How Do You Find <r^2> for Uncertainty in Position?

swain1
Messages
30
Reaction score
0
When trying to work out the uncerainty in position of the expectation value I have read that you have to find <r^2> as well as <r>^2. I have worked out the value of 3a/2 for <r> but what do I have to do to find <r^2>. Do I just sqare the whole function before I integrate?
Also as I am integrating I found, in a book a table that had a general form of integrations between 0 and infinity. I used it without giving it much thought but how would you evaluate this? Thanks:confused:
 
Physics news on Phys.org
Yes to find <r^2> you integrate r^2 p(r) dr instead of rp(r)dr. Or for any function of r, it's just int[f(r)p(r)dr].

I'm guessing the table was for the gaussian interal int[exp(-ax^2)] a>0? The integral can't be evaluated directly, you have to do some weird substitutions, if you want to know how to do it I can show you. It's easiest to remember the the soln is sqrt(pi/a), from 0 to infinity is half this, and for x^2n exp(-ax^2), differentiate both sides wrt a.
 
Ok thanks for that. I have done the inegral and worked out the uncertainty. The value I have got is sqrt(3)a/2. It seems like an awful lot to me. Anyway, do you know if this is right. I am doing this for the ground state of hydrogen. If I have got the correct value. Why is it so large?
 
Not entirely sure, what exactly was the integral? And what is a? I haven't done any quantum, not sure what this is about exactly.
 
\Delta r = \sqrt{&lt;r^2&gt; - &lt;r&gt;^2}
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top