# 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 . . .

## April 13, 2008

### Thesis Defense of the Kaluza-Klein, Intrinsic Spin Hypothesis — EARLY DRAFT

I have been engaged in a number of Usenet and private discussions about the paper Intrinsic Spin and the Kaluza-Klein Fifth Dimension, Rev 3.0 which I posted here on this blog on March 30.

A number of critiques have been raised, which you can see if you check out the recent Usenet threads I started related to intrinsic spin under the heading “Query about intrinsic verus [sic] orbital angular momentum,” over at sci.physics.foundations, sci.physics.relativity and sci.physics.research. These are among the “links of interest” provided in the right-hand pane of this weblog.

I believe that these critiques can be overcome, and that this hypothesis relating to Kaluza-Klein and intrinsic spin and the spatial isotropy of the square of the spin will survive and be demonstrated, ultimately, to be in accord with the physical reality of nature.

I have begun a new paper which is linked at: Thesis Defense of the Kaluza-Klein, Intrinsic Spin Hypothesis, Rev 1.0, which will respond thoroughly and systematically to the various critiques.  What is here so far is the introductory groundwork.  But, I would appreciate continued feedback as this development continues.

Note that the links within the PDF file unfortunately do not work, so to get the intrinsic spin paper, you need to go to Intrinsic Spin and the Kaluza-Klein Fifth Dimension, Rev 3.0.  Also, to get Wheeler’s paper which is referenced, go to Wheeler Geometrodynamics.

## March 30, 2008

### Revised Paper on Kaluza-Klein and Intrinsic Spin, Based on Spatial Isotropy

I have now prepared an updated revision of a paper demonstrating how the compact fifth dimension of Kaluza-Klein is responsible for the observed intrinsic spin of the charged leptons and their antiparticles.  The more global, underlying view, is that all intrinsic spins originate from motion through the curled up, compact $x^5$ dimension.  This latest draft is linked at:

Intrinsic Spin and the Kaluza-Klein Fifth Dimension, Rev 3.0

Thanks to some very helpful critique from Daryl M. on a thread at sci.physics.relativity, particularly post #2, I have entirely revamped section 5, which is the heart of the paper wherein we establish the existence of intrinsic spin in $x^5$, on the basis of “fitting” oscillations around a $4\pi$ loop of the compact fifth dimension (this is to maintain not only orientation but entanglement version).  From this approach, quantization of angular momentum in $x^5$ naturally emerges, it also emerges that the intrinsic $x^5$ angular momentum in the ground state is given by $(1/2) \hbar$.

In contrast to my earlier papers where I conjectured that the intrinsic spin in $x^5$ projects out into the three ordinary space dimensions by virtue of its orthogonal orientation relative to the $x^5$ plane, which was critiqued in several Usenet posts, I now have a much more direct explanation of how the intrinsic spin projects out of $x^5$ to where we observe it.

In particular, we recognize that one of the objections sometimes voiced with regards to a compactified fifth dimension is the question: how does one “bias” the vacuum toward one of four space dimensions, over the other three, by making that dimension compact?  This was at the root of some Usenet objections also raised earlier by DRL.

In this draft — and I think this will overcome many issues — we require that at least as regards intrinsic angular momentum, the square of the $J^5 = (1/2) \hbar$ obtained for the intrinsic angular momentum in $x^5$, must be isotropically shared by all four space dimensions.  That is, we require that there is to be no “bias” toward any of the four space dimensions insofar as squared intrinsic spin is concerned.  Because $J^5 = (1/2) \hbar$ emanates naturally from the five dimensional geometry, we know immediately that $\left(J^{5} \right)^{2} ={\textstyle\frac{1}{4}} \hbar ^{2}$, and then, by the isotropic requirement, that $\left(J^{1} \right)^{2} =\left(J^{2} \right)^{2} =\left(J^{3} \right)^{2} ={\textstyle\frac{1}{4}} \hbar ^{2}$ as well.  We then arrive directly at the Casimir operator $J^{2} =\left(J^{1} \right)^{2} +\left(J^{2} \right)^{2} +\left(J^{3} \right)^{2} ={\textstyle\frac{3}{4}} \hbar ^{2}$ in the usual three space dimensions, and from there, continue forward deductively.

