In the modern interpretation of differential geometry a tangent vector is identified as a derivation: $\frac{\partial}{\partial x}$.

In Quantum Mechanics, momentum - which is classically understood as a vector $p$ is promoted to an operator $-i\hbar\frac{\partial}{\partial x}$.

Is there some connection between these two - on the face of it - different concepts?

That this isn't a straight-forward translation of concepts can be seen from the fact that position, which classically is also a vector is promoted to a multiplication operator and not a derivation.

  • $\begingroup$ See this post: Is the Momentum Operator a Postulate?. $\endgroup$
    – cxx
    Dec 8 '18 at 16:31
  • $\begingroup$ @Hanting Zhang: I don't really see the relevance to my question. What was it that you were intending to point out? $\endgroup$ Dec 8 '18 at 23:37
  • $\begingroup$ Perhaps it would be useful for you if you knew why the momentum operator is defined as it is. From the answers to that post we can see that there are many motivations for the momentum operator, none of which makes a direct connection to tangent vectors $\endgroup$
    – cxx
    Dec 8 '18 at 23:41
  • $\begingroup$ As a side note, not the vector is promoted to an operator but each of the scalar components of the vector is replaced by an operator $\endgroup$
    – Halbeard
    Dec 9 '18 at 1:07
  • $\begingroup$ @Hanting Zhang:That might be useful from a physical point of view, but that link does not go into this at all. In fact, the OP merely states the promotion of the momentum variable to the momentum operator as a given. $\endgroup$ Dec 9 '18 at 1:28

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