Simultaneous diagonalization

1. Dec 22, 2009

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.

2. Dec 22, 2009

Einstein Mcfly

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

3. Dec 23, 2009

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?

4. Dec 23, 2009

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 $d_i$.

5. Dec 28, 2009

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 dont be angry with me. i am slow and take much time to understand things.

6. Dec 28, 2009

Avodyne

Since A is hermitian, the eigenvalues are real.
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.

7. Dec 28, 2009

Mr confusion

understood. thanks.