For those who have followed this development right along, this means in the simplest terms, that rather than use “orthogonality” to get the intrinsic spin out of $x^5$ and into ordinary space, I am instead using “isotropy.”

There is also a new section 8 on positrons and Dirac’s equations which has not been posted before, and I have made other editorial changes throughout the rest of the paper.

## March 29, 2008

### Stepping Back from Kaluza-Klein: Planned Revisions

Those who have followed my Weblog are aware that I have been putting in a lot of work on Kaluza-Klein theory.  This post is to step back from the canvas, lay out the overall picture of what I am pursuing, and summarize what I plan at present to change or correct in the coming days and weeks.  This is in keeping with the concept of this Weblog as “Lab Notes,” or as a public “scientific diary.”

There are really two main aspects to this Kaluza-Klein work:

First, generally, I have found that 5-D Kaluza-Klein theory is most simply approached by starting with (classical) Lorentz force motion, and requiring the Lorentz force motion to be along the geodesics of the five dimensional geometry.  I am far from the first person who recognizes that the Lorentz force can be represented as geodesic motion in a 5-D model.  But I have found, by starting with the Lorentz force, and by requiring the 5-D electromagnetic field strength tensor to be fully antisymmetric, that all of the many “special assumptions” which are often employed in Kaluza-Klein theory energy very naturally on a completely deductive basis, with no further assumptions required.  I also believe that this approach leads to what are perhaps some new results, especially insofar as the Maxwell tensor is concerned, and insofar as QED may be considered in a non-linear context.   The latest draft of this global work on Kaluza-Klein may be seen at Kaluza-Klein Theory and Lorentz Force Geodesics.

Second, specifically, within this broader context, is the hypothesis that the fifth-dimensional “curled” motion is the direct mainspring of intrinsic spin.  More than anything else, the resistance by many physicists to Kaluza-Klein and higher-dimensional theories, rests on the simple fact that this fifth dimension — and any other higher dimensions — are thought to not be directly observable.  In simplest form, “too small” is the usual reason given for this.  Thus, if it should become possible to sustain the hypothesis that intrinsic spin is a directly-observable and universally-pervasive outgrowth of the fifth dimension, this would revitalize Kaluza-Klein as a legitimate and not accidental union of gravitation and electrodynamics, and at the same time lend credence to the higher-dimensional efforts also being undertaken by many researchers.  The latest draft paper developing with this specific line of inquiry is at Intrinsic Spin and the Kaluza-Klein Fifth Dimension.

Now, the general paper at Kaluza-Klein Theory and Lorentz Force Geodesics is very much a work in progress and there are things in this that I know need to be fixed or changed.  If you should review this, please keep in mind the following caveats:

First, sections 1-4 are superseded by the work at Intrinsic Spin and the Kaluza-Klein Fifth Dimension and have not been updated recently.

Second, sections 5-7 are still largely OK, with some minor changes envisioned.  Especially, I intend to derive the “restriction” $\Gamma^u_{55}=0$ from $F^{{\rm M} {\rm N} } =-F^{{\rm N} {\rm M} }$ rather than impose it as an ad hoc condition.

Third, sections 8-11 needs some reworking, and specifically: a) I want to start with an integration over the five-dimensional volume with a gravitational constant $G_{(5)}$ suited thereto, and relate this to the four dimensional integrals that are there at present; and b) I have serious misgivings about using a non-symmetric (torsion) energy tensor and am inclined to redevelop this to impose symmetry on the energy tensor — or at least to explore torsion versus no torsion in a way that might lead to an experimental test.  If we impose symmetry on the energy tensor, then the Maxwell tensor will be the $J^{\mu } =0$ special case of a broader tensor which includes a $J^\mu A^\nu + J^\nu A^\mu$ term and which applies, e.g., to energy flux densities (Poynting components) $T^{0k}$, k=1,2,3 for “waves” of large numbers of electrons.

