# Lab Notes for a Scientific Revolution (Physics)

## May 8, 2008

### How Precisely can we Measure an Electron’s Heisenberg Uncertainty? (or, How Certain is Uncertainty?)

In a May 24 post Heisenberg Uncertainty and Schwinger Anomaly: Two Sides of the Same Coin?, I set forth the hypothesis that the anomalous magnetic moment first characterized by Schwinger, may in fact be a manifestation of the Heisenberg uncertainty relationship, and in particular, that the excess of the uncertainty over $\hbar/2$ may in fact originate from the same basis as the excess of the intrinsic spin magnetic moment g-factor g, over the Dirac value of 2.  This hypothesis is most transparently written as $\Delta x\Delta p=\frac{\left|g\right|}{2} \frac{\hbar }{2} =\left(1+\frac{\alpha}{2\pi } +\ldots \right)\frac{\hbar }{2}$, where $\alpha$ is the running electromagnetic coupling for which $\alpha \left(\mu \right)\to 1/137.036$ at low probe energy $\mu$.  I also pointed out that a crucial next step was to employ a calculation similar to that shown at http://en.wikipedia.org/wiki/Uncertainty_principle#Wave_mechanics, but for a non-Gaussian wavefunction.

I have now concluded a full calculation along these lines, of the precise uncertainty associated with a particle wavefunction of the general form $\psi (x)=Ne^{-\frac{1}{2} A'x^{2} +B'x-V'\left(x\right)}$.  (The primes are a convenience used in calculation where we define $A\equiv A'+A'*$, etc. when calculating expected values, to take into account the possibility of the wavefunction parameters being imaginary.)  While I refer to $V'\left(x\right)$ as an “intrinsic potential,” it is perhaps better to think about this simply as an unspecified, completely-general polynomial in x, which renders the wavefunction completely general.  I have linked an updated draft of my paper which includes this calculation in full and applies it to the hypothesis set forth above, at Heisenberg Uncertainty and the Schwinger Anomaly. While the calculation is lengthy (and took a fair bit of effort to perform, then cross-check), the essence of what is contained in this paper can be summarized quite simply.  So I shall lay out a brief summary below, using the equation numbers which appear in the above-linked paper.

The essence of the results demonstrated in this paper is as follows.  Start with the generalized non-Gaussian wavefunction:

$\psi (x)=Ne^{-\frac{1}{2} A'x^{2} +B'x-V'\left(x\right)}$  (4.1)

Calculate its uncertainty by calculating its Fourier transform $\psi (p)$ (see (6.1)), by calculating each of its variances $(\Delta x)^2$ (5.4) and $(\Delta p)^2$ (7.4), and then by multiplying these together and taking the square root to arrive at the uncertainty.  The calculation is lengthy but straightforward, and it leads to the bottom line result:

$\Delta x\Delta p=\frac{\hbar }{2} \sqrt{1-2A'\left(\frac{dV'}{dB'} \right)^{2} +4B'\frac{dV'}{dB'} } =\frac{\hbar }{2} \sqrt{1-4A'V'\frac{d^{2} V'}{dB'^{2} } +4V'}$.   (8.5)

It is important to emphasize that (8.5) is a mathematical result that is totally independent of the hypothesized relationship of the uncertainty to the intrinsic spin.  So, if you ever been dissatisfied with the inequality of the Heisenberg relationship $\Delta x\Delta p\ge {\tfrac{1}{2}} \hbar$ and wondered what the exact uncertainty is for a given wavefunction, you will find this calculated with precision in sections 4 through 8, and the answer is (8.5) above.  The upshot is that (8.5) above is the precise uncertainty for a wavefunction (4.1) with A’, B’ and V’ all real.  We cannot give a position and momentum with precision, but we can give an uncertainty with precision.  The reasons for having A’, B’ and V’ be real are developed in the paper, but suffice it to say that A’, B’ real is necessary to avert a divergent uncertainty, and if V’ were imaginary rather than real, the uncertainty would always be exactly equal to $\hbar/2$.

Now, with the result (8.5) in hand, we return to the original hypothesis which, if it is true, would require that:

$\frac{\Delta x\Delta p}{\hbar /2} =\sqrt{1+4B'\frac{dV'}{dB'} -2A'\left(\frac{dV'}{dB'} \right)^{2} } =\sqrt{1+4V'-4A'V'\frac{d^{2} V'}{dB'^{2} } } =\frac{\left|g\right|}{2} =1+\frac{a}{2\pi } +\ldots$   (9.1)

Using the series expansion for $\sqrt{1+x}$, we then make the connection:

$V'\equiv \alpha /4\pi$    (9.5)

Now, it behooves us to return to the wavefunction (4.1), and use (9.5) to write:

