# Lab Notes for a Scientific Revolution (Physics)

## April 18, 2008

### Lab Note 6: Operator Decomposition of Intrinsic Spin

I’d like to lay out a nifty little mathematical calculation which allows a “decomposition” of the intrinsic spin matrices $s^{i} ={\tfrac{1}{2}} \hbar \sigma ^{i}$ to include the position and momentum operators $x^{i}$, $p^{i}$, $i=1,2,3$.  To simplify matters, we will employ a Minkowski metric tensor with ${\rm diag}\left(\eta _{\mu \nu } \right)=\left(-1,+1,+1,+1\right)$ so that raising and lowering the space indexes $i=1,2,3$ is simple and at will, and does not entail any sign reversal.  (This lab note is available in PDF form, with a recent update including a new section 2, at intrinsic-spin-decomposition-11.)

We start with the general cross product for two three-vectors A and B.  Written in covariant (index) notation:

$\left(A\times B\right)_{i} \equiv \varepsilon _{ijk} A^{j} B^{k}$.   (1)

One can easily confirm this by taking, for example, $\left(A\times B\right)_{3} \equiv A^{1} B^{2} -A^{2} B^{1}$.  Now, let’s take the triple cross product $\left(A\times B\right)\times C$.  We can apply (1) to itself using $\left(A\times B\right)^{j} \equiv \varepsilon ^{jmn} A_{m} B_{n}$, to write:

$\left[\left(A\times B\right)\times C\right]_{i} =\varepsilon _{ijk} \left(A\times B\right)^{j} C^{k} =\varepsilon _{ijk} \varepsilon ^{jmn} A_{m} B_{n} C^{k}$.  (2)

The fact that the crossing of A and B takes precedence over crossing with C is retained in the fact that $A_{m} B_{n}$ sum with $\varepsilon ^{jmn}$, while $C^{k}$ alone sums into $\varepsilon _{ijk}$.

Let us now expand (2) for the component equation for which $i=3$.  The calculation is as such:

$\begin{array}{l} {\left[\left(A\times B\right)\times C\right]_{3} =\varepsilon _{3jk} \varepsilon ^{jmn} A_{m} B_{n} C^{k} } \\ {=\varepsilon _{312} \varepsilon ^{123} A_{2} B_{3} C^{2} +\varepsilon _{312} \varepsilon ^{132} A_{3} B_{2} C^{2} +\varepsilon _{321} \varepsilon ^{231} A_{3} B_{1} C^{1} +\varepsilon _{321} \varepsilon ^{213} A_{1} B_{3} C^{1} } \\ {=A_{1} B_{3} C^{1} +A_{2} B_{3} C^{2} -A_{3} B_{1} C^{1} -A_{3} B_{2} C^{2} } \\ {=A_{1} B_{3} C^{1} +A_{2} B_{3} C^{2} +A_{3} B_{3} C^{3} -A_{3} B_{1} C^{1} -A_{3} B_{2} C^{2} -A_{3} B_{3} C^{3} } \\ {=A_{1} B_{3} C^{1} +A_{2} B_{3} C^{2} +A_{3} B_{3} C^{3} -A_{3} \left(B\cdot C\right)} \end{array}$,  (3)

where we have added $0=A_{3} B_{3} C^{3} -A_{3} B_{3} C^{3}$ to the fourth line.  Now in the final line, we hit an impasse, because $B_{3}$ is sandwiched between the terms we would like to form into the other dot product $A\cdot C$.  In order to complete the calculation, we must make an assumption that the $A_{i}$commute with $B_{3}$, i.e., that $\left[A_{i} ,B_{3} \right]=0$.  For now, let us make this assumption.

Therefore, we carry out the commutation in (3), and continue along to write:

$\begin{array}{l} {\left[\left(A\times B\right)\times C\right]_{3} =\varepsilon _{3jk} \varepsilon ^{jmn} A_{m} B_{n} C^{k} =A_{1} B_{3} C^{1} +A_{2} B_{3} C^{2} +A_{3} B_{3} C^{3} -A_{3} \left(B\cdot C\right)} \\ {=B_{3} \left(A\cdot C\right)-A_{3} \left(B\cdot C\right)=B_{3} A_{j} C^{j} -A_{3} B_{j} C^{j} } \end{array}$.  (4)

Generalizing fully, we may now write (4) in two equivalent ways as:

$\left\{\begin{array}{c} {\left(A\times B\right)\times C=-A\left(B\cdot C\right)+B\left(A\cdot C\right)\quad \quad } \\ {\varepsilon _{ijk} \varepsilon ^{jmn} A_{m} B_{n} C^{k} =-A_{i} B_{j} C^{j} +B_{i} A_{j} C^{j} } \end{array}\right.$.  (5)

The reader will observe the well-known formula for the cross product.

