Physics Help Forum Need help with algebra

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 May 18th 2018, 05:55 PM #1 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,573 Need help with algebra This should be easy for me but then again that's why I'm brushing up. I.e. if you don't use it then you loose it. See attached picture. How does he do the algebra after the integration is done to get to the end? Many Thanks!!! Attached Thumbnails
 May 18th 2018, 10:21 PM #2 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,573 my apologies. I forgot that I have to show some work that we've done. I'm doing that now and will post it later. Pete is not above the rules here, This is as far as a got Attached Thumbnails   Last edited by Pmb; May 18th 2018 at 11:15 PM.
 May 19th 2018, 01:50 PM #3 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 995 Very quickly and sorry it's so scruffy, but this might help. Considering only the right hand integral and leaving the 1/(b-a)^2 outside the bracket, Expanding and integrating and then substituting a and b into the result brings out the terms in your third line. Note I have taken the factor of 1/4 outside the bracket which means multiplying the terms inside through by 4. The next step is to sparately do this for the left hand integral and you will need to take out the 1/(b-a)^2 outside the bracket, multiplying the result of substituting 0 and a into the left hand integral. This does indeed lead on to line 4 Please ask if you need me to work this out as well. As a matter of interest this looks like an evaluation of the expectation via the state variable in one dimension, between x= a and x = b. Normally in 3 dimensions the integration variable is radial r. It is not shown in your extract why the modulus of A squared is 3/b so I ahve taken this on trust. Perhaps you would like to explain this? Attached Thumbnails
May 20th 2018, 11:14 AM   #4
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 Originally Posted by studiot Very quickly and sorry it's so scruffy, but this might help. Considering only the right hand integral and leaving the 1/(b-a)^2 outside the bracket, Expanding and integrating and then substituting a and b into the result brings out the terms in your third line. Note I have taken the factor of 1/4 outside the bracket which means multiplying the terms inside through by 4. The next step is to sparately do this for the left hand integral and you will need to take out the 1/(b-a)^2 outside the bracket, multiplying the result of substituting 0 and a into the left hand integral. This does indeed lead on to line 4 Please ask if you need me to work this out as well. As a matter of interest this looks like an evaluation of the expectation via the state variable in one dimension, between x= a and x = b. Normally in 3 dimensions the integration variable is radial r. It is not shown in your extract why the modulus of A squared is 3/b so I ahve taken this on trust. Perhaps you would like to explain this?
I don't see anything you mentioned. Here's what I did and got stuck on. In fact here's he entire problem (it was really an algebra problem from where I sit
Attached Files
 Griffiths.pdf (449.9 KB, 6 views)

Last edited by Pmb; May 20th 2018 at 01:42 PM.

 May 21st 2018, 07:27 AM #5 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 270 The mistake is the sign of the third term on the second to last line. It should be $\displaystyle +\frac{b^4}{4}$. With that I get: $\displaystyle = \frac{A^2 a^2}{4} + \frac{A^2}{(b-a)^2}\left[\frac{b^4}{12} - \frac{b^2 a^2}{2} + \frac{2ba^3}{3} - \frac{a^4}{4}\right]$ The rest drops out. The part at the end to yield $\displaystyle \frac{2a+b}{4}$ is tricky. You can try letting $\displaystyle (c_1 a + c_2 b) (b-a)^2 = (b^3 - 3a^2b + 2a^3)$ and then try and find $\displaystyle c_1$ and $\displaystyle c_2$ by equating coefficients. It turns out they are consistent for all coefficients and the factorisation is possible. Pmb likes this.
 May 21st 2018, 11:00 AM #6 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,573 The last line was the least of my concern. It was going from the their to the last to the second to the last. I thought that the second to the last was due to factoring. I proved to myself otherwise. Thanks for trying. In the future when I request a full derivation I mean the whole enchilada. Last edited by Pmb; May 22nd 2018 at 06:53 AM.
 May 26th 2018, 06:47 AM #7 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,573 I'm amazed at how rusty one's skills can get after being unable to do some work for a few years. It's almost scary. Makes me want to stop but I keep pushing myself. That's the hard part.

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