Editing Uncertainty principle (section)

==Position–momentum==
{{Main article|Introduction to quantum mechanics}}

[[File:Sequential superposition of plane waves.gif|360px|right|thumb|The superposition of several plane waves to form a wave packet. This wave packet becomes increasingly localized with the addition of many waves. The Fourier transform is a mathematical operation that separates a wave packet into its individual plane waves. The waves shown here are [[real function|real]] for illustrative purposes only; in quantum mechanics the wave function is generally [[complex function|complex]].]]

It is vital to illustrate how the principle applies to relatively intelligible physical situations since it is indiscernible on the macroscopic<ref>{{cite journal | last1=Jaeger|first1=Gregg|title=What in the (quantum) world is macroscopic?|journal=American Journal of Physics|date=September 2014 | volume=82|issue=9|pages=896–905|doi=10.1119/1.4878358|bibcode = 2014AmJPh..82..896J }}</ref> scales that humans experience. Two alternative frameworks for quantum physics offer different explanations for the uncertainty principle. The [[Schrödinger equation|wave mechanics]] picture of the uncertainty principle is more visually intuitive, but the more abstract [[matrix mechanics]] picture formulates it in a way that generalizes more easily.

Mathematically, in wave mechanics, the uncertainty relation between position and momentum arises because the expressions of the wavefunction in the two corresponding [[orthonormal]] [[basis (linear algebra)|bases]] in [[Hilbert space]] are [[Fourier transforms]] of one another (i.e., position and momentum are [[conjugate variables]]). A nonzero function and its Fourier transform cannot both be sharply localized at the same time.<ref>See Appendix B in {{citation |title=Why photons cannot be sharply localized |first1=Iwo |last1=Bialynicki-Birula |first2=Zofia |last2=Bialynicka-Birula |journal=Physical Review A |date=2009 |volume=79 |issue=3 |pages=7–8|doi=10.1103/PhysRevA.79.032112 |arxiv=0903.3712 |bibcode=2009PhRvA..79c2112B |s2cid=55632217 }}</ref> A similar tradeoff between the variances of Fourier conjugates arises in all systems underlain by Fourier analysis, for example in sound waves: A pure tone is a [[Dirac delta function|sharp spike]] at a single frequency, while its Fourier transform gives the shape of the sound wave in the time domain, which is a completely delocalized sine wave. In quantum mechanics, the two key points are that the position of the particle takes the form of a matter wave, and momentum is its Fourier conjugate, assured by the [[Matter wave|de Broglie relation]] {{math|''p'' {{=}} ''ħk''}}, where {{mvar|k}} is the [[wavenumber]].

In [[matrix mechanics]], the [[mathematical formulation of quantum mechanics#Postulates of quantum mechanics|mathematical formulation of quantum mechanics]], any pair of non-[[commutator|commuting]] [[self-adjoint operator]]s representing [[observable]]s are subject to similar uncertainty limits. An eigenstate of an observable represents the state of the wavefunction for a certain measurement value (the eigenvalue). For example, if a measurement of an observable {{mvar|A}} is performed, then the system is in a particular eigenstate {{mvar|Ψ}} of that observable. However, the particular eigenstate of the observable {{mvar|A}} need not be an eigenstate of another observable {{mvar|B}}: If so, then it does not have a unique associated measurement for it, as the system is not in an eigenstate of that observable.<ref>{{Citation|author1=Claude Cohen-Tannoudji | author2=Bernard Diu | author3=Franck Laloë |title=Quantum mechanics|year=1996|publisher=Wiley|location=Wiley-Interscience | isbn=978-0-471-56952-7 | pages=231–233}}</ref>

===Visualization===
The uncertainty principle can be visualized using the position- and momentum-space wavefunctions for one spinless particle with mass in one dimension.

The more localized the position-space wavefunction, the more likely the particle is to be found with the position coordinates in that region, and correspondingly the momentum-space wavefunction is less localized so the possible momentum components the particle could have are more widespread. Conversely, the more localized the momentum-space wavefunction, the more likely the particle is to be found with those values of momentum components in that region, and correspondingly the less localized the position-space wavefunction, so the position coordinates the particle could occupy are more widespread. These wavefunctions are [[Fourier transform]]s of each other: mathematically, the uncertainty principle expresses the relationship between conjugate variables in the transform.