$\psi (x)=Ne^{-\frac{1}{2} A'x^{2} +B'x-\frac{\alpha }{4\pi } }$,    (9.6)

and to rewrite the uncertainty relationship (9.1) as:

$\frac{\Delta x\Delta p}{\hbar /2} =\sqrt{1+\frac{1}{\pi } B'\frac{d\alpha }{dB'} -\frac{1}{8\pi ^{2} } A'\left(\frac{d\alpha }{dB'} \right)^{2} } =\sqrt{1+\frac{\alpha }{\pi } -A'\frac{\alpha }{4\pi ^{2} } \frac{d^{2} \alpha }{dB'^{2} } } =\frac{\left|g\right|}{2} =1+\frac{\alpha }{2\pi } +\ldots$ (9.7)

Now, let’s get directly to the point: an electron with the wavefunction (9.6), with $A'$ and $B'$ real, will have the uncertainty relationship (9.7), period.  For $\alpha =1/137.036$, the leading uncertainty term $\sqrt{1+\frac{\alpha }{\pi } } =1.00116073607$, while the leading anomaly term $1+\frac{\alpha }{2\pi } =1.00116140973$.  These two terms differ by just under 7 parts in $10^{-7}$.  Therefore, we can state the following:

TheoremFor a wavefunction $\psi (x)=Ne^{-\frac{1}{2} A'x^{2} +B'x-\frac{\alpha }{4\pi } }$ with $A'$ and $B'$ real, the uncertainty ratio $\frac{\Delta x\Delta p}{\hbar /2}$, to leading order in $\alpha$, differs from the intrinsic Schwinger g-factor $g/2$ by less than 7 parts in $10^{-7}$.

We have stated this as a theorem, because this is a simple statement of fact, and involves no interpretation or hypothesis whatsoever.  However, in order to sustain the broader hypothesis

$\Delta x\Delta p=\frac{\left|g\right|}{2} \frac{\hbar }{2} \ge \frac{\hbar }{2}$,   (3.4)

we do need to engage in some interpretation.

First, we define (9.6) as the intrinsic wavefunction of a ground state electron with no orbital angular momentum and no applied external potential.  And, we define (9.7) as the intrinsic uncertainty of this intrinsic wavefunction.  Not every electron will have this wavefunction or this uncertainty or this g-factor, but this wavefunction becomes the baseline electron wavefunction from which any variation is due to extrinsic factors, such as possessing orbital angular momentum or being placed into an external potential, for example, that of a proton.  Thus, our hypothesis (3.4) is a hypothesis about the intrinsic uncertainty associated with the intrinsic wavefunction, and it says that:

Reformulated HypothesisThe intrinsic uncertainty associated with the intrinsic electron wavefunction is identical with the intrinsic g-factor of the anomalous magnetic moment.

The final section 10 of this draft paper linked above, is in progress at this time.  What I am presently trying to do, is make some sense of what appears to be a “new” type of g-factor $\left|g_{{\rm ext}} \right|$, emanating from an extrinsic potential (polynomial) $V_{{\rm ext}}$ in the wavefunction:

$\psi (x)=Ne^{-\frac{1}{2} A'x^{2} +B'x-V_{int} \left(x\right)-V_{{\rm ext}} \left(x\right)} =Ne^{-\frac{1}{2} A'x^{2} +B'x-\frac{\alpha }{4\pi } -V_{{\rm ext}} \left(x\right)}$   (10.1)

This new g-factor is defined in (10.2), and is isolated in (10.3) as such:

$\begin{array}{l} {\frac{\left|g_{{\rm ext}} \right|}{2} =\sqrt{1+\frac{1}{\pi } B'\frac{d\alpha +4\pi dV_{{\rm ext}} }{dB'} -\frac{1}{8\pi ^{2} } A'\left(\frac{d\alpha +4\pi dV_{{\rm ext}} }{dB'} \right)^{2} } -\sqrt{1+\frac{1}{\pi } B'\frac{d\alpha }{dB'} -\frac{1}{8\pi ^{2} } A'\left(\frac{d\alpha }{dB'} \right)^{2} } } \\ {\quad \quad =\sqrt{1+\frac{\alpha +4\pi V_{{\rm ext}} }{\pi } -A'\frac{\alpha +4\pi V_{{\rm ext}} }{4\pi ^{2} } \frac{d^{2} \alpha +4\pi d^{2} V_{{\rm ext}} }{dB'^{2} } } -\sqrt{1+\frac{\alpha }{\pi } -A'\frac{\alpha }{4\pi ^{2} } \frac{d^{2} \alpha }{dB'^{2} } } } \end{array}$.   (10.3)

In section 10, I have provided my “first impression” of where this new g-factor may fit in, in relation to the Paschen-Back effect, but would be interested in the thoughts of the reader regarding what to make of the above g-factor (10.3) and where it might fit into the “scheme of things.”

