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Introduction – What is Heisenberg’s Uncertainty Principle? 
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Why is it Impossible to Measure both Position and Momentum Simultaneously?
In order to illustrate Heisenberg’s uncertainty principle, consider an example where the position of an electron is measured. In order to measure the position of an object, a photon must collide with it and return to the measuring device. Since photons hold some finite momentum, a transfer of momenta will occur when the photon collides with the electron. This transfer of momenta will cause the momentum of the electron to increase. Thus, any attempt at measuring the position of a particle will increase the uncertainty in the value of its momentum.
Applying the same example to a macroscopic object (say a basketball), it can be understood that Heisenberg’s uncertainty principle has a negligible impact on measurements in the macroscopic world. While measuring the position of a basketball, there will still be a transfer of momentum from the photons to the ball. However, the mass of the photon is much smaller than the mass of the ball. Therefore, any momentum imparted by the photon to the ball can be neglected.
If, Δx is the error in position measurement and Δp is the error in the measurement of momentum, then
Δx × Δp ≥ h/4π
Since momentum, p = mv, Heisenberg’s uncertainty principle formula can be alternatively written as
Δx × Δmv ≥ h/4π or ΔX × Δm × Δv ≥ h/4π
Where, ΔV is the error in the measurement of velocity and assuming mass remaining constant during the experiment,
ΔX × ΔV ≥ h/4πm
Accurate measurement of position or momentum automatically indicates larger uncertainty (error) in the measurement of the other quantity.
Applying the Heisenberg principle to an electron in an orbit of an atom, with h = 6.626 ×10^{34}Js and m= 9.11 ×10^{31}Kg,
If the position of the electron is measured accurately to its size (10^{10}m), then the error in the measurement of its velocity will be equal or larger than 10^{6}m or 1000Km.
Heisenberg principle applies to only dualnatured microscopic particles and not to a macroscopic particle whose wave nature is very small.
A striking thought experiment illustrating the uncertainty principle is Bohr’s / Heisenberg’s Gammaray microscope. To observe a particle, say an electron, we shine it with the light ray of wavelength λ and collect the Compton scattered light in a microscope objective whose diameter subtends an angle θ with the electron as shown in the figure below
The precision with which the electron can be located, Delta x, is defined by the resolving power of the microscope,
sinθ = λ/Δx ⇒ Δx = λ/sinθ
It appears that by making λ small, that is why we choose γray, and by making sin θ large, Delta x can be made as small as desired. But, according to the uncertainty principle, we can do so only at the expense of our knowledge of xcomponent of electron momentum.
In order to record the Compton scattered photon by the microscope, the photon must stay in the cone of angle θ and hence its xcomponent of the momentum can vary within ±(h/λ) sin θ. This implies, the magnitude of the recoil momentum of the electron is uncertain by
Δpx = 2h/λ sinθ
The product of the uncertainty yields,
Heisenberg’s principle is applicable to all matter waves. The measurement error of any two conjugate properties, whose dimensions happen to be joule sec, like positionmomentum, timeenergy will be guided by the Heisenberg’s value.
But, it will be noticeable and of significance only for small particles like an electron with very low mass. A bigger particle with heavy mass will show the error to be very small and negligible.
Heisenberg’s uncertainty principle can be considered as a very precise mathematical statement that describes the nature of quantum systems. As such, we often consider two common equations related to the uncertainty principle. They are;
Equation 1: ΔX ⋅ ∆p ~ ħ
Equation 2: ∆E ⋅ ∆t ~ ħ
Where,
ħ = value of the Planck’s constant divided by 2*pi
ΔX = uncertainty in the position
∆p = uncertainty in momentum
∆E = uncertainty in the energy
∆t = uncertainty in time measurement
1. If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h / 4pm × 0.05 nm, is there any problem in defining this value.
Solution:
a) ∆x = 2×1012m; ∆X × ∆mV ≥
∵ ∆mV ≥
b) Momentum mv =
Error in momentum measurement is 10^{10} times larger than the actual momentum. The given momentum will not be acceptable.
2. Position of a chloride ion on a material can be determined to a maximum error of 1μm. If the mass of the chloride ion is 5.86 × 10^{26}Kg, what will be the error in its velocity measurement?
Solution:
∆x = 10^{6} m; ∆X × ∆mV ≥
3. The lifetime of an excited state of an atom is 3 × 10^{3}s. What is the minimum uncertainty in its energy in eV?
Solution: Time and energy are conjugate pairs with Js unit. The product of measurement error is given by Heisenberg’s principle.
Assuming a maximum error in the measurement of lifetime equal to that of lifetime = 3 ×10^{3}s
⸪ 1 Joule = 6.242 × 10^{18}ev,
Uncertainty in the determination of energy of the atom = ∆E = 6.22 × 10^{18} × 1/3×10^{3} × 5.28 ×10^{35}
= 1.1×10^{13}
4. A wet ball weighing 10.1gm has a water of 0.1g on it. The ball is moving with a constant velocity with an uncertainty of momentum of 10^{6} kg m/s. What will be the uncertainty in the measurement of the position of the ball, water and electron in the water molecule?
Solution: ∆X × ∆p ≥ h/4π
Velocity being constant, uncertainty in the measurement of the momentum is associated with the mass of the matter.
Uncertainty in the momentum of the dry ball = mass ×10^{6} = 10×10^{3}×10^{6} Kg m s^{1}.
Uncertainty in the momentum of the water = mass ×10^{6 }= 0.1×10^{3}×10^{6} Kg m s^{1}.
Uncertainty in the momentum of the electron = mass ×10^{6} = 9×10^{31}×10^{6} Kg m s^{1}.
Uncertainty in position measurement is inversely proportional to the uncertainty in momentum
= 10^{8} : 10^{10} : 1.1 × 10^{36 }or
= 1 : 10^{2} : 10^{28}
4. Determine the minimum uncertainties in the positions of the electron if their speeds are known with a precision of 3.0×10^{−3}m/s?
Solution:
Δu=3.0×10^{−3}m/s
Uncertained momentum Δp=m Δu
Uncertainty in position Δx=ℏ/(2Δp)
For electron
Δp=m Δu
= (9.1×10^{−3}^{1}kg × 3.0×10^{−3})
Δp = 2.73 × 10^{33 }kg.m/s
Δx = h/Δp
Δx = 0.12 m
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