19

Let us generalize from four space-time dimensions to a $d$-dimensional Clifford algebra $C$. Define $$\tag{1} p~:=~[\frac{d}{2}], $$ where $[\cdot]$ denotes the integer part. OP's question then becomes Why must the dimension $n$ of a finite dimensional representation $V$ be a multiple of $2^p$? Proof: If $C\subseteq {\rm End}(V)$ and $V$ are both real, ...


16

As previous answers have correctly noted gamma matrices do not forma a basis of $M(4,\mathbb{C})$. Nevertheless you can construct one from them in the following way 1 the identity matrix $\mathbb{1}$ 4 matrices $\gamma^\mu$ 6 matrices $\sigma^{\mu\nu}=\gamma^{[\mu}\gamma^{\nu]}$ 4 matrices $\sigma^{\mu\nu\rho}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho]}$ 1 ...


16

I think the clearest way to think about this is to say that the gamma matrices don't transform. In other words, the fact that they carry a vector index doesn't mean that they form a four vector. This is analogous to how the Pauli matrices work in regular quantum mechanics, so let me talk a little bit about that. Suppose you have a spin $1/2$ particle in ...


14

Although the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}$ is the most famous, there is an expression for the commutator: $$[\gamma^\mu,\gamma^\nu] = 2\gamma^\mu \gamma^\nu - 2 \eta^{\mu\nu}$$ The matrix defined by $[\gamma^\mu,\gamma^\nu]$ actually has a purpose: it forms a representation of the Lorentz algebra. If we define $S^{\mu\nu}$ as $1/4$ the ...


11

To complement V. Moretti's excellent answer, it's worth emphasizing that the dimension of the four-by-four complex matrices $\mathbb C^{4\times 4}$, when seen as a vector space over $\mathbb C$, is $4\!\times\!4=16$. As such, a set of four matrices (i.e. vectors in $\mathbb C^{4\times 4}$) can never be a basis for it. It's also worth saying that the general ...


11

No they do not, due to dimensional reasons, but they are generators of the algebra. That is, $I$ and the products of $\gamma^a$ (products of one, two, three and four matrices) form such a basis. NOTE ADDED. As Emilio Pisanty correctly remarked (also making some further interesting comments) $GL(4, \mathbb C)$ is not a linear space so questions about bases ...


11

Yes. The indices on gamma matrices can be treated like four-vector indices. In particular, indices on gamma matrices are commonly raised and lowered with the Minkowski metric $\eta_{\mu\nu}$ as you indicate; \begin{align} \gamma_\mu = \eta_{\mu\nu}\gamma^\nu. \end{align} Now, as user26143 writes in his comment above, the gamma matrices have the ...


10

You have no other choice than to use $4\times 4$ matrices. All these "representations" are different realizations (related by similarity transformations) of the only possible irreducible representation of the Clifford algebra that is spanned by the abstract $\gamma^\mu$. This representation, in a way, is the definition of what a "Dirac spinor" is, and it is ...


8

The ordering of tensor product factors in matrix representations is ultimately a convention: there is no single canonical way to order the tensor product basis (does $|{\uparrow\downarrow} \rangle$ come before or after $|{\downarrow\uparrow} \rangle$?) and the different orderings will produce different matrix representations. You've found the two ...


7

As the comments explained, you need to know a few properties of the $\gamma$ matrices. First of all, from $$ \{\gamma_\mu, \gamma_\nu\} = 2 \eta_{\mu \nu} \mathbf{1}_4$$ you can infer that (depending on the metric but not on the representation of the dirac algebra!) in (+---) metric $\gamma_0$ is hermitian (hint: look at the $\mu = 0, \nu = 0$ component of ...


7

The reasoning is supposed to go as follows: $\gamma^5$ commutes with all algebra elements, hence with the whole image of the algebra representation. $\gamma^5$ has at least two different eigenvalues, meaning it is not a scalar multiple of the identity. If the representation of the $S^{\mu\nu}$ (that form the Lorentz algebra $\mathfrak{so}(1,3)$) were ...


7

Gamma matrices are defined by the Clifford algebra $$ \{\gamma^\mu, \gamma^\nu\}= 2g^{\mu\nu}\mathbb I_n \,. $$ So, you see the index $\mu$ in $\gamma^\mu$ runs from $0$ upto $D-1$ where $D$ is the number of spacetime dimensions. It does not mean $\gamma^\mu$ is a vector. The $\mu$ index here only tells you how many gamma matrices are there. The ...


7

There are no "two spacetimes". There is a single spacetime $M$, which is a four-dimensional Lorentzian manifold, and there's a bunch of bundles over it. The spin connection is formally a connection form on a $\mathrm{SO}(1,3)$-principal bundle over $M$ that can act on anything that transforms in a representation of the Lorentz algebra. It is not a bivector, ...


6

The term with the derivatives, is really a directional derivative $$ \alpha^k\partial_k=\vec \alpha\cdot \nabla $$ and if applied to a function $f$ it measures its change along that particular direction: $$ \alpha^k\partial_kf=\vec \alpha\cdot \nabla f=|\vec\alpha|\frac{\partial f}{\partial\vec n} =|\vec\alpha|\lim_{\delta x\to0}\frac{f(\vec x+\delta x\:\vec ...


