I'm a little bit confused about how to take the massless limit of the Dirac field: \begin{align} \psi(x)=\int\frac{d^3p}{(2\pi)^2}\frac{1}{\sqrt{E_p}}\sum_{s}\left(a_p^su^s(p)e^{-i p x}+b_p^{s\dagger}v^s(p)e^{ip x}\right)\,,\label{Dirac} \end{align}

EDITED: What are the correct $u^s(p)$ and $v^s(p)$ if the mass is zero? For massive particles, Peskin & Schroeder gives the expression: \begin{align} u^s(p)=\left(\begin{array}{c} \sqrt{p\cdot \sigma}\,\xi^s\\ \sqrt{p\cdot \overline{\sigma}}\,\xi^s \end{array}\right)\,,\qquad v^s(p)=\left(\begin{array}{c} \sqrt{p\cdot \sigma}\,\eta^s\\ -\sqrt{p\cdot \overline{\sigma}}\,\eta^s \end{array}\right) \end{align} If I take the massless limit for $u^s(p)$ with $p$ pointing in the $z$-direction, I find \begin{align} u^1(\vec{p})=\sqrt{E}\left(\begin{array}{c} 0\\ 0\\ 1\\0 \end{array}\right)\,,\quad u^2(\vec{p})=\sqrt{E}\left(\begin{array}{c} 0\\ 1\\ 0\\0 \end{array}\right)\,, \end{align} where I used $\xi^1=(1,0)$ and $\xi^1=(0,1)$. If I do the same for the $v$, I find \begin{align} v^1(\vec{p})=\sqrt{E}\left(\begin{array}{c} 0\\ 0\\ -1\\0 \end{array}\right)\,,\quad v^2(\vec{p})=\sqrt{E}\left(\begin{array}{c} 0\\ 1\\ 0\\0 \end{array}\right)\,. \end{align} This means that we don't have $\overline{u}^i(p)v^j(p)=0$. I thought that for every fixed momentum $p$, the spinors sets $v^i(p)$ and $u^i(p)$ span with $i=1,2$ each span a two-dimensional space, namely $\mathrm{span}(v^1(p),v^2(p))$ and $\mathrm{span}(u^1(p),u^2(p))$ that are orthogonal to each other. However, from above limits it appears as if this were not the case anymore in the massless limit (very large momentum). Did I take the limits incorrectly or did I miss a crucial point?

  • 2
    $\begingroup$ I'm voting to close this question as off-topic because it shows insufficient research. These issues are discussed in practically any QFT book you pick up (I suggest Srednicki). $\endgroup$
    – Prahar
    Commented Sep 17, 2017 at 7:04
  • $\begingroup$ Maybe my question was not very well phrased. I used Peskin & Schroeder and found the discussion of the massless limit, where they explain that the spinors become degenerate. Let me rephrase my question a little bit. It's still a simple question, but not completely trivial (at least not to me right now). $\endgroup$
    – LFH
    Commented Sep 17, 2017 at 15:28
  • $\begingroup$ $u$ and $v$ never mix under proper orthochronous Lorentz transformations, irrespective of whether $p$ massless or massive. $\endgroup$
    – Prahar
    Commented Sep 17, 2017 at 16:03
  • $\begingroup$ Hi Prahar, I guess I didn't explain it super well. Clearly, for different momenta $v$ and $u$ can mix. Given $u^s(p)$ and $v^r(q)$, there might be overlap between them. For the same momenta $p$, $u^s(p)$ and $v^r(p)$ span orthogonal subspaces. However, I somehow couldn't take the massless limit correctly and it looks like as if the spaces mix in the massless limit. That's what my question was mostly about. $\endgroup$
    – LFH
    Commented Sep 18, 2017 at 15:28
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    $\begingroup$ Notice that the "spatial wave functions" corresponding to $u(\textbf{p})$ and $v(\textbf{p})$ are respectively $e^{i\textbf{p} \cdot \textbf{r}}$ and $e^{-i\textbf{p} \cdot \textbf{r}}$. Because the "spatial wave functions" for $u(\textbf{p})$ and $v(-\textbf{p})$ are the same, the corresponding spinor parts, which are $u(\textbf{p})$ and $v(-\textbf{p})$ themselves, have to be orthogonal to each other, as is indeed the case. $\endgroup$
    – higgsss
    Commented Sep 24, 2017 at 10:01

