I am trying to build a simple understanding of what a spinor is, in doing so I have stumbled across two different definitions of spinor:

First definition:
If $|\psi\rangle$ is a generic state of a spin $1/2$ particle then it can be decomposed in the following way: $$|\psi\rangle=c_+|+\rangle+c_-|-\rangle$$ where $|+\rangle,|-\rangle$ are the eigenstates of the third component of the spin ($S_z$), and $c_+,c_-$ are complex numbers. We can also write this as: $$|\psi\rangle=\begin{pmatrix}c_+ \\ c_-\end{pmatrix}$$ and in this form usually we prefer to write $u$ in place of $|\psi\rangle$. This 2D vector representing the state of the particle is called spinor.

Second definition:
Given a particle of spin $1/2$ we can write its wave function (for some reason) in the following way: $$\psi(\vec{x})=\begin{pmatrix}\psi_{1/2}(\vec{x}) \\ \psi _{-1/2}(\vec{x})\end{pmatrix}$$ the wavefunction of the particle, written in this fashion, is called spinor.

I am searching a bridge between this two definitions, as well as a way to remove the ugly "for some reason" from the second statement. Regarding this: in my lecture notes is present the following: $$\psi(\vec{x})=\langle\vec{x}|\psi\rangle=\sum_{l,m}\sum_{s_z=-s}^s\langle \vec{x} | l \ m \ s_z\rangle \langle l \ m \ s_z | \psi\rangle$$ this step is fine since is simply an application of the identity operator, but then: $$\sum_{l,m}\sum_{s_z=-s}^s\langle \vec{x} | l \ m \ s_z\rangle \langle l \ m \ s_z | \psi\rangle=\sum _{l,m,s_z}Y_{l,m}(\theta , \phi)u_{s_z}c^{s_z}_{l,m}(r) \ \ \ \ \ \ (1)$$ but anyway if we take (1) for granted then we have:

$$\psi(\vec{x})=\sum _{l,m} Y_{l,m}(\theta , \phi)\begin{pmatrix}1 \\ 0\end{pmatrix}c^{1/2}_{l,m}(r)+\sum _{l,m} Y_{l,m}(\theta , \phi)\begin{pmatrix}0 \\ 1\end{pmatrix}c^{-1/2}_{l,m}(r)=$$$$=\begin{pmatrix}\sum _{l,m} Y_{l,m}(\theta , \phi) c^{1/2}_{l,m}(r) \\ \sum _{l,m} Y_{l,m}(\theta , \phi) c^{-1/2}_{l,m}(r)\end{pmatrix}=\begin{pmatrix}\psi_{1/2}(r , \theta , \phi) \\ \psi_{-1/2}(r , \theta , \phi) \end{pmatrix}$$

since $u_{s_z}$ is indeed the 2D vector that is the eigenstate of the spin operator.

But I have a huge problem regarding equation (1): I don't get how we can prove that the left hand side is equal to the right hand side. Why is there $Y(\theta , \phi)$? (where the $Y$ are of course the spherical harmonics, so the eigenfunctions of the angular momentum) What is $c^{s_z}_{l,m}(r)$ and why is it $r$ dependent? Why it appears $u_{s_z}$?

  • $\begingroup$ Did your instructor explicitly write down eqn (1)? $\endgroup$ Nov 29, 2020 at 16:41
  • $\begingroup$ (3.6.45) and discussion of Sakurai/Napolitano, 2nd edition addresses the absurdity of expanding your wave function in terms of spherical harmonics: you have conjured up a chimera. $\endgroup$ Nov 29, 2020 at 16:49
  • $\begingroup$ @CosmasZachos Yes he did, why? Is it incorrect? $\endgroup$
    – Noumeno
    Nov 29, 2020 at 16:50
  • $\begingroup$ @CosmasZachos But this expansion seems really useful for practical exercises in QM, as well as being useful in explaining the multiple definitions of spinor.. $\endgroup$
    – Noumeno
    Nov 29, 2020 at 16:52
  • $\begingroup$ @CosmasZachos For example: if I have an exercise that gives me the spinor in terms of the second definition, giving me $\psi_{1/2},\psi_{-1/2}$, and then asks me to find the possible results of a measurement of $L_z,L^2$ then to find the answer I have to expand the spinor/wavefunction using $Y(\theta, \phi)$, otherwise how could I solve an exercise like the one I am mentioning? $\endgroup$
    – Noumeno
    Nov 29, 2020 at 17:00

