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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.

Mozibur Ullah
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  • See this post: Is the Momentum Operator a Postulate?. – cxx Dec 08 '18 at 16:31
  • @Hanting Zhang: I don't really see the relevance to my question. What was it that you were intending to point out? – Mozibur Ullah Dec 08 '18 at 23:37
  • 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 – cxx Dec 08 '18 at 23:41
  • 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 – Halbeard Dec 09 '18 at 01:07
  • @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. – Mozibur Ullah Dec 09 '18 at 01:28
  • @Halberd Rejoyceth: That may be done on a practical level, but physically this does not make sense as this supposes that we fix a basis both in position and momentum space. There are no such natural bases in either of these spaces. – Mozibur Ullah Dec 09 '18 at 01:30
  • My point is that seperate scalar degrees of freedom in phase space are identified with operators since the relevant commutation relation $[x,p] =i\hbar$ only holds for variables of the same direction. The vector components don't have anything to do with the position/momentum space distinction. Only the operator representation does, which you chose as the position space representation in your question. In momentum space the momentum operator would appear as a 'multiplication' and the position as a derivative. – Halbeard Dec 09 '18 at 02:02
  • I don't know if this relevant, but at the classical level the velocity is in the tangent space, the momentum is in the cotangent space. – MBN Dec 09 '18 at 13:37
  • @Halberd Rejoyceth: It's not 'seperate' but 'separate'. – Mozibur Ullah Dec 09 '18 at 15:09

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