Problems about eigenvector in quantum mechanics

[wowowo2006]

I am learning about the basic quantum mechanics
I know that an operator ,call it M^, is generally a matrix
And we also can be represent it b a matrix representation M, associated with certain basis |e>

M^ = sigma ( Mij |e> <e|)
I,j
Where Mij is matrix element of M

So now I wonder which matrix should I use, M^ or M
To find the eigenvector?

 

[ShayanJ]

An operator $\hat M$ is an object that takes a vector and gives another vector. It is not a matrix but it has a matrix representation which depends on the base set used. If the base set is $\{|n\rangle\}_{n=0}^{\infty}$, the matrix elements of $\hat M$ in this basis are $M_{mn}=\langle m | \hat M | n \rangle$. This is the only matrix associated to $\hat M$(up to change of basis, of course).

 

[wowowo2006]

An operator $\hat M$ is an object that takes a vector and gives another vector. It is not a matrix but it has a matrix representation which depends on the base set used. If the base set is $\{|n\rangle\}_{n=0}^{\infty}$, the matrix elements of $\hat M$ in this basis are $M_{mn}=\langle m | \hat M | n \rangle$. This is the only matrix associated to $\hat M$(up to change of basis, of course).

So that means,
When I change the basis, the matrix representation change
So do the eigenvector changes too?

 

[ShayanJ]

So that means,
When I change the basis, the matrix representation change
So do the eigenvector changes too?

I see what's your problem.
You should be able to distinguish between abstract mathematical constructs and their representations. The operators and vectors are entities independent of their representations. They don't change when you change basis, its just that their components w.r.t. different bases are different. Its like vectors in Euclidean space. They are the same no matter you use a particular set i,j,k or another set rotated w.r.t. to the first one. The vector is the same, only the component changes to comply with the change of basis.
So no, eigenvectors do not change, they only have different components w.r.t. to different bases.

 

[vanhees71]

Let's clarify this issue a bit. You have a linear operator on Hilbert space (usually defined on a dense subset). This we denote with $\hat{A}$. Then you can choose any representation you like by taking a complete orthonormal set $|u_{j} \rangle$ of vectors, fulfilling
$$\langle u_j | u_k \rangle=\delta_{jk}, \quad \sum_{j} |u_j \rangle \langle u_j|=\hat{1}.$$
Now you can represent the operator in terms of its matrix elements with respect to this basis,
$$A_{jk}=\langle u_j |\hat{A} u_k \rangle.$$
You get back the operator by inserting two identity operators in terms of the completeness relation for the basis,
$$\hat{A}=\sum_{j,k} |u_j \rangle \langle j|\hat{M} u_k \rangle \langle u_k |=\sum_{jk} M_{jk} |u_j \rangle \langle u_k|.$$
I hope, now at least the formalities are a bit more clear.