[[File:Quantum mechanics travelling wavefunctions wavelength.svg|center|thumb|502px|Position ''x'' and momentum ''p'' wavefunctions corresponding to quantum particles. The colour opacity of the particles corresponds to the [[probability density]] of finding the particle with position ''x'' or momentum component ''p''.<br/>

'''Top:''' If wavelength ''λ'' is unknown, so are momentum ''p'', wave-vector ''k'' and energy ''E'' (de Broglie relations). As the particle is more localized in position space, Δ''x'' is smaller than for Δ''p<sub>x</sub>''.<br/>

'''Bottom:''' If ''λ'' is known, so are ''p'', ''k'', and ''E''. As the particle is more localized in momentum space, Δ''p'' is smaller than for Δ''x''.]]

{{Clear}}

===Wave mechanics interpretation===
{{Main article|Wave packet|Schrödinger equation}}
{{multiple image
| align = right
| direction = vertical
| footer = Propagation of [[matter wave|de Broglie waves]] in 1d—real part of the [[complex number|complex]] amplitude is blue, imaginary part is green. The probability (shown as the colour [[opacity (optics)|opacity]]) of finding the particle at a given point ''x'' is spread out like a waveform, there is no definite position of the particle. As the amplitude increases above zero the [[curvature]] reverses sign, so the amplitude begins to decrease again, and vice versa—the result is an alternating amplitude: a wave.
| image1 = Propagation of a de broglie plane wave.svg
| caption1 = [[Plane wave]]
| width1 = 250
| image2 = Propagation of a de broglie wavepacket.svg
| caption2 = [[Wave packet]]
| width2 = 250
}}
According to the [[Matter wave|de Broglie hypothesis]], every object in the universe is associated with a [[wave]]. Thus every object, from an elementary particle to atoms, molecules and on up to planets and beyond are subject to the uncertainty principle.

The time-independent wave function of a single-moded plane wave of wavenumber ''k''<sub>0</sub> or momentum ''p''<sub>0</sub> is<ref>{{Citation | last = Hall | first = B. C. | title = Quantum Theory for Mathematicians | publisher = Springer | year = 2013 | pages = 60 | bibcode = 2013qtm..book.....H }}</ref>
<math display="block">\psi(x) \propto e^{ik_0 x} = e^{ip_0 x/\hbar} ~.</math>

The [[Born rule]] states that this should be interpreted as a [[probability density function|probability density amplitude function]] in the sense that the probability of finding the particle between ''a'' and ''b'' is
<math display="block"> \operatorname P [a \leq X \leq b] = \int_a^b |\psi(x)|^2 \, \mathrm{d}x ~.</math>

In the case of the single-mode plane wave, <math>|\psi(x)|^2</math> is ''1'' if <math>X=x</math> and ''0'' otherwise. In other words, the particle position is extremely uncertain in the sense that it could be essentially anywhere along the wave packet.

On the other hand, consider a wave function that is a [[superposition principle|sum of many waves]], which we may write as
<math display="block">\psi(x) \propto \sum_n A_n e^{i p_n x/\hbar}~, </math>
where ''A''<sub>''n''</sub> represents the relative contribution of the mode ''p''<sub>''n''</sub> to the overall total. The figures to the right show how with the addition of many plane waves, the wave packet can become more localized. We may take this a step further to the [[continuum limit]], where the wave function is an [[integral]] over all possible modes
<math display="block">\psi(x) = \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^\infty \varphi(p) \cdot e^{i p x/\hbar} \, dp ~, </math>
with <math>\varphi(p)</math> representing the amplitude of these modes and is called the wave function in [[momentum space]]. In mathematical terms, we say that <math>\varphi(p)</math> is the ''[[Fourier transform]]'' of <math>\psi(x)</math> and that ''x'' and ''p'' are [[conjugate variables]]. Adding together all of these plane waves comes at a cost, namely the momentum has become less precise, having become a mixture of waves of many different momenta.<ref name="L&L">{{cite book |first1=Lev Davidovich |last1=Landau|authorlink1=Lev Landau|first2=Evgeny Mikhailovich|last2=Lifshitz|authorlink2= Evgeny Lifshitz|year=1977 |title=Quantum Mechanics: Non-Relativistic Theory |edition=3rd |volume=3 |publisher=[[Pergamon Press]] |isbn=978-0-08-020940-1|url=https://archive.org/details/QuantumMechanics_104}}</ref>

One way to quantify the precision of the position and momentum is the [[standard deviation]]&nbsp;''σ''. Since <math>|\psi(x)|^2</math> is a probability density function for position, we calculate its standard deviation.