Now, let’s take the cross product in which $A=x$, $B=p$ and $C={\bf \sigma }$, where x is the position operator about the center of mass, p is the momentum operator, and ${\bf \sigma }$ are the Pauli spin matrices.  We also take into account the Heisenberg canonical commutation relationship between the position and momentum operators, that is, $\left[x_{\mu } ,p_{\nu } \right]=i\hbar \delta _{\mu \nu }$.  This means that we will have to be careful at the juncture between equations (3) and (4), because the position and momentum operators along the same dimension do not commute.

So, we return to (3) with $A=x$, $B=p$ and $C={\bf \sigma }$, to write:

$\left[\left(x\times p\right)\times {\bf \sigma }\right]_{3} =\varepsilon _{3jk} \varepsilon ^{jmn} x_{m} p_{n} \sigma ^{k} =x_{1} p_{3} \sigma ^{1} +x_{2} p_{3} \sigma ^{2} +x_{3} p_{3} \sigma ^{3} -x_{3} \left(p\cdot {\bf \sigma }\right)$.  (6)

To take the next step, we want to place $p_{3}$ in front of the $x_{i}$.  In so doing, we can commute $p_{3}$ with $x_{i}$ for $i=1,2$.  But, for $i=3$, we must employ $x_{3} p_{3} =p_{3} x_{3} +i\hbar$.  Therefore, (6) now becomes:

$\begin{array}{l} {\left[\left(x\times p\right)\times {\bf \sigma }\right]_{3} =\varepsilon _{3jk} \varepsilon ^{jmn} x_{m} p_{n} \sigma ^{k} =p_{3} x_{1} \sigma ^{1} +p_{3} x_{2} \sigma ^{2} +\left(p_{3} x_{3} +i\hbar \right)\sigma ^{3} -x_{3} \left(p\cdot {\bf \sigma }\right)} \\ {=p_{3} \left(x\cdot {\bf \sigma }\right)-x_{3} \left(p\cdot {\bf \sigma }\right)+i\hbar \sigma _{3} =p_{3} x_{j} \sigma ^{j} -x_{3} p_{j} \sigma ^{j} +i\hbar \sigma _{3} } \end{array}$,  (7)

lowering the index on $i\hbar \sigma ^{3}$ with ${\rm diag}\left(\eta _{ij} \right)=\left(+1,+1,+1\right)$.  Now all of a sudden, $i\hbar \sigma ^{3}$ has made an unexpected appearance.  Generalizing (7), we may write:

$\left\{\begin{array}{c} {\left[\left(x\times p\right)\times {\bf \sigma }\right]=-x\left(p\cdot {\bf \sigma }\right)+p\left(x\cdot {\bf \sigma }\right)+i\hbar {\bf \sigma }\quad \quad \quad } \\ {\varepsilon _{ijk} \varepsilon ^{jmn} x_{m} p_{n} \sigma ^{k} =-x_{i} p_{j} \sigma ^{j} +p_{i} x_{j} \sigma ^{j} +i\hbar \sigma _{i} } \end{array}\right.$,  (8 )

This is the also the well-known formula for the triple-cross product, but with an additional term $i\hbar {\bf \sigma }$ emerging from the canonical commutation relationship.  In fact, moving terms, equation (8 ) gives us a way to decompose the intrinsic spin matrix so as to contain the position and momentum, and as we shall also see, orbital angular momentum operators.

First, we rewrite (8 ) as:

$\left\{\begin{array}{c} {i\hbar s=\left[\left(x\times p\right)\times s\right]+x\left(p\cdot s\right)-p\left(x\cdot s\right)\quad \quad \quad } \\ {i\hbar s_{i} =\varepsilon _{ijk} \varepsilon ^{jmn} x_{m} p_{n} s^{k} +x_{i} p_{j} s^{j} -p_{i} x_{j} s^{j} } \end{array}\right.$,   (9)

where we have multiplied through by ${\tfrac{1}{2}} \hbar$ and then set $s_{i} \equiv {\tfrac{1}{2}} \hbar \sigma _{i}$.  This decomposes the intrinsic spin matrix into an expression involving itself, as well as the position and momentum operators.

Now, using the definition (1) but with $A=x$ and $B=p$, let’s introduce the orbital angular momentum operator :

$l^{j} \equiv \left(x\times p\right)^{j} \equiv l^{j} \equiv \varepsilon ^{jmn} x_{m} p_{n}$  (10)

It is easy to see, for example, that $l^{3} =x_{1} p_{2} -x_{2} p_{1}$.  Using (10), we now rewrite (9) as:

$\left\{\begin{array}{c} {i\hbar s=\left(l\times s\right)+x\left(p\cdot s\right)-p\left(x\cdot s\right)\quad \; \; } \\ {i\hbar s_{i} =\varepsilon _{ijk} l^{j} s^{k} +x_{i} p_{j} s^{j} -p_{i} x_{j} s^{j} } \end{array}\right.$,   (11)

We see that part of this decomposition includes the cross-product $l\times s$ of the orbital angular momentum with the intrinsic spin.  We may also multiply the lower equation (11) through by $\varepsilon ^{mni}$ and then employ the commutation relationship $\left[s^{m} ,s^{n} \right]=i\hbar \varepsilon ^{mni} s_{i}$, to write:

