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In Rydberg Quantum Field Theory page 441 (this edition, unfortunately page 441 is not in the link) it says

If $\xi$ and $\eta$ are Majorana spinors [...] and since $\xi$ and $\eta$ are Grassmann quantities, $$ \xi_i^*\eta_j^* = -\eta_j^*\xi_i^* $$

I know that Grassmann quantities obey the Grassmann algebra, i.e. $\{\xi_i,\eta_j\} = 0$. But why two different spinor fields obey this algebra?

As far as I understood two spinor fields $\xi$ and $\eta$ obey the anticommutation relations between each and its conjugate, $\{\xi_i(\vec{x},t),\xi_j(\vec{y},t)\}=\delta_{ij}\delta^3(\vec{x}-\vec{y})$ and the same for $\eta$. But in this case we have two different fields $\xi$ and $\eta$ which are not the conjugate of each other. So why do they have to satisfy this Grassmann algebra?

For example, in QED there are two fields $\psi$ and $A^\mu$ and each has its commutation/anticommutation relation with its own conjugate (in the case of $A^\mu$ with itself). But there are not any relations between $\psi$ (or $\bar{\psi}$) and $A^\mu$, they are different fields. I mean $[\psi,A^\mu] = \{\psi,A^\mu\} = 0$ and the same for $\bar{\psi}$.

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  • $\begingroup$ Nope, all spinor fields anticommute. Why do you think some anticommute and some don’t? $\endgroup$ – knzhou Aug 25 '18 at 22:35
  • $\begingroup$ Spinors (or, for that matter, any representation of a group) can be defined over both commuting and anti-commuting fields. Either option is fine, but you have to be consistent: it makes no sense to have $\xi$ sometimes being a $c$-number, and sometimes an $a$-number. Once you declare it to be an $a$-number, you have to stick to that choice. And $a$-numbers anti-commute with each other: both with themselves and with other $a$-numbers. $\endgroup$ – AccidentalFourierTransform Aug 25 '18 at 23:06
  • $\begingroup$ Good question. One physical reason to impose anticommutation relations on spinors is to ensure that the energy of a system is bounded from below, else vacuum/particles could keep producing matter incessantly (without violating conservation of energy) and destabilize the system. But, of course, we don't observe this in the real world. $\endgroup$ – Avantgarde Aug 25 '18 at 23:48
  • $\begingroup$ My doubt is about why two different fields, not being related between each other by any conjugation operation, have to satisfy this conditions. Its the same as in non relativistic quantum mechanics, the commutation relations are for $x$ and $p_x$ which are one the conjugate of the other, but $x$ and $p_y$ have nothing to do between each other. I thought the same about fields, $\xi$ and $\bar{\xi}$ must satisfy anticommutation conditions (because they are a conjugate pair), but why $\xi$ and $\eta$? Its the same as saying that in QED $\psi$ have to commute/anticommute with $A^\mu$... $\endgroup$ – user171780 Aug 26 '18 at 2:50
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    $\begingroup$ For your question to make sense, you have to encounter the product of "independent" spinors in real calculations. It is possible if the corresponding fermions interact, so the spinors are not really independent after all. Otherwise (never interacting fermions) both options never appear in meaningful calculations. $\endgroup$ – Vladimir Kalitvianski Aug 26 '18 at 14:49
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OP's title question (v3) reads:

Why two different spinors are Grassmann quantities?

We interpret this as asking why independent spinors anticommute? Here is one line of reasoning:

  1. A general supernumber $z=z_0+z_1$ has a Grassmann-even and a Grassmann-odd component, $z_0$ and $z_1$, respectively.

  2. $z_0$ and $z_1$ are supernumbers of definite Grassmann parity, $|z_0|=0$ and $|z_1|=1$, respectively.

  3. A Grassmann-odd supernumber $z$ with $|z|=1$ squares to zero: $z^2=0$.

  4. A sum $z+w$ of two Grassmann-odd supernumbers $z$ and $w$ is still a Grassmann-odd supernumber. Hence it squares to zero $$ 0~=~(z+w)^2~=~\underbrace{z^2}_{=0} + zw + wz +\underbrace{w^2}_{=0}~=~ zw + wz. $$ Hence two Grassmann-odd supernumbers $z$ and $w$ anticommute, cf. OP's question.

  5. More generally, two supernumbers $z$ and $w$ of definite Grassmann parity supercommute: $$[z,w]_S~:=~zw-(-)^{|z||w|}wz~=~0,$$ where $[\cdot,\cdot]_S$ denotes the supercommutator. In other words, the supercommutator $[z,w]_S=0$ is zero.

  6. When we replace the supernumbers with two superoperators $\hat{z}$ and $\hat{w}$ of definite Grassmann parity, i.e. go from classical to quantum theory, the supercommutator $$ [\hat{z},\hat{w}]_S~=~{\cal O}(\hbar)$$ can acquire non-zero terms at first (and higher) orders in Planck's constant $\hbar$.

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  • $\begingroup$ But why are spinors supernumbers? Aren't the components just complex numbers and thus commuting? $\endgroup$ – md2perpe Aug 27 '18 at 20:48
  • $\begingroup$ That's a different question, which is discussed in this Phys.SE post. $\endgroup$ – Qmechanic Aug 28 '18 at 14:26

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