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How does one derive using, say, the operator formula for reflections

$$ R(r) = (I - 2nn^*)(r),$$

the reflection representation of a vector

$$ R(r) = R(x\hat{i} + y\hat{j} + z\hat{k}) = xR(\hat{i}) + yR(\hat{j}) + zR(\hat{k}) = xs_x + ys_y + zs_z \\ = x \left[ \begin{array}{ c c } 0 & 1 \\ 1 & 0 \end{array} \right] + y\left[ \begin{array}{ c c } 0 & -i \\ i & 0 \end{array} \right] + z \left[ \begin{array}{ c c } 1 & 0 \\ 0 & - 1 \end{array} \right] = \left[ \begin{array}{ c c } z & x - iy \\ x+iy & - z \end{array} \right] $$

that comes up when dealing with spinors in 3-D? Intuitively I can see the matrices are supposed to come from the following geometric picture:

The first Pauli matrix is like a reflection about the "y=x" line. The third Pauli matrix is like a reflection about the "x axis". The second Pauli matrix is like a 90° counterclockwise rotation and scalar multiplication by the imaginary unit https://en.wiktionary.org/wiki/Pauli_matrix

but why and how did we make these choices? I know we're doing it to end up using a basis of $su(2)$, but assuming you didn't know anything about $su(2)$, how could you set this up so that it becomes obvious that what we end up calling $su(2)$ is the right way to represent reflections? The usual ways basically postulates them or show they work through isomorphism or say the come from the fact a vector is associated with the matrix I've written above without explaining where that came from. The closest thing to an explanation is that they come from the quaternionic product whose link to all this, especially something as simple as reflections through lines, escapes me.

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bolbteppa
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  • The operators $;\boldsymbol{-} i (\mathbf{n}\circ\boldsymbol{\sigma});$ and $;\boldsymbol{+} i (\mathbf{n}\circ\boldsymbol{\sigma});$ are special unitary, that is belong to $\Bbb{SU}(2)$, both of them represent a rotation through an angle $+\pi$ or $-\pi$, that is a reflection through the axis $\mathbf{n}$ and are inverses of each other. I think that the geometric picture of the wiki link is wrong. $\mathbf{n}\circ\boldsymbol{\sigma}=n_{1}\sigma_{1}+n_{2}\sigma_{2}+n_{3}\sigma_{3}$. – Frobenius Apr 20 '16 at 07:18

2 Answers2

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Lucubration needs not light with insight. I fear you are expecting to make lemonade with apples. Here is why.

The basic relation is the multiplication law of two Pauli vectors predicated on the abstract properties of the Pauli matrices, not their particular realization, $$(\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma}) = (\vec{a} \cdot \vec{b}) \, I + i ( \vec{a} \times \vec{b} )\cdot \vec{\sigma} ~.$$

Any similarity transformation by an invertible matrix $\vec{\sigma} \mapsto U^{-1} \vec{\sigma} U$ preserving this relation would likewise lead to the fundamental reflection formula you miswrote, $$ - \hat{n}\cdot \vec{\sigma} ~~ \vec{x} \cdot \vec{\sigma} ~~\hat{n}\cdot \vec{\sigma} = ((1\!\!1 -2 \hat{n}\hat{n}) \cdot \vec{x}) \cdot \vec{\sigma}. $$ The missing 2 is crucial in your first formula, if it is to yield a reflection and not a projection. (The first R you wrote is a 3x3 matrix acting on a vector. The second one is a 2x2 matrix representing a different Pauli vector.) This is the Householder formula, of course, acting on a 3-vector properly mapped to the suitable Pauli vector, a 2x2 matrix. (It is used routinely in the adjoint rotation of a Pauli vector, thus seen to amount to the double-angle rotation of the vector rep.)

