Prove: $(Delta x)(Delta lambda) geq frac{lambda^2}{4pi}$

Physics Asked by user276504 on May 8, 2021

Currently I was going through the formula

$$(Delta x)(Delta p)geqfrac{h}{4pi}$$

which is of course the enclosed form of Heisenberg’s Uncertainty Principle. But I also get this formula

$$(Delta x)(Delta lambda)geqfrac{lambda^2}{4pi}.$$

I suppose this is an extension of Heisenberg’s principle but I didn’t get such note in the book. Further I think it maybe came from de Broglie’s hypothesis and may be some calculus but it’s just a guess. So for these reasons I am wishing an answer from you and moreover it will be finer to have some physical interpretation of this formula like as we have for Heisenberg’s principle.

One Answer

Starting from the general uncertainty principle for two arbitrary operators $Omega$ and $Phi$ $$(DeltaOmega)^2(DeltaPhi)^2geqfrac{1}{4}left|langle psi | [Omega,Phi] | psirangleright|^2$$ where $|psirangle$ is an arbitrary state and $[Omega,Phi]=OmegaPhi-PhiOmega$, the commutator. We know from De Broglie that $lambda=frac{h}{p}$; from this, we calculate the Possion bracket of $x$ and $lambda$ as $$begin{align}{lambda,x}&=frac{partial lambda}{partial x}frac{partial x}{partial p}-frac{partial x}{partial x}frac{partiallambda}{partial p} &=frac{h}{p^2} &=frac{lambda^2}{h}end{align}$$ Now, Dirac's popular rule for quantization (works in many cases) is to say that $${omega,phi}=frac{2pi}{i h}[Omega,Phi]$$ so we have $$[Lambda, X]=ifrac{Lambda^2}{2pi}$$. Plugging this into the original general uncertainty principle, $$(DeltaLambda)^2(Delta X)^2geqfrac{langle Lambda^2 rangle^2}{16pi^2}$$ and thus $$(DeltaLambda)(Delta X)geqfrac{langle Lambda^2 rangle}{4pi}$$ Now, from the definition of standard deviation, we know that $(Delta X )^2+langle Xrangle^2=langle X^2rangle$, so we have $$begin{align} (DeltaLambda)(Delta X)&geqfrac{langle Lambda^2 rangle}{4pi} &geqfrac{1}{4pi}((DeltaLambda)^2+langleLambdarangle^2) &geqfrac{langleLambdarangle^2}{4pi} end{align} $$ because standard deviation is positive definite. Thus, denoting the expectation value of $Lambda$ as $lambda$, we finally have $$bbox[5px,border:2px solid black]{(DeltaLambda)(Delta X)geqfrac{lambda^2}{4pi}}$$ This uncertainty princple is different than the normal position-momentum one; instead of the product of uncertainties always being greater than a constant, the product of the standard deviation of position and wavelength is greater than or equal to the mean value of the wavelength divided by $4pi$. Thus, it relates the uncertainties in position and wavelength to the mean value of the wavelength.

Answered by John Dumancic on May 8, 2021

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