Fourth, I am content with section 12, and expect it will survive the next draft largely intact.  Especially important is the covariant derivatives of the electrodynamic potentials being related to the ordinary derivatives of the gravitational potentials, which means that the way in which people often relate electrodynamic potentials to gravitational potentials in Kaluza-Klein theory is valid only in the linear approximation.  Importantly, this gives us a lever in the opposite direction, into non-linear electrodynamics.

Fifth, I expect the development of non-linear QED in section 13 to survive the next draft, but for the fact that the R=0 starting point will be removed as a consequence of my enforcing a symmetric energy tensor in sections 8-11.  Just take out all the “R=0” terms and leave the rest of the equation alone, and everything else is more or less intact.

Finally, the experiment in Section 15, if it stays, would be an experiment to test a symmetric, torsionless energy tensor against a non-symmetric energy tensor with torsion.  (Basically, metric theory versus Cartan theory.)  This is more of a “back of the envelope” section at present, but I do want to pursue specifying an experiment that will test the possible energy tensors which are available from variational principles via this Kaluza-Klein theory.

The paper at Intrinsic Spin and the Kaluza-Klein Fifth Dimension dealing specifically with the intrinsic spin hypothesis is also a work in progress, and at this time, I envision the following:

First, I will in a forthcoming draft explore positrons as well as electrons.  In compactified Kaluza-Klein, these exhibit opposite motions through $x^5$, and by developing the positron further, we can move from the Pauli spin matrices toward the Dirac $\gamma^\mu$ and Dirac’s equation.

Second, I have been engaged in some good discussion with my friend Daryl M. on a thread at sci.physics.relativity.  Though he believes I am “barking up the wrong tree,” he has provided a number of helpful comments, and especially at the bottom of post #2 where he discusses quantization in the fifth dimension using a wavelength $n \lambda = 2 \pi R$.  (I actually think that for fermions, one has to consider orientation / entanglement issues, and so to secure the correct “version,” one should use $n \lambda = 4 \pi R$ which introduces a factor of 2 which then can be turned into a half-integer spin.)  I am presently playing with some calculations based on this approach, which you will recognize as a throwback to the old Bohr models of the atom.

Third, this work of course uses $x^5 = R\phi$ to define the compact fifth dimension.  However, in obtaining $dx^5$, I have taken $R$ to be a fixed, constant radius.  In light of considering a wavelength $n \lambda = 4 \pi R$ per above, I believe it important to consider variations in $R$ rather than fixed $R$, and so, to employ $dx^5 = Rd\phi + \phi dR$.

There will likely be other changes along the way, but these are the ones which are most apparent to me at present.  I hope this gives you some perspective on where this “work in flux” is at, and where it may be headed.

Thanks for tuning in!

Jay.

## March 28, 2008

### Further Considerations on the Energy Tensor: Metric versus Torsion

After reviewing some very helpful discussion on sci.physics.foundations and sci.physics.relativity in the threads I initiated regarding non-symmetric energy tensors and a suggested Kaluza-Klein experiment, and going back to Misner, Thorne and Wheeler’s “Gravitation,” I am starting to shift my viewpoint to be in opposition to the idea of using a non-symmetric (Cartan / Torsion) energy tensor because of the adverse impact this has on formulating a metric theory of gravitation.

There is a non-symmetric energy tensor in equations (15.11) to (15.13) of:

Kaluza-Klein Theory and Lorentz Force Geodesics Rev. 6.0

which is based upon the *non-symmetric* energy tensor of trace matter derived in (11.6).  What I have been turning over, is whether I ought to be comfortable with this result, and my sense runs against it.

However, at the point of original derivation in sections 8-11, there is actually a choice: one can construct the variation of the Lagrangian density of matter with respect to $g_{\mu \nu}$ such that a symmetric tensor will result, or one can choose not to, by creating a symmetric term or not.  This is actually a form of gravitational “symmetry breaking” that occurs in the process of taking the variation of the matter Lagrangian density with respect to the metric tensor.  I think both paths need to be developed, because they lead to on the one hand to a symmetric energy tensor, and on the other to a non-symmetric energy tensor.  In either case, the key term distinguishing this energy tensor from the Maxwell tensor is $J^\mu A^\nu$.

