Ladder Operators - I need help.

Pmb

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Doesn't this arise because the commutator of {a+a-} = 1 ?
 

Pmb

PHF Hall of Fame
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Doesn't this arise because the commutator of {a+a-} = 1 ?
Not that I could see.

FYI: [A, B] is the commutator of A and B while {A, B} is he anticommutator.
 
Apr 2015
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Somerset, England
I downloaded your scans, but haven't had time to wade through them yet to get the hang of Griffith's notation. (I don't have his QM book)
Hopefully I'll be able to do that tomorrow.

As to brackets, let's say it's that shift key at work again.

The number of times I accidentally hit the bloody capslock which is in the wrong place on my laptop and end up with a line of typing I have to redo as it is in the wrong case.
 
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Pmb

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I can place Griffiths text on my website if you want. When you download it let me to so I can delete it. Its the newer second edition too! :)

You should see the massive physics library I have. It takes 2 DVDs to put them all on. I'll mail you a copy if you'd like.
 
Oct 2017
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Can you scan page 47 please?
 

Pmb

PHF Hall of Fame
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Can you scan page 47 please?
Whoops! :D

Done. A friend of mine sent me a description of what's going on. I'll figure it out after I wake up. :)
 
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Oct 2017
536
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Glasgow
Thanks for the extra page.

I think it's easier than it looks:

Eq. 2.57:

\(\displaystyle \hbar \omega \left(a_{\pm}a_{\mp} \pm \frac{1}{2}\right) \psi = E \psi\)

Let \(\displaystyle \psi = \psi_n\) and \(\displaystyle E = E_n = \left(n + \frac{1}{2}\right) \hbar \omega\) from Eq. 2.61.

Therefore

\(\displaystyle \hbar\omega \left(a_{\pm}a_{\mp} \pm \frac{1}{2}\right) \psi_n = E_n \psi_n = \left(n + \frac{1}{2}\right)\hbar\omega \psi_n\)

Cancel \(\displaystyle \hbar \omega\) and expand brackets:

\(\displaystyle a_{\pm}a_{\mp}\psi_n \pm \frac{1}{2}\psi_n = n \psi_n + \frac{1}{2}\psi_n\)

\(\displaystyle \mp\) the term \(\displaystyle \pm \frac{1}{2} \psi_n\) on both sides then yields

\(\displaystyle a_{\pm}a_{\mp}\psi_n = \left(n + \frac{1}{2} \mp \frac{1}{2}\right) \psi_n\)

which expressed separately yields

\(\displaystyle a_{+}a_{-}\psi_n = n \psi_n\) and \(\displaystyle a_{-}a_{+}\psi_n = \left(n+1\right) \psi_n \)
 
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Pmb

PHF Hall of Fame
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Thanks for the extra page.

I think it's easier than it looks:
Thank you very much. Yep. That's what my friend showed me. He added that Griffith wasn't his usual clear self and that's how I got confused. In this particular case it appeared to me from his writing that Eq. (26,5)) followed fom what was before it. And my friend teaches QM at MIT using this text.

I wanted to slap myself once I realized how easy it was.

Thanks for you help. :)
 
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Apr 2015
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Somerset, England
This 'ladder technique' is often presented in QM as something new and cunning but it is really quite old and was originated by messers Newton and Gregory several hundred years ago.

It also goes under the name the shift operator or the displacement operator.
As such it was originally developed as a result of Newton's study of finite differences.

If you think about it is is not so suprising that step up followed by step down leaves you back where you started so if A is the upshift and B is the downshift then

BA = 1.

However AB is more tricky because it depends upon where you start since there is no lower level to descend to if you start with a downshift.

About the Griffiths book

I note it is being offered at reasonable price by lots of vendors, but there seems to be some discrepancies as to whether they are offereing a hardback or paperback, whether the edition is first or second (what's the date of the second please?) or some 'student version' and what the difference between the European and US versions are.

Comments appreciated as I might buy one copy.
 
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