# Ladder operators and matrix elements..

1. Sep 26, 2015

### Activeuser

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.

2. Sep 27, 2015

### 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"?

3. Sep 27, 2015

### 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..

4. Sep 27, 2015

### stevendaryl

Staff Emeritus
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.

5. Sep 27, 2015

### 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, cuz this what I miss and can not find in books; it seems like a bases that I do not have.

6. Sep 27, 2015

### stevendaryl

Staff Emeritus
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:

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

7. Sep 27, 2015

### stevendaryl

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

8. Sep 27, 2015

### Activeuser

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