Uncertainty principle

(Ref [13])

Plane wave

Wave packet

Propagation of de Broglie waves in 1d—real part of the complex amplitude is blue, imaginary part is green. The probability (shown as the colour 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.

According to the de Broglie hypothesis, every object in the universe is a wave, i.e., a situation which gives rise to this phenomenon. The position of the particle is described by a wave function ${\displaystyle \Psi (x,t)}$. 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

$${\displaystyle \operatorname {P} [a\leq X\leq b]=\int _{a}^{b}|\psi (x)|^{2}\,\mathrm {d} x~.}$$

In the case of the single-moded plane wave, ${\displaystyle |\psi (x)|^{2}}$ is a uniform distribution. 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. Click the show button below to see a semi-formal derivation of the Kennard inequality using wave mechanics.

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:

$${\displaystyle {\begin{aligned}g(x)&={\frac {1}{\sqrt {2\pi \hbar }}}\cdot \int _{-\infty }^{\infty }{\tilde {g}}(p)\cdot e^{ipx/\hbar }\,dp\\&={\frac {1}{\sqrt {2\pi \hbar }}}\int _{-\infty }^{\infty }p\cdot \varphi (p)\cdot e^{ipx/\hbar }\,dp\\&={\frac {1}{2\pi \hbar }}\int _{-\infty }^{\infty }\left[p\cdot \int _{-\infty }^{\infty }\psi (\chi )e^{-ip\chi /\hbar }\,d\chi \right]\cdot e^{ipx/\hbar }\,dp\\&={\frac {i}{2\pi }}\int _{-\infty }^{\infty }\left[{\cancel {\left.\psi (\chi )e^{-ip\chi /\hbar }\right|_{-\infty }^{\infty }}}-\int _{-\infty }^{\infty }{\frac {d\psi (\chi )}{d\chi }}e^{-ip\chi /\hbar }\,d\chi \right]\cdot e^{ipx/\hbar }\,dp\\&={\frac {-i}{2\pi }}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }{\frac {d\psi (\chi )}{d\chi }}e^{-ip\chi /\hbar }\,d\chi \,e^{ipx/\hbar }\,dp\\&=\left(-i\hbar {\frac {d}{dx}}\right)\cdot \psi (x),\end{aligned}}}$$

where the canceled term vanishes because the wave function vanishes at infinity. Often the term ${\textstyle -i\hbar {\frac {d}{dx}}}$ is called the momentum operator in position space. Applying Parseval'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.

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

Plugging this into the above inequalities, we get

$${\displaystyle \sigma _{x}^{2}\sigma _{p}^{2}\geq |\langle f\mid g\rangle |^{2}\geq \left({\frac {\langle f\mid g\rangle -\langle g\mid f\rangle }{2i}}\right)^{2}=\left({\frac {i\hbar }{2i}}\right)^{2}={\frac {\hbar ^{2}}{4}}}$$

or taking the square root

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

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.

(Ref [13])

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 B̂, one defines their commutator as

$${\displaystyle [{\hat {A}},{\hat {B}}]={\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}}.}$$

In the case of position and momentum, the commutator is the canonical commutation relation

$${\displaystyle [{\hat {x}},{\hat {p}}]=i\hbar .}$$

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

$${\displaystyle [{\hat {x}},{\hat {p}}]|\psi \rangle =({\hat {x}}{\hat {p}}-{\hat {p}}{\hat {x}})|\psi \rangle =({\hat {x}}-x_{0}{\hat {I}}){\hat {p}}\,|\psi \rangle =i\hbar |\psi \rangle ,}$$

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

$${\displaystyle [{\hat {x}},{\hat {p}}]|\psi \rangle =i\hbar |\psi \rangle \neq 0.}$$

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.

