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The uncertainty principle, also known as Heisenberg's indeterminacy principle, is a fundamental concept in quantum mechanics. It states that there is a limit to the precision with which certain pairs of physical properties, such as position and momentum, can be simultaneously known. In other words, the more accurately one property is measured, the less accurately the other property can be known.

More formally, the uncertainty principle is any of a variety of mathematical inequalities asserting a fundamental limit to the product of the accuracy of certain related pairs of measurements on a quantum system, such as position, x, and momentum, p. Such paired-variables are known as complementary variables or canonically conjugate variables.

First introduced in 1927 by German physicist Werner Heisenberg, the formal inequality relating the standard deviation of position σ_x and the standard deviation of momentum σ_p was derived by Earle Hesse Kennard later that year and by Hermann Weyl in 1928:

${\displaystyle \sigma _{x}\sigma _{p}\geq {\frac {\hbar }{2}}}$

where ${\displaystyle \hbar ={\frac {h}{2\pi }}}$ is the reduced Planck constant.

The quintessentially quantum mechanical uncertainty principle comes in many forms other than position–momentum. The energy–time relationship is widely used to relate quantum state lifetime to measured energy widths but its formal derivation is fraught with confusing issues about the nature of time. The basic principle has been extended in numerous directions; it must be considered in many kinds of fundamental physical measurements.

Position–momentum

It is vital to illustrate how the principle applies to relatively intelligible physical situations since it is indiscernible on the macroscopic scales that humans experience. Two alternative frameworks for quantum physics offer different explanations for the uncertainty principle. The 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 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. 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 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 de Broglie relation p = ħk, where k is the wavenumber.

In matrix mechanics, the mathematical formulation of quantum mechanics, any pair of non-commuting self-adjoint operators representing observables 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 A is performed, then the system is in a particular eigenstate Ψ of that observable. However, the particular eigenstate of the observable A need not be an eigenstate of another observable 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.

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 transforms of each other: mathematically, the uncertainty principle expresses the relationship between conjugate variables in the transform.

Wave mechanics interpretation

According to the 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 k₀ or momentum p₀ is $${\displaystyle \psi (x)\propto e^{ik_{0}x}=e^{ip_{0}x/\hbar }~.}$$

The Born rule states that this should be interpreted as a probability density amplitude function in the sense that the probability of finding the particle between a and b is $${\displaystyle \operatorname {P} =\int _{a}^{b}|\psi (x)|^{2}\,\mathrm {d} x~.}$$

In the case of the single-mode plane wave, ${\displaystyle |\psi (x)|^{2}}$ is 1 if ${\displaystyle X=x}$ 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 sum of many waves, which we may write as $${\displaystyle \psi (x)\propto \sum _{n}A_{n}e^{ip_{n}x/\hbar }~,}$$ where A_n represents the relative contribution of the mode p_n 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 $${\displaystyle \psi (x)={\frac {1}{\sqrt {2\pi \hbar }}}\int _{-\infty }^{\infty }\varphi (p)\cdot e^{ipx/\hbar }\,dp~,}$$ with ${\displaystyle \varphi (p)}$ representing the amplitude of these modes and is called the wave function in momentum space. In mathematical terms, we say that ${\displaystyle \varphi (p)}$ is the Fourier transform of ${\displaystyle \psi (x)}$ 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.

One way to quantify the precision of the position and momentum is the standard deviation σ. Since ${\displaystyle |\psi (x)|^{2}}$ is a probability density function for position, we calculate its standard deviation.

The precision of the position is improved, i.e. reduced σ_x, by using many plane waves, thereby weakening the precision of the momentum, i.e. increased σ_p. Another way of stating this is that σ_x and σ_p 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 variances of position and momentum, defined as $${\displaystyle \sigma _{x}^{2}=\int _{-\infty }^{\infty }x^{2}\cdot |\psi (x)|^{2}\,dx-\left(\int _{-\infty }^{\infty }x\cdot |\psi (x)|^{2}\,dx\right)^{2}}$$ $${\displaystyle \sigma _{p}^{2}=\int _{-\infty }^{\infty }p^{2}\cdot |\varphi (p)|^{2}\,dp-\left(\int _{-\infty }^{\infty }p\cdot |\varphi (p)|^{2}\,dp\right)^{2}~.}$$

