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?

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    $\begingroup$ You didn't change what things are. Instead you've constructed a representation that acts on spinors instead of vectors. $\endgroup$ Dec 8, 2019 at 3:48
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    $\begingroup$ 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, 2019 at 3:54
  • $\begingroup$ Could you elaborate on this? This is surely the thing I'm not understanding $\endgroup$
    – Craig
    Dec 8, 2019 at 3:54
  • $\begingroup$ @Craig $\psi^\dagger\gamma^i\psi$ transforms like a vector $\endgroup$ Dec 8, 2019 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, 2019 at 5:28

1 Answer 1


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.

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