The Robertson–Schrödinger uncertainty relation may be generalized in a straightforward way to describe mixed states.

$${\displaystyle \sigma _{A}^{2}\sigma _{B}^{2}\geq \left|{\frac {1}{2}}\operatorname {tr} (\rho \{A,B\})-\operatorname {tr} (\rho A)\operatorname {tr} (\rho B)\right|^{2}+\left|{\frac {1}{2i}}\operatorname {tr} (\rho [A,B])\right|^{2}.}$$

The Robertson–Schrödinger uncertainty relation can be trivial if the state of the system is chosen to be eigenstate of one of the observable. The stronger uncertainty relations proved by Maccone and Pati give non-trivial bounds on the sum of the variances for two incompatible observables.[30] (Earlier works on uncertainty relations formulated as the sum of variances include, e.g., Ref.[31] due to Huang.) For two non-commuting observables ${\displaystyle A}$ and ${\displaystyle B}$ the first stronger uncertainty relation is given by

$${\displaystyle \sigma _{A}^{2}+\sigma _{B}^{2}\geq \pm i\langle \Psi \mid [A,B]|\Psi \rangle +\mid \langle \Psi \mid (A\pm iB)\mid {\bar {\Psi }}\rangle |^{2},}$$

where ${\displaystyle \sigma _{A}^{2}=\langle \Psi |A^{2}|\Psi \rangle -\langle \Psi \mid A\mid \Psi \rangle ^{2}}$, ${\displaystyle \sigma _{B}^{2}=\langle \Psi |B^{2}|\Psi \rangle -\langle \Psi \mid B\mid \Psi \rangle ^{2}}$, ${\displaystyle |{\bar {\Psi }}\rangle }$ is a normalized vector that is orthogonal to the state of the system ${\displaystyle |\Psi \rangle }$ and one should choose the sign of ${\displaystyle \pm i\langle \Psi \mid [A,B]\mid \Psi \rangle }$ to make this real quantity a positive number.

The second stronger uncertainty relation is given by

$${\displaystyle \sigma _{A}^{2}+\sigma _{B}^{2}\geq {\frac {1}{2}}|\langle {\bar {\Psi }}_{A+B}\mid (A+B)\mid \Psi \rangle |^{2}}$$

where ${\displaystyle |{\bar {\Psi }}_{A+B}\rangle }$ is a state orthogonal to ${\displaystyle |\Psi \rangle }$. The form of ${\displaystyle |{\bar {\Psi }}_{A+B}\rangle }$ implies that the right-hand side of the new uncertainty relation is nonzero unless ${\displaystyle |\Psi \rangle }$ is an eigenstate of ${\displaystyle (A+B)}$. One may note that ${\displaystyle |\Psi \rangle }$ can be an eigenstate of ${\displaystyle (A+B)}$ without being an eigenstate of either ${\displaystyle A}$ or ${\displaystyle B}$. However, when ${\displaystyle |\Psi \rangle }$ is an eigenstate of one of the two observables the Heisenberg–Schrödinger uncertainty relation becomes trivial. But the lower bound in the new relation is nonzero unless ${\displaystyle |\Psi \rangle }$ is an eigenstate of both.

The Robertson–Schrödinger uncertainty can be improved noting that it must hold for all components ${\displaystyle \varrho _{k}}$ in any decomposition of the density matrix given as

$${\displaystyle \varrho =\sum _{k}p_{k}\varrho _{k}.}$$

Here, for the probailities ${\displaystyle p_{k}\geq 0}$ and ${\displaystyle \sum _{k}p_{k}=1}$ hold. Then, using the relation

$${\displaystyle \sum _{k}a_{k}\sum _{k}b_{k}\geq \left(\sum _{k}{\sqrt {a_{k}b_{k}}}\right)^{2}}$$

for ${\displaystyle a_{k},b_{k}\geq 0}$, it follows that[32]

$${\displaystyle \sigma _{A}^{2}\sigma _{B}^{2}\geq \left[\sum _{k}p_{k}L(\varrho _{k})\right]^{2},}$$

where the function in the bound is defined

$${\displaystyle L(\varrho )={\sqrt {\left|{\frac {1}{2}}\operatorname {tr} (\rho \{A,B\})-\operatorname {tr} (\rho A)\operatorname {tr} (\rho B)\right|^{2}+\left|{\frac {1}{2i}}\operatorname {tr} (\rho [A,B])\right|^{2}}}.}$$