1 Answer 1


The 1st equation: $$|\psi\rangle=c_+|+\rangle+c_-|-\rangle$$ is the generic form for any pure state in a two-level system. The space of all normed pairs of ($c_+$, $c_-$) are called the Bloch Sphere.

Being that it applies to any two level system, it does not elucidate the physical nature of a spinor particularly well. In fact, in many teaching scenarios, it may lead to confusion. This is because of the terms "spin-up" and "spin-down". This leads the uninitiated into thinking the spin is either aligned or anti-aligned with magnitude $\frac 1 2$ (in units of $\hbar$), this is not the case.

A spin 1/2 particle has an angular momentum of

$$\sqrt{s(s+1)} = \frac{\sqrt 3} 2$$

and the eigenstates allows to know only the projection on a single axis, which is $\frac 1 2$. That means, that in an $S_z$ eigenstate, there is still:

$$ S_{x \& y} = \sqrt{s(s+1)-s^2} = \frac 1 {\sqrt 2}$$

units of angular momentum that is equally distributed in the $x-y$ plane. That fact, along with the commutations relations:

$$ [S_i, S_j] = 2i\epsilon_{ijk}S_k$$

and the ability to quantize the spin on any axis are critical to understanding spinors.

In short, you can write the coefficients as:

$$ c_+ = \cos{\theta/2}e^{-i\phi/2} $$ $$ c_- = \sin{\theta/2}e^{+i\phi/2} $$

and, while any pair of ($\theta,\phi$) represents a superposition in the original basis, it is a pure spin up state when quantized along an axis pointing in the ($\theta, \phi$) direction.

So that's the spin part of a wave function. If you write the spin wave function as a column vector (your 2nd equations) and combine it with a factored spatial wave function, you get: $$\Psi(\vec{x})=\psi( \vec x)\begin{pmatrix}c_+\\ c_-\end{pmatrix}$$ That represents a particle, with a spatial wave function, and an unentangled spin. In the most general case, the space and spin components can be entangled, and that requires your 3rd equations:

$$\psi(\vec{x})=\begin{pmatrix}\psi_{1/2}( \vec x) \\ \psi _{-1/2}(\vec{x})\end{pmatrix}$$

This is required for systems like the Stern-Gerlach experiment.

If we look at just one spatial component of the wave function, e.g., $\psi_+(\vec x)$, it's written here as a function of a vector, $\vec x$. That can be done in cartesian coordinates:

$$ \psi(\vec x) = \psi(x,y,z)$$

other coordinates, or, what is generally the most useful when angular momentum is conserved: spherical coordinates. In that case, the angular part is captured in spherical harmonics, per that standard projection in your 5th equation:

$$\sum_{l,m}\sum_{s_z=-s}^s\langle \vec{x} | l \ m \ s_z\rangle \langle l \ m \ s_z | \psi\rangle=\sum _{l,m,s_z}Y_{l,m}(\theta , \phi)u_{s_z}c^{s_z}_{l,m}(r) \ \ \ \ \ \ (1)$$

what we see from this equation is that the orbital angular momentum ($(l,m)$) couple to each spin state (per the usual rules of angular momentum addition). Note that eigenstates of total angular momentum quantum numbers ($J^2, J_z$) mix different $(l, m)$ with spin up and spin down, as captured by the $c^{\pm}_{lm}$. This is standard Clebsch-Gordan stuff, e.g.:

$$|\frac 3 2 \frac 1 2\rangle = \begin{pmatrix} \sqrt{\frac 2 3}Y_{1,0}(\theta, \phi) \\ \sqrt{\frac 1 3}Y_{1,1}(\theta, \phi)\end{pmatrix}$$

So that represents and entanglement of the spin and angular coordinates such that total angular momentum quantum numbers are good.

