Spin angular momentum operator queries

[gfd43tg]

Hello,

For the spin angular momentum operator, the eigenvalue problem can be formed into matrix form. I will use $S_{z}$ as my example

$$S_{z} | \uparrow \rangle = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \frac {\hbar}{2} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

$$S_{z} | \downarrow \rangle = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix}= \frac {-\hbar}{2} \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

Why is it that the matrix on the left hand side is assumed to be a 2x2 matrix? Also, why is it when solving for a,b,c, and d the second equation is used. I don't understand how I can just say

$$\begin{pmatrix} a \\ c \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \frac {\hbar}{2} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$
$$\begin{pmatrix} b \\ d \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix}= \frac {-\hbar}{2} \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

I don't understand how I can just isolate the second column of the 2x2 matrix for the spin down case, and isolate the first column for spin up.

 

[blue_leaf77]

Why is it that the matrix on the left hand side is assumed to be a 2x2 matrix?

Those two equations are only for particles with spin 1/2 such as electron. Following the commutation relations for angular momenta, which is satisfied by any angular momentum operator, the z component of an angular momentum L can only be between -L and L with unit increment. That's why spin operator matrix for spin 1/2 particle must be of size 2x2.

I don't understand how I can just isolate the second column of the 2x2 matrix for the spin down case, and isolate the first column for spin up.

To be honest this is also my first time seeing such expressions. But I think you can understand it by viewing the matrix multiplication using coordinate-by-column rule. If you had known this, you can see that such way of writing can only be justified because either for spin up or down vector, one of the element is zero.

 

[gfd43tg]

Could you explain the part about the matrix size in more detail? I still don't understand how one follows from the other

 

[blue_leaf77]

If this is your first exposure in angular momentum theory in quantum mechanics, please just forget "Following the commutation relations for angular momenta, which is satisfied by any angular momentum operator." If you are planning to go deeper in QM you will encounter the formal mathematical proof anyway. For now just keep in mind that the consequence of that statement is that any particle having angular momentum quantum number $s$ ($s$ is necessarily positive and either integer or half-odd integer depending on the particle) will have quantized possible z-components of the angular momentum as given by $-s\hbar,(-s+1)\hbar,\dots,(s-1)\hbar,s\hbar$. Now since electrons is a spin 1/2 particle, its $s$ is 1/2. So the z component of spin angular momentum of electrons can only be -1/2$\hbar$ or 1/2$\hbar$.
Now I assume you are familiar with eigenvalue problem. The eigenvalues of matrix operator $s_z$ for electrons must then be -1/2$\hbar$ or 1/2$\hbar$, so the size of $s_z$ is necessarily 2 by 2. Remember that a square matrix of dimension NxN has N eigenvalues taking any multiplicity into account.

 

[gfd43tg]

Thanks, I was not aware that a square matrix of NxN has N eigenvalues. My linear algebra is too weak to remember that fact, it has been a while since I studied it.