# the operator of spin

I am kind of confused about the operator for spin, all the ones I've seen are written in terms of matrix, but this form can not be applied to things like $1s$ orbital of a hydrogen atom, which is $e^{-(r/a)}$. Or is there actually a way to multiply matirx with functions that Im not aware of. I am taking a quantum chemistry class which does not deal with dirac notation, so I am not sure how to use a matrix as an operator where we have wave functions , not "wave vectors"

Wave functions in ordinary non relativistic quantum mechanics, written $\psi (x)$, only capture the spatial part of the full state of a particle with spin $\not= 0$. The full state $\Psi$ is equal to the tensor product of $\psi (x) \otimes\chi_{s}$ written as the product $\psi(x) \chi_{s}$ where $\chi$ is the 2s+1 dimensional spin state where s is the intrinsic spin of the particle.
For electrons in a hydrogen atom s = 1/2 so $\chi_s$ is a two dimensional column vector.
The spin operator $S$ acts on the spin part of the total state $\Psi$, which is a 2x1 column vector and as such can be acted on by a 2x2 matrix.
• Sorry to confuse you, maybe in words is easier. A particle without spin can be purely describe by its wave function $\psi$. But when a particle has spin it's state can no longer just be described by this wave function, because it has more degrees of freedom than just $x, y, z$ in $R^{3}$. The extra degree of freedom is captured by the spin part of the state. The $\chi_s$ I mentioned. The spin operators S is a vector operator (S_x, S_y, S_yz), their form can be looked up easily on the internet. Just act S_x, or S_y, or S_z, on the spin state $\chi_i$. – CStarAlgebra Sep 22 '16 at 4:45