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I am trying to follow a course on quantum mechanics taught by Leonard Susskind on YouTube.

Link to video: https://youtu.be/JzhlfbWBuQ8?t=4886

At 1:21:30, he introduces the distinction between the way classical mechanics is formulated as compared to quantum mechanics. In classical mechanics all the states of a system are specified as points on a phase space diagram. In QM states are vectors in vector spaces. I am just confused about how in the example of the phase space diagram he drew, he had position on the vertical axis and momentum on the horizontal axis, yet the state was not considered a vector, even thought it is specified by vectors.

Can someone explain my misconception please.

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  • $\begingroup$ With due respect to the formidable Lenny, classical mechanics' difference from quantum mechanics is not in the spaces each is expressed: quantum mechanics can also be expressed in phase space, and classical mechanics in Hilbert space. The two spaces are interconnected invertibly. $\endgroup$ – Cosmas Zachos Nov 24 '19 at 14:56
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Susskind is trying to stress the difference between elements (points) in a phase space and elements (vectors) in a (pre-)Hilbert space, admittedly causing some confusion by making oversimplified claims along the way.

Phase space is not necessarily a vector (affine) space, i.e. it does not necessarily have a linear (affine) structure, respectively. The main case of a phase space is the cotangent bundle $T^{\ast}M$ over a configuration manifold $M$.

Example: The cotangent bundle $T^{\ast}\mathbb{S}^1$ over a circle $\mathbb{S}^1$ is neither a vector space nor an affine space.

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  • $\begingroup$ It might be helpful if you could give an example to illustrate why a phase space need not have the structure of a vector space. I'm thinking that Darboux's theorem guarantees that it has this structure in any neighborhood. Are you talking about a possibility that this can't be extended globally, maybe because the manifold has the wrong topology? $\endgroup$ – user4552 Nov 24 '19 at 22:04
  • $\begingroup$ Yes. I updated the answer. $\endgroup$ – Qmechanic Nov 24 '19 at 22:35
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    $\begingroup$ Nice, thanks. So maybe in less fancy language: -- In quantum mechanics, it always makes sense to talk about mixing together two states, with some complex phase. But in classical mechanics, this need not make sense. For example, suppose our system is an object that can rotate. Then even without worrying about complex numbers, it doesn't necessarily make sense to talk about a state that is the midpoint between orientation A and orientation B. For example, there is no point on the earth's surface that is "the" point half-way between the north and south poles. Does this seem right? $\endgroup$ – user4552 Nov 24 '19 at 22:37
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    $\begingroup$ @Ben Crowell. In QM, linear combinations of kets are meaningful kets; but even though $\vec x_1+ \vec x_2$ is meaningful and so is $\vec p_1+ \vec p_2$, a conjectural 6-tuple $(\vec x_1+ \vec x_2, \vec p_1+ \vec p_2)$ isn't, even though it is ubiquitous in symplectic geometry. Note total angular momentum is not built out of it, but out of the two 3-vectors, $\vec x_1 \times \vec p_1+ \vec x_2 \times \vec p_2$, for example... $\endgroup$ – Cosmas Zachos Nov 25 '19 at 20:25
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(x, p) is not a vector in 3D as it has 6 components. It also is not a vector in 6D as transformations between x and p are not defined.

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    $\begingroup$ canonical transformations will often mix momenta and position, and a significant advantage in going to the Hamiltonian form of the EOM since their form is invariant under canonical transformations. $\endgroup$ – ZeroTheHero Nov 24 '19 at 14:01
  • $\begingroup$ @bencrowell I side with Qmechanic on this. $\endgroup$ – my2cts Nov 24 '19 at 19:49
  • $\begingroup$ I stand corrected on the fact that it need not be a vector space. However, the second sentence still doesn't make much sense. $\endgroup$ – user4552 Nov 24 '19 at 22:05

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