Is the total Spin operator a vector

[BigDig123]

Hello,
I am learning about Excited states of Helium in my undergrad course. I was wondering if the total spin operator

Ŝ

is a vector quantity or not.

Thanks for your help.

 

[BigDig123]

Okay well I have hopefully answered my own question...
I read that the total angular momentum operator is a vector quantity so I have assumed that the total spin operator is one also.

If anybody knows if it is a tensor or anything else please reply.

 

[hilbert2]

The spin operator $\mathbf{S}$ is composed of elements $S_i$ as in $\mathbf{S}=(S_x ,S_y ,S_z )$, and the three components have the same commutation relations as the components of any angular momentum operator. Actually, the commutation relations define what is meant with an angular momentum, so there can't be an angular momentum quantity that wouldn't have three components and be a vector.

 

[blue_leaf77]

If anybody knows if it is a tensor

Any vector is a rank-1 tensor. If it's a vector operator like spin operator then it can also be called a rank-1 tensor operator.

 

[Henryk]

Firstly, definition of a vector. In terms of linear algebra, spin is an operator, not a vector. It is quantum states that are vectors and operators, such a spin, operate on vectors transforming them to another vector. Vectors are entities that can be added, multiplied by a number, etc. You don't do that to operators.

There is another aspect to that: total spin has three components: x, y and z, and in this sense you could talk about it as a vector except for one thing: vectors under coordinate inversion change sign. Spin and angular momentum does not!. In classical physics, angular momentum is a pseudovector because of this invariance under inversion.

 

[hilbert2]

Firstly, definition of a vector. In terms of linear algebra, spin is an operator, not a vector. It is quantum states that are vectors and operators, such a spin, operate on vectors transforming them to another vector. Vectors are entities that can be added, multiplied by a number, etc. You don't do that to operators.

There is another aspect to that: total spin has three components: x, y and z, and in this sense you could talk about it as a vector except for one thing: vectors under coordinate inversion change sign. Spin and angular momentum does not!. In classical physics, angular momentum is a pseudovector because of this invariance under inversion.

Actually, the set of operators acting in $\mathcal{H}$ do form a vector space too, just as the set of $N\times N$ matrices is a $N^2$-dimensional vector space. The way how the operators behave in addition and scalar multiplication satisfies the axioms of a vector space, despite there being a noncommutative operator multiplication too (which doesn't necessary have to exist in a vector space).

There is a difference in the way how vectors are defined in pure mathematics and how they are often defined in physics. The physical definition of a vector is based on how the components of some object transform in rotations (ordinary 3-vector) or in Lorentz transformations (relativistic 4-vector). The spin operator that is discussed in here belongs in the former class, despite the three components being operators rather than real or complex numbers. For instance, if you change your coordinate system in a way where the previous x-axis becomes the new y-axis and the other way around, then also your spin operator changes as $\mathbf{S} = (S_x ,S_y ,S_z ) \rightarrow (S_y , S_x ,S_z)$.