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According to Tobias Hurter’s popular exposition Too Big for a Single Mind (narrated in the present tense):

Dirac makes use of an elegant mathematical tool developed by the Irish mathematician William Hamilton in the nineteenth century.

This is from a passage discussing Dirac in Cambridge in the summer of 1925; see pages 127-128.

What “elegant mathematical tool” is Hurter talking about? Was it quaternions? Unfortunately the notes at the back of the book do not say anything more about this episode.

Mauricio
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James Propp
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2 Answers2

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I'd wager that the ambiguous reference is to the appearance of Hamiltonian quaternions (as an instance of a Clifford algebra) in Dirac's construction of a square root (of a certain sort) of a/the Laplacian, which is nowadays called a/the "Dirac operator". (Not to be confused with Dirac's $\delta$.) It is a linear differential operator with coefficients in a Clifford algebra related to a quadratic form attached to the geometry of the situation.

This is quite different from square roots of positive unbounded self-adjoint operators on Hilbert spaces... For example, the Dirac operator still has a robust distributional sense, while the unbounded-op-on-Hilbert-space version is not distributional.

The Hamiltonian quaternions arise as a Clifford algebra attached to a low-dimensional quadratic form.

paul garrett
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    With the OP specification of the dating to 1925, the Dirac operator is unlikely, it is from 1928. The surrounding passage also talks about a paper sent to Göttingen in 1925, which I think was a draft of this one, where he draws analogies between commutators and Poisson brackets in Hamiltonian dynamics. – Conifold Nov 24 '23 at 06:38
  • @Conifold, ah, yes, with the dates, you're surely right. But then, as in your earlier comment, it is mildly curious that Schroedinger didn't/doesn't get more credit... – paul garrett Nov 24 '23 at 17:34
  • @paulgarrett : "This is quite different from square roots of positive unbounded self-adjoint operators on Hilbert spaces... For example, the Dirac operator still has a robust distributional sense, while the unbounded-op-on-Hilbert-space version is not distributional." Could you please explain what you mean or give a reference? This is not a critique, just trying to understand. – akhmeteli Dec 17 '23 at 06:45
  • @akhmeteli, the functional calculus for unbounded but positive operators such as (a/the self-adjoint extension of...) $-\Delta$ (... restricted to test functions) gives a (positive) square root, yes, but that square root is no longer a differential operator. In contrast, Dirac operators are differential operators, although with coefficients in non-commuting scalars (Hamiltonian quaternions, or, in general, Clifford algebras). – paul garrett Dec 17 '23 at 17:16
  • @paulgarrett : Thank you very much. – akhmeteli Dec 17 '23 at 18:32
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The OP noted in a comment:

I don’t know if Hurter is correct, but I stress that he is writing about Dirac in 1925, not 1927.

Therefore, the text is probably about Dirac's 1925 article , where Dirac uses Hamiltonian function. The equations of motion then feature the Poisson bracket. The Dirac's general approach to quantization involves the replacement of the Poisson brackets by commutators.

akhmeteli
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