$\left[s^{m} ,s^{n} \right]=\varepsilon ^{mni} \varepsilon _{ijk} l^{j} s^{k} +\varepsilon ^{mni} x_{i} p_{j} s^{j} -\varepsilon ^{mni} p_{i} x_{j} s^{j} =l^{m} s^{n} +\varepsilon ^{mni} x_{i} p_{j} s^{j} -\varepsilon ^{mni} p_{i} x_{j} s^{j}$.  (12)

Note, we have also made use of $\varepsilon ^{mni} \varepsilon _{ijk} =\delta ^{mni} _{ijk}$

Equation (11) allows us to decompose the total spin S for a Dirac field $\psi$, as follows: WORDPRESS DOES NOT LIKE THE INTEGRALS — NEED TO FIX

$\left\{\begin{array}{c} {S=\int \left(\overline{\psi }s\psi \right)\, d^{3} x =-{\tfrac{i}{\hbar }} \int \left(\overline{\psi }\left[\left(l\times s\right)+x\left(p\cdot s\right)-p\left(x\cdot s\right)\right]\, \psi \right)\, d^{3} x\quad \; \; } \\ {S_{i} =\int \left(\overline{\psi }s_{i} \psi \right)\, d^{3} x =-{\tfrac{i}{\hbar }} \int \left(\overline{\psi }\left[\varepsilon _{ijk} l^{j} s^{k} +x_{i} p_{j} s^{j} -p_{i} x_{j} s^{j} \right]\psi \right)\, d^{3} x } \end{array}\right.$ (13)

See Ohanian, H., What is spin?,  equation (18).

More to follow . . .

1. I just ran into Ohanian again, in a paper on the Feynman checkerboard model. And I put up a blog post extending the method.

Did you take classes from him? Or how did you run into his paper?

Comment by carlbrannen — April 22, 2008 @ 8:20 pm

2. Hi Carl,

I took classes from him in 1980-82, and asked him questions from time to time after that. But actually, Dr. Thomas Love at USC turned me on to his paper when I first posted about Kaluza Klein and intrinsic spin, suggesting (correctly in my view) that Ohanian’s paper is the most important paper to date on spin.

What I always liked about Ohanian is that he looks for intuitive, physical pictures, and has a certain “reserve” about quantum explanations which is apparent from the way he talks of Pauli’s “pontification” in his paper and insists on a physical, classical picture of spin. He is not a “God does not play Dice” radical, but I suspect he’d get as close to that view as the observed physics would allow.

One day, circa 1982, Ohanian handed me a small sheet of paper with two papers written down on it, and suggested I might enjoy them. One was Reinich’s paper on “already unified field theory” which places source-free electrodynamics onto a totally general relativistic footing. The other was Wheeler’s paper on Geometrodynamics, which popularized and extended Reinich. I was hooked, and ever since that day (aside from taking 18 years off from physics to raise two dynamite kids and start and build my patent law business), I have always used my spare time to pursue the goal of finding a totally geometrodynamic footing for the physical world, including quantum theory.

So, there you have my life story since 1980, in a nutshell.

Going back a bit further, I always loved physics (just talk to anyone who knew me in high school, 1969-1972), but opted out at MIT to study computer science because there seemed to be a better living in that and I sensed that I’d lose any ability to do independent physics work if I went the formal track.

Then, in 1974 I met my wife Debbie. Her dad is recognized as one of the very top EEs in the world, who personally designed the modern power grid in the US. He also has a deep interest in physics, and has a whole library of physics books (most of which I have since borrowed and now have in my home — squatter’s rights ;-). He also had a personal relationship with the late Banesh Hoffman, one of Einstein’s collaborators.

So, when I started law school in 1976 (which I hated with a passion — those were the worst three years of my life), I started delving through his library and reading through such books as “Strange Story of the Quantum” and at least half a dozen books about Einstein and relativity. As well as the Feynman lectures. I was so enamored, that I spent my three years in law school, studying physics on the side.

After I graduated in 1979 and Debbie finished her graduate work in 1980, we came to Schenectady, where I still live. I worked for a couple of years at GE (second worst experience next to law school), and often walked to and from work. Halfway between GE and my then-apartment was Union College where Ohanian used to teach, as well as at RPI. I was in the midst of one physics study or another, which honestly, continued unabated between 1976 and 1986 when my son Joshua was one year old and it was time to focus on him. One day, just on a total whim, I walked into the physics building at Union to see if there was anyone there who would like to talk about physics. Lo and behold, there was only one other person in the building, and it happened to be Ohanian! So, we started to talk, and next thing I knew, I was taking his General Relativity class at RPI.

That beings the nutshell back to 1969. Before that, I suppose was just a typical math/science geek growing up. That gets me back to born 1/15/1954 at New York Hospital overlooking the East River.

Jay.

Comment by Jay R. Yablon — April 24, 2008 @ 2:02 am

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