## Exponentiating 4 by 4 matrix analytically

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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

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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

## 3 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 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. – quid Aug 13 at 14:08

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@Igor: they are a particular matrix representation of a certain Clifford algebra. – Qiaochu Yuan Aug 13 at 16:05

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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.

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(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

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.

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