We can identify Minkowski space-time $M^4$, of metric signature $(1, -1, -1, -1)$, with the (real) space of $2 \times 2$ (complex) Hermitian matrices under the map $(v_0, v_1, v_2, v_3) \mapsto v_0 I + v_1 \sigma_x + v_2 \sigma_y + v_3 \sigma_z$ where $\sigma$'s are Pauli matrices.

Given a Hermitian matrix $X$ and some $A \in SL(2, \mathbb{C})$, the matrix $A X A^{\dagger}$ is again Hermitian, and $\det(AXA^{\dagger}) = \det(X)$, so (Hermitian) conjugation by $A$ is a linear transformation of Minkowski space-time which preserves the metric and is thus a Lorentz transformation. It is orthochronous because $AIA^{\dagger} = AA^{\dagger}$ is a positive operator, and thus $\text{tr}(AA^{\dagger}I) = \text{tr}(AA^{\dagger})$ (which, up to a factor of $1/2$, is the $I$-coefficient of $AA^{\dagger}$ in the $I, \sigma$ basis) is positive.

How can I check it has determinant $1$, though? It seems painful to compute the other coefficients of the matrix corresponding to this transformation. Is there some slick way?

  • $\begingroup$ Maybe I am missing something, but isn't this obvious from the equation you wrote: $det(AXA^{\dagger})=det(X)$. Using the determinant identity $det(ABC)=det(A)det(B)det(C)$, we arrive at $det(A)det(A^{\dagger})=1$, from which it follows that determinant of $A$ is $1$? $\endgroup$ – Anonjohn Oct 16 '20 at 5:16
  • $\begingroup$ It is obvious that its squared determinant is 1 since the conjugate transpose has the same determinant as the original matrix $\endgroup$ – daydreamer Oct 16 '20 at 6:49
  • $\begingroup$ @Anonjohn, the question is not the determinant of $A$, which is 1 by assumption, but the determinant of the linear transformation $X \mapsto AXA^{\dagger}$. $\endgroup$ – Pedro Oct 16 '20 at 16:07

OP has already argued that it is a Lorentz transformation. It is well-known that Lorentz matrices always have determinant $\pm 1$. Let us reformulate OP's question as follows.

How can I check it has determinant $+1$ (as opposed to $-1$)?

Proof: Since $\rho: 𝑆𝐿(2,\mathbb{C})\to O(1,3;\mathbb{R})$ is a continuous map and $𝑆𝐿(2,\mathbb{C})$ is a connected set, the image $\rho(𝑆𝐿(2,\mathbb{C}))$ must again be a connected set. Hence the image $\rho(𝑆𝐿(2,\mathbb{C}))\subseteq SO^+(1,3;\mathbb{R})$ must be a subset of the connected component of $O(1,3;\mathbb{R})$ that contains the identity, i.e. the restricted Lorentz group $SO^+(1,3;\mathbb{R})$. $\Box$

For more details, see this related Phys.SE post.

  • $\begingroup$ Well, that's slick! I somehow hoped an answer would involve some way of figuring out what the matrix for $\rho(A)$ is in terms of the $I, \sigma$ basis, but perhaps it's unwieldy. Do you know how can one see $\rho$ is surjective (onto $SO^{+}(1, 3; \mathbb{R})$)? $\endgroup$ – Pedro Oct 16 '20 at 16:16
  • $\begingroup$ Surjectivity is more elaborate to prove. $\endgroup$ – Qmechanic Oct 16 '20 at 21:07

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