Matrix formulation of an operator

In summary: I have successfully found the N by N matrix corresponding to the operator R.In summary, the conversation discusses finding the N by N matrix for the operator R in a linear vector space with ortho-normal basis states. The operator R is given to operate on basis states as R|b(j) = |b(j+1)> for j=1 to (N-1), and R|bN> = |b1>. However, there is confusion about the result obtained when operating R on |bj> basis vectors, with the individual having trouble verifying the mathematical process and getting incorrect results. The expert suggests that the individual may have a misunderstanding of basis vectors and advises them to practice and master matrices and vectors.
  • #1
tanaygupta2000
208
14
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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  • #2
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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  • #3
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)>
 
  • #4
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##?
 
  • #5
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
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  • #6
Hmm! $$|b_1 \rangle = (1, 0, \dots 0)^T$$
 
  • #7
PeroK said:
Hmm! $$|b_1 \rangle = (1, 0, \dots 0)^T$$
But then R|b1> will be zero?
Because R11 is zero
 
  • #8
tanaygupta2000 said:
But then R|b1> will be zero?
No. It will be ##b_2##.
 
  • #9
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
 
  • #31
tanaygupta2000 said:
Sir please assist me where am I doing mistake.View attachment 278400
That's simply wrong. I can't imagine what process you're using for matrix multiplication of a vector, but it's not the right one.
 
  • #32
tanaygupta2000 said:
I just tried using a hermitian of R since it is behaving like a rotation matrix and I got ... is this good?View attachment 278403
You're matrix multiplication is correct, but you have the wrong idea about what a basis vector is.
 
  • #33
PeroK said:
You're matrix multiplication is correct, but you have the wrong idea about what a basis vector is.
You told it is b1> = transpose(1, 0, 0, ...,)
 
  • #34
tanaygupta2000 said:
You told it is b1> = transpose(1, 0, 0, ...,)
Which is right. You're confusing it with the first component of a vector.
 
  • #35
PS The root of your problem is that you haven't mastered matrices and vectors. You have basic errors and confusions.
 

Related to Matrix formulation of an operator

1. What is the matrix formulation of an operator?

The matrix formulation of an operator is a mathematical representation of an operator in the form of a matrix. It is used to perform operations on vectors and other matrices in linear algebra.

2. How is an operator represented in matrix form?

An operator is represented in matrix form by writing out the coefficients of the operator's action on a basis of vectors. These coefficients are then arranged in a matrix, with each row representing the action of the operator on a different basis vector.

3. What are the advantages of using matrix formulation for operators?

The advantages of using matrix formulation for operators include the ability to easily perform calculations and transformations on vectors and matrices, as well as the ability to use existing techniques and algorithms from linear algebra to solve problems involving operators.

4. How does the matrix formulation of an operator relate to linear transformations?

The matrix formulation of an operator is closely related to linear transformations, as both involve the use of matrices to represent and perform operations on vectors. In fact, linear transformations can be thought of as a special case of operators, where the underlying vector space is the same for both the input and output vectors.

5. What are some applications of the matrix formulation of an operator?

The matrix formulation of an operator has various applications in fields such as physics, engineering, and computer science. It is used to solve systems of linear equations, perform transformations in computer graphics, and model physical systems in quantum mechanics, among others.

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