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

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.