The above relation very often has a bound larger than that of the original Robertson–Schrödinger uncertainty relation. Thus, we need to calculate the bound of the Robertson–Schrödinger uncertainty for the mixed components of the quantum state rather than for the quantum state, and compute an average of their square roots. The following expression is stronger than the Robertson–Schrödinger uncertainty relation

$${\displaystyle \sigma _{A}^{2}\sigma _{B}^{2}\geq \left[\max _{p_{k},\varrho _{k}}\sum _{k}p_{k}L(\varrho _{k})\right]^{2},}$$

where on the right-hand side there is a concave roof over the decompositions of the density matrix. The improved relation above is saturated by all single-qubit quantum states.[32]

With similar arguments, one can derive a relation with a convex roof on the right-hand side[32]

$${\displaystyle \sigma _{A}^{2}F_{Q}[\varrho ,B]\geq 4\left[\min _{p_{k},\Psi _{k}}\sum _{k}p_{k}L(\vert \Psi _{k}\rangle \langle \Psi _{k}\vert )\right]^{2},}$$

where ${\displaystyle F_{Q}[\varrho ,B]}$ denotes the quantum Fisher information and the density matrix is decomposed to pure states as

$${\displaystyle \varrho =\sum _{k}p_{k}\vert \Psi _{k}\rangle \langle \Psi _{k}\vert .}$$

The derivation takes advantage of the fact that the quantum Fisher information is the convex roof of the variance times four.[33][34]

A simpler inequality follows without a convex roof[35]

$${\displaystyle \sigma _{A}^{2}F_{Q}[\varrho ,B]\geq \vert \langle i[A,B]\rangle \vert ^{2},}$$

which is stronger than the Heisenberg uncertainty relation, since for the quantum Fisher information we have

$${\displaystyle F_{Q}[\varrho ,B]\leq 4\sigma _{B},}$$

while for pure states the equality holds.

In the phase space formulation of quantum mechanics, the Robertson–Schrödinger relation follows from a positivity condition on a real star-square function. Given a Wigner function ${\displaystyle W(x,p)}$ with star product ★ and a function f, the following is generally true:[36]

$${\displaystyle \langle f^{*}\star f\rangle =\int (f^{*}\star f)\,W(x,p)\,dx\,dp\geq 0~.}$$

Choosing ${\displaystyle f=a+bx+cp}$, we arrive at

$${\displaystyle \langle f^{*}\star f\rangle ={\begin{bmatrix}a^{*}&b^{*}&c^{*}\end{bmatrix}}{\begin{bmatrix}1&\langle x\rangle &\langle p\rangle \\\langle x\rangle &\langle x\star x\rangle &\langle x\star p\rangle \\\langle p\rangle &\langle p\star x\rangle &\langle p\star p\rangle \end{bmatrix}}{\begin{bmatrix}a\\b\\c\end{bmatrix}}\geq 0~.}$$

Since this positivity condition is true for all a, b, and c, it follows that all the eigenvalues of the matrix are non-negative.

The non-negative eigenvalues then imply a corresponding non-negativity condition on the determinant,

$${\displaystyle \det {\begin{bmatrix}1&\langle x\rangle &\langle p\rangle \\\langle x\rangle &\langle x\star x\rangle &\langle x\star p\rangle \\\langle p\rangle &\langle p\star x\rangle &\langle p\star p\rangle \end{bmatrix}}=\det {\begin{bmatrix}1&\langle x\rangle &\langle p\rangle \\\langle x\rangle &\langle x^{2}\rangle &\left\langle xp+{\frac {i\hbar }{2}}\right\rangle \\\langle p\rangle &\left\langle xp-{\frac {i\hbar }{2}}\right\rangle &\langle p^{2}\rangle \end{bmatrix}}\geq 0~,}$$

or, explicitly, after algebraic manipulation,

$${\displaystyle \sigma _{x}^{2}\sigma _{p}^{2}=\left(\langle x^{2}\rangle -\langle x\rangle ^{2}\right)\left(\langle p^{2}\rangle -\langle p\rangle ^{2}\right)\geq \left(\langle xp\rangle -\langle x\rangle \langle p\rangle \right)^{2}+{\frac {\hbar ^{2}}{4}}~.}$$

