Free Dirac Spinors

Free Dirac solutions have two spin states in each frequency sector. Their explicit components are useful for currents and scattering, but normalization and momentum labels matter as much as the component formula. Covariant spin sums and equal-time Hilbert orthogonality use different conjugations and, for the negative-frequency sector, different momentum labels.

Required background. The Covariant Dirac Equation gives the algebraic constraints; Gamma-Matrix Conventions fixes the Dirac basis and adjoint. Helpful background. The Dirac Hamiltonian as an Operator constructs the energy projectors.

Solving the two block equations

Use natural units $\hbar=c=1$, initially $m>0$, and $p^\mu=(E_{\mathbf p},\mathbf p)$ with $E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}>0$. The two mode conventions are

$$\begin{aligned} \psi_+(x)&=u_s(p)e^{-ip\cdot x},\\ \psi_-(x)&=v_s(p)e^{+ip\cdot x}. \end{aligned}$$

Thus $u$ obeys $(p\!\!\!/-m)u=0$, while $v$ obeys $(p\!\!\!/+m)v=0$. The $v$ mode has eigenvalues $-E_{\mathbf p}$ and $-\mathbf p$ under $i\partial_t$ and $-i\nabla$. The label $p$ in $v(p)$ remains future directed; it is not the mode’s canonical four-momentum.

For $u=(\xi,\zeta)^T$, the Dirac-basis equations are

$$\begin{aligned} (E-m)\xi&=(\boldsymbol\sigma\cdot\mathbf p)\zeta,\\ (E+m)\zeta&=(\boldsymbol\sigma\cdot\mathbf p)\xi. \end{aligned}$$

Choose orthonormal two-spinors $\chi_s$ and, independently, $\eta_s$, with $s=1,2$. A convenient covariant normalization is

$$u_s(p)=\sqrt{E+m} \begin{pmatrix} \chi_s\\ \dfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}\chi_s \end{pmatrix},$$ $$v_s(p)=\sqrt{E+m} \begin{pmatrix} \dfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}\eta_s\\ \eta_s \end{pmatrix}.$$

The Pauli identity $(\boldsymbol\sigma\cdot\mathbf p)^2=\mathbf p^2I$ and $E^2-m^2=\mathbf p^2$ verify both equations directly. At rest $u_s=\sqrt{2m}(\chi_s,0)^T$ and $v_s=\sqrt{2m}(0,\eta_s)^T$. Choosing fixed rest-spin labels and boosting them is one useful convention; helicity labels instead depend on the direction of momentum. No specific charge-conjugation phase relation between $\chi_s$ and $\eta_s$ is required for these results.

Two kinds of normalization

Direct block multiplication gives

$$\begin{aligned} \bar u_r(p)u_s(p)&=2m\delta_{rs},\\ \bar v_r(p)v_s(p)&=-2m\delta_{rs},\\ u_r(p)^\dagger u_s(p)&=2E\delta_{rs},\\ v_r(p)^\dagger v_s(p)&=2E\delta_{rs}. \end{aligned}$$

For example, $u^\dagger u=(E+m)[1+\mathbf p^2/(E+m)^2]=2E$, whereas the minus sign from $\gamma^0$ gives $\bar uu=(E+m)[1-\mathbf p^2/(E+m)^2]=2m$. The negative sign of $\bar vv$ is not a negative probability: the Hilbert density is $v^\dagger v>0$.

For equal future-directed labels, $\bar u_r(p)v_s(p)=0$. However,

$$u_r(p)^\dagger v_s(p)= 2\chi_r^\dagger(\boldsymbol\sigma\cdot\mathbf p)\eta_s$$

need not vanish. The two spinors do not belong to opposite eigenspaces of the same momentum-space Hamiltonian: $v(p)$ belongs to the negative sector at canonical momentum $-\mathbf p$. The correct fixed-momentum orthogonality is

$$u_r(p)^\dagger v_s(-\mathbf p)=0.$$

Here $v_s(-\mathbf p)$ means $v_s(E_{\mathbf p},-\mathbf p)$. This distinction is essential when normalizing Fourier wave packets.

