Estimating Measurement Limits with the Heisenberg Uncertainty Principle

Welcome back. In the previous lesson, de Broglie’s relation connected a particle’s momentum to its wavelength:

$$\lambda =\frac{h}{p}.$$

That connection explains why electrons display diffraction and interference at atomic length scales. We now take the next step: if an electron is described by a wave, what does it mean to say where it is and how fast it is moving?

This lesson develops the Heisenberg uncertainty relation as a quantitative limit on simultaneous knowledge of position and momentum. You will learn to calculate the smallest possible uncertainty in one quantity when the uncertainty in the other is specified, and to judge why the effect is central for electrons but negligible for everyday objects.

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A limit built into quantum states

For motion along one direction, conventionally called the $x$-direction, the Heisenberg uncertainty relation is

$$\Delta x\Delta p_x\ge \frac{\hslash }{2}.$$

Because

$$\hslash =\frac{h}{2\pi },$$

the same relation is often written as

$$\Delta x\Delta p_x\ge \frac{h}{4\pi }.$$

The symbols have precise meanings:

• $\Delta x$: uncertainty, or spread, in position along $x$
• $\Delta p_x$: uncertainty, or spread, in momentum along $x$
• $h=6.626\times 10^{-34}\text{J s}$: Planck’s constant
• $\hslash =1.055\times 10^{-34}\text{J s}$: reduced Planck’s constant

The relation does not say that position and momentum cannot be measured at all. It says that a quantum state cannot have both an arbitrarily narrow position distribution and an arbitrarily narrow momentum distribution at the same time.

If position is made more definite, momentum must become less definite. If momentum is made more definite, position must become less definite.

A crucial detail is the inequality sign. The product can be larger than $\hslash /2$, but it cannot be smaller:

$$\Delta x\Delta p_x<\frac{\hslash }{2}$$

is physically impossible.

When an exercise asks for the minimum uncertainty, set the two sides equal:

$$\Delta x\Delta p_x=\frac{\hslash }{2}.$$

That equality represents the smallest allowed product. It is not automatically the uncertainty product for every quantum state.

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Why waves produce uncertainty

The uncertainty relation follows from the wave nature of matter introduced in the previous lesson.

A single plane wave has one wavelength and therefore one momentum. But a plane wave extends over all space: it provides no localized position. In this idealized case,

$$\Delta p_x=0,$$

while position is completely indefinite.

To localize a particle, quantum mechanics combines plane waves with several slightly different wavelengths. Because de Broglie’s relation connects wavelength and momentum, several wavelengths mean several possible momenta. The more tightly the combined wave is localized in space, the broader its momentum range must be.

This is not primarily a story about imperfect laboratory equipment. Even a perfect apparatus cannot prepare a particle in a state with both exactly defined position and exactly defined momentum. The limitation belongs to the quantum state itself.

The language of “uncertainty” can be misleading if it is treated as ordinary experimental error. In a rigorous treatment, $\Delta x$ and $\Delta p_x$ are standard deviations: they quantify the widths of the distributions obtained from many identically prepared particles. A position measurement still gives one localized detection event, but repeated measurements display a spread. Momentum measurements on the same prepared state likewise display a spread.

This is why quantum mechanics replaces the classical picture of an electron following an exactly known path with a probabilistic description. The next lesson will make that description explicit through the wavefunction and the Born rule.

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Turning the relation into a calculation

For a particle of fixed mass,

$$p_x=m{v}_x.$$

Therefore, if the uncertainty comes from uncertainty in velocity,

$$\Delta p_x\approx m\Delta v_x.$$

The approximation is appropriate when the particle’s mass is known precisely compared with the velocity uncertainty, as it normally is in introductory chemistry problems.

Substituting this into the uncertainty relation gives a useful form:

$$\Delta xm\Delta v_x\ge \frac{\hslash }{2}.$$

To estimate the minimum position uncertainty from a known velocity uncertainty, rearrange:

$$\Delta x_{\min }=\frac{\hslash }{2m\Delta v_x}.$$

Conversely, if the position uncertainty is given, the minimum momentum uncertainty is

$$\Delta p_{x,\min }=\frac{\hslash }{2\Delta x}.$$

A reliable solution process

1. Identify what uncertainty is given: $\Delta x$, $\Delta p_x$, or $\Delta v_x$.

2. If velocity uncertainty is supplied, calculate momentum uncertainty with

   $$\Delta p_x=m\Delta v_x.$$

3. Rearrange the uncertainty relation for the unknown quantity.

4. Use equality only because the question asks for a minimum possible uncertainty.

5. State the result as a lower bound, using $\ge$, unless the question specifically requests the minimum numerical value.