Since the Robertson and Schrödinger relations are for general operators, the relations can be applied to any two observables to obtain specific uncertainty relations. A few of the most common relations found in the literature are given below.

• For position and linear momentum, the canonical commutation relation ${\displaystyle [{\hat {x}},{\hat {p}}]=i\hbar }$ implies the Kennard inequality from above:
  $${\displaystyle \sigma _{x}\sigma _{p}\geq {\frac {\hbar }{2}}.}$$

  
• For two orthogonal components of the total angular momentum operator of an object:
  $${\displaystyle \sigma _{J_{i}}\sigma _{J_{j}}\geq {\frac {\hbar }{2}}{\big |}\langle J_{k}\rangle {\big |},}$$

  
  where i, j, k are distinct, and J_i denotes angular momentum along the x_i axis. This relation implies that unless all three components vanish together, only a single component of a system's angular momentum can be defined with arbitrary precision, normally the component parallel to an external (magnetic or electric) field. Moreover, for ${\displaystyle [J_{x},J_{y}]=i\hbar \varepsilon _{xyz}J_{z}}$, a choice ${\displaystyle {\hat {A}}=J_{x}}$, ${\displaystyle {\hat {B}}=J_{y}}$, in angular momentum multiplets, ψ = |j, m⟩, bounds the Casimir invariant (angular momentum squared, ${\displaystyle \langle J_{x}^{2}+J_{y}^{2}+J_{z}^{2}\rangle }$) from below and thus yields useful constraints such as j(j + 1) ≥ m(m + 1), and hence j ≥ m, among others.
• In non-relativistic mechanics, time is privileged as an independent variable. Nevertheless, in 1945, L. I. Mandelshtam and I. E. Tamm derived a non-relativistic time–energy uncertainty relation, as follows.[37][38] For a quantum system in a non-stationary state ψ and an observable B represented by a self-adjoint operator ${\displaystyle {\hat {B}}}$, the following formula holds:
  $${\displaystyle \sigma _{E}{\frac {\sigma _{B}}{\left|{\frac {d\langle {\hat {B}}\rangle }{dt}}\right|}}\geq {\frac {\hbar }{2}},}$$

  
  where σ_E is the standard deviation of the energy operator (Hamiltonian) in the state ψ, σ_B stands for the standard deviation of B. Although the second factor in the left-hand side has dimension of time, it is different from the time parameter that enters the Schrödinger equation. It is a lifetime of the state ψ with respect to the observable B: In other words, this is the time interval (Δt) after which the expectation value ${\displaystyle \langle {\hat {B}}\rangle }$ changes appreciably.
  An informal, heuristic meaning of the principle is the following: A state that only exists for a short time cannot have a definite energy. To have a definite energy, the frequency of the state must be defined accurately, and this requires the state to hang around for many cycles, the reciprocal of the required accuracy. For example, in spectroscopy, excited states have a finite lifetime. By the time–energy uncertainty principle, they do not have a definite energy, and, each time they decay, the energy they release is slightly different. The average energy of the outgoing photon has a peak at the theoretical energy of the state, but the distribution has a finite width called the natural linewidth. Fast-decaying states have a broad linewidth, while slow-decaying states have a narrow linewidth.[39]
  The same linewidth effect also makes it difficult to specify the rest mass of unstable, fast-decaying particles in particle physics. The faster the particle decays (the shorter its lifetime), the less certain is its mass (the larger the particle's width).
• For the number of electrons in a superconductor and the phase of its Ginzburg–Landau order parameter[40][41]
  $${\displaystyle \Delta N\,\Delta \varphi \geq 1.}$$

