Matrix formulation of an operator

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Homework Statement
Consider an N-dimensional linear vector space spanned by the ortho-normal basis
states, {|b1>, |b2>, ...., |bN>}. An operator R is given to operate on these basis stats as: R|b(j) = |b(j+1)> for j=1 to (N-1), and R|bN> = |b1>.
Write R in the matrix form in the given basis.
Relevant Equations
<ψ|R|ψ> = Σ(i) Σ(j) <ψ|bj> <bj|R|bi> <bi|ψ>
where
R operator = <bj|R|bi> matrix
IMG_20210220_095529.jpg


I have successfully found the N by N matrix corresponding to the operator R.
But the problem is, whenever I try to operate R on |bj> basis vectors, I am not getting |b(j+1)> as it should be.
Instead, I am getting result as given in the question only by <bj|R = <b(j+1)|

Matrix is not working with kets.
Please help !
 
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tanaygupta2000 said:
Homework Statement:: Consider an N-dimensional linear vector space spanned by the ortho-normal basis
states, {|b1>, |b2>, ..., |bN>}. An operator R is given to operate on these basis stats as: R|b(j) = |b(j+1)> for j=1 to (N-1), and R|bN> = |b1>.
Write R in the matrix form in the given basis.
Relevant Equations:: <ψ|R|ψ> = Σ(i) Σ(j) <ψ|bj> <bj|R|bi> <bi|ψ>
where
R operator = <bj|R|bi> matrix

View attachment 278350

I have successfully found the N by N matrix corresponding to the operator R.
But the problem is, whenever I try to operate R on |bj> basis vectors, I am not getting |b(j+1)> as it should be.
Instead, I am getting result as given in the question only by <bj|R = <b(j+1)|

Matrix is not working with kets.
Please help !
The matrix looks right to me. What do you get for ##R|b_1 \rangle##?
 
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PeroK said:
The matrix looks right to me. What do you get for ##R|b_1 \rangle##?
It is behaving like R|b(j) = |b(j-1)> instead of R|b(j) = |b(j+1)>
 
PeroK said:
The question was what is ##R|b_1 \rangle##?
It is |bN>
IMG_20210220_183817.jpg
 

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PeroK said:
Hmm! $$|b_1 \rangle = (1, 0, \dots 0)^T$$
But then R|b1> will be zero?
Because R11 is zero
 
PeroK said:
Try with ##N = 3##.
For N=2, R = [0, 0 first row and 1, 0 second row]
Is it right?
 
PeroK said:
Try ##N = 3## instead.
For N=3, R = [0, 0, 0; 1, 0, 0; 0, 1, 0]
R|b1> is still 0
since first row is zero
 
PeroK said:
Well, no. ##R## has a ##1## at the end of the first row.
Sir the end has <b1|b4>
How it can be 1?
 
tanaygupta2000 said:
Sir the end has <b1|b4>
How it can be 1?
In accordance with <b1|R|b3>
 
PeroK said:
There is no ##b_4## if ##N = 3##.
So what will be <b1|R|b3>?
I am getting confused
 
PeroK said:
I can see that. From your post #5, we have:
$$R(N=3) = [0, 0, 1; 1, 0, 0; 0, 1, 0]$$
Sir are the last entries of rows II and III belong from the last column of R(N, N)?
 
I think I am getting ...
 
IMG_20210220_191406.jpg
IMG_20210220_191406.jpg
 
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Thank You so much !
 
PeroK said:
Okay, but note that I've spotted an error now.
Yes sir my mistake it should be 0, 0, 1, 0
 
PeroK said:
No. It will be ##b_2##.
Sir please assist me where am I doing mistake.
IMG_20210221_074525.jpg
 
In Shankar it is given like this:
Shankar.PNG


But I'm having trouble verifying it mathematically as given in the question.
 
tanaygupta2000 said:
In Shankar it is given like this:
View attachment 278401

But I'm having trouble verifying it mathematically as given in the question.
I just tried using a hermitian of R since it is behaving like a rotation matrix and I got ... is this good?
IMG_20210221_083300.jpg