Without loss of generality, we will assume that the 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 $${\displaystyle \sigma _{x}^{2}=\int _{-\infty }^{\infty }x^{2}\cdot |\psi (x)|^{2}\,dx}$$ $${\displaystyle \sigma _{p}^{2}=\int _{-\infty }^{\infty }p^{2}\cdot |\varphi (p)|^{2}\,dp~.}$$

The function ${\displaystyle f(x)=x\cdot \psi (x)}$ can be interpreted as a 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: $${\displaystyle \langle u\mid v\rangle =\int _{-\infty }^{\infty }u^{*}(x)\cdot v(x)\,dx,}$$ where the asterisk denotes the complex conjugate.

With this inner product defined, we note that the variance for position can be written as $${\displaystyle \sigma _{x}^{2}=\int _{-\infty }^{\infty }|f(x)|^{2}\,dx=\langle f\mid f\rangle ~.}$$

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

The term ${\textstyle -i\hbar {\frac {d}{dx}}}$ is called the momentum operator in position space. Applying Plancherel's theorem, we see that the variance for momentum can be written as $${\displaystyle \sigma _{p}^{2}=\int _{-\infty }^{\infty }|{\tilde {g}}(p)|^{2}\,dp=\int _{-\infty }^{\infty }|g(x)|^{2}\,dx=\langle g\mid g\rangle .}$$

The Cauchy–Schwarz inequality asserts that $${\displaystyle \sigma _{x}^{2}\sigma _{p}^{2}=\langle f\mid f\rangle \cdot \langle g\mid g\rangle \geq |\langle f\mid g\rangle |^{2}~.}$$

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

All that remains is to evaluate these inner products.

$${\displaystyle {\begin{aligned}\langle f\mid g\rangle -\langle g\mid f\rangle &=\int _{-\infty }^{\infty }\psi ^{*}(x)\,x\cdot \left(-i\hbar {\frac {d}{dx}}\right)\,\psi (x)\,dx-\int _{-\infty }^{\infty }\psi ^{*}(x)\,\left(-i\hbar {\frac {d}{dx}}\right)\cdot x\,\psi (x)\,dx\\&=i\hbar \cdot \int _{-\infty }^{\infty }\psi ^{*}(x)\left\,dx\\&=i\hbar \cdot \int _{-\infty }^{\infty }\psi ^{*}(x)\left\,dx\\&=i\hbar \cdot \int _{-\infty }^{\infty }\psi ^{*}(x)\psi (x)\,dx\\&=i\hbar \cdot \int _{-\infty }^{\infty }|\psi (x)|^{2}\,dx\\&=i\hbar \end{aligned}}}$$

Plugging this into the above inequalities, we get $${\displaystyle \sigma _{x}^{2}\sigma _{p}^{2}\geq |\langle f\mid g\rangle |^{2}\geq \left({\frac {\langle f\mid g\rangle -\langle g\mid f\rangle }{2i}}\right)^{2}=\left({\frac {i\hbar }{2i}}\right)^{2}={\frac {\hbar ^{2}}{4}}}$$ and taking the square root $${\displaystyle \sigma _{x}\sigma _{p}\geq {\frac {\hbar }{2}}~.}$$

with equality if and only if p and x are linearly dependent. Note that the only physics involved in this proof was that ${\displaystyle \psi (x)}$ and ${\displaystyle \varphi (p)}$ 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

In matrix mechanics, observables such as position and momentum are represented by self-adjoint operators. When considering pairs of observables, an important quantity is the commutator. For a pair of operators  and ${\displaystyle {\hat {B}}}$, one defines their commutator as $${\displaystyle ={\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}}.}$$ In the case of position and momentum, the commutator is the canonical commutation relation $${\displaystyle =i\hbar .}$$

The physical meaning of the non-commutativity can be understood by considering the effect of the commutator on position and momentum eigenstates. Let ${\displaystyle |\psi \rangle }$ be a right eigenstate of position with a constant eigenvalue x₀. By definition, this means that ${\displaystyle {\hat {x}}|\psi \rangle =x_{0}|\psi \rangle .}$ Applying the commutator to ${\displaystyle |\psi \rangle }$ yields $${\displaystyle |\psi \rangle =({\hat {x}}{\hat {p}}-{\hat {p}}{\hat {x}})|\psi \rangle =({\hat {x}}-x_{0}{\hat {I}}){\hat {p}}\,|\psi \rangle =i\hbar |\psi \rangle ,}$$ where Î is the identity operator.

