# Dirac equation in QFT vs relativistic QM

How does the Dirac equation in quantum field theory solve the existing problems in the interpretation Dirac equation (as a single-particle wave equation) in relativistic quantum mechanics?

EDIT: The question was not clear. So I ask again. Why is the Dirac equation not meaningful as a single particle wave equation? Does many-particle interpretation of Dirac equation make it meaningful? If yes, how?

• Specifically which problem are you asking about? This question is quite vague. Mar 1, 2014 at 1:52
• @user1504- I have edited the question.
– SRS
Mar 1, 2014 at 4:34

Your problem is not related only with the Dirac equation, but with whatever field equation you have. Your question is about the interpretation of the field as a wave function of one particle. There are two concepts that I would like to show here that might answer your question: fields and wave functions (they are different things).

• A field is an object $$\phi_{\alpha} (x)$$, where $$x$$ behaves as a event in space time and $$\alpha$$ is an index that carry a finite representation of Lorentz Symmetry. They obey some equations that related the $$\alpha$$ indices and the $$x$$, but this is irrelevant here. What is relevant is that they also obey the Klein Gordon equation, which is given by $$\partial^2\phi_{\alpha}= \pm m^2\phi_{\alpha},$$ for the metric signature ($$\pm,\mp, \cdots,\mp$$).

• A wave function is a representation of a quantum state in a particular bases. $$\psi (x)=\langle x|\psi\rangle$$ where $$|x\rangle$$ behave as a event (in non-relativistic quantum mechanics this is $$(t,\vec {x})$$). Turns out that this state $$|x\rangle$$ in relativistic quantum mechanics is: $$|x\rangle=\int\frac{d^3p}{(2\pi)^{3/2}}\sqrt{\frac{k^0}{p^0}}e^{-ipx}|p\rangle$$ You can prove yourself as an exercise. Note that there is a square root, which prevents this integral to be a simple Fourier transform of $$|p\rangle$$, the eigenvector of P (generator of translations). The set of states $$|x\rangle$$ does not constitute an orthogonal basis since $$\langle x|y \rangle \neq 0$$ and the square root in the formula above is the responsible for that. So, if we work with a wave function $$\langle x|\psi\rangle$$ in a relativistic QM we are dealing with am over-complete basis representation.

Now if we upgrade our language and move to the second-quantization framework we have $$|x\rangle=a^{\dagger}_x|0\rangle$$ Defining linear combinations of creation and annihilation operators of the form: $$\phi(x)=Aa_x+Ba_x^{\dagger}$$ such that $$[\phi(x),\phi^{\dagger}(y)]= 0$$ at space like separations is a useful construction to model local interactions of relativistic QM. This leads to the Quantum Field Theory enterprise.

The field is not interpreted as a wave function but as an operator $\hat{\psi}$ which creates/annihilates particles. This quantisation procedure is done by expanding the field in its Fourier components which depend on the momentum, spin, etc, $a_s(p)$, $a_s^\dagger(p)$, $b_s(p)$, $b_s^\dagger(p)$ (the fact that these $\hat{\psi}$ operators are not self conjugate requires the introduction of these two kinds) and interpreting them as creation and annihilation operators of particles/antiparticles with such momentum & spin. Then after imposing the right (anti)commutation relations for this operators and defining the vacuum state, $|0\rangle$ as the state annihilated by al the $a$s, we build the single and multi particle states as the $a^\dagger$s acting on the vacuum, just as we do with the simple harmonic oscillator. For instance $$|p_1,s_1;p_2,s_2\rangle = a_{s_1}^\dagger(p_1) a_{s_2}^\dagger(p_2) |0\rangle$$ (I'm ignoring normalisation in case someone asks) And the same with the $b^\dagger$s for antiparticles.

Now the Dirac equation gives the evolution of such operators and the Hamiltonian itself, being a combination of this operators, acting on those states give positive energies.

In case you wonder states of definite position are constructed with the field operators acting on the vacuum, so for instance the wave function of a definite momentum and spin state would be something like

$$\psi(x) = \langle x | p,s\rangle = \langle 0 |\hat{\psi}\,a_s^\dagger(p)|0\rangle$$ again ignoring normalisation.

• I dropped the hats on the $a$s and $b$s to make the notation lighter but after quantisation they are operators.
– jpm
Feb 28, 2014 at 18:48
– SRS
Feb 28, 2014 at 19:59
• Ok, I think I got it know. You were not asking about the negative energy problem but about the many particles interpretation. Then the point is that the field being a quantum dynamical entity in itself expanded infinite number of creation/annihilation operators allows the construction of the Hilbert Space in terms of those.
– jpm
Feb 28, 2014 at 20:08

As showed by Dirac and others QFT is a disjoint theory of RQM; as a consequence, QFT cannot solve the problems of RQM. You must want to revise the section "8.3 Does QFT solve the problems of relativistic QM?" of the Phys. Found. paper Quantum mechanics: Myths and facts. This is the conclusion:

Thus, instead of saying that QFT solves the problems of relativistic QM, it is more honest to say that it merely sweeps them under the carpet.

There is some little historical reference.

The Dirac equation has two linearly solutions which are the eigenstates of Dirac hamiltonian: first refers to the positive values of energy while the second refers to the negative ones. In classical field theory we can violate this problem by setting the trivial initial conditions $\psi (\mathbf r , t)$ for negative values, but in quantum theory it doesn't predict the absense of "negative" states, because there is possible discrete transitions between "positive" and "negative" states. The problem of negative energies is very important, because the absense of restrictions on the negative energy states promotes things like perpetuum mobile. It is experimentally unobserved.

Dirac phenomenologically solved this problem by adding the Dirac sea conceprion to the free field theory. This gives the infinite number of bonded particles with negative energies, which makes possible the absence of free solutions with negative energies due to Pauli principle. But simultaneously it determines the processes of annihilating or processing particles. For example, if there is the idle state in the vacuum with negative energy $-E$, the free electron with energy $2E$ can take this state with emission of energy $2E$ (it is equal to annihilation of particles). So there is possible of decreasing of total number and charge of free particles.

But there's nothing critical for interpreting the Dirac theory as one-particle, because the bonded electrons can't interact between each other: interaction means transition from one bonded state to another one, but this is impossible because of the Pauli principle. So in the case of free Dirac particles this infinite number is unobservable due to isotropy and homogeneity of the space-time, and even after assuming the Dirac sea conception you can still interpret the Dirac equation as single-particle one. But lets introduce some external field. It can interact with the Dirac sea and pull out the particles from it. So the Dirac theory stops being one-particle one when the external field may give ene energy more than $2mc^2$.

All of these thinking leaded to the quantum field theory. All the ideas and problems which I described are still present in quantization formalizm (infinite vacuum energy, creation and destruction operators, scattering processes with pairs creation in an external field etc.), but it is important to distinguish the reasons and consequences. In our world there aren't free particles with negative energies - this is a reason why we introduce the quantum field theory.