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I'm perplexed by the following non numbered equation at page 54 of Peskin & Schroeder, right between $(3.92)$ and $(3.93)$

$$ [\psi_a(x),\overline{\psi}_b(x)]=\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}\sum_s\left(u_a^s(p) \overline{u}_b^s(p)e^{-ip\cdot(x-y)}+v_a^s(p) \overline{v}_b^s(p)e^{ip\cdot(x-y)}\right)=\\=\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}\sum_s\left((\not p+m)_{ab}e^{-ip\cdot(x-y)}+(\not p-m)_{ab}e^{ip\cdot(x-y)}\right)=\\=(i\not\partial_x+m)_{ab}\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}\left(e^{-ip\cdot(x-y)}-e^{ip\cdot(x-y)}\right)=\\=(i\not\partial_x+m)_{ab}[\phi(x),\phi(y)]$$

Where $\psi_a$ are Dirac fields. I have questions both about the notation and the actual content.

  1. What is meant by the subscript $ab$ on operators? E.g. $(\not p +m)_{ab}.$ On the fields I interpreted it as if we had multiple fields with the same Lagrangian, i.e. a total Lagrangian density given by $$\mathcal{L}=\sum_a \overline{\psi}_a(i\not \partial -m)\psi_a$$ but the two subscripts don't make sense to me written that way.
  2. Looking at this computation, it seems like $$u_a^s(p) \overline{u}_b^s(p)=(\not p+m)_{ab}$$ (whatever that means) and similar for the antiparticles, but I would've expected $$ u_a^s(p) \overline{u}_b^s(p)=2m\delta_{ab}.$$ How can I resolve this?
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(1) Recall that the written out form of

$${\not p } + m = \gamma^\mu p_\mu + m 1_4$$

and so

$$({\not p } + m )_{ab} = (\gamma^\mu p_\mu + m 1_4)_{ab} $$

is just denoting the $ab$ entry of the 4$\times$ 4 matrix.

(2)

$$u_a^s(p) \overline{u}_b^s(p)=(\not p+m)_{ab}$$

Not quite. Rather,

$$\sum_s u_a^s(p) \overline{u}_b^s(p)=(\not p+m)_{ab}.$$

This is shown on page 34 of Fun with Spinor indices. Your concern about thinking that $"u_a^s(p) \overline{u}_b^s(p)=2m\delta_{ab}"$ is also addressed on that page.

Edit So in short $ab$ are, in general, spinor indices meaning that $\psi_a$ is the $a^{th}$ component of the dirac spinor $\psi$, and, say, $\bar{\psi}_a\gamma^\mu_{ab}\psi_b = \bar{\psi} \gamma^\mu \psi$ where the latter is found by usual matrix multiplication. That is, $ab$ represent the components of a matrix in spinor space.

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  • $\begingroup$ Thanks a lot! For (2), all clear, I was stupid. For (1), I gather that my supposition about the meaning of the indices $a$ and $b$ for the field was incorrect, and it means the $a$ or $b$ component of the spinor, correct? $\endgroup$ – user2723984 Jan 3 at 20:53
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    $\begingroup$ Yes. Please see my edit and let me know if that answered your question. $\endgroup$ – InertialObserver Jan 3 at 21:00

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