Suppose, for the sake of proof by contradiction, that ${\displaystyle |\psi \rangle }$ is also a right eigenstate of momentum, with constant eigenvalue p₀. If this were true, then one could write $${\displaystyle ({\hat {x}}-x_{0}{\hat {I}}){\hat {p}}\,|\psi \rangle =({\hat {x}}-x_{0}{\hat {I}})p_{0}\,|\psi \rangle =(x_{0}{\hat {I}}-x_{0}{\hat {I}})p_{0}\,|\psi \rangle =0.}$$ On the other hand, the above canonical commutation relation requires that $${\displaystyle |\psi \rangle =i\hbar |\psi \rangle \neq 0.}$$ 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, $${\displaystyle \sigma _{x}={\sqrt {\langle {\hat {x}}^{2}\rangle -\langle {\hat {x}}\rangle ^{2}}}}$$ $${\displaystyle \sigma _{p}={\sqrt {\langle {\hat {p}}^{2}\rangle -\langle {\hat {p}}\rangle ^{2}}}.}$$

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

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: $${\displaystyle {\hat {x}}={\sqrt {\frac {\hbar }{2m\omega }}}(a+a^{\dagger })}$$ $${\displaystyle {\hat {p}}=i{\sqrt {\frac {m\omega \hbar }{2}}}(a^{\dagger }-a).}$$

Using the standard rules for creation and annihilation operators on the energy eigenstates, $${\displaystyle a^{\dagger }|n\rangle ={\sqrt {n+1}}|n+1\rangle }$$ $${\displaystyle a|n\rangle ={\sqrt {n}}|n-1\rangle ,}$$ the variances may be computed directly, $${\displaystyle \sigma _{x}^{2}={\frac {\hbar }{m\omega }}\left(n+{\frac {1}{2}}\right)}$$ $${\displaystyle \sigma _{p}^{2}=\hbar m\omega \left(n+{\frac {1}{2}}\right)\,.}$$ The product of these standard deviations is then $${\displaystyle \sigma _{x}\sigma _{p}=\hbar \left(n+{\frac {1}{2}}\right)\geq {\frac {\hbar }{2}}.~}$$

In particular, the above Kennard bound is saturated for the ground state n=0, for which the probability density is just the normal distribution.

Quantum harmonic oscillators with Gaussian initial condition

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 x₀ as $${\displaystyle \psi (x)=\left({\frac {m\Omega }{\pi \hbar }}\right)^{1/4}\exp {\left(-{\frac {m\Omega (x-x_{0})^{2}}{2\hbar }}\right)},}$$ where Ω describes the width of the initial state but need not be the same as ω. Through integration over the propagator, we can solve for the full time-dependent solution. After many cancelations, the probability densities reduce to $${\displaystyle |\Psi (x,t)|^{2}\sim {\mathcal {N}}\left(x_{0}\cos {(\omega t)},{\frac {\hbar }{2m\Omega }}\left(\cos ^{2}(\omega t)+{\frac {\Omega ^{2}}{\omega ^{2}}}\sin ^{2}{(\omega t)}\right)\right)}$$ $${\displaystyle |\Phi (p,t)|^{2}\sim {\mathcal {N}}\left(-mx_{0}\omega \sin(\omega t),{\frac {\hbar m\Omega }{2}}\left(\cos ^{2}{(\omega t)}+{\frac {\omega ^{2}}{\Omega ^{2}}}\sin ^{2}{(\omega t)}\right)\right),}$$ where we have used the notation ${\displaystyle {\mathcal {N}}(\mu ,\sigma ^{2})}$ to denote a normal distribution of mean μ and variance σ. Copying the variances above and applying trigonometric identities, we can write the product of the standard deviations as $${\displaystyle {\begin{aligned}\sigma _{x}\sigma _{p}&={\frac {\hbar }{2}}{\sqrt {\left(\cos ^{2}{(\omega t)}+{\frac {\Omega ^{2}}{\omega ^{2}}}\sin ^{2}{(\omega t)}\right)\left(\cos ^{2}{(\omega t)}+{\frac {\omega ^{2}}{\Omega ^{2}}}\sin ^{2}{(\omega t)}\right)}}\\&={\frac {\hbar }{4}}{\sqrt {3+{\frac {1}{2}}\left({\frac {\Omega ^{2}}{\omega ^{2}}}+{\frac {\omega ^{2}}{\Omega ^{2}}}\right)-\left({\frac {1}{2}}\left({\frac {\Omega ^{2}}{\omega ^{2}}}+{\frac {\omega ^{2}}{\Omega ^{2}}}\right)-1\right)\cos {(4\omega t)}}}\end{aligned}}}$$

