# Exponentiating 4 by 4 matrix analytically

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

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What is the result you are alluding to for $2\times 2? matrices? – Igor Rivin Aug 12 '12 at 23:22 By "analytically" you mean "explicitely"? Put your matrix$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)$which is quite explicit to compute... – Qfwfq Aug 12 '12 at 23:47 @Qfwfq: that is computationally tractable, but NOT explicit (try writing a formula in terms of matrix elements of$X$) – Igor Rivin Aug 13 '12 at 0:12 Also, why all the votes to close? – Igor Rivin Aug 13 '12 at 0:12 ## 2 Answers Perhaps I misunderstand the question. When you say you have Dirac matrices, does that mean that you are computing the exponential of a liner combination of Dirac matrices? If so, then there is a very simple analytical formula in any dimension: just use the Clifford relations in the exponential series. More concretely, suppose that you would like to compute the exponential of a matrix$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\$.

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For us heathens: what are Dirac matrices? – Igor Rivin Aug 13 '12 at 13:35
@Igor Rivin: as far as I understand (which is not to say much) the first display (Clifford relation) is sort-of the defintion. – user9072 Aug 13 '12 at 14:08
@Igor: they are a particular matrix representation of a certain Clifford algebra. – Qiaochu Yuan Aug 13 '12 at 16:05

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

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(There seems to be a problem with the link.) – Andrés Caicedo Aug 13 '12 at 0:12
@Andres: should be fixed now... – Igor Rivin Aug 13 '12 at 0:35
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 '12 at 1:29