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?

  • 1
    $\begingroup$ Use \langle and \rangle instead of < and >. $\endgroup$ Sep 3, 2021 at 17:18
  • $\begingroup$ @Jakob Thank you $\endgroup$
    – Frank
    Sep 3, 2021 at 17:38

2 Answers 2


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 $$\hat{S}_z|1/2,1/2\rangle = \hbar\frac{1}{2}|1/2,1/2\rangle. $$

  • $\begingroup$ So is $\langle \alpha | \alpha \rangle = \delta_{i, j}$ really implying that the wavefunction, regardless of the other parameters $(n, l, m_l)$, is kronecker delta? And really they just left the other parameters $n, l, m_l$ out of the math? Should I assume $| \alpha \rangle$ is really $| n, l, m_l, \alpha \rangle$ for ANY $n, l, m_l$ in the identity $\langle \alpha | \alpha \rangle = \delta_{i, j}$? $\langle n,l,m, \alpha | n,l,m, \alpha \rangle = \delta_{i, j}$ $\endgroup$
    – Frank
    Sep 4, 2021 at 15:29
  • $\begingroup$ $\langle\alpha|\alpha\rangle=1$, not a kronecker delta, because these are the same state. The full hydrogen wave function is $|nlm\alpha\rangle$ which can be interpreted as a product $|nlm\rangle|\alpha\rangle$, where spin operators only hit the $|\alpha\rangle$ part and spatial operators only hit the $|nlm\rangle$ part. Your use of $\delta_{ij}$ is incorrect. $\langle nlm\alpha|nlm\alpha\rangle=1$, but $\langle n'l'm'\alpha'|nml\alpha\rangle = \delta_{ll'}\delta_{nn'}\delta_{mm'}\delta_{\alpha\alpha'}$. I.e., if any two of the quantum numbers are different, the inner product is zero. $\endgroup$
    – march
    Sep 4, 2021 at 15:50
  • $\begingroup$ What is the actual function defined by $| \alpha \rangle$ ? I think we are really saying there is some function $f(s, m_s)$ that has the property $\langle f(s, m) | f(s, m')\rangle = \delta_{mm'}$. But I have never actually seen anyone write out this function ever. $\endgroup$
    – Frank
    Sep 4, 2021 at 15:56
  • $\begingroup$ You can kind of think of $|\alpha\rangle$ as a $\alpha(\sigma)$, where $\sigma$ takes on only two possible values (1/2 and -1/2), and $\alpha(\sigma)$ is the amplitude for the state being in state $|1/2\rangle$. We're abusing an mixing up notation a little bit. Generally speaking, $|\alpha\rangle = c_1|1/2,{-1/2}\rangle+c_2|1/2,{1/2}\rangle$, and the relationship to the function language is $\alpha(-1/2) = c_1$ and $\alpha(1/2)\c_2$. It's easier to just think of $|alpha\rangle$ as a vector in 2D vector space. $\endgroup$
    – march
    Sep 4, 2021 at 18:14

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:

$$ \langle \chi|\chi\rangle = \big[ \alpha^*\langle \frac 1 2,+\frac 1 2|+\beta^*\langle \frac 1 2,-\frac 1 2| \big] \big[\alpha|\frac 1 2, +\frac 1 2\rangle+\beta|\frac 1 2, -\frac 1 2\rangle\big]\equiv 1$$


$$\alpha^*\alpha\langle \frac 1 2, +\frac 1 2|\frac 1 2, +\frac 1 2\rangle + \alpha^*\beta\langle \frac 1 2, +\frac 1 2|\frac 1 2, -\frac 1 2\rangle + \beta^*\alpha\langle \frac 1 2, -\frac 1 2|\frac 1 2, +\frac 1 2\rangle + \beta^*\beta\langle \frac 1 2, -\frac 1 2|\frac 1 2, -\frac 1 2\rangle=1$$

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.

  • $\begingroup$ If hydrogen radial function is $Re(n, l, r)$ and spherical harmonics is $Y(l, m_l, \theta, \Phi)$ then is $|n, l, m_l, m_s \rangle = Re * Y * |s, m_s> $? If so, what is the function $|s, m_s>$ ? I think I need to see a proof of how $\langle \frac{1}{2}, m | \frac{1}{2}, m' \rangle = \delta_{mm'}$ before I really understand this. $\endgroup$
    – Frank
    Sep 4, 2021 at 15:38
  • $\begingroup$ @Frank The Hilbert space is the Bloch Sphere, which describes any two state quantum system. There are no functions of position, as there is no position. $\endgroup$
    – JEB
    Sep 4, 2021 at 16:30
  • 1
    $\begingroup$ $a$ and $b$ aren't coordinates, they're labels. If they label an orthonormal basis, then: $\langle a,b|a',b'\rangle = \delta_{aa'}\delta_{bb'}$, by definition. E.g in a 2d infinite square well, you can label eigenstates with $a=(1,2,\ldots)$ and $b=(1,2,\ldots)$, then instead of writing $\psi(x,y)=\frac 1 {N_{ab}} \sin(a\pi x/L)\sin(b\pi y/L)$ you just call it $|a,b\rangle$. $\endgroup$
    – JEB
    Sep 9, 2021 at 16:42
  • $\begingroup$ Hi I deleted that because this video explained it to me. I can consider spin up like (1, 0) and spin down like (0, 1) vectors: youtube.com/watch?v=BWM0RXg-uvI $\endgroup$
    – Frank
    Sep 9, 2021 at 16:43
  • $\begingroup$ No to delete questions. I had added the vector form already. I think it's better to learn them 1st, before going abstract with bra-ket. Nevertheless, the vector space in which those column vectors live is pretty abstract. $\endgroup$
    – JEB
    Sep 9, 2021 at 16:54

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