From the relations $${\displaystyle {\frac {\Omega ^{2}}{\omega ^{2}}}+{\frac {\omega ^{2}}{\Omega ^{2}}}\geq 2,\quad |\cos(4\omega t)|\leq 1,}$$ we can conclude the following (the right most equality holds only when Ω = ω): $${\displaystyle \sigma _{x}\sigma _{p}\geq {\frac {\hbar }{4}}{\sqrt {3+{\frac {1}{2}}\left({\frac {\Omega ^{2}}{\omega ^{2}}}+{\frac {\omega ^{2}}{\Omega ^{2}}}\right)-\left({\frac {1}{2}}\left({\frac {\Omega ^{2}}{\omega ^{2}}}+{\frac {\omega ^{2}}{\Omega ^{2}}}\right)-1\right)}}={\frac {\hbar }{2}}.}$$

Coherent states

A coherent state is a right eigenstate of the annihilation operator, $${\displaystyle {\hat {a}}|\alpha \rangle =\alpha |\alpha \rangle ,}$$ which may be represented in terms of Fock states as $${\displaystyle |\alpha \rangle =e^{-{|\alpha |^{2} \over 2}}\sum _{n=0}^{\infty }{\alpha ^{n} \over {\sqrt {n!}}}|n\rangle }$$

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, $${\displaystyle \sigma _{x}^{2}={\frac {\hbar }{2m\omega }},}$$ $${\displaystyle \sigma _{p}^{2}={\frac {\hbar m\omega }{2}}.}$$ Therefore, every coherent state saturates the Kennard bound $${\displaystyle \sigma _{x}\sigma _{p}={\sqrt {\frac {\hbar }{2m\omega }}}\,{\sqrt {\frac {\hbar m\omega }{2}}}={\frac {\hbar }{2}}.}$$ with position and momentum each contributing an amount ${\textstyle {\sqrt {\hbar /2}}}$ 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

Consider a particle in a one-dimensional box of length ${\displaystyle L}$. The eigenfunctions in position and momentum space are $${\displaystyle \psi _{n}(x,t)={\begin{cases}A\sin(k_{n}x)\mathrm {e} ^{-\mathrm {i} \omega _{n}t},&0<x<L,\\0,&{\text{otherwise,}}\end{cases}}}$$ and $${\displaystyle \varphi _{n}(p,t)={\sqrt {\frac {\pi L}{\hbar }}}\,\,{\frac {n\left(1-(-1)^{n}e^{-ikL}\right)e^{-i\omega _{n}t}}{\pi ^{2}n^{2}-k^{2}L^{2}}},}$$ where ${\textstyle \omega _{n}={\frac {\pi ^{2}\hbar n^{2}}{8L^{2}m}}}$ and we have used the de Broglie relation ${\displaystyle p=\hbar k}$. The variances of ${\displaystyle x}$ and ${\displaystyle p}$ can be calculated explicitly: $${\displaystyle \sigma _{x}^{2}={\frac {L^{2}}{12}}\left(1-{\frac {6}{n^{2}\pi ^{2}}}\right)}$$ $${\displaystyle \sigma _{p}^{2}=\left({\frac {\hbar n\pi }{L}}\right)^{2}.}$$