I Understanding spinor transformation law

pellis
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According to Misner, Thorne & Wheeler, the transformation law for spinors is given as:
ξ => ξ’ = Rξ …where R(θ) = cos(θ/2) – isin(θ/2)[σ.n]
Does this mean a resulting spinor rotation is, in effect, equivalent to a reflection?
REMOVED pending revision
 
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pellis said:
Or have I got it wrong? Advice greatly appreciated - Paul
I don't have the MTW book, so I hope this clarifies things.

The spin of a particle in QM is represented by a spin state, which can be thought of as a vector in an abstract vector space. This vector space of spin states is not Euclidean 3D space, but a 2D complex vector space.

If we study the effect of a spatial rotation of the system in 3D Euclidean space, then we can use the mathematical machinery of QM to find the effect of this spatial rotation on the spin state of spin 1/2 particle. And what we find is that the spin state changes according to a operator whose parameter is half the spatial rotation angle. To be precise we find that the spin state is changed according to the following operator:
$$\exp(-\frac{(\vec \sigma \cdot \hat n)\theta}{2})$$I'm sure MTW has the matrix expansion of this, so I'll save myself typing it in.

The perhaps surprising result is that for a spin 1/2 particle a 360 degree spatial rotation transforms the spin state to the negative of the original state (*). This means that the expected value of every observable is unchanged (the factor of ##-1## is simply a phase factor). Nevertheless, the spin state is not the same as the original. And this may be demonstrated in a neutron interferometry experiment to study ##2\pi## rotations, as referenced in Modern QM by Sakurai, page 162.

(*) PS $$\exp(-\frac{(\vec \sigma \cdot \hat n)2\pi}{2}) = -I$$Where ##I## is the identity transformation.
 
Thank you for your reply.

Although I posted the original on Quantum Physics, and mentioned spinors as 2 complex-component objects, the question wasn’t directed at fermion spin, but at an aspect of spinor transformations more generally. (MTW, as you probably know, is about gravitation, and even goes so far as to say “…there is nothing that limits the usefulness of the spinor concept to rotations.”)

Later in MTW it refers to Lorentz transformations of spinors: “The spinor has acquired a significance of its own through one's having pulled out half of the transformation formula…”

That, and my understanding of how rotations can be viewed as pairs of reflections, motivated me to ask, originally, if R(θ)ξ rotates ξ by θ or θ/2, but checking just after posting the question I realized that the inversion that defines spinors requires it to be θ/2, as also for a single side of the vector “sandwich product”, hence my modification to the Summary: “Does this mean a resulting spinor rotation is, in effect, equivalent to a reflection?”

And I will also appreciate any comments on the revised version to follow.
 
pellis said:
“Does this mean a resulting spinor rotation is, in effect, equivalent to a reflection?”
How could it be? A rotation has an axis and a parameter ##\theta##. And, a reflection has only an axis.
 
From MTW (there's a PDF version on line, wjhich I acquired after originally purchasing a hardcopy of fthe book):

Why do half-angles put in an appearance? And what is behind the law of combination
of rotations? The answer to both questions is the same: a rotation through
the angle 0 about a given axis may be visualized as the consequence of successive
reflections in two planes that meet along that axis at the angle 0/2 (Figure 41.2).
Two rotations therefore. imply four reflections. However, it can be arranged that
reflections no. 2 and no. 3 take place in the same plane, the plane that includes
the two axes of rotation. Then reflection no. 3 exactly undoes reflection no. 2. By
now there remain only reflections no. 1 and no. 4, which together constitute one
rotation: the net rotation that was desired (Figures 41.3 and 41.4).

.
.
.
1636401111346.png

x
Figure 41.3.
Composition of two rotations seen in terms of reflections. The first rotation (for instance, 90· about
02 in the example of Figure 41.1.) is represented in terms of reflection I followed by reflection 2 (the
planes of the two reflections being separated by 90·/2 = 45· in the example). The second reflection
appears as the resultant of reflections 3 and 4. But the reflections 2 and 3 take place in the common
plane 20X. Therefore one reflection undoes the other. Thus the sequence of four operations 1234
collapses to the two reflections I and 4. Their place in tum is taken by a single rotation about the axis
OA.

Neat!

I'll repost my original, which explains how this is relevant to my query, but with some modifications,

Thanks for your interest..
 
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