Thanks for listening, and for your thoughts.

Jay.

## May 1, 2008

### Heisenberg Uncertainty and Schwinger Anomaly Continued: Draft Paper

I have been writing a paper to rigorously develop the hypothesis I presented last week, in a post linked at Heisenberg Uncertainty and Schwinger Anomaly: Two Sides of the Same Coin?.  I believe there is enough developed now, and I think enough of the kinks are now out, so you all may take a sneak preview.  Thus, I have linked my latest draft at:

Heisenberg Uncertainty and the Schwinger Anomaly

Setting aside the hypothesized connection between the magnetic anomaly and uncertainty, Sections 4 through 7, which have not been posted in any form previously, stand completely by themselves, irrespective of this hypothesis.  These sections are strictly mathematical in nature, and they provide an exact measure for how the uncertainty associated with a wavefunction varies upwards from $\hbar/2$ as a function of the potential, and the parameters of the wavefunction itself.  The wavefunction employed is completely general, and the uncertainty relation is driven by a potential $V$.

This is still under development, but this should give you a very good idea of where this is headed.

Of course, I welcome comment, as always.

Best regards,

Jay.

## April 24, 2008

### Heisenberg Uncertainty and Schwinger Anomaly: Two Sides of the Same Coin?

In section 3 of Heisenberg Uncertainty and Schwinger Anomaly: Two Sides of the Same Coin?, I have posted a calculation which shows why the Schwinger magnetic anomaly may in fact be very tightly tied to the Heisenberg inequality $\Delta x\Delta p\ge {\tfrac{1}{2}} \hbar$.  The bottom line result, in (3.11) and (3.12), is that the gyromagnetic “g-factor” for a charged fermion wave field with only intrinsic spin (no angular momentum) is given by:

$\left|g\right|=2\frac{\left(\Delta x\Delta p\right)}{\hbar /2} \ge 2$  (3.11)

It is also helpful to look at this from the standpoint of the Heisenberg principle as:

$\Delta x\Delta p=\frac{\left|g\right|}{2} \frac{\hbar }{2} \ge \frac{\hbar }{2}$  (3.12)

First, if (3.11) is true, then the greater than or equal to inequality of Heisenberg says, in this context, that the magnitude of the intrinsic g-factor of a charged wavefunction is always greater than or equal to 2.  That is, the inequality $\Delta x\Delta p\ge {\tfrac{1}{2}} \hbar$ becomes another way of stating a parallel inequality $\left|g\right|\ge 2$.  We know this to be true for the charged leptons, which have $g_{e} /2=1.0011596521859$, $g_{\mu } /2=1.0011659203$, and $g_{\tau } /2=1.0011773$ respectively. [The foregoing data is extracted from W.-M. Yao et al., J. Phys. G 33, 1 (2006)]

Secondly, the fact that the charged leptons have g-factors only slightly above 2, suggests that these a) differ from perfect Gaussian wavefunctions by only a very tiny amount, b) the electron is slightly more Gaussian than the muon, and the muon slightly more-so than the tauon.  The three-quark proton, with $g_{P} /2=2.7928473565$, is definitively less-Gaussian the charged leptons.  But, it is intriguing that the g-factor is now seen as a precise measure of the degree to which a wavefunction differs from a perfect Gaussian.

Third, (3.11) states that the magnetic moment anomaly via the g-factor is a precise measure of the degree to which $\Delta x\Delta p$ exceeds $\hbar /2$.  This is best seen by writing (3.11) as (3.12).

Thus, for the electron, $\left(\Delta x\Delta p\right)_{e} =1.0011596521859\cdot \left(\hbar /2\right)$, to give an exact numerical example.  For a different example, for the proton, $\left(\Delta x\Delta p\right)_{P} =2.7928473565\cdot \left(\hbar /2\right)$.

Fourth, as a philosophical and historical matter, one can achieve a new, deeper perspective about uncertainty.  Classically, it was long thought that one can specify position and momentum simultaneously, with precision.  To the initial consternation of many and the lasting consternation of some, it was found that even in principle, one could at best determine the standard deviations in position and momentum according to $\Delta x\Delta p\ge {\tfrac{1}{2}} \hbar$.  There are two aspects of this consternation:  First, that one can never have$\Delta x\Delta p=0$ as in classical theory.  Second, that this is merely an inequality, not an exact expression, so that even for a particle with $\Delta x\Delta p\ge {\tfrac{1}{2}} \hbar$, we do not know for sure what is its exact value of $\Delta x\Delta p$.  This latter issue is not an in-principle limitation on position and momentum measurements; it is a limitation on the present state of human knowledge.

