Can Two Hermitian Matrices be Simultaneously Diagonalized if They Commute?

[Mr confusion]

hello,
i am having some trouble understanding simultaneous diagonalization. i have understood the proof which tells us that two hermitian matrices can be simultaneously diagonalized by the same basis vectors if the two matrices commute. but my book then shows a proof for the case when the matrices are both degenerate. it is basically going over my head, will anyone please help me with a good proof/ sketch?
my book uses block diagonal elements for the proof which i cannot understand...
thank you.

 

[Einstein Mcfly]

What do you mean by "the matricies are both degenerate"?

 

[Mr confusion]

i mean that multiple eigenvectors for the same eigenvalue.
ie. the characteristic equation has multiple roots like 3,3,1 etc.
the characteristic equation of both matrices have this property.

am i clear now,sir?

 

[Avodyne]

Let |i\rangle be an eigenstate of A with eigenvalue a_i. Then, sandwiching AB-BA=0 between \langle i| and |j\rangle, we get

(a_i-a_j)\langle i|B|j\rangle = 0.

So, if a_i\ne a_j, then B_{ij}\equiv \langle i|B|j\rangle = 0. If all the eigenvalues of A are different, then a_i\ne a_j whenever i\ne j, and so all the off-diagonal elements of B_{ij} are zero. Thus B is diagonal in the same basis as A.

But if two or more eigenvalues of A are the same (say a_1=a_2=\ldots=a_N) then we do not know anything about B_{ij} for i,j = 1,\ldots,N.

Let b be the N\times N hermitian matrix with matrix elements B_{ij}, i,j = 1,\ldots,N. We can diagonalize this matrix with a unitary transformation, b=UdU^\dagger, where d is diagonal. (It does not matter whether or not any of the diagonal elements of d are equal.) Now define new basis states |\tilde i\rangle; these are the same as the old basis states for i>N, and for 1\le i\le N,

|i'\rangle = \sum_{j=1}^N U_{ij}|j\rangle.

Now the |i'\rangle states are eigenstates of both A and B. For i=1,\ldots,N, the eigenvalue of A is a_1, and the eigenvalue of B is d_i.

 

[Mr confusion]

many many thanks avodyne. i am just starting this subject, so i am inexperienced . but i thought that the first step will give us (ai*-aj).
also please tell me how i can be sure that the new basis,' i ' prime will still remain the basis of A after the unitary passive transformation that diagonalises B?
please don't be angry with me. i am slow and take much time to understand things.

 

[Avodyne]

i thought that the first step will give us (ai*-aj).

Since A is hermitian, the eigenvalues are real.

also please tell me how i can be sure that the new basis,' i ' prime will still remain the basis of A after the unitary passive transformation that diagonalises B?

We are making the transformation only among the N states that all have the same eigenvalue of A. In this subspace, A can be written as the number a_1 (the eigenvalue of A) times the N\times N identity matrix I. A unitary transformation in this subspace therefore leaves A unchanged.

 

[Mr confusion]

understood. thanks.