# Why we know spin vector is an axial vector?

1. Apr 26, 2012

### ndung200790

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.

2. Apr 29, 2012

### Jano L.

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.

3. Apr 29, 2012

### tiny-tim

hi ndung200790!
we can't add axial vectors and non-axial vectors (if forget the correct name )

if spin wasn't axial, we wouldn't be able to interchange spin angular momentum and orbital angular momentum

4. Apr 29, 2012

### vanhees71

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.

5. Apr 29, 2012

### strangerep

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".)