The precision of the position is improved, i.e. reduced ''σ''<sub>''x''</sub>, by using many plane waves, thereby weakening the precision of the momentum, i.e. increased ''σ''<sub>''p''</sub>. Another way of stating this is that ''σ''<sub>''x''</sub> and ''σ''<sub>''p''</sub> have an [[inverse relationship]] or are at least bounded from below. This is the uncertainty principle, the exact limit of which is the Kennard bound.

===Proof of the Kennard inequality using wave mechanics===

We are interested in the [[variance]]s of position and momentum, defined as
<math display="block">\sigma_x^2 = \int_{-\infty}^\infty x^2 \cdot |\psi(x)|^2 \, dx - \left( \int_{-\infty}^\infty x \cdot |\psi(x)|^2 \, dx \right)^2</math>
<math display="block">\sigma_p^2 = \int_{-\infty}^\infty p^2 \cdot |\varphi(p)|^2 \, dp - \left( \int_{-\infty}^\infty p \cdot |\varphi(p)|^2 \, dp \right)^2~.</math>

[[Without loss of generality]], we will assume that the [[expected value|means]] vanish, which just amounts to a shift of the origin of our coordinates. (A more general proof that does not make this assumption is given below.) This gives us the simpler form
<math display="block">\sigma_x^2 = \int_{-\infty}^\infty x^2 \cdot |\psi(x)|^2 \, dx</math>
<math display="block">\sigma_p^2 = \int_{-\infty}^\infty p^2 \cdot |\varphi(p)|^2 \, dp~.</math>

The function <math>f(x) = x \cdot \psi(x)</math> can be interpreted as a [[vector space|vector]] in a [[function space]]. We can define an [[inner product]] for a pair of functions ''u''(''x'') and ''v''(''x'') in this vector space:
<math display="block">\langle u \mid v \rangle = \int_{-\infty}^\infty u^*(x) \cdot v(x) \, dx,</math>
where the asterisk denotes the [[complex conjugate]].

With this inner product defined, we note that the variance for position can be written as
<math display="block">\sigma_x^2 = \int_{-\infty}^\infty |f(x)|^2 \, dx = \langle f \mid f \rangle ~.</math>

We can repeat this for momentum by interpreting the function <math>\tilde{g}(p)=p \cdot \varphi(p)</math> as a vector, but we can also take advantage of the fact that <math>\psi(x)</math> and <math>\varphi(p)</math> are Fourier transforms of each other. We evaluate the inverse Fourier transform through [[integration by parts]]:
<math display="block">\begin{align}
g(x) &= \frac{1}{\sqrt{2 \pi \hbar}} \cdot \int_{-\infty}^\infty \tilde{g}(p) \cdot e^{ipx/\hbar} \, dp \\
&= \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^\infty p \cdot \varphi(p) \cdot e^{ipx/\hbar} \, dp \\
&= \frac{1}{2 \pi \hbar} \int_{-\infty}^\infty \left[ p \cdot \int_{-\infty}^\infty \psi(\chi) e^{-ip\chi/\hbar} \, d\chi \right] \cdot e^{ipx/\hbar} \, dp \\
&= \frac{i}{2 \pi} \int_{-\infty}^\infty \left[ \cancel{ \left. \psi(\chi) e^{-ip\chi/\hbar} \right|_{-\infty}^\infty } - \int_{-\infty}^\infty \frac{d\psi(\chi)}{d\chi} e^{-ip\chi/\hbar} \, d\chi \right] \cdot e^{ipx/\hbar} \, dp \\
&= -i \int_{-\infty}^\infty \frac{d\psi(\chi)}{d\chi} \left[ \frac{1}{2 \pi}\int_{-\infty}^\infty \, e^{ip(x - \chi)/\hbar} \, dp \right]\, d\chi\\
&= -i \int_{-\infty}^\infty \frac{d\psi(\chi)}{d\chi} \left[ \delta\left(\frac{x - \chi }{\hbar}\right) \right]\, d\chi\\
&= -i \hbar \int_{-\infty}^\infty \frac{d\psi(\chi)}{d\chi} \left[ \delta\left(x - \chi \right) \right]\, d\chi\\
&= -i \hbar \frac{d\psi(x)}{dx} \\
&= \left( -i \hbar \frac{d}{dx} \right) \cdot \psi(x) ,
\end{align}</math>
where <math>v=\frac{\hbar}{-ip}e^{-ip\chi/\hbar}</math> in the integration by parts, the cancelled term vanishes because the wave function vanishes at both infinities and <math>|e^{-ip\chi/\hbar}|=1</math>, and then use the [[Dirac delta function#History|Dirac delta function]] which is valid because <math>\dfrac{d\psi(\chi)}{d\chi}</math> does not depend on ''p'' .

