38-7 Heisenberg’s Uncertainty Principle#

Prompts

State Heisenberg’s uncertainty principle for position $x$ and momentum $p_x$. What is $\hbar$ (h-bar) in terms of $h$?
Why can’t we measure position and momentum simultaneously with unlimited precision? Give a physical argument involving the act of detection (e.g., scattering light off an electron).
For a free particle with definite momentum ($\Delta p_x \approx 0$), what does the uncertainty principle say about $\Delta x$? How does this connect to the constant $|\psi|^2$ in section 38-6?
“Measure $p_x$ precisely first, then measure $x$—don’t we get both?” Why does this fail?
An electron has speed $2.05\times 10^6$ m/s known to 0.50%. What is the minimum $\Delta x$ allowed by the uncertainty principle?

Lecture Notes#

Overview#

Heisenberg’s uncertainty principle (1927): It is impossible to assign measured values to a particle’s position and momentum simultaneously with unlimited precision.
The product of uncertainties satisfies $\Delta x\,\Delta p_x \geq \hbar/2$ (and similarly for $y$, $z$), where $\hbar = h/(2\pi)$.
The principle arises from the wave nature of matter—measurements involve probabilities, not certainties. The act of detection itself disturbs the system.
A free particle with definite momentum has $\Delta p_x \approx 0$ and thus $\Delta x \to \infty$: the particle could be found anywhere (consistent with constant $|\psi|^2$).

The uncertainty relations#

Werner Heisenberg proposed in 1927 that the probabilistic nature of quantum physics places a fundamental limit on simultaneous measurements of position and momentum. We cannot assign both $\vec{r}$ and $\vec{p}$ with unlimited precision at the same time.

In terms of $\hbar$ (h-bar), defined as

(373)#\[ \hbar = \frac{h}{2\pi} \approx 1.055 \times 10^{-34}\ \text{J}\cdot\text{s} \]

the Heisenberg uncertainty principle states

(374)#\[ \Delta x\,\Delta p_x \geq \frac{\hbar}{2}, \quad \Delta y\,\Delta p_y \geq \frac{\hbar}{2}, \quad \Delta z\,\Delta p_z \geq \frac{\hbar}{2} \]

Here $\Delta x$ and $\Delta p_x$ are the uncertainties in the $x$ components of position and momentum. Even with ideal instruments, each product is greater than or equal to $\hbar/2$—never less.

Interpretation

We can view $\Delta x$ and $\Delta p_x$ as the spread (standard deviations) in repeated measurements. The principle is not about clumsy apparatus; it reflects the wave nature of matter and the fact that position and momentum are complementary—sharpening one blurs the other.

Physical argument: detection disturbs the system#

Classically, we locate an object by watching it—light scatters off the object and we detect the scattered light. For a car or a pool ball, that scattering barely affects the object’s motion.

For an electron, the act of detection does alter its state. To determine position very precisely, we need short-wavelength light (high resolution). But a short-wavelength photon carries large momentum; when it scatters from the electron, it transfers momentum and changes the electron’s motion. The more precisely we fix $\Delta x$ (smaller $\Delta x$), the more we disturb the momentum—the larger $\Delta p_x$ becomes.

Conversely, to measure momentum very precisely, we need a long-wavelength probe (small momentum transfer). But a long-wavelength wave is spread out—it cannot localize the electron. So a small $\Delta p_x$ implies a large $\Delta x$. The uncertainties are linked: decreasing one increases the other.

Connection to the free particle#

In section 38-6 we found that a free particle with $\psi = Ae^{ikx}$ has a definite momentum $p_x = \hbar k$ (since $k = 2\pi/\lambda = 2\pi p/h = p/\hbar$). Thus $\Delta p_x \approx 0$. By the uncertainty principle, $\Delta x \geq \hbar/(2\Delta p_x) \to \infty$: the position is completely uncertain. The particle could be detected anywhere—exactly what we found: $|\psi|^2 = A^2$ is constant along the $x$ axis.

Why not measure twice?#

“Measure $p_x$ precisely first, then measure $x$ wherever the electron shows up—don’t we get both?” The flaw: the second measurement necessarily alters the momentum. If we measure position very precisely (e.g., by scattering a short-wavelength photon), we disturb the momentum. After the position measurement, we no longer know $p_x$. We cannot have both values simultaneously with arbitrary precision.

Example#

Example

Electron with $v_x = 2.05\times 10^6$ m/s, known to 0.50%: The momentum is $p_x = m_e v_x \approx 1.87\times 10^{-24}$ kg·m/s. The uncertainty is $\Delta p_x = 0.0050\times p_x \approx 9.35\times 10^{-27}$ kg·m/s. The minimum position uncertainty is $$ \Delta x = \frac{\hbar}{2\Delta p_x} \approx \frac{1.055\times 10^{-34}}{2\times 9.35\times 10^{-27}} \approx 5.6\times 10^{-9}\ \text{m} \approx 6\ \text{nm} $$ —tens of atomic diameters. Even with a 0.5% speed measurement, we cannot simultaneously know the electron's position to better than ~6 nm. (Some texts use $\Delta x,\Delta p_x \geq \hbar$ and obtain ~11 nm.)

Summary#

Heisenberg’s principle: $\Delta x\,\Delta p_x \geq \hbar/2$ (and $y$, $z$); $\hbar = h/(2\pi)$.
Fundamental limit: Not due to instrument quality; arises from wave nature and complementarity.
Detection disturbs: Precise position measurement requires short $\lambda$ light → large momentum transfer → large $\Delta p_x$.
Free particle: Definite $p$ $\Rightarrow$ $\Delta p \approx 0$ $\Rightarrow$ $\Delta x \to \infty$; consistent with constant $|\psi|^2$.