When one sees this in, say, the Hydrogen atom, the radial wave function is entirely factored out. Your final equation, however, allows radial dependence in the $c^{\pm}_{lm}(r)$. This introduces spatial entanglement between spin and the space coordinate in the radial direction. (I cannot think of a system off hand where that comes up...since, if you're using spherical harmonics, the problem should be separable in spherical coordinates). Nevertheless, it is the most general spinor wave function in spherical coordinates.

Addendum: Note that the $\theta$ and $\phi$ dependence on the RHS of (1) is directly related to $\vec x$ on the LHS. Here I am going to $\vec r$ for a 3D position vector, as $\vec x$ will cause confusion:

In cartesian coordinates $(x, y, z)$ and the cartesian basis:

$${\bf r} = x{\bf\hat x}+y{\bf\hat y}+z{\bf\hat z} $$

which is, in spherical coordinates $(r, \theta, \phi)$ and the cartesian basis:

$${\bf r} = r\sin\theta\cos\phi{\bf\hat x}+r\sin\theta\sin\phi{\bf\hat y}+r\cos\theta{\bf\hat z} $$

Now we can switch from the cartesian basis vectors to spherical basis vectors (spherical basis vectors are the 3 eigenvectors of rotations about the z-axis, they are not the spherical-coordinate basis vectors ($\hat r, \hat\theta, \hat\phi)$, and understanding them is very helpful in understanding spin, and tensor operators, etc):

$${\bf r} = r\sin\theta\cos\phi\frac 1 {\sqrt 2}({\bf\hat e^+}+{\bf\hat e^-})+r\sin\theta\sin\phi \frac i {\sqrt 2}({\bf\hat e^+}-{\bf\hat e^-})+r\cos\theta{\bf\hat e^0} $$

which can be rearranged:

$${\bf r} = \frac 1 {\sqrt 2}r\sin\theta e^{+i\phi}{\bf\hat e^+}+\frac i {\sqrt 2}r\sin\theta e^{-i\phi} {\bf\hat e^-}+r\cos\theta{\bf\hat e^0} $$

note that the angular coefficients are the $l=1$ spherical harmonics:

$${\bf r} = r\sqrt{\frac{4\pi} 3}\big[Y_1^1(\theta,\phi){\bf\hat e^+} + Y_1^0(\theta,\phi){\bf\hat e^0} +Y_1^{-1}(\theta,\phi){\bf\hat e^-} \big ]$$

That formula may take some getting used to. In the spherical basis, the spherical harmonics are the coordinates. It's strange, how can a function be a coordinate? Well, look at $Y_1^0(\theta,\phi)$. On the unit sphere, it is $z$. Well $z$ is a function (it's a polynomial), and so are $x$ and $y$. So we use polynomial coefficients in the straight cartesian rep, we just don't usually think of them as such, we think of them as labeled numbers.

  • $\begingroup$ I was following you until you re-written my equation (1). I feel like you breeze through it without properly showing why we can say that the equation is true, and this was one of the main points of my question. Don't get me wrong your answer is useful but I still have some perplexities, like: Y is dependent upon $\theta$ and $\phi$, but in the left hand side of the equation there is noting dependent on $\theta$ and $\phi$, so where does the dependence come from? I would expect something like $\langle \theta \ \phi| l \ m\rangle$. I think that you could expend on the derivation of equation (1). $\endgroup$
    – Noumeno
    Nov 30, 2020 at 15:12
  • $\begingroup$ That's why I mention coordinates. $\vec x$ depends on theta and phi. I will add an addendum that will convince you...because it is what convince me. $\endgroup$
    – JEB
    Nov 30, 2020 at 16:04
  • $\begingroup$ Wonderful. For an eventual future reader I think this link may be helpful: it expands on spherical basis vectors and how they work. $\endgroup$
    – Noumeno
    Nov 30, 2020 at 20:07

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