Why we know spin vector is an axial vector?

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The discussion centers on the classification of spin vectors as axial vectors in the context of angular momentum. Participants clarify that both orbital angular momentum and spin are derived from similar mathematical principles, specifically the product of position and momentum vectors. The axial nature of these vectors is emphasized, particularly in relation to their behavior under coordinate transformations. The conversation also touches on the implications of relativistic quantum mechanics, where the distinction between orbital and spin angular momentum becomes complex.

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Please teach me this:
We know that orbit angular momentum is the product of coordinate operator vector and momentum operator,so when we reflect the coordinate system the angular momentum is unchanging(axial vector).But I do not understand why spin vector is axial vector.
Thank you very much in advance.
 
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Good question. I think that spin was introduced just as another moment of momentum, in addition to the orbital moment of momentum. So all the mathematics is very similar, including the axialness.
One way to think of spin moment of momentum is that the particle performs fast circular motion around some average position, and then spin is again product of position vector and momentum. This picture has however the problems that the particle should then radiate electromagnetic waves, which was not recognized in experiments.
 
hi ndung200790! :smile:
ndung200790 said:
… But I do not understand why spin vector is axial vector.

we can't add axial vectors and non-axial vectors (if forget the correct name :redface:)

if spin wasn't axial, we wouldn't be able to interchange spin angular momentum and orbital angular momentum :wink:
 
Unfortunately this argument is not true since the weak-interaction current is precisely of (V-A) form, i.e., you subtract the axial-vector from the vector current. Of course, your argument is true, when considering only parity-conserving interactions (strong, electromagnetic).

Further, in relativistic quantum mechanics there is no unique frame-independent splitting of orbital angular momentum and spin. Thus it is not easy to answer the question within relativistic theory, but of course total angular momentum must be an axial vector.
 
ndung200790 said:
PWe know that orbit angular momentum is the product of coordinate operator vector and momentum operator,so when we reflect the coordinate system the angular momentum is unchanging(axial vector). But I do not understand why spin vector is axial vector.
Let's write the (classical) orbital angular momentum tensor (components) as
$$
J_{ij} ~=~ x_i p_j - x_j p_i ~.
$$
As you said, this does not change under a coordinate reflection.

But this is often written in the form ##L = x \times p## or, with explicit indices, as
$$
L_i ~=~ \varepsilon_{ijk} x^j p^k
$$
where ##\varepsilon_{ijk}## is the Levi-Civita antisymmetric symbol, and the Einstein summation convention is used.

Now, ##\varepsilon_{ijk}## are components of a pseudo-tensor, since it also changes sign under a reflection. Therefore the ##L_i## form a pseudo-vector (which is another name for "axial vector".)

Here's some Wiki links for more information:

http://en.wikipedia.org/wiki/Levi-Civita_tensor
http://en.wikipedia.org/wiki/Axial_vector
 
Hi Vanhees71!
Do you mean that in relativistic QM the sum of orbit angular momentum and spin is conservable,then spin must be axial vector?
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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