Quantum mechanics MOC

Uncertainty principle

For any observables 𝐴 and 𝐵,

𝜎2𝐴𝜎2𝐵≥(12𝑖⟨[ˆ𝐴,ˆ𝐵]⟩)2

which is the generalized uncertainty principle.

Proof of inequality

Let ˆ𝐴 and ˆ𝐵 be Hermitian operators corresponding to observables. Define

|𝑓⟩=(ˆ𝐴−⟨𝐴⟩)|𝜓⟩|𝑔⟩=(ˆ𝐵−⟨𝐵⟩)|𝜓⟩

so that 𝜎2𝐴 =⟨𝑓|𝑓⟩ and 𝜎2𝐵 =⟨𝑔|𝑔⟩. Then by the Cauchy-Schwarz inequality

𝜎2𝐴𝜎2𝐵=⟨𝑓|𝑓⟩⟨𝑔|𝑔⟩≥|⟨𝑓|𝑔⟩|2≥ℑ(⟨𝑓|𝑔⟩)2=(12𝑖(⟨𝑓|𝑔⟩−⟨𝑔|𝑓⟩))2

Now

⟨𝑓|𝑔⟩=⟨𝜓|(ˆ𝐴−⟨𝐴⟩)(ˆ𝐵−⟨𝐵⟩)|𝜓⟩=⟨𝜓|(ˆ𝐴ˆ𝐵−ˆ𝐴⟨𝐵⟩−⟨𝐴⟩ˆ𝐵+⟨𝐴⟩⟨𝐵⟩)|𝜓⟩=⟨𝜓|ˆ𝐴ˆ𝐵|𝜓⟩−⟨𝐴⟩⟨𝐵⟩−⟨𝐴⟩⟨𝐵⟩+⟨𝐴⟩⟨𝐵⟩=⟨ˆ𝐴ˆ𝐵⟩−⟨𝐴⟩⟨𝐵⟩

and by similar calculation ⟨𝑔|𝑓⟩ =⟨ˆ𝐵ˆ𝐴⟩ −⟨𝐵⟩⟨𝐴⟩ Hence

𝜎2𝐴𝜎2𝐵≥(12𝑖(⟨ˆ𝐴ˆ𝐵⟩−⟨ˆ𝐵ˆ𝐴⟩))2=(12𝑖⟨[ˆ𝐴,ˆ𝐵]⟩)2

as claimed.

Minimum uncertainty wavepacket

In order for the inequalities above to become equalities, we require |𝑔⟩ =𝑐|𝑓⟩ and ℜ(⟨𝑓|𝑔⟩) =ℜ(𝑐⟨𝑓|𝑓⟩) =0, implying ℜ(𝑐) =0. Thus a minimum uncertainty |𝜓⟩ has

|𝑔⟩=𝑖𝑎|𝑓⟩(ˆ𝐵−⟨ˆ𝐵⟩)|𝜓⟩=𝑖𝑎(ˆ𝐴−⟨ˆ𝐴⟩)|𝜓⟩

with 𝑎 real. For ˆ𝐴 =ˆ𝑥 and ˆ𝐵 =ˆ𝑝 in 1D position-space we get a Gaußian¹

𝜓(𝑥)=𝐴exp⁡(−𝑎(𝑥−⟨𝑥⟩)2)2ℏ)exp⁡(𝑖⟨𝑝⟩𝑥ℏ)

in which case 𝜎𝑥𝜎𝑝 =ℏ2. Hence the uncertainty principle is correct, giving the greatest lower bound on the uncertainty of arbitrary observables.

Intuition: The uncertainty principle for classical waves

Consider a classical wave, e.g. a sinusoidal wave traveling down a long rope. It is difficult to say where the wave is, since it is distributed throughout the rope, but it is possible to give the wavelength. Now consider a localized bump travelling down the rope. In this case, it is possible to describe the position of the bump, but giving its wavelength is difficult, since it has no periodicity. Hence we see that the more precise a wave’s position is, the less precise its wavelength, and vice versa. In quantum mechanics, wavelengths correspond to momenta.¹

Heisenberg uncertainty princple

The Heisenberg uncertainty principle is the special case of

𝜎2𝑥𝜎2𝑝≥ℏ2

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Footnotes

1. 2018. Introduction to Quantum Mechanics, §3.5.2, pp. 108–109 ↩ ↩²