# Lab Notes for a Scientific Revolution (Physics)

## 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.