Ladder operators and matrix elements...

[Activeuser]

Please I need your help in such problems..
in terms of ladder operators to simplify the calculation of matrix elements... calculate those
i) <u+2|P²|u>
ii) <u+1| X³|u>
If u is different in both sides, then the value is 0? is it right it is 0 fir both i and ii?
when exactly equals 0, please explain slowly, my background is chemistry and not physics..
Thank you.

 

[strangerep]

You'll need to supply a bit more context...

I guess P and X are momentum and position operators respectively? But what is your "u"?

 

[Activeuser]

Well, in terms of ladder operators, X= (h/2mw)^1/2(a+a⁺) you can say = constant (a+a⁺) it is a real
P=i(hwm/2)^1/2 (a-a⁺) or constant (a-a⁺)
about u and u+1 are functions; some books write them like this and others write PSI or PHI of subscript u and u+1;I couldn't type that way as there is no symbols here.
a⁺a=n occupation number operator , [a,a⁺]=1

If needed I can give an example I know..

 

[stevendaryl]

Well, in terms of ladder operators, X= (h/2mw)^1/2(a+a⁺) you can say = constant (a+a⁺) it is a real
P=i(hwm/2)^1/2 (a-a⁺) or constant (a-a⁺)
about u and u+1 are functions; some books write them like this and others write PSI or PHI of subscript u and u+1;I couldn't type that way as there is no symbols here.
a⁺a=n occupation number operator , [a,a⁺]=1

If needed I can give an example I know..

I'm assuming that |u\rangle means the state such that the number operator N = a^\dagger a has value u? In that case, it's completely straight-forward, if tedious, to compute expressions such as

\langle u+2 | P^2 | u \rangle

Just write P in terms of a and a^\dagger, and apply the rules:

a |u\rangle = \sqrt{u} |u - 1\rangle
a^\dagger |u\rangle = \sqrt{u+1} |u+1\rangle
\langle n | m \rangle = 0 if n \neq m
\langle n | n \rangle = 1

So P^2 = A (a^2 - a a^\dagger - a^\dagger a + (a^\dagger)^2) (for some constant A)
P^2 | u \rangle = A (a^2 |u \rangle - a a^\dagger |u\rangle - a^\dagger a |u\rangle + (a^\dagger)^2 |u\rangle)

Then you just work out what a^2 |u\rangle is, etc.

 

[Activeuser]

ok great. I know till this step, my question is to separate these operations and get the final out put like this
<u+2|aa|u> - <u+2|aa⁺|u> - <u+1|a⁺a|u>... so on
and solve each one on its own, when I get any different matrix output, the term equals 0; like <u+1|u> or <u+2|u-1>... right? Cuz my confusion is about this part.

the other part is how to work a³,, if aaa, the 1st one from left operates with the left , and the 1st a from right operates with the right function of the matrix, the middle a operator works with which function.. as it gives different values..
I am sorry for my silly questions, because this what I miss and can not find in books; it seems like a bases that I do not have.

 

[stevendaryl]

Well, just look at the term

\langle u+2|a\ a|u \rangle

Use the rule: a |u\rangle = \sqrt{u} |u - 1\rangle. So that term simplifies(?) to:

\langle u+2|a\ a|u \rangle = \sqrt{u} \langle u + 2 | a | u-1\rangle

Use the same rule again: a |u-1\rangle = \sqrt{u-1} |u - 2\rangle. So we have:

\langle u+2|a\ a|u \rangle = \sqrt{u(u-1)}| \langle u + 2 | u-2\rangle

which is zero.

The only term you're going to get a nonzero result for is:

\langle u+2|a^\dagger\ a^\dagger|u \rangle

 

[stevendaryl]

By the way, this should probably be in the Advanced Physics Homework section.

 

[Activeuser]

Thank you so much for this clarification.. it is very helpful.
Next time I will follow the right section.. new here.