  

Suppose we consider a quantum particle on a ring, where the wave function depends on an angular variable ${\displaystyle \theta }$, which we may take to lie in the interval ${\displaystyle [0,2\pi ]}$. Define "position" and "momentum" operators ${\displaystyle {\hat {A}}}$ and ${\displaystyle {\hat {B}}}$ by

$${\displaystyle {\hat {A}}\psi (\theta )=\theta \psi (\theta ),\quad \theta \in [0,2\pi ],}$$

and

$${\displaystyle {\hat {B}}\psi =-i\hbar {\frac {d\psi }{d\theta }},}$$

where we impose periodic boundary conditions on ${\displaystyle {\hat {B}}}$. The definition of ${\displaystyle {\hat {A}}}$ depends on our choice to have ${\displaystyle \theta }$ range from 0 to ${\displaystyle 2\pi }$. These operators satisfy the usual commutation relations for position and momentum operators, ${\displaystyle [{\hat {A}},{\hat {B}}]=i\hbar }$.[42]

Now let ${\displaystyle \psi }$ be any of the eigenstates of ${\displaystyle {\hat {B}}}$, which are given by ${\displaystyle \psi (\theta )=e^{2\pi in\theta }}$. These states are normalizable, unlike the eigenstates of the momentum operator on the line. Also the operator ${\displaystyle {\hat {A}}}$ is bounded, since ${\displaystyle \theta }$ ranges over a bounded interval. Thus, in the state ${\displaystyle \psi }$, the uncertainty of ${\displaystyle B}$ is zero and the uncertainty of ${\displaystyle A}$ is finite, so that

$${\displaystyle \sigma _{A}\sigma _{B}=0.}$$

Although this result appears to violate the Robertson uncertainty principle, the paradox is resolved when we note that ${\displaystyle \psi }$ is not in the domain of the operator ${\displaystyle {\hat {B}}{\hat {A}}}$, since multiplication by ${\displaystyle \theta }$ disrupts the periodic boundary conditions imposed on ${\displaystyle {\hat {B}}}$.[26] Thus, the derivation of the Robertson relation, which requires ${\displaystyle {\hat {A}}{\hat {B}}\psi }$ and ${\displaystyle {\hat {B}}{\hat {A}}\psi }$ to be defined, does not apply. (These also furnish an example of operators satisfying the canonical commutation relations but not the Weyl relations.[43])

For the usual position and momentum operators ${\displaystyle {\hat {X}}}$ and ${\displaystyle {\hat {P}}}$ on the real line, no such counterexamples can occur. As long as ${\displaystyle \sigma _{x}}$ and ${\displaystyle \sigma _{p}}$ are defined in the state ${\displaystyle \psi }$, the Heisenberg uncertainty principle holds, even if ${\displaystyle \psi }$ fails to be in the domain of ${\displaystyle {\hat {X}}{\hat {P}}}$ or of ${\displaystyle {\hat {P}}{\hat {X}}}$.[44]

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[6] is saturated for the ground state n=0, for which the probability density is just the normal distribution.

Position (blue) and momentum (red) probability densities for an initial Gaussian distribution. From top to bottom, the animations show the cases Ω=ω, Ω=2ω, and Ω=ω/2. Note the tradeoff between the widths of the distributions.

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 σ². 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) .

$${\displaystyle \sigma _{x}\sigma _{p}\geq {\frac {\hbar }{4}}{\sqrt {3+{\frac {1}{2}}\left({\frac {\Omega ^{2}}{\omega ^{2}}}+{\frac {\omega ^{2}}{\Omega ^{2}}}\right)-\left({\frac {1}{2}}\left({\frac {\Omega ^{2}}{\omega ^{2}}}+{\frac {\omega ^{2}}{\Omega ^{2}}}\right)-1\right)}}={\frac {\hbar }{2}}.}$$

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,

$${\displaystyle \sigma _{x}^{2}={\frac {\hbar }{2m\omega }},}$$