Proof of the generalized Uncertainty Principle?

[patric44]

Homework Statement

i am trying to proof the generalized uncertainty principle , and i am stuck at some point

Relevant Equations

(ΔC)^2 = <φ|A^2|φ>

hi guys
i am trying to follow a proof of the generalized uncertainty principle and i am stuck at the last step :

i am not sure why he put these relations in (4.20) :
$$(\Delta\;C)^{2} = \bra{\psi}A^{2}\ket{\psi}$$
$$(\Delta\;D)^{2} = \bra{\psi}B^{2}\ket{\psi}$$
i tried to prove these using the difination of the expectation value and had no success! any help will be appreciated, thanks.

 

[vela]

Inequality (4.20) holds for any two hermitian operators. In particular, it holds for C and D. So you have
$$\lvert \langle [C,D] \rangle \rvert^2 \le 4 \langle C^2 \rangle \langle D^2 \rangle.$$ Then you can use (4.22) and (4.23) to write the righthand side in terms of $\Delta A$ and $\Delta B$. And since [A,B] = [C,D], you end up with
$$\lvert \langle [A,B] \rangle \rvert^2 \le 4 (\Delta A)^2 (\Delta B)^2 .$$ Divide by 4 and take the square root to finish the derivation.

I'm wondering if the author meant to use A and B in (4.24) instead of C and D. There's also a pretty obvious typo in (4.23).

 

[patric44]

Inequality (4.20) holds for any two hermitian operators. In particular, it holds for C and D. So you have
$$\lvert \langle [C,D] \rangle \rvert^2 \le 4 \langle C^2 \rangle \langle D^2 \rangle.$$ Then you can use (4.22) and (4.23) to write the righthand side in terms of $\Delta A$ and $\Delta B$. And since [A,B] = [C,D], you end up with
$$\lvert \langle [A,B] \rangle \rvert^2 \le 4 (\Delta A)^2 (\Delta B)^2 .$$ Divide by 4 and take the square root to finish the derivation.

I'm wondering if the author meant to use A and B in (4.24) instead of C and D. There's also a pretty obvious typo in (4.23).

thanks so much it becomes clear now , i guess the author meant $A,B$