# Some questions about derivation of uncertainty principle

In Introduction to Quantum Mechanics by Griffiths and Schroeter, they derive the Uncertainty principle in the following way:

First, they define $$f=\left(\hat A-\langle A\rangle\right)|\Psi\rangle$$ $$g=\left(\hat B-\langle B\rangle\right)|\Psi\rangle$$

Where $$\hat A, \hat B$$ are two Hermitian Operators.

Then they use the definition of standard deviation and Schwarz inequality to show that

$$(\sigma_A\sigma_B)^2\geq\vert\langle f\vert g\rangle\vert^2 \tag{1}$$

Now since in general $$\langle f\vert g\rangle$$ is complex, they define $$z=\langle f\vert g\rangle$$ and use the fact that $$\vert z\vert^2\geq [Im(z)]^2=[(z-z^*)/2i]^2\tag{2}$$

Thus

$$(\sigma_A\sigma_B)^2\geq \left[\frac 1{2i}\left(\langle f\vert g\rangle -\langle g\vert f\rangle\right) \right]\tag{3}$$

Question 1: Since we are interested in the minimum possible value of $$\sigma_A\sigma_B$$ wouldn't it make more sense to talk about (1) rather than (3)? After all, the limit put on $$\sigma_A\sigma_B$$ by (1) is greater than (3).

Question 2: To me, it seems arbitrary, at this stage, to use $$\vert z\vert^2\geq [Im(z)]^2$$ in (2), when one could've just used $$\vert z\vert^2\geq [Re(z)]^2$$. So what's the rationale to choose imaginary over real?

• ...which definition of the standard deviation? Commented Mar 14 at 17:44

It is true that by taking step (2) we are 'unnecessarily' making the uncertainty principle weaker. Furthermore, it's possible to use $$|z|^2=Re(z)^2+Im(z)^2$$ to derive the stronger Robertson-Schrödinger uncertainty principle: $$\sigma_A^2\sigma_B^2\geq \left|\frac{1}{2}\langle \{ \hat{A},\hat{B}\}\rangle-\langle \hat{A} \rangle \langle \hat{B} \rangle \right|^2 + \left|\frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle \right|^2$$ And of course, leaving out the first term, we get the well-known uncertainty principle (3): $$\sigma_A^2\sigma_B^2\geq \left|\frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle \right|^2$$ So why do we generally see the latter over the former? It's largely a matter of practicality. Note that the expected value of any operator $$\hat{O}$$ is $$\langle \hat{O} \rangle=\langle\psi|\hat{O}|\psi\rangle$$, which is dependent on the state $$|\psi \rangle$$. Thus, $$\langle \{ \hat{A},\hat{B}\}\rangle$$, $$\langle \hat{A} \rangle$$ and $$\langle \hat{B} \rangle$$ are generally not known. On the other hand, there are many cases where $$\langle[\hat{A},\hat{B}]\rangle$$ is either the same regardless of $$|\psi\rangle$$, such as: $$\langle[\hat{x},\hat{p}]\rangle=\langle i\hbar \hat{I} \rangle = i\hbar \Longrightarrow \sigma_x\sigma_p\geq \frac{\hbar}{2}$$ or otherwise noteworthy, such as: $$\langle[\hat{L_x},\hat{L_y}]\rangle=\langle i\hbar \hat{L_z} \rangle\Longrightarrow \sigma_{L_x}\sigma_{L_y}\geq \left|\frac{\hbar}{2}\langle L_z \rangle\right|$$ and: $$\langle[\hat{O},\hat{H}]\rangle=i\hbar \frac{d\langle\hat{O}\rangle}{dt} \Longrightarrow \frac{\sigma_{O}}{\left|\frac{d\langle\hat{O}\rangle}{dt}\right|}\sigma_{H}\geq \frac{\hbar}{2}$$
At the end of the day, the uncertainty principle has little utility for doing exact calculations (if you have $$|\psi\rangle$$ you can just explicitly calculate $$\sigma_A$$ and $$\sigma_B$$), but it does provide some insight into the relationship between the observables of a system. In that regard, the term $$\left|\frac{1}{2}\langle \{ \hat{A},\hat{B}\}\rangle-\langle \hat{A} \rangle \langle \hat{B} \rangle \right|^2$$ simply isn't very useful and is therefore left out.