And that's that. Edit: since you asked for an illustration, taking the axis perpendicular to the plane of reflexion (x,z) to be along y, so n=(0,1,0), the above formula reduces to just $$ -\sigma_y ~~\vec{x} \cdot \vec{\sigma} ~~\sigma_y= x \sigma_x -y \sigma_y +z\sigma_z, $$ the celebrated conjugation of the pseudoreal rep of SU(2).

The above generic 3d reflections have nothing to do with the somewhat fanciful mnemonic rules of the wiktionary you quote for the representation-dependent three Pauli matrices in terms of 2d, plane reflections across lines, not planes, as any similarity transform of the standard Paulis would do, as seen above! The (somewhat frivolous) mnemonic summarizes, "for desert island use", plane transformations that reflect a 2-vector according to the standard convention Pauli matrices. It is no more than a convention though---admittedly the easiest one. You could permute them cyclically and so get the cyclically permuted mnemonic. Since this is aggressively representation-dependent, it amounts to a case where one could safely reassure oneself that there is no deep geometrical connection involved---a rare occasion, indeed.

Cosmas Zachos
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  • Thanks for your answer and for pointing out the missing 2, unfortunately I think you are missing just the crux of my question but you get the gist! I was trying my best to trying to ask:

    "where does $ \vec{a} \cdot \vec{\sigma} = R(r) = xs_x + ys_y + zs_z = ...$ come from? I think you use Householder formula, not sure... When I use it, it doesn't work, e.g. when I compute $R(r)$ on the standard orthogonal basis vector $\hat{i}$ it does not give the Pauli matrix $s_x$ I have above. How do I get them & thus the matrix I gave?"

    In trying to make sure I am on the right track:

     – bolbteppa May 25 '16 at 21:02
  • I am thinking all of this comes from noting we want to find a 'reflection' representation of a vector, motivated by the fact that a rotation decomposes into a product of reflections, so maybe it amounts to understanding the reflection representation of the basis vectors, which give the Pauli matrices, but how do you actually get the expressions from those? With this Householder formula? I can't get it using this formula, why do we get these expressions if the formula doesn't give it? – bolbteppa May 25 '16 at 21:06
  • It seems like your answer to all this is that the multiplication rule

    $XY = \frac{1}{2}(XY + XY+YX-YX) = \frac{1}{2}(XY+YX)+\frac{1}{2}( XY-YX) = (\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma})$

    $= (\vec{a} \cdot \vec{b}) , I + i ( \vec{a} \times \vec{b} )\cdot \vec{\sigma}$

    explains everything, but this is just a reflection of the fact that a rotation decomposes into a product of reflections, and is the intuitive constraint on what we're trying to do. I am asking how you get one of the reflections in that product, by asking what the reflection rep of the basis vectors is.

     – bolbteppa May 25 '16 at 21:09
  • I derived the householder formula for vectors by the well-known reflection formula for Pauli vectors. Of course it works: try it. Take n=(1,0,0), plug in, and confer that this reflects a 3 vector on an x-y mirror. I believe the first formula you wrote is the Householder formula and the second is a generic Pauli vector, which, as I described, for unit n=(x,y,z) produces the Pauli vector used twice, not once, of course, in my demonstration. – Cosmas Zachos May 25 '16 at 22:05
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A rotation is of the form $$\begin{bmatrix} \cos(\theta) & - \sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$$ A reflection is of the form $$\begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & - \cos(2\theta) \end{bmatrix}$$

If we want to find a $2-$dimensional representation of a $3-$dimensional rotation then we can assume $$\mathbb{R}^3 = \mathbb{R} \times \mathbb{C}$$ so that $$(x,y,z) \mapsto (z,x+iy)$$

Thus in a space where all vectors are of the form $(z,x+iy)$, the column vectors of a rotation matrix will be of this form, with the second column orthogonal to the first, thus a reflection matrix would be $$\begin{bmatrix} z & x-iy \\ x+iy & - z \end{bmatrix}$$

Mapping $(1,0,0) \mapsto (0,1)$ etc... gives the Pauli matrices.

bolbteppa
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