So, I'm trying to read the section on superspace from the book on string theory by Becker, Becker and Schwarz, and I realized that I've been stuck on something simple for a while. Some relevant equations are: $$Y^\mu(\sigma,\theta)=X^\mu(\sigma)+\bar\theta\psi^\mu(\sigma)+\frac{1}{2}\bar\theta\theta B^\mu(\sigma)\tag{4.19}$$ $$Q_A=\frac{\partial}{\partial\bar\theta_A}-(\rho^\alpha\theta)_A\partial_\alpha\tag{4.20}$$ Here $Y$ is a superfield, $Q$ the SUSY generator, $\theta$ a Grassmann spinor, and $\{\rho^\alpha\}$ the two-dimensional Dirac matrices.

  1. The book defines the supercharge $Q_A$ in equation 4.20 and goes on to state $$\delta\theta_A = [\bar{\epsilon}Q, \theta^A] = \epsilon^A. \tag{4.21}$$ $$\delta\sigma^\alpha = [\bar{\epsilon}Q, \sigma^\alpha] = -\bar{\epsilon}\rho^\alpha\theta = \bar{\theta}\rho^\alpha \epsilon. \tag{4.22} $$ Are these definitions, or can they be justified using equation 4.20? I assumed the latter and tried to "derive" them by sticking a test function of the worldsheet supercoordinates, but I got stuck because of the second term in the commutator. Why does it vanish?

  2. Also, why is the following a superfield transformation given by $$\delta Y^\mu = [\bar\epsilon Q, Y^\mu] = \bar{\epsilon}QY^\mu. \tag{4.23} $$ Specifically, the commutator also contains a second term, but somehow dropping it still gives the correct answer.

  3. Finally, if one plugs in the general expression for the superfield which is equation 4.19 of the book, one does recover the correct worldsheet supersymmetry transformations, provided one takes the derivative of $\bar{\theta}\theta$ with respect to $\bar{\theta}$ to be -2. How does one justify that? The transformations are: $$\delta X^\mu=\bar\epsilon \psi^\mu\tag{4.25}$$ $$\delta\psi^\mu=\rho^\alpha\partial_\alpha X^\mu\epsilon+B^\mu\epsilon\tag{4.26}$$ $$\delta B^\mu=\bar\epsilon\rho^\alpha\partial_\alpha\psi^\mu\tag{4.27}$$

  • $\begingroup$ Check out exercise 4.4 on p. 117. $\endgroup$
    – Ryan Unger
    May 1, 2015 at 9:58
  • 1
    $\begingroup$ I already did...that's why I posed this question. Also the solution pretty much states this without explanation. $\endgroup$ May 1, 2015 at 10:20

1 Answer 1

  1. The variation $\delta F$ for any field (or degree of freedom) $F$, given an infinitesimal transformation, is always calculated as the commutator $$ \delta F = [ \bar\epsilon Q, F ] $$ where $\bar \epsilon$ is a parameter ("angle" or "shift" or some generalization) of the transformation and $Q$ is the generator. (Those may be replaced by other letters.)

This is the usual Lie-algebra-based way how operators transform. The finite (but very close to identity) transformation may be said to be $$ U = \exp(\bar\epsilon Q) = 1 + \bar\epsilon Q + o(\epsilon) $$ and the difference of the conjugated $F$ from the original one is the variation $$\delta F = U F U^{-1} - F $$ So these totally general rules that are already taught in undergraduate quantum mechanics etc. are just applied to the generator $Q$, the infinitesimal "superangle" $\bar\epsilon$, and operators like $\theta^A$, $\sigma$, and $Y$...

Note that the product $\bar\epsilon$ is "bosonic", so its commutators, and not anticommutators, enter the formulae. However, they may be decomposed to anticommutators.

This explains the first "equation" in (4.21) and (4.22). The following ones are the actual calculations, using (4.20). The second term in $Q$ according to (4.20), one which contrains $\partial_\alpha$, doesn't contribute anything to (4.21) because $\theta^A$ and $\sigma^\alpha$ are independent coordinates of the superspace (super world sheet), so the partial derivative of one with respect to the other vanishes.

Analogously, the first term vanishes and only the second term contributes in (4.22).

  1. In (4.23), the expression $QY^\mu$ simply means the same as $[Q,Y^\mu]$: it is the differential operators in $Q$, with all the right coefficients, acting on $Y^\mu$. It is similar to differentiating functions of positions in ordinary quantum mechanics. Imagine that you have a function $V(x)$ of the operator $x$. Then you may write $V'(x)$, a different, differentiated function of the same operator $x$, as $i/\hbar$ times $[p,V(x)]$. The commutator of $p$ (the $x$-derivative) with the operator does the differentiating of the functions. On state vectors, derivatives act simply from the left, but the analogous action on the operators has to be written as commutators.

The other term $[\bar\epsilon,Y^\mu]=$ doesn't contribute, it is zero, because $\bar\epsilon$ is a (Grassmannian but still) $c$-number. So this analogous is zero much like the commutator $[5,x]$ in quantum mechanics.

  1. The derivative of $\bar\theta^A \theta$ with respect to $\bar\theta^A$ is $+\theta$, like division, and one may get a factor of $2$ there is a sum over $A$. It is, up to possible signs, the same claim as that the $x$-derivative of $xy$ is $y$. You must have missed some prefactors $\theta$ in some terms when you decided about the incorrect result of the derivative.

The name of the male co-author is John Schwarz, not Schwartz.

  • $\begingroup$ Thank you for the very detailed response Luboš. Apologies fro the typos in my original post, which I think I have fixed (some of which can be attributed to text completion I overlooked, on a friend's computer). I didn't have internet access for the past several hours, so I couldn't respond sooner to the requests from other users about including all the equations explicitly. $\endgroup$ May 2, 2015 at 9:09
  • $\begingroup$ I just have one more query, regarding point 3: the derivative of $\bar{\theta}^A \theta$ is $+\theta$. But when the authors write a term like $\bar{\theta}\theta$, do I assume they mean $\bar{\theta}^A \theta_A$, since $\theta$ is a Majorana spinor, or simply $\bar{\theta} \theta$, a formal product of two Grassmann numbers? I think my confusion about the factor of 2 stems from a misunderstanding of the notation. Thank you again for the detailed responses to the other questions; the analogy between [Q, Y] and [p, V(x)] in quantum mechanics was very helpful. $\endgroup$ May 2, 2015 at 9:16
  • $\begingroup$ Dear @leastaction, thanks for reading. Concerning $\bar\theta\theta$, I believe that there is a mistake in your formula. It seems like a chiral superfield that only depends on $\bar \theta$, look at the $\psi$ term, so that should be written as an argument on the left hand side and the last $B$ term should actually be $\bar\theta\bar\theta$, and that's meant to represent $\epsilon_{AB}\bar \theta^A\bar\theta^B$, perhaps with a factor of $1/2$ or $i/2$ or whatever is their convention. But these chiral fields should only contain $\theta$ or only $\bar\theta$. $\endgroup$ May 2, 2015 at 12:24
  • $\begingroup$ @LubošMotl I, not OP, wrote the equation in question. I took it directly from the text, and your correction does not appear in the official list of errata. Note that $Y^\mu$ is not chiral. $\endgroup$
    – Ryan Unger
    May 4, 2015 at 21:25
  • $\begingroup$ In that case, $\theta$ and $\bar\theta$ must be treated together and be equivalent to $\theta_A$ with different components $A$. It's still true that the derivative of $\theta$ bilinear is $\theta$ linear. $\endgroup$ May 5, 2015 at 4:46

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