Question about Braket notation

1. Apr 23, 2009

chemstudent09

Ok, here is my question.

When you have < r | i >, this equals Sri. So logically if that is that case, if you had SriSaj this would equal < r | j >< a | j >, right?

If so, then what does < r | j ><a | j > equal? I'm working a problem where I am trying to get a final answer of < r | h | a >. Is that even possible from what I have here?

By the way, S = Kronecker delta.

2. Apr 23, 2009

Pengwuino

In an orthonormal basis, yes, $$\left\langle {r} \mathrel{\left | {\vphantom {r i}} \right. \kern-\nulldelimiterspace} {i} \right\rangle = \delta _{ri}$$.

However, what you wrote, $$\delta _{ri} \delta _{aj} = \left\langle {r} \mathrel{\left | {\vphantom {r i}} \right. \kern-\nulldelimiterspace} {i} \right\rangle \left\langle {a} \mathrel{\left | {\vphantom {a j}} \right. \kern-\nulldelimiterspace} {j} \right\rangle$$ is not true. I suspect a typo? Did you mean to put 'i' instead of the first 'j'?

EDIT: Ok until latex is back up, the first equation is <r|i> = S_ri. The second is S_ri*S_aj = <r|i><a|j>.

3. Apr 23, 2009

chemstudent09

Yes sorry, that was a typo on my part. That first j was supposed to be an i.

4. Apr 23, 2009

Pengwuino

Ok well, to answer your question, what you have with <r|j><a|j> is S_rj*S_aj.

5. Apr 23, 2009

chemstudent09

Ok, now, is it possible for <r|i><a|j> = <r|h|a> ?

6. Apr 24, 2009

Pengwuino

H is your Hamiltonian i assume? I can't imagine you'll get that manipulation ever if it is.

Last edited: Apr 24, 2009
7. Apr 24, 2009

Staff Emeritus
Those don't have the same units. The left hand side is a pure number, and the right hand side has dimensions of energy.

8. Apr 24, 2009

chemstudent09

Ok, I messed up in my problem. I have my problem worked down to this:

(I have to make E = Capital sigma (summation) because the LaTeX isn't working for some reason.)

E<i|h|j>[<r|i><a|j> - <r|j><a|i>]

(Sigma is summed over ij, I just don't know how to make it show on here)

Now is it possible for this to equal <r|h|a> ? In other words:

E<i|h|j>[<r|i><a|j> - <r|j><a|i>] = <r|h|a> ?

9. Apr 24, 2009