How do we solve for equation 1.8.7 on page 33 of Shankar's QM 2nd Edition?

[bugatti79]

Hi Folks,

How do we arrive at equation 1.8.7 page 33.

We have $<i|\Omega-\omega I|V>=0$, given $I=\Sigma_{j=1}|i><i|$ we can write

$<i|\Omega-\omega \Sigma_{j=1}|i><i||V>=0$

not sure how to proceed from here...

 

[andresB]

You can decompose

Hi Folks,

How do we arrive at equation 1.8.7 page 33.

We have $<i|\Omega-\omega I|V>=0$, given $I=\Sigma_{j=1}|i><i|$ we can write

$<i|\Omega-\omega \Sigma_{j=1}|i><i||V>=0$

not sure how to proceed from here...

You can write your vector V as a linear combination of the basis vectors

##|V>=Sigmav_j|j>##

and then using the fact that the basis vector are orthogonal with each other you can evaluate the bracket, taking into account that ω_ij are the matrix element of Ω in the |i> basis (<i|Ω|j>=ω_ij)

 

[bugatti79]

I can get as far as

$<I| \Omega - \omega \Sigma_{I} |I><I| \Sigma_{j} v_{j} |j>=0$

$\Sigma_{j} \Omega - \omega \Sigma_{I} |I><I| v_{j} \delta_{ij}=0$

How is the identity dealt with?

 

[andresB]

Don't put the identity in the first place, and remember that <i|Ω|j>=Ω_ij (eq 1.6.1) (I wrote it wrong in my last post)

 

[bugatti79]

I do not follow regarding the identity. I do not know how to remove it. Can u clarify?

 

[andresB]

I could show you the whole process but I don't know how to type it.

 

[andresB]

Ok here it is

$\left\langle i\right|\Omega-\omega\left|v\right\rangle = \left\langle i\right|\Omega-\omega(\sum_{j}v_{j}\left|j\right\rangle )$
$= \sum_{j}v_{j}\left\langle i\right|\Omega\left|j\right\rangle -\omega\sum_{j}v_{j}\left\langle i\right.\left|j\right\rangle$
$=\sum_{j}v_{j} \Omega_{ij}-\omega\sum_{j}v_{j}\delta_{ij}$
$= \sum_{j}v_{j}(\Omega_{ij}-v_{j}\delta_{ij})$

 

[bugatti79]

Thanks for your efforts, just some questions:

1) How do you justify omitting the identity I?
Are you treating this equal to 1 because the action of the identity operator on a ket is just the same ket?
Additionally, Shankar specifically instructed to apply the identity operator to the left of |V>

2) How did you arrive with a 2nd bra <i| on the second line? There is only one.

regards

 

[andresB]

1) Yeah, the identity is implied, it action of a ket and a bra is to leave it as they are. (reading Shankar I don't know why one would introduce the representation of the identity though)

2)Well, the actual equation is
$\left\langle i\right|(\Omega-\omega\left)|v\right\rangle=0$

Look at 1.8.3 and multiply by the left by the i bra.

 

[andresB]

To clarify, You can't really have

$(\left\langle i\right|\Omega)-(\omega\left|v\right\rangle)$

because that would be adding a bra to a ket and you can't do that since they belong to different vector spaces.