Minimising for Lambda; Differentiation?

[RadonX]

Hi, so first post on this forum so I hope I'm doing everything good-as-gold!

I've been twisting my head around a derivation of the uncertainty principle and I'm a little stumped on something that I feel I really should know. Hope someone would like to help.

Certain things are defined at the start,
[tex]\hat{Q}'=\hat{Q}-\left\langle\hat{Q}\right\rangle[/tex]
[tex]\hat{R}'=\hat{R}-\left\langle\hat{R}\right\rangle[/tex]
[tex]\phi(x)=(\hat{Q}'+\textit{i}\lambda\hat{R}')\psi(x)[/tex]
[tex]I(\lambda)=\intdx\phi^{*}(x)\phi(x)\geq0[/tex]

I've followed the manipulation of [tex]I(\lambda)[/tex] all the way to the following line;
[tex]I(\lambda)=(\Delta Q)^{2}+\lambda^{2}(\Delta R)^{2}+\textit{i}\lambda\left\langle\left[\hat{Q},\hat{R}\right]\right\rangle\geq0[/tex]

That's all fine. In the notes he then says "Minimise for [tex]\lambda[/tex]" following it with;
[tex]2 \lambda(\Delta R)^{2}+\textit{i}\left\langle\left[\hat{Q},\hat{R}\right]\right\rangle=0[/tex]

He then subs it back into [tex]I(\lambda)[/tex]
It looks like he's differentiated wrt lambda. But I'm not entirely sure what he means when he says minimise for lambda.
Also, I haven't seen anywhere where 'I' has been defined. Not sure what it is!
Has anyone got any ideas?

MUCH appreciated!

 

[eljose79]

Hi there! It's great to see someone delving into the uncertainty principle and trying to understand it at a deeper level. I'm happy to help you with your question.

First off, I is defined as the integral of the squared wavefunction, which is always greater than or equal to zero. This is a fundamental property of probability and ensures that the wavefunction is normalized.

Now, when your notes say to "minimise for lambda", they mean to find the value of lambda that minimizes the expression for I. In other words, they are looking for the smallest possible value of I. This is because the uncertainty principle states that the product of the uncertainties in position and momentum (or any two non-commuting observables) must be greater than or equal to a certain value, which is given by the expression for I.

To find this minimum value, we can take the derivative of I with respect to lambda and set it equal to zero. This is what your notes are doing in the next step. By setting the derivative equal to zero, we can solve for the value of lambda that minimizes I.

I hope this helps clarify things for you. Keep exploring and questioning, that's what science is all about!