27 May 2025

Heisenberg Uncertainty Principle and Its Consequences

uncertainty-principle zero-point-energy atomic-size nuclear-confinement

For a normalized state and a Hermitian observable $\hat A$, define the centred operator and standard deviation by

\[\delta\hat A=\hat A-\langle A\rangle, \qquad (\Delta A)^2=\langle(\delta\hat A)^2\rangle.\]

Apply the Cauchy-Schwarz inequality to the vectors $\delta\hat A\lvert\psi\rangle$ and $\delta\hat B\lvert\psi\rangle$:

\[(\Delta A)^2(\Delta B)^2 \geq\left\lvert\langle\delta\hat A\,\delta\hat B\rangle\right\rvert^2.\]

Split the product into Hermitian and anti-Hermitian parts,

\[\delta\hat A\,\delta\hat B =\frac12\{\delta\hat A,\delta\hat B\} +\frac12[\hat A,\hat B].\]

The expectation of the anticommutator is real, while the expectation of the commutator is purely imaginary. Therefore the squared modulus is at least the square of its imaginary part:

\[(\Delta A)^2(\Delta B)^2 \geq\frac14\left\lvert\langle[\hat A,\hat B]\rangle\right\rvert^2.\]

Taking the non-negative square root gives Robertson’s relation,

\[\boxed{\Delta A\,\Delta B \geq\frac12\left\lvert\langle[\hat A,\hat B]\rangle\right\rvert}.\]

Since $[\hat x,\hat p_x]=i\hbar$,

\[\boxed{\Delta x\,\Delta p_x\geq\frac{\hbar}{2}}.\]

Both sides have units of action: $\mathrm{m\,kg\,m\,s^{-1}}=\mathrm{J\,s}$. This spread is a property of identically prepared states, not merely a disturbance caused by an imperfect measuring instrument.

The path of an object

A classical path requires a definite position and momentum at every instant. If $\Delta x\to0$, the uncertainty relation requires $\Delta p_x\to\infty$; if $\Delta p_x\to0$, it requires $\Delta x\to\infty$. Thus an exact microscopic trajectory cannot be assigned. A wave-packet centre can approximate a classical path only when $\Delta x$ and $\Delta p$ are both negligible compared with the macroscopic length and momentum scales of the motion.

Zero-point energy

For the one-dimensional oscillator,

\[\hat H=\frac{\hat p^2}{2m}+\frac12m\omega^2\hat x^2.\]

The minimum-energy state is centred at the equilibrium point with $\langle x\rangle=\langle p\rangle=0$. Hence

\[\langle H\rangle =\frac{(\Delta p)^2}{2m}+\frac12m\omega^2(\Delta x)^2.\]

Using $\Delta p\geq\hbar/(2\Delta x)$ gives the lower bound

\[E\geq\frac{\hbar^2}{8m(\Delta x)^2} +\frac12m\omega^2(\Delta x)^2.\]

Differentiate the right-hand side with respect to the positive width $\Delta x$:

\[-\frac{\hbar^2}{4m(\Delta x)^3} +m\omega^2\Delta x=0,\]

so

\[(\Delta x)^2=\frac{\hbar}{2m\omega},\qquad (\Delta p)^2=\frac{m\hbar\omega}{2}.\]

Substitution yields

\[\boxed{E_{\min}=\frac12\hbar\omega}.\]

The two positive terms each contribute $\hbar\omega/4$. The oscillator therefore cannot have both $x=0$ and $p=0$, even at its lowest energy.

Size of an atom

Localizing the electron in hydrogen to a scale $r$ implies a momentum scale $\Delta p\sim\hbar/r$. This order-of-magnitude estimate deliberately suppresses numerical factors that depend on the three-dimensional trial state. The competing kinetic and Coulomb energies are

\[E(r)\sim \frac{\hbar^2}{2m_er^2} -\frac{e^2}{4\pi\varepsilon_0r}.\]

The kinetic term is positive and varies as $r^{-2}$; the attractive potential is negative and varies as $r^{-1}$. Setting $dE/dr=0$ gives

\[-\frac{\hbar^2}{m_er^3} +\frac{e^2}{4\pi\varepsilon_0r^2}=0,\]

and hence the atomic length scale

\[\boxed{r\sim a_0 =\frac{4\pi\varepsilon_0\hbar^2}{m_e e^2}}.\]

The expression has units of length, and its value is $a_0\simeq5.29\times10^{-11}\,\mathrm m$. As $r\to0$, the $r^{-2}$ localization energy dominates the $-r^{-1}$ attraction, preventing classical collapse into the proton.

Existence of an electron inside the nucleus

If an electron were confined inside a nucleus of radius $R$, then

\[\Delta p\gtrsim\frac{\hbar}{2R},\qquad pc\gtrsim\frac{\hbar c}{2R}.\]

Because $\hbar c\simeq197.3\,\mathrm{MeV\,fm}$, confinement to $R=1\,\mathrm{fm}$ requires $pc\gtrsim98.7\,\mathrm{MeV}$. This is far larger than the electron rest energy $m_ec^2=0.511\,\mathrm{MeV}$, so the non-relativistic expression $p^2/(2m_e)$ is invalid. The relativistic kinetic energy is

\[K=\sqrt{p^2c^2+m_e^2c^4}-m_ec^2 \gtrsim98\,\mathrm{MeV}.\]

Such an energy is incompatible with a low-energy electron pre-existing inside an ordinary nucleus. Electrons emitted in beta decay are created in the decay process rather than released from nuclear confinement.

The stationary-width, zero-point-energy, and atomic-scale substitutions are verified in the Maxima worksheet.