Talk:Orientation entanglement

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Is it correct to say (just before Formal details) "a spinor can be represented as a vector whose head is a flag lying on one side of a Möbius strip"?

This representation is a version of the "flagpole+flag+orientation-entanglement relation" image that I believe originated with Penrose, and is described in some detail in Misner, Thorne and Wheeler's "Gravitation" (MTW).

While MTW says on p. 1157 that "A spinor consists of this combination of (1) null flagpole plus (2) flag plus (3) the orientation-entanglement relation between the flag and its surroundings", nevertheless the next paragraph starts "One goes from a spinor ξ, a mathematical object with two complex components ξ1 and ξ2, to the geometric object of the "flagpole plus flag plus orientation-entanglement relation" in two steps:..." and then goes on to describe in detail how this combination is built up from 2-spinors.

In which case, might one try to diagram the 2-spinors themselves (or projections thereof), rather than the composite object described? If a spinor is an irreducible representation of a symmetry group, then shouldn't it have a geometry corresponding to its symmetry class in a similar way to that which gives the basic symmetry types of e.g. molecular structures, for which the geometries corresponding to the irreps are usually easily visualisable as the "basic" symmetries of the molecules?

If so, then the nearest I can get to a visualisation of a 2-spinor is something like a pair of co-axial helices (as complex exponential terms) that might somehow combine to form a vector, but I'm sure that's not (quite) right. 163.119.193.40 00:09, 11 November 2007 (UTC)[reply]

Question[edit]

If the cup is now brought up through the center of one coil of this helix, and passed onto its other side, the twist disappears. Can somebody explain this idea!!! Gvozdet (talk) 20:41, 10 March 2010 (UTC)[reply]

I don't know what you mean. The "idea" is already explained quite carefully, including a diagram that illustrates the same general principle with a ribbon. Sławomir Biały (talk) 15:14, 12 March 2010 (UTC)[reply]

I just don't understant some words - bring up through and pass onto Gvozdet (talk) 15:26, 13 March 2010 (UTC)[reply]

I see. I had considered the possibility that it might be a question of the actual meaning of the sentence. This is probably more easily illustrated with a belt. Putting two full twists in a belt creates two coils of a double helix. Now holding both ends of the belt fixed, the twist can be removed by performing a suitable ambient isotopy (that "passes" one end of the belt "through" one loop of the double helix). A native English speaker (or just a better writer) might be helpful in writing this differently. Sławomir Biały (talk) 18:12, 13 March 2010 (UTC)[reply]

Thank you. I think that I have understood. I have just watched the video. Gvozdet (talk) 20:06, 13 March 2010 (UTC)[reply]

The dynamic picture at [Plate trick] is clearer, as it shows the sides of the bands, and therefore the (lack of) twisting. Why not use it for this article as well?114.248.221.180 (talk) 04:07, 16 November 2017 (UTC)[reply]

Merge[edit]

Yes, I think the Plate_trick page should be merged into the bigger Orientation_entanglement page.. it's the same topic and the latter page does not mention "Feynman's plate trick" or "Dirac's belt trick" which seem to be common names/descriptions of this fascinating spinor behavior. Another name is "Dirac's spinor spanner" which I will edit in. —Preceding unsigned comment added by TBond (talk • contribs) 09:49, 29 August 2010 (UTC)[reply]

Support. SpiralSource (talk) 23:51, 28 March 2022 (UTC)[reply]

This makes no sense![edit]

Please assume I am not an idiot. Thanks. The article says that the cup is attached with two bands to two opposite walls. OK. The article says that rotating the cup 360° about its vertical axis results in it being "twisted". This makes no sense - each point of the cup is in exactly the same position it was prior to the rotation. Certainly the bands are twisted. Not the cup. The article goes on to claim imply that the floor has some importance here. I am GUESSING that the article is trying to explain that the two bands, rather than being wrapped around the circumference of the cup will be twisted around each other? Clearly an additional rotation by 360° (in the same direction) will double the twists. This has NOTHING to do with what happens in the first video (as far as I can see). This article is badly flawed and needs to be fixed. (The video shows rotation about a horizontal (not vertical) axis. Is this the problem?) I have better than average spatial intelligence, so if I am not "getting" this, then most people won't. No offense to the artist of the two coffee cup diagrams, but they are fairly useless. If it IS the bands that are the subject, then their twists ought to be clearly illustrated.173.189.79.137 (talk) 00:52, 15 October 2013 (UTC)[reply]

Green belt[edit]

If you look at the top or bottom of the green belt in the first gif, it appears to be perpetually twisting in the same direction around its long axis, with no relaxation. — Preceding unsigned comment added by 173.108.107.34 (talk) 06:06, 6 April 2019 (UTC)[reply]

Formal details[edit]

The paragraph starting “A unit quaternion ...” is telling in words the simple formulas appearing in Euler–Rodrigues formula, subsection Rotation. And the point is not about the quaternion (they are just a formal alternative to SU(2) matrices). So I would suggest to state in this last paragraph the obvious “formal detail”.

A unit quaternion has the cosine of half the rotation angle as its scalar part and the sine of half the rotation angle multiplying a unit vector along some rotation axis (here assumed fixed) as its vector part (also called imaginary part, see Euler–Rodrigues formula). If the initial orientation of a rigid body (with unentangled connections to its fixed surroundings) is identified with a unit quaternion having zero vector part and +1 for the scalar part, then after one complete rotation (2π rad) the vector part returns to zero and the scalar part has become −1 (entangled). After two complete rotations (4π rad) the vector part again returns to zero and the scalar part returns to +1 (unentangled), completing the cycle.

Replace by

The SU(2) matrix M for a rotation about the unit vector n (axis) by an angle θ can be written

{\begin{aligned}M&=\cos({\frac {\theta }{2}})\ {\begin{pmatrix}1&0\\0&1\end{pmatrix}}+{\rm {i}}\,\sin({\frac {\theta }{2}})\,n_{1}\ {\begin{pmatrix}0&1\\1&0\end{pmatrix}}+{\rm {i}}\,\sin({\frac {\theta }{2}})\,n_{2}\ {\begin{pmatrix}0&-{\rm {i}}\\{\rm {i}}&0\end{pmatrix}}+{\rm {i}}\,\sin({\frac {\theta }{2}})\,n_{3}\ {\begin{pmatrix}1&0\\0&-1\end{pmatrix}}\\&=\cos(\theta /2)\,I+{\rm {i}}\,\sin(\theta /2)\,n_{1}\,\sigma _{1}+{\rm {i}}\,\sin(\theta /2)\,n_{2}\,\sigma _{2}+{\rm {i}}\,\sin(\theta /2)\,n_{3}\,\sigma _{3},\end{aligned}}

where the $σ i$ are the Pauli spin matrices. If the initial orientation of the body (with unentangled connections to its fixed surroundings) is identified with the angle θ = 0, then after one complete rotation (θ = 2π rad) the matrix M has become −I (entangled). After two complete rotations (θ = 4π rad) the matrix returns to +I (unentangled), completing the cycle. DieHenkels (talk) 10:29, 24 October 2023 (UTC)[reply]