Monday, 30 September 2019

quantum field theory - Grassmann paradox weirdness



I'm running into an annoying problem I am unable to resolve, although a friend has given me some guidance as to how the resolution might come about. Hopefully someone on here knows the answer.


It is known that a superfunction (as a function of space-time and Grassmann coordinates) is to be viewed as an analytic series in the Grassmann variables which terminates. e.g. with two Grassmann coordinates $\theta$ and $\theta^*$, the expansion for the superfunction $F(x,\theta,\theta^*)$ is


$$F(x,\theta)=f(x)+g(x)\theta+h(x)\theta^*+q(x)\theta^*\theta.$$


The product of two Grassmann-valued quatities is a commuting number e.g. $\theta^*\theta$ is a commuting object. One confusion my friend cleared up for me is that this product need not be real or complex-valued, but rather, some element of a 'ring' (I don't know what that really means, but whatever). Otherwise, from $(\theta^*\theta)(\theta^*\theta)=0$, I would conclude necessarily $\theta^*\theta=0$ unless that product is in that ring.


But now I'm superconfused (excuse the pun). If Dirac fields $\psi$ and $\bar\psi$ appearing the QED Lagrangian $$\mathcal{L}=\bar\psi(i\gamma^\mu D_\mu-m)\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ are anticommuting (Grassmann-valued) objects, whose product need not be real/complex-valued, then is the Lagrangian no longer a real-valued quantity, but rather takes a value which belongs in my friend's ring??? I refuse to believe that!!



Answer



A supernumber $z=z_B+z_S$ consists of a body $z_B$ (which always belongs to $\mathbb{C}$) and a soul $z_S$ (which only belongs to $\mathbb{C}$ if it is zero), cf. Refs. 1 and 2.


A supernumber can carry definite Grassmann parity. In that case, it is either $$\text{Grassmann-even/bosonic/a $c$-number},$$ or $$\text{Grassmann-odd/fermionic/an $a$-number},$$ cf. Refs. 1 and 2.$^{\dagger}$ The letters $c$ and $a$ stand for commutative and anticommutative, respectively.


One can define complex conjugation of supernumbers, and one can impose a reality condition on a supernumber, cf. Refs. 1-4. Hence one can talk about complex, real and imaginary supernumbers. Note that that does not mean that supernumbers belong to the set of ordinary complex numbers $\mathbb{C}$. E.g. a real Grassmann-even supernumber can still contain a non-zero soul.


An observable/measurable quantity can only consist of ordinary numbers (belonging to $\mathbb{C}$). It does not make sense to measure a soul-valued output in an actual physical experiment. A soul is an indeterminate/variable, i.e. a placeholder, except it cannot be replaced by a number to give it a value. A value can only be achieved by integrating it out!



In detail, a supernumber (that appears in a physics theory) is eventually (Berezin) integrated over the Grassmann-odd (fermionic) variables, say $\theta_1$, $\theta_2$, $\ldots$, $\theta_N$, and the coefficient of the fermionic top monomial $\theta_1\theta_2\cdots\theta_N$ is extracted to produce an ordinary number (in $\mathbb{C}$), which in principle can be measured.


E.g. the Grassmann-odd (fermionic) variables $\psi(x,t)$ in the QED Lagrangian should eventually be integrated over in the path integral.


References:




  1. planetmath.org/supernumber.




  2. Bryce DeWitt, Supermanifolds, Cambridge Univ. Press, 1992.





  3. Pierre Deligne and John W. Morgan, Notes on Supersymmetry (following Joseph Bernstein). In Quantum Fields and Strings: A Course for Mathematicians, Vol. 1, American Mathematical Society (1999) 41–97.




  4. V.S. Varadarajan, Supersymmetry for Mathematicians: An Introduction, Courant Lecture Notes 11, 2004.




--


$^{\dagger}$ In this answer, the words bosonic (fermionic) will mean Grassmann-even (Grassmann-odd), respectively.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...