After quantizing the Dirac field, we have:

$$\psi(x) = \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_p}} \sum_s a^s_p u^s (p) e^{-ip\cdot x} + b^{s \dagger}_p v^s (p) e^{ip\cdot x}$$ $$\bar{\psi}(x) = \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_p}} \sum_s b^s_p \bar{v}^s (p) e^{-ip\cdot x} + a^{s \dagger}_p \bar{u}^s (p) e^{ip\cdot x}.$$

What is the relationship between the Dirac field, and the single-electron state? Is a single electron state of definite momenta and spin given by: $$|\vec{p}_{e^{-}} , s\rangle = \sqrt{2 E_p}a^{s \dagger}_p | 0 \rangle$$ or instead by: $$|\vec{p}_{e^-} , s\rangle = \sqrt{2 E_p}a^{s \dagger}_p u^s (p)| 0 \rangle?$$ In other words, does the $a^\dagger$ "create" the spinor part already?

For the real Klein Gordon field, we have that a particle in a state of definite position $x$ is given by: $$\phi(\vec{x}) |0 \rangle = \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{2 E_p}e^{-ip\cdot x} |p\rangle. $$ Does the analogy hold for the Dirac field? In other words, would an anti- electron in a state of definite position $x$ be: $$\psi(x) |0 \rangle = \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{2 E_p}\sum_s e^{ip\cdot x}|\vec{p}_{e^+},s \rangle$$

In general, what are the single electron/positron states, and how is the spinor part treated in the labeling?


1 Answer 1


The single particle states are really in the space spanned by $|\mathbf{p},s\rangle = a^\dagger(\mathbf{p},s)|0\rangle$. This is, perhaps, most transparent in Weinberg's presentation in which one actually starts from relativistic particle states and only later talks about fields.

Single particle states are, by definition, states in a Hilbert space carrying one unitary irreducible representation of the universal cover of the Poincaré group. The possible Hilbert spaces are determined by Wigner's classification, explained in Weinberg's "The Quantum Theory of Fields", Chapter 2. For the electron one would pick the irreducible representation with mass $m = m_e$ and spin $j = \frac{1}{2}$ which is spanned by the states $|\mathbf{p},s\rangle$ where $s=\pm \frac{1}{2}$. Starting from such single particle Hilbert space one can then carry out the Fock space construction and build a Hilbert space whose states are all possible multiparticle states. One then introduces $a^\dagger(\mathbf{p},s)$ and $a(\mathbf{p},s)$.

In this picture it is made crystal clear that the field is constructed out of $a^\dagger(\mathbf{p},s)$ and $a(\mathbf{p},s)$ later to facilitate the construction of one interaction Hamiltonian which leads to a Lorentz invariant theory satisfying the cluster decomposition principle.

Observe that in this description of the states we don't have the spinors. The spin is captured by the label $s=\pm \frac{1}{2}$ of the basis states.

It turns out, though, that there is one alternative and equivalent characterization of the states, in which the spin label is actually traded for the spinors. This is one "position space" description of the states, but it is important to understand that it is different from the "position space" we have in non-relativistic QM. In QFT we do not have a position observable and therefore we don't have eigenstates of position operators.

The way in which this second representation arises is when we study LSZ reduction. In that case we find, for example, that the operator $a^\dagger_{\rm in}(\mathbf{p},s)$ creating the incoming state $|\mathbf{p},s\rangle_{\rm in}$, can be obtained by taking the Dirac inner product between the field operator $\psi(x)$ and the "position space wavefunction" $\psi_{s,p}(x)=u_s(p)e^{-ip\cdot x}$.

These "position space wavefunctions" are actually positive frequency solutions to the field equation and can be seem as one alternative representation of the single-particle states. In these ones the spinors appear. So in a sense we have a correspondence $$\psi_{s,p}(x)= u_s(p)e^{-ip\cdot x} \sim |\mathbf{p},s\rangle$$ but you should not mix the two things. I also urge you to observe that $\psi_{s,p}(x)$ is not a probability amplitude to find a particle at $x$, but rather the wavefunction such that the Dirac inner product of the field operator $\psi(x)$ with it recovers the creation operator for the state.

For more on these kinds of wavefunctions see this Phys.SE thread " Why is it useful to characterize relativistic particle states by this kind of wavefunction? ". I also encourage you to see the discussion by Itzykson & Zuber on LSZ reduction.


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