# Phase Factor of Spinors (what they represent)

1. Jun 22, 2012

### lzydesmond

Hi all,

I am troubled by the flag and flagpole analogy for two-spinors and would like some clarification.

Please refer to the post by Hans de Vries.

Am I right to say that the usage of spin rotation operators (eg exp(-i($\phi$/2)σ(x)), which is a $\phi$ degrees rotation in the 3D space about the positive x axis would affect the phase factor of the spinor?

spinor s is given by
s = s*exp(-i($\alpha$/2)) (cos($\vartheta$/2))exp(-i($\phi$/2)), sin($\vartheta$/2))exp(i($\phi$/2))) (column vector here)

note, alpha, theta and phi represent angle of the flag about the flagpole, angle downwards from the z-axis and angle from the x-axis. phi here is not the same as the arbitrary phi for the spin rotation operator.

My question is whether it is true that when the spin rotation operator is used, it affects the term exp(-i($\alpha$/2) in a way such that a rotation of 360 degrees makes the spinor negative and thus a rotation of 720 degrees is needed for the spinor to return to its original state. Ie, when the flagpole rotates, the flag is also rotating but while the flagpole has a period of 360 degrees(like a usual vector in 3D), the flag has a period of 720 degrees.

I mean is this the only way to see how spin rotation operators affect the spinor? (in other words the state of electron) This is because I can't see directly how rotation operators affect the phase factor other than this way. Please enlighten me. Thank you.

2. Jun 22, 2012

### lzydesmond

hi there, it seems that no one has an answer/understood what I was saying.

I gave some though to the issue...
what I think is that the rotation of the flag (angle alpha) has no effect on rotations. it is what is called global phase factor I believe? exp(-i(alpha)/2)

I think I got muddled up with the spin rotation operator which does produce a phase factor when acting on an eigenstate. This phase factor exp(-i(phi)/2) is crucial in determining the state of spin (since phi = 2pi gives a phase factor of -1).

SU(2) is a group homomorphism to SO(3) (2:1) precisely because we need 2 turns of 360 degrees to get back to the original state. But the first turn and the second turn are essentially the same thing in SO(3), a rotation of 360 degrees.

3. Jun 22, 2012

### PhilDSP

Hi lzydesmond,

I'll take an initial swing at this and others can correct or augment... SU(2) is not simply a homomorphism of SO(3) but is a double cover or 2-to-1 homomorphism of SO(3). There are 2 ways within SU(2) to get to any result in SO(3). What is considered an axis in SO(3) is really a pseudo-vector or axial vector in contrast to a polar vector. The algebraic rules for polar and pseudo-vectors are

$i r = -r~~~~$ when $r$ is a polar vector
$i r = r~~~~~~$ when $r$ is a pseudo-vector

The cross product $r \times s$ is a pseudo-vector when $r$ and $s$ are of the same type and a polar vector when there are of mixed types. The cross product is really an alternate description of a rotation in SO(3) or SU(2).

If you flip the rotation axis in SO(3) the rotation appears to switch directions. Another (180 degree) flip will return the original rotation direction. But in SU(2), you need to consider the effect that a flip of the rotation axis (or flagpole) has in terms of conjugates of the vectors or pseudo-vectors as described above. That is, SU(2) forces the distinction above.

This seems to mean that if a rotation can be described in SO(3) without recourse to SU(2) then you won't observe a 360 degree to 720 degree difference, while if your physical situation can be adequately described only within SU(2), spin $\frac{1}{2}$ particles for example, then the difference can potentially be observed in certain circumstances.

Last edited: Jun 22, 2012
4. Jun 22, 2012

### lzydesmond

okay thanks for your reply. The part on the global phase and the phase introduced by spin rotations is still not resolved though. I am still unsure whether the flag actually rotates during a rotation.

5. Jun 22, 2012

### PhilDSP

The flag/flagpole terminology was introduced by Penrose and Rindler I think. I suspect it has special meaning only within their particular conceptualization. More generally, a spinor represents a rotation so I don't think you need to ask if it is itself rotating.

You can compound or perform an integration of any number of infinitesimal rotations to arrive at a particular physical rotation. It doesn't matter in what order you perform the integration, the total rotation is invariant. That is the rationale for developing spinors as opposed to dealing with the non-commuting nature of normal 3D axial space rotations.

Last edited: Jun 22, 2012
6. Jun 22, 2012

### lzydesmond

yes I kind of get what you mean by a spinor represents a rotation. Actually I would think it is a spin state, a state whereby we can express in terms of eigenspinors. I mean I found their conceptualization quite useful, except for the flag rotation part. Is there a better conceptualization? or anyway whereby I can conceptualize rotation of spinors in three dimensions.

7. Jun 22, 2012

### PhilDSP

Elie Cartan's original conceptual development of spinors as isotropic vectors is probably an excellent way to map normal 3 vectors to spinors and vice versa. I particularly like Hladik's more recent treatment of the subject.

