I've been working on some quantum information theory problems and I've revisited Griffith's Quantum Mechanics. On page 109, he derives the uncertainty principle. He goes through the steps:
- For operators $\hat{A}$ and $\hat{B}$, defines : $|f\rangle = ( \hat{A} - \langle \hat{A}\rangle)|\psi\rangle$ and $|g\rangle = (\hat{B} - \langle \hat{B}\rangle)|\psi\rangle$
- Defines variance : $\sigma^{2}_{A} = \langle f | f\rangle$ and $\sigma^{2}_{B} = \langle g | g\rangle$
- Invokes Cauchy-Schwarz inequality: $\sigma^{2}_{A} \sigma^{2}_{B} = \langle f | f \rangle \langle g | g \rangle \geq |\langle f | g \rangle|^{2}$
- Defines $z$ as a complex number : $z = \langle f | g \rangle $
- Utilizes the magnitude of $z$ and discarding the real component (see equation 3.136) : $|z|^{2} = (\text{Re}(z))^{2} + (\text{Im}(z))^{2} \geq (\text{Im}(z))^{2} = [\frac{1}{2i}(z-z^{*}]^{2}$
- He keeps only the imaginary component and plugs the result from equation 3.136 (step 5) on the right hand side of the Cauchy-Schwarz inequality (step 3). I realize that by discarding the real component he his not violating his inequality in equation 3.135 (step 3), so he can technically do this. Doing additional commutator math, he gets: $\sigma_A^2\sigma_B^2 \geq \left|\frac{1}{2i} \langle[ \hat{A}, \hat{B}] \rangle\right|^2$
Question:
- In the Griffith's derivation, why did he discard the Real component? It seems by doing so you now no longer need to know anything about the wave function. I.e. you don't need to calculate $\langle \psi | \hat{A} | \psi \rangle = \langle \hat{A} \rangle$)? This would make the math easier from this aspect.
If you kept the real component, it seems like you'd have a stronger inequality. In fact Wikipedia gives such a derivation yielding : $\sigma_A^2\sigma_B^2 \geq \left| \frac{1}{2}\langle\{\hat{A}, \hat{B}\}\rangle - \langle \hat{A} \rangle\langle \hat{B}\rangle \right|^2+ \left|\frac{1}{2i} \langle[ \hat{A}, \hat{B}] \rangle\right|^2$ )