Matrix formulation of an operator

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Homework Help Overview

The discussion revolves around the matrix formulation of an operator R acting on an N-dimensional linear vector space spanned by ortho-normal basis states. The original poster expresses difficulty in obtaining the expected results when applying the operator R to the basis vectors, specifically noting discrepancies in the outcomes compared to the defined behavior of the operator.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the matrix representation of the operator R and its application to basis vectors. Questions arise regarding the correctness of the matrix entries and the expected results when operating on specific basis states. There is also discussion about the implications of the matrix structure and its relation to the operator's defined behavior.

Discussion Status

The discussion is ongoing, with various participants attempting to clarify the matrix representation and its application. Some have offered insights into potential errors in understanding the operator's action, while others are questioning specific entries in the matrix and their implications. There is a recognition of confusion regarding the relationship between matrix elements and basis vectors.

Contextual Notes

Participants are working under the constraints of the problem statement, which defines the operator's action on the basis states. There is an acknowledgment of potential misunderstandings regarding matrix multiplication and the representation of basis vectors.

tanaygupta2000
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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)>
 
tanaygupta2000 said:
It is behaving like R|b(j) = |b(j-1)> instead of R|b(j) = |b(j+1)>
The question was what is ##R|b_1 \rangle##?
 
PeroK said:
The question was what is ##R|b_1 \rangle##?
It is |bN>
IMG_20210220_183817.jpg
 

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  • IMG_20210220_183817.jpg
    IMG_20210220_183817.jpg
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Hmm! $$|b_1 \rangle = (1, 0, \dots 0)^T$$
 
PeroK said:
Hmm! $$|b_1 \rangle = (1, 0, \dots 0)^T$$
But then R|b1> will be zero?
Because R11 is zero
 
tanaygupta2000 said:
But then R|b1> will be zero?
No. It will be ##b_2##.
 
Try with ##N = 3##.
 
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  • #10
PeroK said:
Try with ##N = 3##.
For N=2, R = [0, 0 first row and 1, 0 second row]
Is it right?
 
  • #11
tanaygupta2000 said:
For N=2, R = [0, 0 first row and 1, 0 second row]
Is it right?
Try ##N = 3## instead.
 
  • #12
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
 
  • #13
tanaygupta2000 said:
For N=3, R = [0, 0, 0; 1, 0, 0; 0, 1, 0]
R|b1> is still 0
Well, no. ##R## has a ##1## at the end of the first row.
 
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  • #14
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?
 
  • #15
tanaygupta2000 said:
Sir the end has <b1|b4>
How it can be 1?
There is no ##b_4## if ##N = 3##.
 
  • #16
tanaygupta2000 said:
Sir the end has <b1|b4>
How it can be 1?
In accordance with <b1|R|b3>
 
  • #17
PeroK said:
There is no ##b_4## if ##N = 3##.
So what will be <b1|R|b3>?
I am getting confused
 
  • #18
tanaygupta2000 said:
So what will be <b1|R|b3>?
I am getting confused
I can see that. From your post #5, we have:
$$R(N=3) = [0, 0, 1; 1, 0, 0; 0, 1, 0]$$
 
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  • #19
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)?
 
  • #20
I think I am getting ...
 
  • #21
tanaygupta2000 said:
Sir are the last entries of rows II and III belong from the last column of R(N, N)?
If ##N = 3##, then the ##1## at the end is in the 3rd column. ##N## is a variable in this case.
 
  • #22
IMG_20210220_191406.jpg
IMG_20210220_191406.jpg
 
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  • #23
Not exactly, the last row is wrong on ##R_4##!
 
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  • #24
Thank You so much !
 
  • #25
tanaygupta2000 said:
Thank You so much !
Okay, but note that I've spotted an error now.
 
  • #26
PeroK said:
Okay, but note that I've spotted an error now.
Yes sir my mistake it should be 0, 0, 1, 0
 
  • #27
tanaygupta2000 said:
Yes sir my mistake it should be 0, 0, 1, 0
That's right now.
 
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  • #28
PeroK said:
No. It will be ##b_2##.
Sir please assist me where am I doing mistake.
IMG_20210221_074525.jpg
 
  • #29
In Shankar it is given like this:
Shankar.PNG


But I'm having trouble verifying it mathematically as given in the question.
 
  • #30
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
 

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