I read that the state of a pair of particles is the tensor product of the single states of both, and you will get a wavefunction with the parameters of both, if you swap the parameters you will get a change of sign on the function if they are fermions. First, is that correct?

And now, can i do the same thing with two solutions of the Dirac equation? I mean make the tensor product, and get a 2-rank spinor (i don't know what's the formal name) where the components are the product of the components of each spinor and make a conjugate/traspose swaping the indices like the other function named before, and get a change of sign? Will it be 0 if the two solutions are the same?

Or how i do it? What's the relation between the two aproaches?

  • 1
    $\begingroup$ Another user suggests A Quantum Theory of Bi-Spinor Fields as a useful reference. $\endgroup$
    – rob
    Mar 29, 2018 at 0:15
  • $\begingroup$ The first paragraph is correct; solutions of the Dirac equation are generally called 4-component spinors; these are complex solutions; with a reality condition they are called Majorana-Weyl spinors; and 2-component spinors are otherwise known as Weyl spinors; mathematically, they are representations of double covers of the SO(n) or just representations of Spin(n). $\endgroup$ May 2, 2018 at 22:12

2 Answers 2


The first paragraph as people have pointed out in the comments, is indeed correct.

About the second one, the short answer is no you can't do the same, but more over the proper statement is you don't need to do the same. Within the framework of field theory, in contrast to classical quantum mechanics, multi-particle states are already taken into account. That is the construction of the Fock space (see Second quantization) is done by acting with creation operators on the vacuum. Acting with creation operators labeled by different momenta or coming from different fields, leads to states that represent multi-particle states, so there is no need to take tensor products of spinors.

Nonetheless, fermionic fields (solutions to Dirac's equation) are subject to canonical anti-commutation relations (as opposed to the bosonic case where these become commutation relations): $$\{\psi_a(\vec{x}),\psi^\dagger_b(\vec{y}) \}\big|_{t_x=t_y} = \delta(\vec{x}-\vec{y})$$ which induce the correct commutation relations on the creation and annihilation operators $a,a^\dagger$ of the usual mode decomposition.

So the difference of both approaches in the end is what the fundamental object of study is under each description. In quantum mechanics wave-functions describe a single particle, in quantum field theory, fields describe the collective interactions of particles.

More over, in quantum field theory one must also respect Lorentz symmetry among other possible symmetries of a theory. So any interesting object must be a Lorentz invariant object. Additionally one imposes also locality so for operators within the Lagrangian of your theory, there is the requirement of depending on a single space-time point.

One could consider a two point function, or correlation function, this one is actually of interest within QFT: $$\langle \psi^\dagger(x) \psi(y) \rangle$$ Which is just the propagator of the field $\psi$ in a free theory and tells you how the field evolves between the space-time points $x$ and $y$ in the absence of interactions.


Looking at the expression of a quantized Dirac field $$ \psi = \sum_s \int a_s(p) u_s(p) e^{-ipx} + b_s^{\dagger}(p) v_s(p) e^{ipx} \frac{d^3 p}{(2\pi)^2\sqrt{2E}} , $$ one finds that it contains annihilation and creation operators $a_s(p)$ and $b_s^{\dagger}(p)$, as well as spinors $u_s(p)$ and $v_s(p)$. The anticommuting properties of the field is provided by the annihilation and creation operators, where is the solution of the Dirac equaltions are provided by the spinors. As a result, the spinors on their own won't give you the anticommuting properties.


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