8. Jun 23, 2012

### Hans de Vries

Yes, the reverse is also true.

The U(1) operator $e^{i\phi}$ has the beautiful fundamentally important property that it
always rotates a spinor around its own axis entirely independent of the direction
of the spinor. For a spinor $\xi_s$ pointing in the $\vec{s}$ direction we can therefor write.

$$e^{i\vec{s}\cdot\vec{\sigma}}\xi_s~~=~~e^{is}\xi_s$$

The reason for this is that we define the spinor $\xi_s$ as being an eigenvector of
the boost matrix $\vec{s}\cdot\vec{\sigma}$ in the $\vec{s}$ direction, because if you boost a spin in its own
direction then its changes magnitude but the direction stays the same.
We therefor may replace the matrix $\vec{s}\cdot\vec{\sigma}$ at the left hand side by the eigenvalue
$s$ at the right hand side.

Hans.

9. Jun 23, 2012

### lzydesmond

ok so in short, whenever a spin rotation matrix is used (around any axis), angle alpha measured in the orientation of the flag changes so the flag rotates and the flag has a period of 720 degrees (which corresponds to 2 full circles). This term exp(-i(alpha/2) boosts the magnitude of the spinor. (it multiplies the spinor by a complex number). When the flag achieves a rotation of 360 degrees, it becomes obvious when the spinor gains a negative sign (as compared to a similar rotation, with ref to your figure 16.7). So the spinor has an overall sign change.

Am I right in all these?

10. Jun 23, 2012

### Hans de Vries

That's all correct except that the magnitude stays equal when applying $e^{-i\alpha/2}$.

(The term $-i\tfrac12\vec{s}\cdot\vec{\sigma}$ is the generator of the rotation operator while $\pm\tfrac12\vec{s}\cdot\vec{\sigma}$ is the
generator of boost operations)

Hans.

11. Jun 23, 2012

### lzydesmond

Okay, I think I kind of get what you mean. Because exp(-i(alpha/2) has modulus 1, the magnitude remains unchanged. Could you clarify on this part (as quoted above)? What are boost operations?

12. Jun 23, 2012

### Hans de Vries

A boost is just another word for a one-time acceleration (change of one velocity to
another one) in the context of special relativity and Lorentz transforms.

If you give a particle with spin a velocity then, in general the spin-pointer will change
direction, except if you accelerate the particle in the direction parallel to the spin.

Hans.

13. Jun 23, 2012

### lzydesmond

Hi, I have tried to search for a general boost operator but to no avail. I understand that exp(-i($\vartheta$/2)(n$\cdot$$\sigma$)) is the spin operator around axis n, where n is a unit vector. May I ask what is the equivalent for boost operator? Also, I don't see how exp(is) changes the magnitude of the spinor. Do you mean exp(s) instead?

14. Jun 23, 2012

### Hans de Vries

The boost operator is very similar. Just remove the i from the argument. See also here in 16.16

It doesn't. The i makes the difference between a boost and a rotation (there's also a
sign depending on which of the two chiral components you are operating on).

Hans.

15. Jun 23, 2012

### lzydesmond

okay I kind of understood what you have said. You used the boost matrix (which is also the spin matrix) $\vec{s}\cdot\vec{\sigma}$ by taking its eigenvalue, (s here) to show the above equation for spin rotation operator. But the equivalent for boost operator would be without the i right? The state of the spin would not change in direction but receive a boost in magnitude, exp(s) instead. Am I right on this now?

for boost operator
$$e^{\vec{s}\cdot\vec{\sigma}}\xi_s~~=~~e^{s}\xi_s$$

16. Jun 23, 2012

### Hans de Vries

Yes,

Hans

17. Jul 8, 2012

### PhilDSP

Looking into this further a bit, I gather that flags and flagpoles are a development that lead to the Twistor formalization. Penrose and Rindler in "Spinors and space-time" on p. 127 state:

A flag manifold can then be defined wherein the space of Twistors is associated with the family of flag manifolds.

The rationale for developing Twistor theory seems to be to overcome a limitation in working with SU(n) groups. Working purely in SU(n), as you might for modeling basic characteristics of elementary particles, you essentially work with dynamics. That is, you make no presumptions about the application of a time-space metric ($E^4$ and $E^3$ are implicit). If you wish to apply a space-time metric, you then work kinematically. However, it is known that there exist no non-trivial SU(n) anti-self-dual gauge fields in Minkowski space. Twistor formalism gives mappings between Twistor space and Minkowski space. Twistors provide the self-dual and anti-self-dual structures compatible with SU(n).

Last edited: Jul 8, 2012
18. Jul 8, 2012

### Spinnor

So if I have an electron that is equally likely to be spin up as spin down and I boost the electron in the up direction it will now be more likely to be a spin up electron?

Thanks for any help!

19. Jul 8, 2012

### lzydesmond

You need to define the axis for the spin and determine the axis for the boost as well. After that, you just need to do some calculations for the amplitude and square it for the probability.