Prove: $(\Delta x)(\Delta \lambda) \geq \frac{\lambda^2}{4\pi}$ Currently I was going through the formula
$$(\Delta x)(\Delta p)\geq\frac{h}{4\pi}$$
which is of course the enclosed form of Heisenberg’s Uncertainty Principle. But I also get this formula
$$(\Delta x)(\Delta \lambda)\geq\frac{\lambda^2}{4\pi}.$$
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.
 A: Starting from the general uncertainty principle for two arbitrary operators $\Omega$ and $\Phi$ $$(\Delta\Omega)^2(\Delta\Phi)^2\geq\frac{1}{4}\left|\langle \psi | [\Omega,\Phi] | \psi\rangle\right|^2$$
where $|\psi\rangle$ is an arbitrary state and $[\Omega,\Phi]=\Omega\Phi-\Phi\Omega$, the commutator. We know from De Broglie that $\lambda=\frac{h}{p}$; from this, we calculate the Possion bracket of $x$ and $\lambda$ as
$$\begin{align}\{\lambda,x\}&=\frac{\partial \lambda}{\partial x}\frac{\partial x}{\partial p}-\frac{\partial x}{\partial x}\frac{\partial\lambda}{\partial p}\\
&=\frac{h}{p^2}\\
&=\frac{\lambda^2}{h}\end{align}$$
Now, Dirac's popular rule for quantization (works in many cases) is to say that $$\{\omega,\phi\}=\frac{2\pi}{i h}[\Omega,\Phi]$$ so we have $$[\Lambda, X]=i\frac{\Lambda^2}{2\pi}$$. Plugging this into the original general uncertainty principle,
$$(\Delta\Lambda)^2(\Delta X)^2\geq\frac{\langle \Lambda^2 \rangle^2}{16\pi^2}$$
and thus $$(\Delta\Lambda)(\Delta X)\geq\frac{\langle \Lambda^2 \rangle}{4\pi}$$
Now, from the definition of standard deviation, we know that $(\Delta X )^2+\langle X\rangle^2=\langle X^2\rangle$, so we have
$$\begin{align}
(\Delta\Lambda)(\Delta X)&\geq\frac{\langle \Lambda^2 \rangle}{4\pi}\\
&\geq\frac{1}{4\pi}((\Delta\Lambda)^2+\langle\Lambda\rangle^2) \\
&\geq\frac{\langle\Lambda\rangle^2}{4\pi}
\end{align}
$$
because standard deviation is positive definite. Thus, denoting the expectation value of $\Lambda$ as $\lambda$, we finally have
$$\bbox[5px,border:2px solid black]{(\Delta\Lambda)(\Delta X)\geq\frac{\lambda^2}{4\pi}}$$
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 $4\pi$. Thus, it relates the uncertainties in position and wavelength to the mean value of the wavelength.
