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Does there exist an analytical method by which i can exponentiate a 4 by 4 matrix, in the same way as the general 2 by 2 matrix case in pauli matrix basis. I have dirac matrices (which are composed of direct products of pauli matrices) as my basis for 4 by 4 matrices. I need an analytical way ! Any reply is appreciated. regards

flag	asked Aug 12 at 23:18 ad_tls 61●1

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What is the result you are alluding to for $2\times 2? matrices? – Igor Rivin Aug 12 at 23:22

By "analytically" you mean "explicitely"? Put your matrix $X$ in Jordan form. Then $X=S+N$ where $S$ is diagonal and $N$ nilpotent and $SN=NS$, and $\exp(X)=\exp(S)\cdot\exp(N)$ which is quite explicit to compute... – Qfwfq Aug 12 at 23:47

@Qfwfq: that is computationally tractable, but NOT explicit (try writing a formula in terms of matrix elements of $X$) – Igor Rivin Aug 13 at 0:12

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Also, why all the votes to close? – Igor Rivin Aug 13 at 0:12

José Figueroa-O'Farrill · Answer 1 · 2012-08-13 18:41:46Z UTC

Perhaps I misunderstand the question. When you say you have Dirac matrices, does that mean that you are computing the exponential of a liner combination of Dirac matrices? If so, then there is a very simple analytical formula in any dimension: just use the Clifford relations in the exponential series.

More concretely, suppose that you would like to compute the exponential of a matrix $X := \sum_i x^i \Gamma_i$, where the Dirac matrices $\Gamma_i$ obey the Clifford relation $$ \Gamma_i \Gamma_j + \Gamma_j \Gamma_i = - 2 g_{ij} I~, $$ with $I$ the identity matrix. Then it follows from this relation that $$ X^2 = - x^2 I~, $$ where I have introduced the (indefinite, if $g_{ij}$ has indefinite signature) "squared norm" $$ x^2 = \sum_{i,j} x^i x^j g_{ij}~. $$

If $x^2$ = 0, then $$ \exp X = I + X $$ and if $x^2 \neq 0$, then letting $x = \sqrt{x^2}$ (which could be imaginary), $$ \exp X = \cos x I + \frac{\sin x}{x} X~. $$

Added (for the "heathens")

Quiaochu's comment is correct. Here are some more details.

Let $V$ be a finite-dimensional real vector space with a non-degenerate inner product $\left<-,-\right>$. Let $Cl(V)$ be the corresponding Clifford algebra. Let $\rho: Cl(V) \to \operatorname{End}(M)$ be an irreducible representation of $Cl(V)$. Let $(e_i)$ be a basis for $V$. Then $\Gamma_i := \rho(e_i)$ are called Dirac matrices of $CL(V)$ in the representation $M$.

Igor Rivin · Answer 2 · 2012-08-13 00:51:06Z UTC

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There is a completely explicit formula in this paper of Bensauod and Mouline (rendicotti Palermo, 2005), which is quite compact for low dimensions.

edited Aug 13 at 0:51

answered Aug 13 at 0:09

Igor Rivin
31.1k●3●37●71

(There seems to be a problem with the link.) – Andres Caicedo Aug 13 at 0:12

@Andres: should be fixed now... – Igor Rivin Aug 13 at 0:35

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It's explicit in terms of the solution of a differential equation related to the characteristic polynomial. Of course, to solve that differential equation explicitly you need the eigenvalues... – Robert Israel Aug 13 at 1:29

ad_tlse · Answer 3 · 2012-08-14 09:13:21Z UTC

Thanks all,

Igor, are you sure about the Dirac matrices satisfy this particular Clifford relation ?? I guess it's +/- depending upon the indices.I tried this derivation before and got stuck at the anticommutator relations particularly.

Exponentiating 4 by 4 matrix analytically

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