Additional: Heisenberg's Uncertainty Principle
Uncertainty in Position and Momentum ($ \Delta x \Delta p \ge \hbar/2 $)
The wave-particle duality of matter, highlighted by the de Broglie hypothesis and experiments like Davisson-Germer, has profound consequences for how we describe the properties of particles at the quantum level. Classical physics assumes that a particle can have a definite position and a definite momentum simultaneously. However, in quantum mechanics, this is not possible. This is expressed by Heisenberg's Uncertainty Principle, formulated by Werner Heisenberg in 1927.
Statement of the Uncertainty Principle
Heisenberg's Uncertainty Principle states that it is impossible to simultaneously determine with perfect accuracy certain pairs of physical properties of a particle, such as its position and momentum. The more precisely you know one property, the less precisely you can know the other.
For the position ($x$) and momentum ($p_x$) of a particle along the same direction, the principle states that the product of the uncertainty in position ($\Delta x$) and the uncertainty in momentum ($\Delta p_x$) is always greater than or equal to a small constant related to Planck's constant ($\hbar = h/(2\pi)$).
$ \Delta x \Delta p_x \ge \frac{\hbar}{2} $
Similar relations hold for uncertainties in other pairs of conjugate variables:
$ \Delta y \Delta p_y \ge \frac{\hbar}{2} $
$ \Delta z \Delta p_z \ge \frac{\hbar}{2} $
Where $\Delta x, \Delta y, \Delta z$ are the uncertainties in position along the x, y, z axes, and $\Delta p_x, \Delta p_y, \Delta p_z$ are the uncertainties in the corresponding momentum components. The uncertainties refer to the standard deviation of measurements if the experiment were repeated many times.
Physical Interpretation
The uncertainty principle is not a statement about the limitations of our measuring instruments or techniques. It is a fundamental property of nature at the quantum scale, arising from the wave nature of particles.
Consider trying to measure the position and momentum of a particle simultaneously.
-
If we want to measure the position of a particle very precisely (small $\Delta x$), we would need to use a probe (like a photon or another particle) with a very short wavelength (since resolution is limited by wavelength). According to de Broglie and photon theory ($p=h/\lambda$), a short wavelength implies a high momentum for the probe. When this high-momentum probe interacts with the particle (e.g., scattering off it), it transfers a significant amount of momentum to the particle, unpredictably changing the particle's momentum. Thus, a precise measurement of position introduces a large uncertainty in the particle's momentum.
-
Conversely, if we want to measure the momentum of a particle very precisely (small $\Delta p_x$), we need to interact with it in a way that minimises the disturbance to its momentum. This would typically involve using a probe with low momentum, which, by de Broglie's relation, has a long wavelength. A long wavelength limits the precision with which we can locate the particle. Thus, a precise measurement of momentum introduces a large uncertainty in the particle's position.
The uncertainty principle is a fundamental limit on the precision with which we can simultaneously know these conjugate variables.
The constant $\hbar/2$ is very small (of the order of $10^{-35} \, J \cdot s$). For macroscopic objects, the uncertainties in position and momentum are negligible compared to the scale of measurements. For example, measuring the position and velocity of a cricket ball, the uncertainties are far below what the principle requires for them to be significant relative to the classical trajectory. However, for microscopic particles like electrons confined to atomic scales, the uncertainties become significant and determine the fundamental behaviour described by quantum mechanics.
Example 1. The uncertainty in the position of an electron is $1.0 \times 10^{-10}$ m (approximately the size of an atom). Calculate the minimum uncertainty in its momentum. ($h = 6.63 \times 10^{-34} \, J \cdot s$).
Answer:
Given:
Uncertainty in position, $\Delta x = 1.0 \times 10^{-10} \, m$
Planck's constant, $h = 6.63 \times 10^{-34} \, J \cdot s$. Reduced Planck constant $\hbar = h/(2\pi) = 6.63 \times 10^{-34} / (2 \times 3.14159) \approx 1.055 \times 10^{-34} \, J \cdot s$.
According to Heisenberg's Uncertainty Principle for position and momentum:
$ \Delta x \Delta p_x \ge \frac{\hbar}{2} $
We are looking for the minimum uncertainty in momentum, so we use the equality:
$ \Delta p_x = \frac{\hbar}{2 \Delta x} $
Substitute the given values:
$ \Delta p_x = \frac{1.055 \times 10^{-34} \, J \cdot s}{2 \times (1.0 \times 10^{-10} \, m)} $
$ \Delta p_x = \frac{1.055}{2} \times 10^{-24} \, kg \cdot m/s $ (Unit check: $J \cdot s / m = (N \cdot m) \cdot s / m = N \cdot s = (kg \cdot m/s^2) \cdot s = kg \cdot m/s$)
$ \Delta p_x = 0.5275 \times 10^{-24} \, kg \cdot m/s $
The minimum uncertainty in the electron's momentum is approximately $5.275 \times 10^{-25} \, kg \cdot m/s$. This is a significant uncertainty at the atomic scale.
Uncertainty in Energy and Time ($ \Delta E \Delta t \ge \hbar/2 $)
Besides position and momentum, other pairs of physical quantities are also subject to the uncertainty principle. A particularly important pair is energy ($E$) and time ($t$).
