In Euclidean space with metric $\delta_{ij}$ a Galilean vector rotates via $\hat{V} = \Lambda(\theta) V$ where $\Lambda$ is a member of $O(3)$. This can be represented by having the $V$ be column vectors and $\Lambda$ a $3\times 3$ matrix such that $\Lambda^T=\Lambda^{-1}$. A second rank symmetric tensor $T$ will rotate thus: $\hat{T} = \Lambda(\theta) T\Lambda^T(\theta)$. However we can also represent our vectors in a Clifford algebra (as opposed to the representation in the coordinate bases which is typically used).

Given $\gamma_1,\gamma_2,\gamma_3$ that satisfy $$ \gamma_j\gamma_k+\gamma_k\gamma_j=2\delta_{jk}. $$ And with $$V= v^i\gamma_i $$ The rotation of $V$ through an angle $\theta$ will be $$ \hat{V}= \exp\left(\frac{\theta}{2}\gamma_1\gamma_2\right) V \exp\left(-\frac{\theta}{2}\gamma_1\gamma_2\right). $$ However this looks an awful lot like the equation for the rotation of a tensor through an angle as well. A similar question arises as to why spinor transformations now look like vector transformations:

$$ \hat{\psi}= \exp\left(\frac{\theta}{2}\gamma_1\gamma_2\right) \psi $$ Does employing geometric algebra somehow change our space so that spinors are vectors, and physical vectors, are higher rank tensors in this space?

  • 1
    $\begingroup$ You didn't change what things are. Instead you've constructed a representation that acts on spinors instead of vectors. $\endgroup$ – Gabriel Golfetti Dec 8 '19 at 3:48
  • 1
    $\begingroup$ Of course, this is just rediscovering tensor products. A rank 2 tensor is the product of two vectors, so its transformation involves two factors of the vector transformation. Similarly, a vector is contained in the product of two spinors. $\endgroup$ – knzhou Dec 8 '19 at 3:54
  • $\begingroup$ Could you elaborate on this? This is surely the thing I'm not understanding $\endgroup$ – Craig Dec 8 '19 at 3:54
  • $\begingroup$ @Craig $\psi^\dagger\gamma^i\psi$ transforms like a vector $\endgroup$ – Gabriel Golfetti Dec 8 '19 at 3:55
  • $\begingroup$ @knzhou can you suggest more readings on this. All my sources sort of take this for granted. I suppose my question is how this "change of basis" changes our space we're acting on? $\endgroup$ – Craig Dec 8 '19 at 5:28

In the Clifford algebra framework, the transformation law of spinor $\psi$ (a rotation along $z$ axis as an example) $$ \psi \rightarrow e^{\frac{\theta}{2}\gamma_1\gamma_2} \psi $$ is fundamental, while all other transformations are derived.

Accordingly, a vector $V^\mu$ such as fermion current $$ V^\mu = \bar{\psi}\gamma^\mu\psi =\psi^\dagger\gamma^0\gamma^\mu\psi, $$ transforms as \begin{align} V^\mu &\rightarrow (e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi)^\dagger\gamma^0\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\ &=\psi^\dagger(e^{\frac{\theta}{2}\gamma_1\gamma_2})^\dagger\gamma^0\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\ &=\psi^\dagger e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^0\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\ &=\psi^\dagger \gamma^0(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2})\psi \\ &=\psi^\dagger \gamma^0(\Lambda^\nu_\mu(\theta)\gamma^\mu) \psi \\ &=\Lambda^\nu_\mu(\theta)\psi^\dagger \gamma^0\gamma^\mu \psi \\ &=\Lambda^\nu_\mu(\theta)V^\mu. \end{align}

Note that

  • The fermion current $V^\mu$ is defined as $\bar{\psi}\gamma^\mu\psi= \psi^\dagger\gamma^0\gamma^\mu\psi$ not as $\psi^\dagger\gamma^\mu\psi$. Bear in mind that for Lorentz boost $(e^{\frac{\theta}{2}\gamma_0\gamma_i})^\dagger\gamma^0 = e^{+\frac{\theta}{2}\gamma_0\gamma_i}\gamma^0 = \gamma^0e^{-\frac{\theta}{2}\gamma_0\gamma_i}$. The additional $\gamma^0$ in $\bar{\psi} = \psi^\dagger\gamma^0$ is essential in guaranteeing the Lorentz covariance of the vector $V^\mu$.
  • $\Lambda^\nu_\mu(\theta)$ are just real numbers, so that they can be moved freely around Clifford algebra elements.

And you can also derive that according to the spinor transformation rule, an antisymmetric tensor $T^{\mu\nu}$ such as $$ T^{\mu\nu} = \bar{\psi}\gamma^\mu\gamma^\nu\psi =\psi^\dagger\gamma^0\gamma^\mu\gamma^\nu\psi, $$ transforms as \begin{align} T^{\mu\nu} &\rightarrow (e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi)^\dagger\gamma^0\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\ &=\psi^\dagger(e^{\frac{\theta}{2}\gamma_1\gamma_2})^\dagger\gamma^0\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\ &=\psi^\dagger e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^0\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\ &=\psi^\dagger \gamma^0(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2})\psi \\ &=\psi^\dagger \gamma^0(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2} )(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2})\psi \\ &=\psi^\dagger \gamma^0(\Lambda^{\mu'}_\mu(\theta)\gamma^\mu)(\Lambda^{\nu'}_\nu(\theta)\gamma^\nu) \psi \\ &=\Lambda^{\mu'}_\mu(\theta)\Lambda^{\nu'}_\nu(\theta)\psi^\dagger \gamma^0\gamma^\mu\gamma^\nu \psi \\ &=\Lambda^{\mu'}_\mu(\theta)\Lambda^{\nu'}_\nu(\theta)T^{\mu\nu} . \end{align}

For related discussions,

  • See here for how to embed vectors and tensors in a spinor.
  • See here for how to correctly define covariant derivatives with respect to single-sided vs double-sided gauge transformations.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.