The term <math display="inline">-i \hbar \frac{d}{dx}</math> is called the [[momentum operator]] in position space. Applying [[Plancherel theorem|Plancherel's theorem]], we see that the variance for momentum can be written as
<math display="block">\sigma_p^2 = \int_{-\infty}^\infty |\tilde{g}(p)|^2 \, dp = \int_{-\infty}^\infty |g(x)|^2 \, dx = \langle g \mid g \rangle.</math>

The [[Cauchy–Schwarz inequality]] asserts that
<math display="block">\sigma_x^2 \sigma_p^2 = \langle f \mid f \rangle \cdot \langle g \mid g \rangle \ge |\langle f \mid g \rangle|^2 ~.</math>

The [[modulus squared]] of any complex number ''z'' can be expressed as
<math display="block">|z|^{2} = \Big(\text{Re}(z)\Big)^{2}+\Big(\text{Im}(z)\Big)^{2} \geq \Big(\text{Im}(z)\Big)^{2} = \left(\frac{z-z^{\ast}}{2i}\right)^{2}. </math>
we let <math>z=\langle f|g\rangle</math> and <math>z^{*}=\langle g\mid f\rangle</math> and substitute these into the equation above to get
<math display="block">|\langle f\mid g\rangle|^2 \geq \left(\frac{\langle f\mid g\rangle-\langle g \mid f \rangle}{2i}\right)^2 ~.</math>

All that remains is to evaluate these inner products.

<math display="block">\begin{align}
\langle f\mid g\rangle-\langle g\mid f\rangle  &= \int_{-\infty}^\infty \psi^*(x) \, x \cdot \left(-i \hbar \frac{d}{dx}\right) \, \psi(x) \, dx - \int_{-\infty}^\infty \psi^*(x) \, \left(-i \hbar \frac{d}{dx}\right) \cdot x \, \psi(x) \, dx \\
&= i \hbar \cdot \int_{-\infty}^\infty \psi^*(x) \left[ \left(-x \cdot \frac{d\psi(x)}{dx}\right) + \frac{d(x \psi(x))}{dx} \right] \, dx \\
&= i \hbar \cdot \int_{-\infty}^\infty \psi^*(x) \left[ \left(-x \cdot \frac{d\psi(x)}{dx}\right) + \psi(x) + \left(x \cdot \frac{d\psi(x)}{dx}\right)\right] \, dx \\
&= i \hbar \cdot \int_{-\infty}^\infty \psi^*(x) \psi(x) \, dx \\
&= i \hbar \cdot \int_{-\infty}^\infty |\psi(x)|^2 \, dx \\
&= i \hbar
\end{align}</math>

Plugging this into the above inequalities, we get
<math display="block">\sigma_x^2 \sigma_p^2 \ge |\langle f \mid g \rangle|^2 \ge \left(\frac{\langle f\mid g\rangle-\langle g\mid f\rangle}{2i}\right)^2 = \left(\frac{i \hbar}{2 i}\right)^2 = \frac{\hbar^2}{4}</math>
and taking the square root
<math display="block">\sigma_x \sigma_p \ge \frac{\hbar}{2}~.</math>

with equality if and only if ''p'' and ''x'' are linearly dependent. Note that the only ''physics'' involved in this proof was that <math>\psi(x)</math> and <math>\varphi(p)</math> are wave functions for position and momentum, which are Fourier transforms of each other. A similar result would hold for ''any'' pair of conjugate variables.