A quick unit check confirms that the result has units of length:

$$\frac{\hslash }{\Delta p_x}=\frac{{\text{kg m}}^2{\text{s}}^{-1}}{{\text{kg m s}}^{-1}}=\text{m}.$$

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Example: an electron with a fairly precise velocity

Suppose an electron’s velocity along $x$ is known with uncertainty

$$\Delta v_x=1.0\times 10^5{\text{m s}}^{-1}.$$

Use the electron mass,

$$m_e=9.11\times 10^{-31}\text{kg}.$$

First calculate the momentum uncertainty:

$$\Delta p_x=m_e\Delta v_x.$$ $$\Delta p_x=\left(9.11\times 10^{-31}\text{kg}\right)\left(1.0\times 10^5{\text{m s}}^{-1}\right).$$ $$\Delta p_x=9.11\times 10^{-26}{\text{kg m s}}^{-1}.$$

Now calculate the smallest allowed position uncertainty:

$$\Delta x_{\min }=\frac{1.055\times 10^{-34}\text{J s}}{2\left(9.11\times 10^{-26}{\text{kg m s}}^{-1}\right)}.$$ $$\Delta x_{\min }=5.79\times 10^{-10}\text{m}.$$

In nanometers,

$$\Delta x_{\min }=0.579\text{nm}.$$

This is a chemically large distance. It is several times the approximate diameter of a hydrogen atom, which is on the order of $0.1\text{nm}$. Thus, if an electron’s velocity is constrained to this degree of precision, its position cannot simultaneously be confined to a tiny atomic-scale region.

The interpretation is not that the electron is “somewhere inside a $0.579\text{nm}$ box” in a classical sense. Instead, the electron’s state must have a spatial spread at least this large under the stated momentum constraint.

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Reversing the calculation: tight localization requires broad momentum

Now imagine that an electron is localized along one direction within

$$\Delta x=50\text{pm}.$$

Convert picometers to meters:

$$\Delta x=50\times 10^{-12}\text{m}=5.0\times 10^{-11}\text{m}.$$

The smallest allowed momentum uncertainty is

$$\Delta p_{x,\min }=\frac{\hslash }{2\Delta x}.$$ $$\Delta p_{x,\min }=\frac{1.055\times 10^{-34}}{2\left(5.0\times 10^{-11}\right)}.$$ $$\Delta p_{x,\min }=1.06\times 10^{-24}{\text{kg m s}}^{-1}.$$

For an electron, this corresponds to a velocity uncertainty of

$$\Delta v_{x,\min }=\frac{\Delta p_{x,\min }}{m_e}.$$ $$\Delta v_{x,\min }=\frac{1.06\times 10^{-24}}{9.11\times 10^{-31}}.$$ $$\Delta v_{x,\min }=1.16\times 10^6{\text{m s}}^{-1}.$$

An electron localized on a scale comparable with an atom must therefore have a very broad range of possible velocities. This is one reason the classical image of an electron as a tiny object moving on a sharply defined circular orbit is inadequate. Atomic electrons are described by spatial probability distributions, not by simultaneously exact positions and momenta.

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Why the uncertainty principle is invisible in everyday mechanics

The relation applies to every object, including a baseball or a laboratory instrument. Its practical importance depends on mass and on the scales being studied.

Consider a $0.100\text{kg}$ ball whose velocity along $x$ is known to within

$$\Delta v_x=0.010{\text{m s}}^{-1}.$$

Its momentum uncertainty is

$$\Delta p_x=m\Delta v_x=\left(0.100\right)\left(0.010\right)=1.0\times 10^{-3}{\text{kg m s}}^{-1}.$$

The minimum position uncertainty is then

$$\Delta x_{\min }=\frac{1.055\times 10^{-34}}{2\left(1.0\times 10^{-3}\right)}.$$ $$\Delta x_{\min }=5.28\times 10^{-32}\text{m}.$$

That length is absurdly smaller than any experimental or physical feature relevant to a ball. Quantum uncertainty still exists, but it is overwhelmed by practical measurement limits and is irrelevant to everyday motion.

The contrast can be summarized as follows:

System	Given velocity uncertainty	Minimum position uncertainty	Physical significance
Electron	$1.0\times 10^5{\text{m s}}^{-1}$	$5.79\times 10^{-10}\text{m}$	Comparable with atomic dimensions
$0.100\text{kg}$ ball	$0.010{\text{m s}}^{-1}$	$5.28\times 10^{-32}\text{m}$	Negligible in practice

The small value of $\hslash$ makes the uncertainty bound effectively invisible for massive objects. Electrons have such low mass that even modest uncertainty in their velocity can imply a position spread relevant to chemistry.

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Key takeaways

The Heisenberg uncertainty relation for position and momentum along one direction is

$$\Delta x\Delta p_x\ge \frac{\hslash }{2}=\frac{h}{4\pi }.$$

For constant mass,

$$\Delta p_x\approx m\Delta v_x.$$

Use equality only to calculate the minimum permitted uncertainty. The resulting answer is a lower bound, so phrase it as

$$\Delta x\ge \Delta x_{\min }$$

or

$$\Delta p_x\ge \Delta p_{x,\min }.$$

Most importantly, this is not merely a limitation of measurement devices. Localizing a matter wave requires combining a range of wavelengths, which necessarily means a range of momenta. For atomic electrons, that tradeoff is large enough to rule out a classical trajectory with simultaneously exact position and momentum.

Next, we will turn from uncertainty ranges to the wavefunction itself and learn how the Born rule connects $\psi$ to measurable probabilities.