Question about Braket notation

  • #1

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
Pengwuino
Gold Member
4,989
15
In an orthonormal basis, yes, [tex]\left\langle {r}
\mathrel{\left | {\vphantom {r i}}
\right. \kern-\nulldelimiterspace}
{i} \right\rangle = \delta _{ri} [/tex].

However, what you wrote, [tex]\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 [/tex] 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
Yes sorry, that was a typo on my part. That first j was supposed to be an i.
 
  • #4
Pengwuino
Gold Member
4,989
15
Ok well, to answer your question, what you have with <r|j><a|j> is S_rj*S_aj.
 
  • #5
Ok, now, is it possible for <r|i><a|j> = <r|h|a> ?
 
  • #6
Pengwuino
Gold Member
4,989
15
H is your Hamiltonian i assume? I can't imagine you'll get that manipulation ever if it is.
 
Last edited:
  • #7
Vanadium 50
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Ok, now, is it possible for <r|i><a|j> = <r|h|a> ?
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
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
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> ?

Well, if |i> and |j> form a complete set of basis vectors (and i believe they do in this case), then the above expression reduces to this,
E<i|h|j>[<r|i><a|j> - <r|j><a|i>] = <r|h|a> - <a|h|r> = 2*iota*Im( <r|h|a> )
where Im() represents the imaginary part.
 

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