# Distinguishing between eigenstates

1. Feb 22, 2009

### james477

Hi, i am having trouble understanding a section of a QM course concerning degenerate eigenstates.

Suppose that some operator B is compatible with A (so A and B have a common eigenbasis). My notes say that this means that some r and r+pi/2 give eigenstates of both A and B in the form |a(r)> = (cos(r)|a1> + sin(r)|a2>) where |a1> and |a2> are two orthogonal degenerate eigenstates of A. I don't understand why the eigenstates have this form?

It goes on to say that in order to find the appropriate value of r we must solve for the eigenvectors of:

<a1|B|a1> <a1|B|a2>
<a2|B|a1> <a2|B|a2>

I don't understand this either... what is the significance of the eigenvectors of this matrix? (i believe the eigenvalues are the quantised measurables of B)

Any help would really be appreciated, Thanks

2. Feb 22, 2009

### xepma

So the idea with degenerate states is that when an operator A is said to be degenerate means that for a particular eigenvalue the corresponding eigenspace has a dimension larger than one. For simplicity we take this dimension to be two.

This means that when we construct an eigenbasis for A, there will always be two eigenvectors of this basis which share the same eigenvalue, i.e.

$A|\lambda_1\rangle = A|\lambda_2\rangle = \lambda|\lambda_1\rangle$

Note that $|\lambda_1\rangle$ en $|\lambda_2\rangle$ are linearly independent. Also, we may choose any linear combination of these two vectors and we still end up with an eigenvector of A:

$|\lambda_3\rangle = a|\lambda_1\rangle + b |\lambda_2\rangle \Longrightarrow A|\lambda_3\rangle = \lambda|\lambda_3\rangle$

What is important to realize is that there does not exist a natural basis for A within this degenerate eigenspace. Any two linearly independent eigenvectors will do.

So, now we get to the second point. Suppose we have a second operator B, which commutes with A (i.e. they are compatible). This means that there exists an eigenbasis of B which is also an eigenbasis of A. However, it is in general not true that any eigenbasis of A is also an eigenbasis for B. The reason for this is that the degenerate eigenspace of A might not be degenerate for B. In that case the operator B "distinguishes" among these eigenvectors of A.

So: $B|\lambda_1\rangle \neq B|\lambda_2\rangle$

In fact, the eigenvectors of A $|\lambda_1\rangle$ and $|\lambda_2\rangle$ will in general not even be eigenvectors of B (remember: any linear combination of these two eigenvectors is also an eigenvector of A). So the aim is now to find a new set of states $|\lambda_1'\rangle$ and $|\lambda_2'\rangle$ which also form an eigenbasis of B. By looking for the eigenvalues of the matrix you wrote down you are precisely determining that. Namely, these eigenvalues correspond to two eigenvectors. And these eigenvectors in turn diagonalize this matrix.

In other words, we need to find the basis $|\lambda_1'\rangle$ and $|\lambda_2'\rangle$ such that we have:

$\left( \begin{matrix} \langle\lambda_1'|B|\lambda_1'\rangle & \langle\lambda_1'|B|\lambda_2'\rangle\\ \langle\lambda_2'|B|\lambda_1'\rangle & \langle\lambda_2'|B|\lambda_2'\rangle \end{matrix}\right) = \left( \begin{matrix} \lambda_1' & 0\\ 0 & \lambda_2' \end{matrix}\right)$

If the matrix is of this form, then B is diagonal, which is precisely what we want.

3. Feb 23, 2009

### james477

Thanks, thats much clearer :)

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