Eigenketes and Eigenvalues of operators

[jasonchiang97]

Homework Statement

Again, consider the two-dimensional vector space, with an orthonormal basis consisting of kets |1> and |2>, i.e. <1|2> = <2|1> = 0, and <1|1> = <2|2> = 1. Any ket in this space is a linear combination of |1> and |2>. a) [2pt] The operator A acts on the basis kets as A|1> = |1>, A|2> = −|2>. Find the eigenkets (=eigenvectors) of A, and the corresponding eigenvalues. Find the matrix which represents the operator A in the basis of |1> and |2>. Find the matrix which represents the operator A in the basis of eigenkets of A.

Homework Equations

A|a> = c|a>

The Attempt at a Solution

Well for the first part, I just said that if A|a> = c|a> where c is a constant, then |a> is the eigenket or eigenvector and c is the eigenvalue so I got

|1> is the eigenvector and 1 is the eigenvalue for A|1> and |2> is the eigenket and -1 is the eigenvector for A|2>. I am unsure of how to find the corresponding matrix that represents A in the basis of |1> and |2>

Is it just (1 -1) but as a column matrix?

 

[kuruman]

The matrix is 2x2 and its ijth element is $A_{ij} =<i|A|j>$. Can you construct the matrix now?

 

[jasonchiang97]

Yes I figured it out. Thanks!

 

[jasonchiang97]

Okay, so I have my matrix in in the basis of |1> and |2> as (1,0)
(0 1)

Sorry I am not sure how to write matrices on this forum. I am wondering how would I apply the change of basis to get them into the basis of the eigenkets of A? Since the eigenkets is just |1> and |2> wouldn't they be the same basis?

 

[kuruman]

To write matrices and other math expressions, click on the LaTeX link, bottom left and to the right of the question mark.

Constructing the matrix in the eigenket representation, $|V_1>$, $|V_2>$, use the same procedure. The ijth element is $A_{ij}=<V_i|A|V_j>$. If you do it correctly, you will see something that should have been obvious in retrospect.

 

[jasonchiang97]

So if my eigenket is |1> and |2> then would my result not be the same? Or is my eigenkets not |1> and |2>

 

[Orodruin]

|1> is the eigenvector and 1 is the eigenvalue for A|1> and |2> is the eigenket and -1 is the eigenvector for A|2>.

I believe you have a slight, but important misunderstanding. There is no such thing as an eigenvector of A|1> or an eigenvector of A|2>. Instead, A|1> = |1> means that |1> is an eigenvector of A with eigenvalue 1 and similarly |2> is an eigenvector of A with eigenvalue -1.

A is an operator from a 2D space to a 2D space. As such it should be represented by a 2x2 matrix, not a column or row matrix.

 

[jasonchiang97]

I believe you have a slight, but important misunderstanding. There is no such thing as an eigenvector of A|1> or an eigenvector of A|2>. Instead, A|1> = |1> means that |1> is an eigenvector of A with eigenvalue 1 and similarly |2> is an eigenvector of A with eigenvalue -1.

A is an operator from a 2D space to a 2D space. As such it should be represented by a 2x2 matrix, not a column or row matrix.

I see. Just to make sure, the |1> and |2> are used as symbols here to denote the vector in A|1> and A|2> correct? As in it's the same notation but different vector if I were to write B|x> and B|y>

 

[kuruman]

I see. Just to make sure, the |1> and |2> are used as symbols here to denote the vector in A|1> and A|2> correct? As in it's the same notation but different vector if I were to write B|x> and B|y>

Although A|1> is a vector, it is not the same as vector |1>. It is the vector you get when you operate on vector |1> with operator A. It is analogous to this idea. If you have a function $f(x)$, then $g(x) = df(x)/dx$ is a new function that you get when you operate on $f(x)$ with operator $d/dx$. For the same reason you cannot say that $f(x)$ is used as a "symbol to denote" the function $df(x)/dx$, you cannot say that vector |1> is used as a symbol to denote vector A|1>.

 

[jasonchiang97]

I understand much better now. Thanks!