# Is Schwinger's derivation of the two classes bosons and fermions a no-go theorem for supersymmetry?

In his book "Quantum Kinematics and Dynamics", Julian Schwinger gives a mathematical argument which shows that only two classes of dynamical variables can exist in quantum mechanics. He also derives the commutation relations for the two classes, thus establishing by theoretical means the existence and properties of bosonic and fermionic dynamical variables.

In the course of the proof, Schwinger introduces a group of transformations of the dynamical variables. It turns out that the transformations must be block diagonal in the two classes of dynamical variables. In other words the transformations only act within each class and do not mix variables between the two classes.

Schwinger's proof that transformations must not mix bosonic and fermionic dynamical variables seems to contradict the notion of supersymmetry where transformations are introduced which mix bosonic and fermionic dynamical variables. This cannot in fact be the case because there must be many physicists who work in supersymmetry and are also familiar with Schwinger's work.

So, my question is, "How does on one reconcile Schwinger's proof that transformations don't mix bosonic and fermionic dynamical variables with the notion of supersymmetry where these variables are transformed into each other?"

Finally, in order to make this question clearer, here is a sketch of Schwinger's argument which is in section 3.7 of "Quantum Kinematics and Dynamics".

Schwinger starts from his quantum mechanical action principle. He assumes the action operator is, $$\hat{S}=\int (\frac{1}{4}(\hat{x}_{a}A_{ab}d\hat{x}_{b}-d\hat{x}_{a}A_{ab}\hat{x}_{b})-\hat{H}dt)$$ where the $\hat{x}_{a}$ are the dynamical variables and the $A_{ab}$ is a matrix which has to be skew-Hermitian $A^{\dagger}=-A$ and $\hat{H}$ is the Hamiltonian. The form of the action seems motivated by symplectic geometry because, classically, $A_{ab}$ is the symplectic 2-form.

The action is varied by changing the dynamical variables by $\delta \hat{x}_{a}$. Schwinger assumes that displacing a variation across a dynamical variable induces a linear transformation on these variables, $$\delta \hat{x}_{a}\hat{x}_{b}=(k_{a}\hat{x})_{b}\delta\hat{x}_{a}$$ where $k_{a}$ is a matrix depending the variable $\hat{x}_{a}$ in the variation. By assuming the dynamical variables are Hermitian, Schwinger argues that the matrices obey $k_{a}k_{a}=1$.

Schwinger then says there must be freedom to transform to new dynamical variables by $\hat{x}'=l\hat{x}$ where $l$ is a transformation matrix. By applying the transformation to the rule for displacing a variation across a variable, Schwinger gets a consistency condition, $$\sum_{b}l_{ab}k_{b} (l^{-1})_{bc}=\delta_{ac}k_{a} \ .$$ This equation mixes the matrices $k_{a},k_{b}$ with components of the transformation such as $l_{ab}$. This equation cannot hold for arbitrary transformation $l$ unless the matrices $k_{b}$ are identical for all values of $b$ that can be connected by the linear transformation and linear transformations only within on each class are permissible. Schwinger goes on to use $k_{a}k_{a}=1$ to show that only two classes of dynaical variables can exist and to derive the commutation properties by showing that the matrices $k_{a}$ must have a particularly simple form. However, the point has been made, the transformations $l$ must not mix variables of different classes.

• The 2nd and 3rd full-line equations you have each contain mismatching indices on the left and righthand sides; it would be easier to follow if you clean that up. I'm not familiar enough with Schwinger's argument to give a complete authoritative answer, but my guess would be that he made the assumption that the elements of the $l$ transformation matrices had to be real or complex valued, whereas in supersymmetry they can be Grassmann valued, opening up more possibilities. – reductionista Oct 26 '16 at 22:33
• @JeffLJones : I fixed an error in the second equation, but the third is the same as the book. Note $k_{a},k_{b}$ are matrices with components $(k_{a})_{ef}, (k_{b})_{ef}$. – Stephen Blake Oct 27 '16 at 4:01
• +1 to JeffLJones comment, and also note that supersymmetric theories exist, even those which have been solved to an excruciating level of detail, so it is unlikely that any generic internal inconsistency in SUSY theories have been missed. – Peter Kravchuk Oct 27 '16 at 5:29

There is no such contradiction between Schwinger's statement and supersymmetry. What Schwinger shows is that a symmetry cannot change the spin statistic of a field. Supersymmetry does not change the spin statistic of a field.

As an example, we can check Wess & Bagger supersymmetry book, chapter 3. Their simplest Lagrangian is

${\mathcal L}={\rm i}\partial_n \bar\psi \bar\sigma^n \psi+A^*\Box A +F^*F+m\left(AF +A^* F^*-\frac{1}{2}\psi\psi-\frac{1}{2}\bar\psi\bar\psi\right)$

$\mathcal L$ is invariant under

$\delta_\xi A=\sqrt{2} \xi \psi\\ \delta_\xi \psi= \sqrt{2}\sigma^n \bar\xi\partial_n A+\sqrt{2}\xi F \\ \delta_\xi F= {\rm i} \bar\xi \bar\sigma^m \partial_m \psi$

which is a susy transformation.

The use of a spinor $\xi$ ensures us that, even though the symmetry transformation relates bosons $(A,F)$ with spinors $\psi$, it does not transform a boson into a spinor. Indeed, $\xi \psi$ and $\bar\xi \bar\sigma^m \partial_m \psi$ have the spin statistic of a boson, while $\delta_\xi \psi= \sqrt{2}\sigma^n \bar\xi\partial_n A+\sqrt{2}\xi F$ have the spin statistic of a Fermion. If $\xi$ was a C-number, or a bosonic type of field, then we would get a conflict between the Schwinger derivations and supersymmetry.

Summarizing: SUSY is a symmetry between boson and fermions, which relates them, but it does not change their spin-statistics.

• When you say "spin statistic", I guess you mean "transforms as" so that $\delta_{\xi}A=\sqrt{2}\xi^{\alpha}\psi_{\alpha}$ and both sides of the formula are scalars? – Stephen Blake Nov 3 '16 at 18:57
• "spin statistic" means either bosonic or ferminic. A scalar field is bosonics, with spin 0. Both the LHS and the RHS of the transformation $\delta_\xi A=\sqrt{2} \xi^\alpha \psi_\alpha$ are spin 0 fields and none of them is a grassmannian variable, as expected. – CGH Nov 4 '16 at 1:10