# Quantum Question

1. Mar 8, 2009

### latentcorpse

Just reading through my notes and found a step I can't follow:

If you look at p3 of the follwing document, one of the lines in the proof of the Generalised Uncertainty Principle has a (2) next to it. I can't get from the line before it to that line.

Can anyone help me out?

http://www.ph.ed.ac.uk/teaching/course-notes/documents/64/786-lecture5.pdf

thanks

2. Mar 8, 2009

### gabbagabbahey

Use, equation (1) on that page to show that $$[\hat{X},\hat{Y}]=[\hat{A},\hat{B}]$$, then use the definition of expectation value

3. Mar 8, 2009

### latentcorpse

hey. thanks. i have another question about it though.

in the line before (2), how does the $(\Delta \hat{A}_t)^2$ term work? Surely when we square $\hat{X}$, we get $(\Delta \hat{A}_t)^2$ as well as other stuff arising from the cross terms???

also in the very last line, where does he get the $i$ in the RHS from - i'm assuming it's so that we end up with a $\geq$ not a $\leq$ but i don't follow it...

thanks

4. Mar 8, 2009

### gabbagabbahey

By definition, $$(\Delta \hat{A}_t)^2=(\hat{A}-\langle\hat{A}\rangle_t)^2=\hat{X}^2$$

And the $i$ is just a way to account for the negative sign since $$\langle i [\hat{A},\hat{B}]\rangle_t^2=i^2\langle [\hat{A},\hat{B}]\rangle_t^2=-\langle[\hat{A},\hat{B}]\rangle_t^2$$

5. Mar 8, 2009

### latentcorpse

but $\Delta \hat{A}_t = \sqrt{\langle \hat{A^2}_t \rangle - \langle \hat{A}_t \rangle^2}$

so why do you get what you've written?

6. Mar 8, 2009

### gabbagabbahey

The two definitions are equivalent:

$$(\hat{A}-\langle\hat{A}\rangle_t)^2=\hat{A}^2-2\hat{A}\langle\hat{A}\rangle_t+\langle\hat{A}\rangle_t^2$$

$$\implies \int_{-\infty}^{\infty} \Psi^{*}(x,t)(\hat{A}-\langle\hat{A}\rangle_t)^2\Psi(x,t)dx=\int_{-\infty}^{\infty} \Psi^{*}(x,t)\hat{A}^2\Psi(x,t)dx-2\langle\hat{A}\rangle_t \int_{-\infty}^{\infty} \Psi^{*}(x,t)\hat{A}\Psi(x,t)dx+\langle\hat{A}\rangle_t^2\int_{-\infty}^{\infty} \Psi^{*}(x,t)\Psi(x,t)dx$$

$$=\langle\hat{A}^2\rangle_t-2\langle\hat{A}\rangle_t^2+\langle\hat{A}\rangle_t^2=\langle\hat{A}^2\rangle_t-\langle\hat{A}\rangle_t^2=(\Delta\hat{A}_t)^2$$

$$\implies (\Delta\hat{A}_t)^2=(\hat{A}-\langle\hat{A}\rangle_t)^2$$

since they both integrate to the same thing