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I have an expression and the transformation rules, and I wonder if this qualifies as a spinor. Can the following expression written with complex Clifford algebra be seen as a spinor? In any case, it certainly reproduces quantum mechanics of spin wave vectors.

The expression for a single spin-1/2 particle given by Euler angles $\theta,\phi$ is $$\Psi=\frac{1}{\sqrt{2}}\left((e_1+if_1)\cos\frac{\theta}{2}+(e_2+if_2)\sin\frac{\theta}{2}e^{i\phi}\right)$$ where $e_1,f_1,e_2,f_2$ are orthonormal basis vectors which all square to $e_i^2=f_i^2=+1$ and $i$ is the imaginary unit. This can be recognized as a way to write the usual spin wave vector $\psi=(\cos\frac{\theta}{2},\sin\frac{\theta}{2}e^{i\phi})$. For $n$-dimensional wave vectors it would be $\Psi=\frac{1}{\sqrt2}\sum_k(e_k+if_k)\psi_k$.

If I take the expression $$\Omega=i(\Psi\Psi^\dagger-1)$$ where the multiplication is the clifford multiplication of vectors, then for a single particle I get $$ \Omega=T+X\sin\theta\cos\phi+Y\sin\theta\sin\phi+Z\cos\theta $$ with $$ \begin{align*} X&=\frac{1}{2}\left(e_1f_2+e_2f_1\right)\\ Y&=\frac{1}{2}\left(e_1e_2+f_1f_2\right)\\ Z&=\frac{1}{2}(e_2f_2-e_1f_1)\\ T&=\frac{1}{2}(e_1f_1+e_2f_2) \end{align*} $$ which looks like the Bloch sphere.

The probability to measure a state $\Omega^{(A)}$ in an observable $\Omega^{(B)}$ is the inner product $$ P(A\to B)=\langle \Omega^{(A)}\Omega^{(B)\dagger}\rangle $$ which is conventionally expressed by the Born rule $P=|\langle \psi^{(A)}|\psi^{(B)}\rangle|^2$.

Spatial rotations around a normalized axis $(r_x, r_y, r_z)$ and an angle $\alpha$ can be performed with the equations $$ \begin{align*} \Omega'&=R\,\Omega\, R^\dagger\\ \Psi'&=R\Psi \end{align*} $$ with the rotor $$ \begin{align*} R&=\exp\left(\frac{\alpha}{2}(T+Xr_x+Yr_y+Zr_z)\right)\\ &=\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2}(T+Xr_x+Yr_y+Zr_z) \end{align*} $$

With all that information, and assuming the algebra is correct (which it is), does the initial expression $\Psi$ qualify as a spinor? Is the relation between $\Omega$ and $\Psi$ what is informally called the "square root of a vector"?

(You could also do the whole calculation with real clifford algebra if you use $i=e_0f_0$. Also it works for a wave vector of any dimension)

...It's been a while and no answer :( With so many spinor questions in this forum, I thought checking if transformation rules apply should be easy. Maybe, someone can leave some thoughts in the comments. It's a new formulation and it surely reproduces QM and rotations, but I'm not familiar with spinor theory to tell how they relate. However, it looks suspiciously close the what I've seen about spinors and it looks like an explicit expression for spinors.

A lightning intro to clifford algebra: think of the basis vectors as matrices, because they similarly do not commute. All you need now is that each basis vector squared is $a^2=+1$, and if you have two different basis vectors they anti-commute under multiplication $ab=-ba$. $\dagger$ means reversing the order in a basis vector product (and taking the complex conjugate of the scalar coefficient).

Also note the relation: $\hat a^\dagger_\uparrow=e_1+if_1$ and $\hat a^\dagger_\downarrow=e_2+if_2$ for a connection to creation operators.