===Matrix mechanics interpretation===
{{Main article|Matrix mechanics}}
In matrix mechanics, observables such as position and momentum are represented by self-adjoint operators.<ref name="L&L"/> When considering pairs of observables, an important quantity is the ''[[commutator]]''. For a pair of operators {{mvar|Â}} and <math>\hat{B}</math>, one defines their commutator as
<math display="block">[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}.</math>
In the case of position and momentum, the commutator is the [[canonical commutation relation]]
<math display="block">[\hat{x},\hat{p}]=i \hbar.</math>

The physical meaning of the non-commutativity can be understood by considering the effect of the commutator on position and momentum [[eigenstate]]s. Let <math>|\psi\rangle</math> be a right eigenstate of position with a constant eigenvalue {{math|''x''<sub>0</sub>}}. By definition, this means that <math>\hat{x}|\psi\rangle = x_0 |\psi\rangle.</math> Applying the commutator to <math>|\psi\rangle</math> yields
<math display="block">[\hat{x},\hat{p}] | \psi \rangle = (\hat{x}\hat{p}-\hat{p}\hat{x}) | \psi \rangle = (\hat{x} - x_0 \hat{I}) \hat{p} \, | \psi \rangle = i \hbar | \psi \rangle,</math>
where {{mvar|Î}} is the [[identity matrix|identity operator]].

Suppose, for the sake of [[proof by contradiction]], that <math>|\psi\rangle</math> is also a right eigenstate of momentum, with constant eigenvalue {{mvar|''p''<sub>0</sub>}}. If this were true, then one could write
<math display="block">(\hat{x} - x_0 \hat{I}) \hat{p} \, | \psi \rangle = (\hat{x} - x_0 \hat{I}) p_0 \, | \psi \rangle = (x_0 \hat{I} - x_0 \hat{I}) p_0 \, | \psi \rangle=0.</math>
On the other hand, the above canonical commutation relation requires that
<math display="block">[\hat{x},\hat{p}] | \psi \rangle=i \hbar | \psi \rangle \ne 0.</math>
This implies that no quantum state can simultaneously be both a position and a momentum eigenstate.

When a state is measured, it is projected onto an eigenstate in the basis of the relevant observable. For example, if a particle's position is measured, then the state amounts to a position eigenstate. This means that the state is ''not'' a momentum eigenstate, however, but rather it can be represented as a sum of multiple momentum basis eigenstates. In other words, the momentum must be less precise. This precision may be quantified by the standard deviations,
<math display="block">\sigma_x=\sqrt{\langle \hat{x}^2 \rangle-\langle \hat{x}\rangle^2}</math>
<math display="block">\sigma_p=\sqrt{\langle \hat{p}^2 \rangle-\langle \hat{p}\rangle^2}.</math>

As in the wave mechanics interpretation above, one sees a tradeoff between the respective precisions of the two, quantified by the uncertainty principle.

===Quantum harmonic oscillator stationary states===
{{Main article|Quantum harmonic oscillator|Stationary state}}
Consider a one-dimensional quantum harmonic oscillator. It is possible to express the position and momentum operators in terms of the [[creation and annihilation operators]]:
<math display="block">\hat x = \sqrt{\frac{\hbar}{2m\omega}}(a+a^\dagger)</math>
<math display="block">\hat p = i\sqrt{\frac{m \omega\hbar}{2}}(a^\dagger-a).</math>

Using the standard rules for creation and annihilation operators on the energy eigenstates,
<math display="block">a^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle</math>
<math display="block">a|n\rangle=\sqrt{n}|n-1\rangle, </math>
the variances may be computed directly,
<math display="block">\sigma_x^2 = \frac{\hbar}{m\omega} \left( n+\frac{1}{2}\right)</math>
<math display="block">\sigma_p^2 = \hbar m\omega \left( n+\frac{1}{2}\right)\, .</math>
The product of these standard deviations is then
<math display="block">\sigma_x \sigma_p = \hbar \left(n+\frac{1}{2}\right) \ge \frac{\hbar}{2}.~</math>

In particular, the above Kennard bound<ref name="Kennard" /> is saturated for the [[ground state]] {{math|''n''{{=}}0}}, for which the probability density is just the [[normal distribution]].

