In this video lecture, the lecturer wrote spin up is $|\alpha\rangle = [\frac{1}{2}, \frac{1}{2}]$ and spin-down is $|\beta\rangle= [\frac{1}{2}, -\frac{1}{2}]$.
He then wrote that $\langle\alpha|\alpha\rangle = 1$ and $\langle\alpha|\beta\rangle = 0$
But I'm thinking that $\langle\alpha|\alpha\rangle = \alpha \cdot \alpha = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ ?
Above I'm basically taking the dot product.
Can someone explain what I'm getting wrong?
The terminology $|\alpha\rangle = |1/2,1/2\rangle$ does not mean that this is a two-element vector with components 1/2 and 1/2. Instead, the first element refers to the total spin $s=1/2$, and the second element refers to the $z$-component of the spin. These are just labels for the state that correspond to eigenvalues of the spin operators. Thus, these are the quantum numbers of the state. The state as written is assumed to be the normalized eigenvector of those operators with the corresponding eigenvalues, i.e. $$\hat{S}^2|1/2,1/2\rangle = \hbar^2\left(\frac{1}{2}\left(\frac{1}{2}+1\right)\right)|1/2,1/2\rangle, $$ and $$\hat{S}_z|1/2,1/2\rangle = \hbar\frac{1}{2}|1/2,1/2\rangle. $$
How are you taking a dot product? Regarding the ket $|s, m_s\rangle$ the quantity $s^2 + m_s^2$ does mean anything.
The Hilbert space for the spin of a spin $\frac 1 2$ fermion is two dimensional. The basis states are:
$$|s=\frac 1 2,m_s=+\frac 1 2\rangle $$ $$|s=\frac 1 2,m_s=-\frac 1 2\rangle $$
which are commonly called "spin up" and "spin down", respectively. They are orthonormal, so that:
$$ \langle \frac 1 2, m|\frac 1 2, m'\rangle= \delta_{mm'}$$
(note that $s$ is always $\frac 1 2$).
A general pure state can be written as:
$$ |\chi\rangle = \alpha|\frac 1 2, +\frac 1 2\rangle+\beta|\frac 1 2, -\frac 1 2\rangle$$
where $\alpha,\beta$ are complex numbers in the unit disk. Note that $\chi$ is normalized so that:
$$ \langle \chi|\chi\rangle = \big[ \alpha^*\langle \frac 1 2,+\frac 1 2|+\beta^*\langle \frac 1 2,-\frac 1 2| \big] \big[\alpha|\frac 1 2, +\frac 1 2\rangle+\beta|\frac 1 2, -\frac 1 2\rangle\big]\equiv 1$$
or
$$\alpha^*\alpha\langle \frac 1 2, +\frac 1 2|\frac 1 2, +\frac 1 2\rangle + \alpha^*\beta\langle \frac 1 2, +\frac 1 2|\frac 1 2, -\frac 1 2\rangle + \beta^*\alpha\langle \frac 1 2, -\frac 1 2|\frac 1 2, +\frac 1 2\rangle + \beta^*\beta\langle \frac 1 2, -\frac 1 2|\frac 1 2, -\frac 1 2\rangle=1$$
Apply the orthonormality conditions:
$$|\alpha|^2 +|\beta|^2= 1$$
so that you can write them in terms of real angles:
$$ \alpha= \cos{\frac{\theta}2}$$ $$ \beta= \sin{\frac{\theta}2}e^{i\phi}$$
(note an overall global phase has been factored out make $\alpha$ real). In this form, the Hilbert space is referred to as The Bloch Sphere.
If the bra-ket notation is too abstract, one can write a mathematical representation of the general state as:
$$ \chi = \left[\begin{array}{c}\alpha\\\beta\end{array}\right]$$
The spin up/down states are eigenvectors of:
$$ \hat S_z = \left[\begin{array}{cc}+\frac 1 2 & 0 \\0&-\frac 1 2\end{array}\right]$$
with eigenvalue $\pm \frac 1 2 $. The orthonormality relations are then straight forward.
All states are eigenstates of total angular momentum squared ($\hbar=1$):
$$ \hat S^2 = \left[\begin{array}{cc}\frac 3 4 & 0 \\0&\frac 3 4\end{array}\right]$$
with eigenvalue:
$$\lambda = s(s+1) = \frac 1 2(\frac 1 2 + 1)= \frac 3 4$$
Note the state $\chi$ is the "wave function", though it's usually called a "state" because, unlike $\psi(x)$, it's not defined over a continuous (position) variable, which is how we usually think of function.