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    $\begingroup$ A spinor is an element of the even subrepresentation of the clifford algebra. So, for a usual spinor, the clifford algebra is four dimensional and the spinor can be a sum of a scalar, a bivector, or the clifford product of four elements. I'm pretty sure that your "spinor" is not an even subrepresentation and so not a spinor. $\endgroup$ Feb 18 at 11:25
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    $\begingroup$ @tonetillo4 Does it satisfy the transformation laws though? $\endgroup$
    – Gerenuk
    Feb 18 at 12:39
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    $\begingroup$ If I look at physics.stackexchange.com/q/639161 then my $\Omega$ corresponds to his $V$ and my $\Psi$ corresponds to his $\xi$. And the question is to show that both version do or do not obey the same transformation laws. $\endgroup$
    – Gerenuk
    Feb 18 at 13:06
  • $\begingroup$ Actually, one could premultiply the expression by a new vector e3 to make it an even element. It wouldn't change the results and be unnecessary baggage, but it would make it formally even. And since it's complex clifford algebra one could premultiply it in a non-trivial way, too. $\endgroup$
    – Gerenuk
    Feb 18 at 16:07
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    $\begingroup$ I deleted my earlier wrong answer. I took vector multiplication such as $e_1f_2$ as the conventional $|e_1><f_2|$, whereas actually the vector basis vectors live in the Clifford space, therefore the $|e_1><f_2|$ interpretation is not correct. $\endgroup$
    – MadMax
    Mar 15 at 13:36

2 Answers 2

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... what is called informally the square root of a vector

A spinor is often described by physicists as a double valued representation of the orthogonal group. Mathematicians, who don't like double valued anything rephrase this and say a k-valued representation of a group is actually a representation of a k-covering of the group. Now double coverings of any orthogonal groups exist and are called, unsuprisingly, spin groups. Thus, by the preceding, a spinor is a representation of a spin group and usually this is taken to be a fundamental representation and so the smallest rep.

So where are the 'square roots'?

Well, one informal way of thinking about it is to think of the square root in the complex plane. This organises itself into a double covering of the complex plane branched at the origin. But we can extend this to a double covering of the complex sphere branched at two points (the other branch point is the additional infinite point). And this rather looks like our double coverimg above, but isn't of course. Firstly because it is branched at two points, but also because its a double cover of the sphere rather than of its rotation group.

Now, not only can spinors be defined over flat space, they can be defined over manifolds. But it turns out that not every manifold can support a spinor, those that do are called spin. And this in turn means they must support what is called a spin bundle, otherwise known as the bundle of spin frames or spin structures. This is the manifold equivalent of a spin group and we form a spinor bundle by using a spin representation. A spinor here, is more precisely a section of this bundle, and is more properly called a spinor field.

Now, it turns out that over a Kahler manifold, spin bundles are isomorphic to the square root of the top holomorphic cotangent bundle. Such bundles are vector bundles. So their sections are vector fields. So in some sense, spinors are not just "informally a square root of a vector field", they actually are one.

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    $\begingroup$ How exactly does this answer OP's question about whether $\Psi$ is a spinor? Or whether $\Psi$ and $\Omega$ are related by the square root of a vector? $\endgroup$
    – Kyle Kanos
    Feb 23 at 15:45
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    $\begingroup$ It seems to me that you are answering a haphazardly-related very general question instead of the very specific question that was asked. There's no mention of $\Psi$, $\Omega$, $R$, $e_i$, $f_i$, etc in here, which is something I'd expect when answering about those particular variables. $\endgroup$
    – Kyle Kanos
    Feb 23 at 20:26
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    $\begingroup$ I use haphazard loosely here, don't get hung up on that. But it seems to me that the question that you have answered here is a truncated version of OPs actual question. That's why I'm asking how this answers the question posted, because for all intents and purposes, it doesn't really address anything OP asked. $\endgroup$
    – Kyle Kanos
    Feb 23 at 22:05
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    $\begingroup$ (By "truncated", I mean, OP's question is of the form, "Do X & Y satisfy condition Z" and you answered the question "What is condition Z?" It is certainly part of what was asked, but not actually what was asked). $\endgroup$
    – Kyle Kanos
    Feb 23 at 22:08
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    $\begingroup$ I appreciate the effort in contributing information which is related to the topic. Yet, you did not address anything important from the question. You did not even answer half the question, because even if someone shows $\Psi$ is a spinor (which is the only interesting part in the question), your text does not show anything about the relation between $\Psi$ and $\Omega$. You completely discarded everything about the question. It's honorable that you contribute, but it would be wise if you'd read more carefully and listen to comments. $\endgroup$
    – Gerenuk
    Feb 24 at 7:33
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A couple of observations:

  1. Yes, $\Omega$ and $\Psi$ transform as vector and spinor. Note that the vector basis $e_1,f_1,e_2,f_2$ generates Clifford algebra $Cl(4,0)$. The spinor $\Psi$ can take a more general form such as the whole $Cl(4,0)$ space (16 components), or even part of $Cl(4,0)$ (8 components, as advocated by David Hestenes). OP's choice is limited to vector portion of $Cl(4,0)$ (4 components).

  2. For Clifford algebra $Cl(4,0)$, the symmetry group spanned by $Cl(4,0)$ bi-vectors is $SO(4)/Spin(4)$. The OP's $SU(2)$ (spanned by $X, Y, Z$) is a special case of Chris Doran's approach of embedding $SU(m)$ in $SO(2m)/Spin(2m)$ (with m=2). The OP's $T,X,Y,Z$ correspond to equations 6.1, 6.4a, 6.4b, and 6.5 in Doran's paper, respectively.

  3. The most general transformations on $\Omega$ and $\Psi$ are spanned by generators corresponding to all the $Cl(4,0)$ 6 bi-vectors and 4 tri-vectors, which constitute the de Sitter group $SO(4,1)$. In this case, the spinor $\Psi$ has to live in the whole Clifford algebra $Cl(4,0)$ space (8 odd + 8 even components), since the tri-vector-related operators would transform between Clifford-odd and Clifford-even sectors of the spinor $\Psi$.

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  • $\begingroup$ Thanks. It is interesting to see how the individual expressions fit into group theory! It's just beyond my knowledge at the moment whereas my expressions are simple algebra. The main point of my formalism is though, that I have an explicit equation $\Omega=i(\Psi\Psi^\dagger-1)$ and also the physical equation for the probability $P=\langle \Omega_1\Omega_2\rangle$. Is there a reference which mentions this? $\endgroup$
    – Gerenuk
    Mar 16 at 8:05
  • $\begingroup$ Also, I approached C. Doran with an early version of my idea, and he only briefly commented that when he tried something like that, he couldn't get rigid body rotation to work. It may be a misunderstanding though. Unfortunately, he didn't get back to me after I sent him my final version. But now I'm not sure C. Doran ever went this particular route. $\endgroup$
    – Gerenuk
    Mar 16 at 8:08
  • $\begingroup$ For all this to work with actual quantum mechanics, it may actually be crucial to not have the most general algebra, but the particular subset I write. At least, it accurately reproduces common multi-spin quantum mechanics and quantum computing, and does not work with more generalized expressions. $\endgroup$
    – Gerenuk
    Mar 16 at 8:15
  • $\begingroup$ My observation 3 on de Sitter group goes beyond Doran's construction, since Doran's approach is limited to the bi-vector-related transformations. BTW, your original transformation like $\Psi'=R\Psi$ could rotate your vector-based spinor into tri-vectors. $\endgroup$
    – MadMax
    Mar 16 at 13:37
  • $\begingroup$ Yes, technically $\Psi'=R\Psi$ could create expressions which are not trivially mapped to wave vectors. In a way, because it should actually be $\Psi'=R\Psi R^\dagger$ which still would yield identical results. This alternative equation just requires more computation and the additional right multiplications cancels in $\Omega$ anyway. $\endgroup$
    – Gerenuk
    Mar 16 at 13:58

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