Matrix formulation of an operator

[tanaygupta2000]

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

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 !

 

[PeroK]

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 $R|b_1 \rangle$?

 

[tanaygupta2000]

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]

It is behaving like R|b(j) = |b(j-1)> instead of R|b(j) = |b(j+1)>

The question was what is $R|b_1 \rangle$?

 

[tanaygupta2000]

The question was what is $R|b_1 \rangle$?

It is |bN>

 

[PeroK]

Hmm! $$|b_1 \rangle = (1, 0, \dots 0)^T$$

 

[tanaygupta2000]

Hmm! $$|b_1 \rangle = (1, 0, \dots 0)^T$$

But then R|b1> will be zero?
Because R11 is zero

 

[PeroK]

But then R|b1> will be zero?

No. It will be $b_2$.

 

[PeroK]

Try with $N = 3$.

 

[tanaygupta2000]

Try with $N = 3$.

For N=2, R = [0, 0 first row and 1, 0 second row]
Is it right?

 

[PeroK]

For N=2, R = [0, 0 first row and 1, 0 second row]
Is it right?

Try $N = 3$ instead.

 

[tanaygupta2000]

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]

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.

 

[tanaygupta2000]

Well, no. $R$ has a $1$ at the end of the first row.

Sir the end has <b1|b4>
How it can be 1?

 

[PeroK]

Sir the end has <b1|b4>
How it can be 1?

There is no $b_4$ if $N = 3$.

 

[tanaygupta2000]

Sir the end has <b1|b4>
How it can be 1?

In accordance with <b1|R|b3>

 

[tanaygupta2000]

There is no $b_4$ if $N = 3$.

So what will be <b1|R|b3>?
I am getting confused

 

[PeroK]

So what will be <b1|R|b3>?
I am getting confused

I can see that. From your post #5, we have:
$$R(N=3) = [0, 0, 1; 1, 0, 0; 0, 1, 0]$$

 

[tanaygupta2000]

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)?

 

[tanaygupta2000]

I think I am getting ...

 

[PeroK]

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.

 

[tanaygupta2000]

 

[PeroK]

Not exactly, the last row is wrong on $R_4$!

 

[tanaygupta2000]

Thank You so much !

 

[PeroK]

Thank You so much !

Okay, but note that I've spotted an error now.

 

[tanaygupta2000]

Okay, but note that I've spotted an error now.

Yes sir my mistake it should be 0, 0, 1, 0

 

[PeroK]

Yes sir my mistake it should be 0, 0, 1, 0

That's right now.

 

[tanaygupta2000]

No. It will be $b_2$.

Sir please assist me where am I doing mistake.

 

[tanaygupta2000]

In Shankar it is given like this:

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

 

[tanaygupta2000]

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?

 

[PeroK]

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.

 

[PeroK]

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.

 

[tanaygupta2000]

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, ...,)

 

[PeroK]

You told it is b1> = transpose(1, 0, 0, ...,)

Which is right. You're confusing it with the first component of a vector.

 

[PeroK]

PS The root of your problem is that you haven't mastered matrices and vectors. You have basic errors and confusions.