was in fact also Dirac who discovered this in 1927 when introducing annihilation and creation operators to describe photon emission and absorption; quantum field theory was also formulated by Jordan an Pauli in 1928.

In quantum field theory there is also no problem with bosons, be they charged (complex field or non-selfadjoint field operator) or uncharged (real field or selfadjoint field operator). The problem with bosons is that you cannot argue anymore that you simply occupy the unwanted negative-particle states and reinterpret the holes as antiparticles, because arbitrarily many bosons can occupy the ground state, but with negative-energy states the free-particle Hamiltonian would be unbounded and thus there was no stable ground state, implying that there could be no stable bosonic state at all. This contradicts of course observation, because nowadays we know that there are bosons (also scalar bosons like the pions) in nature. So QFT is really the much more convincing concept. The trick is mathematical and even a bit inspired by Dirac's hole theory. It's due to Stueckelberg and (most probably independently) to Feynman: While the positive frequency modes can be used to write down the mode decomposition of the field operators (it's important that we have field operators rather than c-number fields now) with annihilation operators as "amplitudes" (as known from non-relativistic QFT based on the Schrödinger or Pauli equation), for the negative-frequency mode you have to write down a creation operator. Then you only have single-particle states with positive energies in the game.

You need the creation operators in this field-operator decomposition in order to build quantum fields that behave under Lorentz transformations in the local way as do classical fields, and you can easily build locally interacting particle theories with it (in terms of local Lagrangians when formulating the whole theory based on the action principle, which is the most natural way to study its symmetries and to get the equal-time commutator relations within the heuristics of canonical field quantization). Now you also have to demand that the theory, which now describes many-body theory based on the usual occupation-number representation (Fock states), has a Hamiltonian that is bounded from below, and then a very remarkable feature occurs: You are forced to quantize the scalar fields as bosons and the Dirac field, which describes spin-1/2 particles, as fermions. Then and only then your energy-density and thus the total energy (Hamiltonian) is positive semidefinite. At the same time the quantized conserved current $\hat{j}^{\mu} = :\overline{\hat{\psi}} \gamma^{\mu} \hat{\psi}:$ is not the quantum version of some probability current but the quantized version of a true charge-like current. For free particles in terms of the particle-number operator the corresponding charge operator reads
$$\hat{Q}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \sum_{\sigma=\pm 1/2} [\hat{N}_a(\vec{p},\sigma)-\hat{N}_b(\vec{p},\sigma)],$$
where $a$ and $b$ denote the particle and the antiparticle states, respectively. That means that the particle and antiparticle carry opposite charges, but the same positive energy $E=\sqrt{m^2+\vec{p}^2}$. This doesn't matter, of course, because now the fields are no longer wave functions, and you don't need to interpret them as probability amplitudes anymore. Rather they are building blocks to define the operators representing observables in the Hilbert space of states, which is now constructed as the corresponding Fock space, and this has the appropriate Hilbert-space structure.