Statement of the Uncertainty Principle for Energy and Time
Heisenberg's Uncertainty Principle for energy and time states that the product of the uncertainty in the energy ($\Delta E$) of a system and the uncertainty in the time interval ($\Delta t$) during which the system is in that energy state is always greater than or equal to a small constant related to Planck's constant:
$ \Delta E \Delta t \ge \frac{\hbar}{2} $
Physical Interpretation
This relation means that if a system exists in a certain energy state for a very short duration ($\Delta t$ is small), then its energy cannot be precisely determined ($\Delta E$ will be large). Conversely, if a system is known to have a very precisely defined energy ($\Delta E$ is small), then it must exist in that state for a long duration ($\Delta t$ must be large, ideally infinite for a perfectly defined energy state).
This principle has implications in various areas:
-
Energy Levels of Atoms: Electrons in excited states of atoms exist for a very short lifetime before decaying to lower energy states by emitting a photon. The uncertainty in the lifetime ($\Delta t$) leads to an uncertainty in the energy of the excited state ($\Delta E$). This energy uncertainty manifests as a natural broadening of the spectral lines emitted by atoms (the lines are not perfectly sharp). Shorter lived states have broader spectral lines.
-
Virtual Particles: In quantum field theory, the energy-time uncertainty principle allows for the temporary violation of energy conservation for very short durations. This concept is used to explain the existence of virtual particles, which can pop into existence from the vacuum and disappear very quickly, provided that their energy and lifetime satisfy $\Delta E \Delta t \approx \hbar$.
-
Particle Lifetimes: For unstable particles, their lifetime ($\Delta t$) determines the uncertainty in their mass/energy ($\Delta E$). Particles with very short lifetimes have a large uncertainty in their mass, appearing as a "width" in their mass distribution.
Derivation (Conceptual)
A rigorous derivation of this uncertainty relation requires advanced quantum mechanics. However, a hand-waving argument can be made by considering a wave packet.
A wave packet, representing a localised particle, is formed by superposing waves of different wavelengths and frequencies. To form a wave packet that is sharply localised in space (small $\Delta x$), you need to superpose a wide range of wavelengths (large uncertainty in wavelength $\Delta\lambda$). By de Broglie's relation $p=h/\lambda$, a large $\Delta\lambda$ implies a large uncertainty in momentum $\Delta p$, leading to $\Delta x \Delta p \sim h$ or $\hbar$.
Similarly, to form a pulse that is sharply localised in time (small $\Delta t$), you need to superpose waves covering a wide range of frequencies (large uncertainty in frequency $\Delta\nu$). The energy of a quantum is $E=h\nu$, so a large $\Delta\nu$ implies a large uncertainty in energy $\Delta E$. The relation between duration ($\Delta t$) and frequency spread ($\Delta\nu$) for a wave packet is approximately $\Delta t \Delta\nu \sim 1/(2\pi)$. Multiplying by $h$, we get $\Delta t (h\Delta\nu) \sim h/(2\pi)$, i.e., $\Delta t \Delta E \sim \hbar$.
The uncertainty principle is a fundamental aspect of quantum mechanics, highlighting that position and momentum, and energy and time, are not simultaneously precisely knowable properties of a quantum system.
Example 1. An electron in an excited state of an atom remains in that state for an average lifetime of about $10^{-8}$ seconds before decaying to the ground state. Estimate the minimum uncertainty in the energy of this excited state. Express the result in eV. ($\hbar \approx 1.055 \times 10^{-34} \, J \cdot s$, $1 \, eV = 1.602 \times 10^{-19} \, J$).
Answer:
Given:
Uncertainty in the lifetime (time interval), $\Delta t \approx 10^{-8} \, s$
Reduced Planck constant, $\hbar = 1.055 \times 10^{-34} \, J \cdot s$
According to Heisenberg's Uncertainty Principle for energy and time:
$ \Delta E \Delta t \ge \frac{\hbar}{2} $
We are looking for the minimum uncertainty in energy, so we use the equality:
$ \Delta E = \frac{\hbar}{2 \Delta t} $
Substitute the given values:
$ \Delta E = \frac{1.055 \times 10^{-34} \, J \cdot s}{2 \times (10^{-8} \, s)} $
$ \Delta E = \frac{1.055}{2} \times 10^{-26} \, J $ (Unit check: $J \cdot s / s = J$)
$ \Delta E = 0.5275 \times 10^{-26} \, J $
Now, convert this energy uncertainty from Joules to electron volts (eV). $1 \, J = 1 / (1.602 \times 10^{-19}) \, eV$.
$ \Delta E \, (in \, eV) = \frac{0.5275 \times 10^{-26} \, J}{1.602 \times 10^{-19} \, J/eV} $
$ \Delta E \, (in \, eV) = \frac{0.5275}{1.602} \times 10^{-7} \, eV $
$ \Delta E \, (in \, eV) \approx 0.329 \times 10^{-7} \, eV = 3.29 \times 10^{-8} \, eV $
The minimum uncertainty in the energy of this excited state is approximately $3.29 \times 10^{-8}$ eV. This energy uncertainty is related to the natural width of the spectral line emitted when the electron transitions from this state.
