structural:summarization, quality:correction
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 k or momentum p is ψ (x) ∝ e i k 0 x = e i p 0 x / ℏ. {\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 P = ∫ a b | ψ (x) | 2 d x. {\displaystyle \operatorname {P} =\int _{a}^{b}|\psi (x)|^{2}\,\mathrm {d} x~.}

In the case of the single-mode plane wave, | ψ (x) | 2 {\displaystyle |\psi (x)|^{2}} is 1 if X = x {\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 ψ (x) ∝ ∑ n A n e i p n x / ℏ , {\displaystyle \psi (x)\propto \sum _{n}A_{n}e^{ip_{n}x/\hbar }~,} where A represents the relative contribution of the mode p 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 ψ (x) = 1 2 π ℏ ∫ − ∞ ∞ φ (p) ⋅ e i p x / ℏ d p , {\displaystyle \psi (x)={\frac {1}{\sqrt {2\pi \hbar }}}\int _{-\infty }^{\infty }\varphi (p)\cdot e^{ipx/\hbar }\,dp~,} with φ (p) {\displaystyle \varphi (p)} representing the amplitude of these modes and is called the wave function in momentum space. In mathematical terms, we say that φ (p) {\displaystyle \varphi (p)} is the Fourier transform of ψ (x) {\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 | ψ (x) | 2 {\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, by using many plane waves, thereby weakening the precision of the momentum, i.e. increased σ. Another way of stating this is that σ and σ 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.

We are interested in the variances of position and momentum, defined as σ x 2 = ∫ − ∞ ∞ x 2 ⋅ | ψ (x) | 2 d x − (∫ − ∞ ∞ x ⋅ | ψ (x) | 2 d x) 2 {\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}} σ p 2 = ∫ − ∞ ∞ p 2 ⋅ | φ (p) | 2 d p − (∫ − ∞ ∞ p ⋅ | φ (p) | 2 d p) 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 σ x 2 = ∫ − ∞ ∞ x 2 ⋅ | ψ (x) | 2 d x {\displaystyle \sigma _{x}^{2}=\int _{-\infty }^{\infty }x^{2}\cdot |\psi (x)|^{2}\,dx} σ p 2 = ∫ − ∞ ∞ p 2 ⋅ | φ (p) | 2 d p. {\displaystyle \sigma _{p}^{2}=\int _{-\infty }^{\infty }p^{2}\cdot |\varphi (p)|^{2}\,dp~.}

The function f (x) = x ⋅ ψ (x) {\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: ⟨ u ∣ v ⟩ = ∫ − ∞ ∞ u ∗ (x) ⋅ v (x) d x, {\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 σ x 2 = ∫ − ∞ ∞ | f (x) | 2 d x = ⟨ f ∣ f ⟩. {\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 g ~ (p) = p ⋅ φ (p) {\displaystyle {\tilde {g}}(p)=p\cdot \varphi (p)} as a vector, but we can also take advantage of the fact that ψ (x) {\displaystyle \psi (x)} and φ (p) {\displaystyle \varphi (p)} are Fourier transforms of each other. We evaluate the inverse Fourier transform through integration by parts: g (x) = 1 2 π ℏ ⋅ ∫ − ∞ ∞ g ~ (p) ⋅ e i p x / ℏ d p = 1 2 π ℏ ∫ − ∞ ∞ p ⋅ φ (p) ⋅ e i p x / ℏ d p = 1 2 π ℏ ∫ − ∞ ∞ ⋅ e i p x / ℏ d p = i 2 π ∫ − ∞ ∞ ⋅ e i p x / ℏ d p = − i 2 π ∫ − ∞ ∞ ∫ − ∞ ∞ d ψ (χ) d χ e − i p χ / ℏ d χ e i p x / ℏ d p = − i ℏ 2 π ∫ − ∞ ∞ d ψ (χ) d χ e − i p χ / ℏ d χ = (− i ℏ d d x) ⋅ ψ (x), {\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\\&={\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\\&=-i{\sqrt {\frac {\hbar }{2\pi }}}\int _{-\infty }^{\infty }{\frac {d\psi (\chi)}{d\chi }}e^{-ip\chi /\hbar }\,d\chi \\&=\left(-i\hbar {\frac {d}{dx}}\right)\cdot \psi (x),\end{aligned}}} where v = ℏ − i p e − i p χ / ℏ {\displaystyle v={\frac {\hbar }{-ip}}e^{-ip\chi /\hbar }} in the integration by parts, the cancelled term vanishes because the wave function vanishes at infinity, and the final two integrations re-assert the Fourier transforms. Often the term − i ℏ d d x {\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 σ p 2 = ∫ − ∞ ∞ | g ~ (p) | 2 d p = ∫ − ∞ ∞ | g (x) | 2 d x = ⟨ g ∣ g ⟩. {\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 σ x 2 σ p 2 = ⟨ f ∣ f ⟩ ⋅ ⟨ g ∣ g ⟩ ≥ | ⟨ f ∣ g ⟩ | 2. {\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 | z | 2 = (Re (z)) 2 + (Im (z)) 2 ≥ (Im (z)) 2 = (z − z ∗ 2 i) 2. {\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 z = ⟨ f | g ⟩ {\displaystyle z=\langle f|g\rangle } and z ∗ = ⟨ g ∣ f ⟩ {\displaystyle z^{*}=\langle g\mid f\rangle } and substitute these into the equation above to get | ⟨ f ∣ g ⟩ | 2 ≥ (⟨ f ∣ g ⟩ − ⟨ g ∣ f ⟩ 2 i) 2. {\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.

⟨ f ∣ g ⟩ − ⟨ g ∣ f ⟩ = ∫ − ∞ ∞ ψ ∗ (x) x ⋅ (− i ℏ d d x) ψ (x) d x − ∫ − ∞ ∞ ψ ∗ (x) (− i ℏ d d x) ⋅ x ψ (x) d x = i ℏ ⋅ ∫ − ∞ ∞ ψ ∗ (x) d x = i ℏ ⋅ ∫ − ∞ ∞ ψ ∗ (x) d x = i ℏ ⋅ ∫ − ∞ ∞ ψ ∗ (x) ψ (x) d x = i ℏ ⋅ ∫ − ∞ ∞ | ψ (x) | 2 d x = i ℏ {\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 σ x 2 σ p 2 ≥ | ⟨ f ∣ g ⟩ | 2 ≥ (⟨ f ∣ g ⟩ − ⟨ g ∣ f ⟩ 2 i) 2 = (i ℏ 2 i) 2 = ℏ 2 4 {\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 σ x σ p ≥ ℏ 2. {\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 ψ (x) {\displaystyle \psi (x)} and φ (p) {\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.