6

It is actually possible, and not too difficult, to prove this without expanding the exponentials to first order only. What you are trying to prove is $S^\dagger \gamma^0 = \gamma^0 S^{-1}$, this is equivalent to $$ \gamma^0 S^\dagger \gamma^0 = S^{-1} $$ because $( \gamma^0 )^2 = 1$. Expand $S^\dagger = \sum_n \frac{1}{n!} \left( \frac i 4 \omega_{\mu\nu} \...


5

Follow these guidelines: 1) Use the notations $\sigma^\mu= (1, \vec \sigma)$, $\tilde \sigma^\mu= (1, -\vec \sigma)$. You will get expressions with lower indices $\mu$, by using the Minkowski metrics $\eta_{\mu\nu} = (1, -1, -1, -1)$. 2) Use the chiral / Weyl basis, and express, for the LHS and RHS expressions of the equality, the matrices between the ...


5

It is not the case because the gamma matrices do not commute and neither do the covariant derivatives. We can always write a 2-index tensor as the sum of its antisymmetric and its symmetric part, $$D^\mu D^\nu = \frac{1}{2}(D^\mu D^\nu + D^\nu D^\mu) + \frac{1}{2}(D^\mu D^\nu - D^\nu D^\mu).$$ The latter part vanishes precisely when the curvature (= field ...


5

Try using the definition of $\gamma^5$ and just apply the conjugation. Remember that conjugation flips the order of the matrices, which means you want to change their order before applying the conjugation. Then realize that the anticommutation relations give you a way of interchanging two gamma matrices, giving only a minus sign (provided the indices are ...


5

One option is to start out with the matrix representation for two sets of conjugate Grassmann numbers (see previous thread), $\theta_i, \pi_i$ with $i=1,...,N$, such that $\{\theta_i,\theta_j\}=0,\quad\{\pi_{i},\pi_{j}\} = 0, \quad \{\theta_i,\pi_j\} = \delta_{ij}$ Then a $2N$-dimensional Clifford algebra can be built by $\gamma_{i}=\theta_{i}+\pi_{i}\\ \...


5

I) Two square matrices $A$ and $B$ are similar matrices if they are connected via a relation $$\tag{1}AP~=~PB$$ for some invertible matrix $P$. II) Two square matrices $A$ and $B$ are unitarily similar matrices if $P$ in eq. (1) is a unitary matrix.


5

Regular numbers could never fulfill $$ AB + BA = 0, \quad AA = 1 = BB. $$ The only way to fulfill the first eq. is to have either $A = 0 $ or $B = 0$ but this violates the second equation. Matrices on the other hand can fulfill such equations and since Dirac knew about matrices he did not discard his idea after finding something that's impossible with ...


5

It gets easier if you use the result from part 1. Then you also don't have to deal with the $\mathcal O(\omega^2)$ (see my answer to your other question). In your calculation, you transformed $\bar\psi$ and $\psi$, but not $\gamma^\lambda$. This is correct, as I will show in the end, but I will take another point of view which is really helpful here: $\...


5

Expanding on my comment. The basic idea is that what you mean by adjoint depends on the vector space being considered. For example, we might have $\mathbb{C}^n$ as our space, with the usual inner product; in that case, the adjoint of a vector or matrix is the transpose conjugate. Note that technically, taking adjoint of a vector doesn't return a vector, ...


4

Intuitive explanation Preliminary: A vector has a many components as elements of the vector space basis. A Clifford algebra basis is generated by all (independent) products of the generators (in Dirac's equation case these are the $\gamma$'s). The counting There are as many $\gamma$'s as the dimension of the spacetime, and according to the definition ...


4

From Wikipedia: The number 5 is a relic of old notation in which $\gamma^0$ was called "$\gamma^4$".


4

In order to appreciate the periodicity of graphene one has to recognize that it consists out of two interpenetrating hexagonal Bravais sublattices, A and B, which together make up the honeycomb lattice. The two sublattices are like two degrees of freedom, and the electron can have an amplitude to be on sublattice A, and an amplitude to be on the sublattice B....


4

The Lorentz group $O(3,1)$ has spinor representations (actually $SL(2,\mathbb C)$, that is the universal cover of $O(3,1)$), as well known. The problem is that now, in general relativity, we want to deal with generic transformations. So we are working with G$L(4)$. Roughly speaking, the associated Lie Algebra $\mathfrak{gl}(4)$ doesn't admit spinor ...


4

In his "Recollections of an Exciting Era" (History of Twentieth Century Physics: Proceedings of the International School of Physics "Enrico Fermi". Course LVII.-New York; London: Academic Press, 1977. P. 109-146), Dirac wrote that, to derive what is now known as the Dirac equation, he had played with mathematical formulas, and he needed the Pauli matrices $\...


4

The Clifford algebra $\mathrm{Cl}(\mathbb{R}^{1,3})$ is the algebra generated by endowing the vectors in $\mathbb{R}^{1,3}$ with a free algebra multiplication and then imposing the constraint given by $v \circ v = \eta^{\mu\nu}v_\mu v_\nu$ with $\eta$ as the Minkowski metric on $\mathbb{R}^{1,3}$. Any Clifford algebra has a natural connection to the ...


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