Spin sums and completeness

Summing over the two spin states uses $\sum_s\chi_s\chi_s^\dagger=\sum_s\eta_s\eta_s^\dagger=I_2$. The $u$ sum is $\sum_su_s(p)\bar u_s(p)=p\!\!\!/+m$, whose explicit blocks are

$$p\!\!\!/+m= \begin{pmatrix} (E+m)I_2&-\boldsymbol\sigma\cdot\mathbf p\\ \boldsymbol\sigma\cdot\mathbf p&-(E-m)I_2 \end{pmatrix}.$$

Similarly,

$$\sum_sv_s(p)\bar v_s(p)=p\!\!\!/-m.$$

These are basis-independent identities once the stated spinor normalization is fixed. They are not themselves unit-normalized orthogonal Hilbert projectors. Multiplying by $\gamma^0$ and dividing by $2E$ gives

$$\begin{aligned} \frac1{2E}\sum_su_s(p)u_s(p)^\dagger&=P_+(\mathbf p),\\ \frac1{2E}\sum_sv_s(-\mathbf p)v_s(-\mathbf p)^\dagger&=P_-(\mathbf p). \end{aligned}$$

Their sum is $I_4$. The scalar denominators $2m$ in covariant normalization and $2E$ in equal-time projectors have different roles.

Packets and the massless limit

With the present spinors, write a free square-integrable packet as

$$\begin{aligned} \psi(x) &=\sum_s\int\frac{d^3p}{(2\pi)^{3/2}\sqrt{2E_{\mathbf p}}} \,F_s(p,x),\\ F_s(p,x)&=a_s(\mathbf p)u_s(p)e^{-ip\cdot x}\\ &\quad+b_s(\mathbf p)v_s(p)e^{+ip\cdot x}. \end{aligned}$$

The spatial delta function in a mixed term enforces opposite labels; the orthogonality just derived then gives

$$\int d^3x\,\psi^\dagger\psi =\sum_s\int d^3p\, \left(|a_s(\mathbf p)|^2+|b_s(\mathbf p)|^2\right).$$

Both signs enter positively. These coefficients describe a c-number one-particle equation; replacing them by field operators and assigning particle statistics is a further construction.

At $m\to0$ with fixed nonzero $\mathbf p$, the component formulas and the $2E$ Hilbert norms remain finite. The scalar bilinears $\bar uu$ and $\bar vv$ vanish, so a convention that divides spinors or projectors by $m$ has no direct massless continuation. For positive-frequency spinors choose $(\boldsymbol\sigma\cdot\widehat{\mathbf p})\chi_\lambda =\lambda\chi_\lambda$, $\lambda=\pm1$. Then $u_\lambda=\sqrt E(\chi_\lambda,\lambda\chi_\lambda)^T$ is an eigenvector of $\gamma^5$ with eigenvalue $\lambda$ and of helicity with eigenvalue $\lambda/2$. For massive spinors this equality of chirality and twice helicity no longer holds. Negative-frequency mode labels and physical antiparticle helicity require their own conjugation conventions.

Exercises

1. For $\mathbf p=p\widehat{\mathbf z}$ and $\chi=\eta=(1,0)^T$, write $u(p)$ and $v(-\mathbf p)$ and check their Hilbert orthogonality.

Solution

They are $\sqrt{E+m}(1,0,p/(E+m),0)^T$ and $\sqrt{E+m}(-p/(E+m),0,1,0)^T$. Their scalar product is $-p+p=0$. Using $v(+\mathbf p)$ would instead give $2p$.

2. Rescale $u$ to unit Hilbert norm. What becomes of $\bar uu$ and the spin sum?

Solution

With $w=u/\sqrt{2E}$, $w^\dagger w=1$, $\bar ww=m/E$, and $\sum_sw_s\bar w_s=(p\!\!\!/+m)/(2E)$. The energy factor is frame dependent; the invariant $2m$ convention has intentionally been changed.

3. For $m>0$, take the trace of the positive-frequency spin sum. Why does it match two spin states rather than four?

Solution

$\operatorname{tr}(p\!\!\!/+m)=4m$. On the other side, $\operatorname{tr}(u_s\bar u_s)=\bar u_su_s=2m$ for each of two states. A four-component spinor constrained to one on-shell energy sector has only two independent spin amplitudes.

References

• J. D. Bjorken and S. D. Drell, Relativistic Quantum Mechanics, McGraw–Hill, 1964 — explicit free spinors; translate normalization before combining formulas.
• C. Itzykson and J.-B. Zuber, Quantum Field Theory, McGraw–Hill, 1980 — covariant spinor normalization and completeness.
• M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014, doi:10.1017/9781139540940 — spin sums and helicity.