If we impose symmetry on the energy tensor, then the Maxwell tensor will be the $J^u=0$ special case of a broader tensor which includes a $J^\mu A^\nu + J^\nu A^\mu$ term and which applies, e.g., to energy flux densities (Poynting components) $T^{0k}$, $k=1,2,3$ for “waves” of large numbers of electrons.

Then, the experiment becomes — not a test of the torsion tensor — but a test as between an energy tensor with and without torsion.  That is, the experiment as reformulated, becomes a test of metric-style versus Cartan-style theories of gravitation.

Dealing with the currents $J^\mu$ is clear.  Regarding how to deal with the potential $A^\mu$ in doing the experiment, think about a beam of electrons. They of course will all repel, so the beam will emerge conically from the electron gun if nothing is done to force them onto a parallel path. Now, take a circular cross section of electrons from the beam striking an energy flux detector.  One can think of the cross-sectional surface where the electron stream meets the detector as a “disk,” not unlike a charged, flat, frisbee, which is also productively thought about as a dielectric.  I would submit that one can assign a “zero” potential to the center of the cross section, and a varying non-zero potential to the periphery.  That is, if one were to take a circular dielectric disk and fill it with electric charge, then float some positive charge nearby, the positive charge — I believe — would be attracted toward and seat itself at the center of the disk, and so that would be a natural place to define the zero of potential.

This would also mean that regional detections of flux toward the fringes of the detector will be different than toward the center, assuming uniformity of charge distribution, because the energy created by the potentials among the electrons are different in different regions.  So, there is a way to assign potentials even without applying an external voltage, though someone conducting this experiment may want to also apply an external voltage simply to vary the range of experiments.

Now, to the main point: one should do the experiment with random, unpolarized electrons, and then again with spins aligned with and against the direction of propagation, merely to test the symmetric versus non-symmetric energy tensors one to the other.   One will win, the other not.  Metric versus torsion.

I am planning a restructuring of the paper at the above link.  In the near future, I will outline the proposed restruturing — what I would plan to keep and what I plan to change.

Jay.

## March 22, 2008

### A Possible Kaluza-Klein Experiment

It has been suggested — appropriately so — that I consider whether there might be one or more experiments which can be designed to validate or falsify some of the Klauza-Klein results which I have been posting of late. I believe that one possible experiment resides in the non-symmetric energy tensor of trace matter derived in (11.6) of my latest posted paper. Thus, I have added a new section 15 to this paper, and reposted the entire paper, with this new section 15, at Kaluza-Klein Theory and Lorentz Force Geodesics Rev. 6.0 Because this is of particular interest as it may open some new experimental windows, I have posted section 15 below as well. Please note: the specific discussion of the connection between the compactified fifth dimension, and intrinsic spin, is not updated in this paper, and the latest discusssion I have written up on this topic, is at Intrinsic Spin and the Kaluza-Klein Fifth Dimension.

Section 15: At this juncture, we have enough information to propose an experiment to validate or falsify some of the results derived thus far.  We turn for this purpose to the stress energy tensor of matter (11.6), which we raise into contravariant notation as follows:

$\kappa T^{\nu \mu } =-\kappa \left(F^{\mu \tau } F^{\nu } _{\tau } -{\textstyle\frac{1}{4}} g^{\mu \nu } F^{\sigma \tau } F_{\sigma \tau } \right)+{\textstyle\frac{\sqrt{2} }{2}} \overline{\kappa }g^{5\mu } J^{\nu } =\kappa T^{\mu \nu } _{Maxwell} +{\textstyle\frac{\sqrt{2} }{2}} \overline{\kappa }g^{5\mu } J^{\nu }$. (15.1)

The Maxwell tensor $T^{\mu \nu } _{Maxwell} =T^{\nu \mu } _{Maxwell}$ is, of course, a symmetric tensor.  But the added trace matter term $g^{5\mu } J^{\nu }$ is not necessarily symmetric, that is, there is no a priori reason why $g^{5\mu } J^{\nu }$ must be equal to $g^{5\nu } J^{\mu }$.  The origin of this non-symmetry was discussed earlier in Section 9.