=== Quantum harmonic oscillators with Gaussian initial condition ===
{{multiple image
| align = right
| direction = vertical
| footer =
Position (blue) and momentum (red) probability densities for an initial Gaussian distribution. From top to bottom, the animations show the cases {{nowrap|1=Ω = ''ω''}}, {{nowrap|1=Ω = 2''ω''}}, and {{nowrap|1=Ω = ''ω''/2}}. Note the tradeoff between the widths of the distributions.
| width1 = 360
| image1 = Position_and_momentum_of_a_Gaussian_initial_state_for_a_QHO,_balanced.gif
| width2 = 360
| image2 = Position_and_momentum_of_a_Gaussian_initial_state_for_a_QHO,_narrow.gif
| width3 = 360
| image3 = Position_and_momentum_of_a_Gaussian_initial_state_for_a_QHO,_wide.gif
}}

In a quantum harmonic oscillator of characteristic angular frequency ''ω'', place a state that is offset from the bottom of the potential by some displacement ''x''<sub>0</sub> as
<math display="block">\psi(x)=\left(\frac{m \Omega}{\pi \hbar}\right)^{1/4} \exp{\left( -\frac{m \Omega (x-x_0)^2}{2\hbar}\right)},</math>
where Ω describes the width of the initial state but need not be the same as ''ω''. Through integration over the [[Propagator#Basic examples: propagator of free particle and harmonic oscillator|propagator]], we can solve for the {{Not a typo|full time}}-dependent solution. After many cancelations, the probability densities reduce to
<math display="block">|\Psi(x,t)|^2 \sim \mathcal{N}\left( x_0 \cos{(\omega t)} , \frac{\hbar}{2 m \Omega} \left( \cos^2(\omega t) + \frac{\Omega^2}{\omega^2} \sin^2{(\omega t)} \right)\right)</math>
<math display="block">|\Phi(p,t)|^2 \sim \mathcal{N}\left( -m x_0 \omega \sin(\omega t), \frac{\hbar m \Omega}{2} \left( \cos^2{(\omega t)} + \frac{\omega^2}{\Omega^2} \sin^2{(\omega t)} \right)\right),</math>
where we have used the notation <math>\mathcal{N}(\mu, \sigma^2)</math> to denote a normal distribution of mean ''μ'' and variance ''σ''<sup>2</sup>. Copying the variances above and applying [[list of trigonometric identities|trigonometric identities]], we can write the product of the standard deviations as
<math display="block">\begin{align}
\sigma_x \sigma_p&=\frac{\hbar}{2}\sqrt{\left( \cos^2{(\omega t)} + \frac{\Omega^2}{\omega^2} \sin^2{(\omega t)} \right)\left( \cos^2{(\omega t)} + \frac{\omega^2}{\Omega^2} \sin^2{(\omega t)} \right)} \\
&= \frac{\hbar}{4}\sqrt{3+\frac{1}{2}\left(\frac{\Omega^2}{\omega^2}+\frac{\omega^2}{\Omega^2}\right)-\left(\frac{1}{2}\left(\frac{\Omega^2}{\omega^2}+\frac{\omega^2}{\Omega^2}\right)-1\right) \cos{(4 \omega t)}}
\end{align}</math>

From the relations
<math display="block">\frac{\Omega^2}{\omega^2}+\frac{\omega^2}{\Omega^2} \ge 2, \quad |\cos(4 \omega t)| \le 1,</math>
we can conclude the following (the right most equality holds only when {{nowrap|1=Ω = ''ω''}}):
<math display="block">\sigma_x \sigma_p \ge \frac{\hbar}{4}\sqrt{3+\frac{1}{2} \left(\frac{\Omega^2}{\omega^2}+\frac{\omega^2}{\Omega^2}\right)-\left(\frac{1}{2} \left(\frac{\Omega^2}{\omega^2}+\frac{\omega^2}{\Omega^2}\right)-1\right)} = \frac{\hbar}{2}. </math>

===Coherent states===
{{Main article|Coherent state}}
A coherent state is a right eigenstate of the [[annihilation operator]],
<math display="block">\hat{a}|\alpha\rangle=\alpha|\alpha\rangle,</math>
which may be represented in terms of [[Fock state]]s as
<math display="block">|\alpha\rangle =e^{-{|\alpha|^2\over2}} \sum_{n=0}^\infty {\alpha^n \over \sqrt{n!}}|n\rangle</math>

In the picture where the coherent state is a massive particle in a quantum harmonic oscillator, the position and momentum operators may be expressed in terms of the annihilation operators in the same formulas above and used to calculate the variances,
<math display="block">\sigma_x^2 = \frac{\hbar}{2 m \omega},</math>
<math display="block">\sigma_p^2 = \frac{\hbar m \omega}{2}.</math>
Therefore, every coherent state saturates the Kennard bound
<math display="block">\sigma_x \sigma_p = \sqrt{\frac{\hbar}{2 m \omega}} \, \sqrt{\frac{\hbar m \omega}{2}} = \frac{\hbar}{2}. </math>
with position and momentum each contributing an amount <math display="inline">\sqrt{\hbar/2}</math> in a "balanced" way. Moreover, every [[squeezed coherent state]] also saturates the Kennard bound although the individual contributions of position and momentum need not be balanced in general.