Now, while ${\tfrac{1}{2}} \hbar$ is a lower bound in principle, the question remains open to the present day, whether there is a way, for a given particle, to specify the precise degree to which its $\Delta x\Delta p$ exceeds ${\tfrac{1}{2}} \hbar$, and how this would be measured.  For example, one might ask, is there any particle in the real world that is a perfect Gaussian, and therefore can be located in spacetime and conjugate momentum space, down to exactly ${\tfrac{1}{2}} \hbar$.  Equation (3.12) above suggests that if such a particle exists, it must be a perfect Gaussian, and, that we would know it was a perfect Gaussian, if its g-factor was experimentally determined to be exactly equal to the Dirac value of 2.  Conversely, (3.12) tells us that it is the g-factor itself, which is the direct experimental indicator of the magnitude of $\Delta x\Delta p$ for any given particle wavefunction.  The classical precision of $\Delta x\Delta p=0$ comes full circle, and while it will never return, there is the satisfaction of being able to replace this with the quantum  mechanical precision of (3.12), $\Delta x\Delta p=\left|g\right|\hbar /4$, rather than the weaker inequality of $\Delta x\Delta p\ge {\tfrac{1}{2}} \hbar$.

Fifth, if (3.12) is correct, then since it is independently known from Schwinger that $\frac{g}{2} =1+\frac{a}{2\pi } +\ldots$, this would mean that we would have to have:

$\Delta x\Delta p=\frac{\left|g\right|}{2} \frac{\hbar }{2} =\left(1+\frac{a}{2\pi } +\ldots \right)\frac{\hbar }{2}$  (3.13)

Thus, from the perturbative viewpoint, the degree to which $\Delta x\Delta p$ exceeds ${\tfrac{1}{2}} \hbar$ would have to be a function of the running coupling strength $\alpha =e^{2} /4\pi$ in Heaviside-Lorentz units.  As Carl Brannnen has explicitly pointed out to me, this means that a Gaussian wavepacket is by definition non-interacting; as soon as there is an interaction, one concurrently loses the exact Gaussian.

Sixth, since deviation of the g-factor above 2 would arise from a non-Gaussian wavefunction such as $\psi (x)=N\exp \left(-{\tfrac{1}{2}} Ax^{2} +Bx\right)$, the rise of the g-factor above 2 would have to stem from the $Bx$ term in this non-Gaussian wavefuction.  In this regard, we note to start, that $N\int \exp \left(-{\tfrac{1}{2}} Ax^{2} +Bx\right)dx= \sqrt{2\pi /A} \exp \left(B^{2} /2A\right)$, for a non-Gaussian wavefunction, versus $N\int \exp \left(-{\tfrac{1}{2}} Ax^{2} \right)dx= \sqrt{2\pi /A}$ for a perfect Gaussian.

Finally, to calculate this all out precisely, one would need to employ a calculation similar to that shown at http://en.wikipedia.org/wiki/Uncertainty_principle#Wave_mechanics, but for the non-Gaussian $N\int \exp \left(-{\tfrac{1}{2}} Ax^{2} +Bx\right)dx= \sqrt{2\pi /A} \exp \left(B^{2} /2A\right)$ rather than the Gaussian$N\int \exp \left(-{\tfrac{1}{2}} Ax^{2} \right)dx= \sqrt{2\pi /A}$, to arrive at the modified bottom line equation of this Wiki section.  That is the next calculation I plan, but this is enough, I believe, to post at this time.

## 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 . . .

## March 20, 2008

### Derivation of Heisenberg Uncertainty from Kaluza Klein Geometry

For those who have followed my Kaluza-Klein (KK) work, I believe that it is now possible to derive not only intrinsic spin, but Heisenberg uncertainty directly from a fifth, compactified dimension in Kaluza Klein.  This would put canonical quantum mechanics on a strictly Riemannian geometric foundation which — as a side benefit — unites gravitation and electromagnetism.

I need to consolidate over the next few days and will of course make a more expanded post when I am ready, but here is the basic outline.  First, take a look at:

Intrinsic Spin and the Kaluza-Klein Fifth Dimension

where I show how intrinsic spin is a consequence of the compactified fifth dimension.  This paper, at present, goes so far as to show how the Pauli spin matrices emerge from KK.

Next, go to the two page file:

Spin to Uncertainty

This shows how one can pop Heisenberg out of the spin matrices.

Finally, go to the latest draft paper on KK generally, at:

Kaluza-Klein Theory and Lorentz Force Geodesics with Non-linear QED

This lays out the full context in which I am developing this work.  Please note that the discussion on intrinsic spin in the third link is superseded by the discussion thereof in the first link.

More to follow . . .

Jay.

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