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Heisenberg Uncertainty Principle
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The Planck constant occurs in statements of Werner Heisenberg's uncertainty
principle. Given a large number of particles prepared in the same state, the
uncertainty in their position, Δx, and the uncertainty in their momentum
(in the same direction), Δp, obey
ΔxΔp >= h/2
where the uncertainty is given as the standard deviation of the measured value from
its expected value. There are a number of other such pairs of physically measurable
values which obey a similar rule. One example is time vs. energy. The either-or nature
of uncertainty forces measurement attempts to choose between trade offs, and given
that they are quanta, the trade offs often take the form of either-or (as in Fourier
analysis), rather than the compromises and gray areas of time series analysis.
In addition to some assumptions underlying the interpretation of certain values in
the quantum mechanical formulation, one of the fundamental cornerstones to the entire
theory lies in the commutator relationship between the position operator and the
momentum operator:
[p_{i},x_{j}] = -ihδ_{ij}
where δ_{ij} is the Kronecker delta.
The principle can be derived in several ways. One way is to look at things in terms
of wave functions which relate to the probability of finding the electron at any point
in space. A perfect plane wave with a wavelength λ implies that the momentum
is precisely known from p = h/λ but the probability of finding the particle is
spread over all space - this is because the wavefunction e^{ikx} stretches
from + to - infinity so the probability of finding it is the same everywhere.
To represent a localized particle it is necessary to construct a wave packet by
superpositioning free particle eigenstates (i.e. plane waves with different values
of momentum).
ψ(x) = Σ_{p}A_{p}exp(ipx/h) ... Fourier series
Note that if ψ(x) is a solution to the Schrodinger equation, any linear combination
of plane waves is also a solution.
The position and momentum wavefunctions are Fourier Transforms of each other.
ψ(x) = (1/√2π)∫φ(p)exp(ipx/h)dp
φ(p) = (1/√2π)∫ψ(x)exp(-ipx/h)dx
It is a property of the Fourier Transform that if ψ(x) is very narrow, then
φ(p) is very broad and vice versa. This is illustrated in the following example.
Consider the following wavefunction:
ψ(x) = exp(-α|x|)
It can be easily seen that the width of this function is controlled by α. As
α increases Δx decreases. Therefore, Δx ∝ 1/α
The normalized wavefunction can be shown to be:
ψ(x) = √α exp(-α|x|)
Now take the Fourier Transform of the function and integrate to get φ(p).
By comparison, φ(p) is equivalent to the amplitude, A_{p} in the Fourier
series expansion above. Thus,
φ(p) ≡ A_{p} = √(α/L)∫exp(-α|x|)exp(-ipx/h) dx
Where √(1/L) is the normalization factor for exp(-ipx/h).
After some math this produces a solution:
A_{p} = √(α/L)[2α/(α^{2} + p^{2}/h^{2})]
So,
P(p) = |A_{p}|^{2} = (4α^{3}/L)[1/(α^{2} + p^{2}/h^{2})^{2}]
P(p) is peaked at p = 0 and falls off rapidly. If we pick a point such that:
α^{2} = p^{2}/h^{2} or α = p/h
Since α is inversely proportional to Δx we can write,
1/Δx = Δp/h
or
ΔxΔp = h Q.E.D.
Thus as the width in p-space is varied, the width in x-space varies to keep the product
constant and vice versa. In other words, While the superpositiom leads to increased
certainty in position, it causes uncertainty in the momentum.
Consider a Gaussian wavepacket. The envelope function in x and p space is given by:
P(x) = (1/√2πσ_{x}^{2})exp(-[x - x_{o}]^{2}/2σ_{x}^{2})
P(k) = (1/√2πσ_{k}^{2})exp(-[k - k_{o}]^{2}/2σ_{k}^{2})
where σ is the width of the distribution,
Construct a wavepacket ψ(x):
ψ(x) = (1/2πα)^{1/4}exp(-x^{2}/4α)exp(ik_{o}x)
where (1/2πα)^{1/4} is the normalization factor and 4α = 2σ_{x}^{2}.
Probability density for x:
P(x) = ψ(x)^{*}ψ(x) = (1/√2πα)exp(-x^{2}/2α) where x_{o} = 0
=> 2σ_{x}^{2} = 2α ∴ σ_{x} = √α
Now, using the Fourier Transform:
φ(k) = (1/√2π)∫ψ(x)exp(-ikx)dx
= (1/√2π)∫(1/2πα)^{1/4}exp(-x^{2}/4α)exp(ik_{o}x)exp(-ikx)dx
= (2α/π)^{1/4}exp(-α[k - k_{o}]^{2})
Probability density for k:
P(k) = φ(k)^{*}φ(k) = √(2α/π)exp(-2α[k - k_{o}]^{2})
=> 2σ_{k}^{2} = 1/2α ∴ σ_{k} = 1/2√α
Therefore,
σ_{x}σ_{k} = ΔxΔk = √α(1/2√α) = 1/2
p = hk ∴ Δp = hΔk
ΔxΔk = h/2 Q.E.D.
Yet another way to derive the Uncertainty Principle is via the single slit
experiment. Consider:
At the slit:
slit width = d
deflection angle = θ ( = θ' )
Δy ~ d (The y position at the slit is known to a certainty d)
Δp_{y} = 0 (at the slit)
δ = dsinθ = λ for n = 1 ∴ sinθ = λ/d
At the screen:
Component of momentum in y direction, Δp_{y} = psinθ = pλ/d
∴ Δp_{y}d = pλ
∴ Δp_{y}Δy since d ~ Δy
∴ Δp_{y}Δy = h since p = h/λ
Narrow the slit => better position but greater uncertainty in momentum.
NOTE: The uncertainty in the momentum and position applies to a single direction
(i.e. ΔxΔp_{x} = h/2). It is possible to measure position and momentum precisely
for orthogonal components. This is expressed in the following commutator:
[Δx_{i},Δp_{j}] = ihδ_{ij}