1 Answer 1


By the definition, the coefficient $u_{s}(p)$ is the spinor in front of $e^{-ipx}$, while the coefficient $v_{s}(p)$ is the spinor in front of $e^{ipx}$. There are no continuous Lorentz transformations which change the sign of energy, and therefore your statement about mixing of $u_{s}(p)$ and $v_{s}(p)$ is not clear.

From the other side, note that for any mass, the existence of the term $v_{s}(p)e^{ipx}$ is required by the causality principle: for any two operators $A(x), B(y)$ composed from the relativistic (fermionic) fields $\psi$ the mean value $[A(x),B(y)]$ is zero for space-like intervals, which is reflected in the requirement $[\psi(x),\psi^{\dagger}(y)]_{+} = 0$ for space-like $(y-x)^{2}$'s. The latter is impossible without the $v_{s}$ term.

Note also that $u_{s}(\mathbf p)$ and $v_{s}(\mathbf p)$ are related to each other because the field $\psi(x)$ must be transformed in the definite way, as representing the particle with given spin (helicity) and mass, under the Lorentz transformations. For massive case, for example, such relation reads $$ u_{s}(\mathbf p) = (-1)^{s+\frac{1}{2}}\gamma_{5}v_{-s}(\mathbf p) $$ Since the Dirac equation has correct massless limit, such relation holds in the massless case.


I don't understand why you are confused. Massless limit doesn't require $u, v$ to span different sub-spaces; this is because the antiparticle obviously doesn't disappear with $m\to 0$. Rather the different sub-spaces are spanned by $u_{s}, u_{-s}$ and $v_{s}, v_{-s}$ correspondingly, which is simply the statement that the Dirac equation is splitted on two Weyl equations $\sigma^{\mu}p_{\mu}\psi(p) = 0$ and $\tilde{\sigma}^{\mu}p_{\mu}\psi(p) = 0$.

  • $\begingroup$ Thank you for your answer. I completely agree with you on the level of the full solution containing the wavefunction part. When I was talking about spinor, I was only talking about the 4-component spinor. I now understand that $v$ and $u$ don't mix for the same momenta (in other words: the $u(p)$ and $v(p)$ subspaces are invariantly orthogonal under the action of the little group associated to $p$). I also clarified in my question what I'm still confused about... $\endgroup$
    – LFH
    Commented Sep 18, 2017 at 15:30
  • $\begingroup$ @LFH : I've updated my answer. $\endgroup$
    – Name YYY
    Commented Sep 18, 2017 at 19:09
  • $\begingroup$ What do you mean with $u_{-s}$? I assume you mean that $u^s(\vec{p})$ always spans an orthogonal subspace to $v^s(-\vec{p})?$? I suppose that's what I mixed up, because for my work I focused on massive spinors with $\vec{p}=0$, so there it didn't matter. I think this clarifies it for me. Thank you! $\endgroup$
    – LFH
    Commented Sep 18, 2017 at 22:01
  • $\begingroup$ @LFH : I meaned that $u_{1/2}(p)$ and $u_{-1/2}(p)$ are from different sub-spaces that aren't mixed in the massless case. But not $u_{s},v_{s}$. $\endgroup$
    – Name YYY
    Commented Sep 19, 2017 at 6:03
  • $\begingroup$ I'm not sure if I understand your clarification. What do you mean by $u_{-1/2}(p)$. What I understood so far: For a given 3-momentum $\vec{p}$, there are four basis vectors that span the 4-dimensional spinor space: namely, first $u_s(\vec{p})$ for $s=1,2$ and $v_s(-\vec{p})$ for $s=1,2$. Moreover, these span two 2-dimensional subspaces that are orthogonal to each other. I don't know what you mean if an index running over 1,2 gets a minus sign. $\endgroup$
    – LFH
    Commented Sep 20, 2017 at 4:04

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