===Particle in a box===
{{Main article|Particle in a box}}
Consider a particle in a one-dimensional box of length <math>L</math>. The [[Particle in a box#Wavefunctions|eigenfunctions in position and momentum space]] are
<math display="block">\psi_n(x,t) =\begin{cases}
A \sin(k_n x)\mathrm{e}^{-\mathrm{i}\omega_n t}, & 0 < x < L,\\
0, & \text{otherwise,}
\end{cases}</math>
and
<math display="block">\varphi_n(p,t)=\sqrt{\frac{\pi L}{\hbar}}\,\,\frac{n\left(1-(-1)^ne^{-ikL} \right) e^{-i \omega_n t}}{\pi ^2 n^2-k^2 L^2},</math>
where <math display="inline">\omega_n=\frac{\pi^2 \hbar n^2}{8 L^2 m}</math> and we have used the [[de Broglie relation]] <math>p=\hbar k</math>. The variances of <math>x</math> and <math>p</math> can be calculated explicitly:
<math display="block">\sigma_x^2=\frac{L^2}{12}\left(1-\frac{6}{n^2\pi^2}\right)</math>
<math display="block">\sigma_p^2=\left(\frac{\hbar n\pi}{L}\right)^2. </math>

The product of the standard deviations is therefore
<math display="block">\sigma_x \sigma_p = \frac{\hbar}{2} \sqrt{\frac{n^2\pi^2}{3}-2}.</math>
For all <math>n=1, \, 2, \, 3,\, \ldots</math>, the quantity <math display="inline">\sqrt{\frac{n^2\pi^2}{3}-2}</math> is greater than 1, so the uncertainty principle is never violated. For numerical concreteness, the smallest value occurs when <math>n = 1</math>, in which case
<math display="block">\sigma_x \sigma_p = \frac{\hbar}{2} \sqrt{\frac{\pi^2}{3}-2} \approx 0.568 \hbar > \frac{\hbar}{2}.</math>

===Constant momentum===
{{Main article|Wave packet}}
[[File:Guassian Dispersion.gif|360 px|thumb|right|Position space probability density of an initially Gaussian state moving at minimally uncertain, constant momentum in free space]]
Assume a particle initially has a [[momentum space]] wave function described by a normal distribution around some constant momentum ''p''<sub>0</sub> according to
<math display="block">\varphi(p) = \left(\frac{x_0}{\hbar \sqrt{\pi}} \right)^{1/2} \exp\left(\frac{-x_0^2 (p-p_0)^2}{2\hbar^2}\right),</math>
where we have introduced a reference scale <math display="inline">x_0=\sqrt{\hbar/m\omega_0}</math>, with <math>\omega_0>0</math> describing the width of the distribution—cf. [[nondimensionalization]]. If the state is allowed to evolve in free space, then the time-dependent momentum and position space wave functions are
<math display="block">\Phi(p,t) = \left(\frac{x_0}{\hbar \sqrt{\pi}} \right)^{1/2} \exp\left(\frac{-x_0^2 (p-p_0)^2}{2\hbar^2}-\frac{ip^2 t}{2m\hbar}\right),</math>
<math display="block">\Psi(x,t) = \left(\frac{1}{x_0 \sqrt{\pi}} \right)^{1/2} \frac{e^{-x_0^2 p_0^2 /2\hbar^2}}{\sqrt{1+i\omega_0 t}} \, \exp\left(-\frac{(x-ix_0^2 p_0/\hbar)^2}{2x_0^2 (1+i\omega_0 t)}\right).</math>

Since <math> \langle p(t) \rangle = p_0</math> and <math>\sigma_p(t) = \hbar /(\sqrt{2}x_0)</math>, this can be interpreted as a particle moving along with constant momentum at arbitrarily high precision. On the other hand, the standard deviation of the position is
<math display="block">\sigma_x = \frac{x_0}{\sqrt{2}} \sqrt{1+\omega_0^2 t^2}</math>
such that the uncertainty product can only increase with time as
<math display="block">\sigma_x(t) \sigma_p(t) = \frac{\hbar}{2} \sqrt{1+\omega_0^2 t^2}</math>