With an eye toward conducting an experiment, let us now consider (15.1) in the linear approximation of (13.6) where ${\rm L}_{QCD} \approx -A^{\beta } J_{\beta } -{\textstyle\frac{1}{4}} F^{\sigma \tau } F_{\sigma \tau }$.  In the linear approximation, as used to reach (13.3), (12.11) reduces to $g^{5\mu } \approx \overline{\kappa }\left(\frac{\phi ^{5\mu } -{\textstyle\frac{1}{2}} bA^{\mu } }{1+{\textstyle\frac{1}{2}} \overline{\kappa }\phi } \right)\approx -{\textstyle\frac{1}{2}} \overline{\kappa }bA^{\mu }$, and (15.1) becomes:

$T^{\nu \mu } \approx -\left(F^{\mu \tau } F^{\nu } _{\tau } -{\textstyle\frac{1}{4}} g^{\mu \nu } F^{\sigma \tau } F_{\sigma \tau } \right)-2J^{\nu } A^{\mu } =T^{\mu \nu } _{Maxwell} -2J^{\nu } A^{\mu }$, (15.2)

where we have also used $b^{2} =8$ and $2\kappa =\overline{\kappa }^{2}$, and divided out $\kappa$.  The transpose of this non-symmetric energy tensor is:

$T^{\mu \nu } \approx -\left(F^{\mu \tau } F^{\nu } _{\tau } -{\textstyle\frac{1}{4}} g^{\mu \nu } F^{\sigma \tau } F_{\sigma \tau } \right)-2J^{\mu } A^{\nu } =T^{\mu \nu } _{Maxwell} -2J^{\mu } A^{\nu }$, (15.3)

Now, it is known that a non-symmetric energy tensor, physically, is indicative of a non-zero spin density.  In particular, using (15.2) and (15.3), the non-symmetry of the energy tensor is related to a non-zero spin density tensor $S^{\mu \nu \alpha }$ according to: [A good, basic discussion of the spin tensor is at http://en.wikipedia.org/wiki/Spin_tensor.]

$S^{\mu \nu \alpha } _{;\alpha } =T^{\mu \nu } -T^{\nu \mu } =-2J^{\mu } A^{\nu } +2J^{\nu } A^{\mu }$. (15.4)

For such a non-symmetric tensor, the “energy flux” is not identical to the “momentum density, as these differ by (15.4), for $\mu =0$, $\nu =k=1,2,3$ and vice versa.  If the spin density $S^{\mu \nu \alpha } =0$, then  in this special case, (15.4) yields:

$J^{\mu } A^{\nu } =J^{\nu } A^{\mu }$. (15.5)

So, for $S^{\mu \nu \alpha } =0$, (15.3) may be written using (15.5) as the explicitly-symmetric tensor:

$T^{\mu \nu } \approx -\left(F^{\mu \tau } F^{\nu } _{\tau } -{\textstyle\frac{1}{4}} g^{\mu \nu } F^{\sigma \tau } F_{\sigma \tau } \right)-J^{\mu } A^{\nu } -J^{\nu } A^{\mu } =T^{\mu \nu } _{Maxwell} -J^{\mu } A^{\nu } -J^{\nu } A^{\mu }$. (15.6)

Now, let’s consider a experiment which is entirely classical.  The $T^{0k}$ “Poynting” components of (15.4), (15.6) represent the energy flux across a two-dimensional area, for a flux of matter which we will take to be a stream of electrons, while the $T^{k0}$ components represent the momentum density.  The proposed experiment, then, will be to fire a stream of a very large number of electrons thereby constituting an electron “wave,” and to detect the aggregate flux of energy across a two-dimensional surface under various spin preparations, in precisely the same manner that one might test the flow of luminous energy across a surface when using light waves rather than electron waves.  Specifically, we propose in test I to fire electrons without doing anything to orient their spins, so that, statistically, the number of electrons flowing through the flux surface with positive helicity is equal to the number with negative helicity and so the spin density is zero, and (15.6) applies.  In test II, we fire electrons, but apply a magnetic field before detecting the flux, to ensure that all of the electrons are aligned to positive helicity.  In this event, the spin density, by design, is non-zero, and one of (15.2) or (15.3) will apply.  In test III, we do the same, but now apply the magnetic field to ensure that all of the electrons have negative helicity, before detecting the flux.

## 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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