# I Common eigenfuctions for degeneracy

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1. Jan 11, 2017

### goodphy

Hello.

I read the textbook and found that common eigenfunctions are even possible for degenerate eigenvalues.

Let's say operators A and B commutes and eigenvalue a of operator A is N-fold degenerate, means that there are N linearly independent eigenfunctions having same eigenvalue a. These eigenfunctions are not necessarily common eigenfunctions of operator B, unlikely to non-degenerate eigenfunctions that are always common for commuting operators. However, the textbook said that we can always find a set of new N linearly independent eigenfunctions of eigenvalue a that are common for operator B by linear combination of old eigenfunctions of a. It is really fantastic theorem but I couldn't find its proof.

Could you please give me the proof of this? If I fully understand the proof, then I will be really confident that commuting operators have common eigenfunctions for all their eigenvalues, even some eigenvalues are degenerate.

2. Jan 11, 2017

### Strilanc

You're missing some of the criteria for proving the theorem. As stated, there are trivial counter-examples. For example, set $$N=2, A=H^{\otimes 2}, B=|0\rangle\langle 0| + i |1\rangle\langle 1| - |2\rangle\langle 2| - i |3\rangle \langle 3|$$.

Where $H$ is the Hadamard gate.

3. Jan 11, 2017

### PeroK

Here's is an outline of what's happening.

You start with operator A. As it's Hermitian it has a complete spectrum of eigenvectors, but may be degenerate, in the sense that it has an $N$-dimensional eigenspace for one or more of its eigenvalues. A is represented by a diagonal matrix in a basis of its eigenvectors

If you take any commuting operator, B, then it must be diagonal in this basis, except on the degenerate eigenspace.

Now, B might also be degenerate to some extent on that degenerate eigenspace, and if we have picked a B that is just as degenerate as A on that space, then we are no further forward. But, assume that B is less degenerate than A on that eigenspace. B then splits that N-dimensional space into several smaller eigenspaces on which A and B are both degenerate. Where these eigenspaces are 1-dimensional, we are finished, otherwise we must find an operator C and continue.

Note: at each stage you redefine the basis on the degenerate eigenspace to make all your operators diagonal.

To give an example:

If A is degenerate on a 5-dimensional eigenspace, then we find B that is 1-2-2 degenerate on that eigenspace. Now, we have two 2-dimensional eigenspaces on which A and B are both degenerate. Then, we find C that is non-degenerate on one of these and, say, D that is non-degenerate on the other.

In the finite dimensional case, these operators clearly exist, as you can simply define them by their matrix representation in the original basis. For infinite dimensional operators, you would need to be more careful.

The fact that operators representing useful observables are always available is perhaps another matter.

The point is that there is nothing fantastic going on here if you simply look at the operators in a suitable matrix representation.

4. Jan 11, 2017

### goodphy

Hello.

Your answer is far more complicated than I expected. It was a long time ago that I studied matrix representation of operators. But of course, I'll re-study this for reminding to understand your comments.

Before doing that, I just want to gett an answer of one question: is statement of "for compatible (commuting) operators {A, B}, we can always find a set of common eigenfunctions of all eigenvalues of both operators" true?

5. Jan 11, 2017

### Staff: Mentor

Yes.

6. Jan 11, 2017

### PeroK

That question doesn't really make sense. Instead:

Let $v$ be an eigenvector of $A$ corresponding to eigenvalue $\lambda$. If $A$ and $B$ commute, then:

$ABv = BAv = \lambda Bv$

Hence, $Bv$ is an eigenvector of $A$ corresponding to eigenvalue $\lambda$.

Now, if $A$ is not degenerate in $\lambda$, then $Bv = \mu v$ for some $\mu$. As, only scalar multiples of $v$ are eigenvectors of $A$ corresponding to $\lambda$.

And, if $A$ is degenerate in $\lambda$, then $Bv$ lies somewhere in that degenerate eigenspace.

In summary, $B$ maps eigenspaces of $A$ to themselves. When these eigenspaces are 1-dimensional (non-degenerate) it's simple, and you have a common eigenvector.

When these spaces are n-dimensional ($n > 1$), then you need to look at the spectrum of $B$ on that eigenspace.

Let's assume, for simplicity, that $B$ is not degenerate on that eigenspace. There we have $n$ distinct eigenvalues for $B$ on that space each with a linearly independent eigenvector. This forms a basis for the eigenspace on which the combination of eigenvalues of $A$ and $B$ is unique.

That completes the decomposition of the whole space using shared eigenvectors of $A$ and $B$.

7. Jan 11, 2017

### PeroK

Just to give an example, in a 4D space, you might get for eigenvalues and vectors for $A$

$\lambda_1, \ v_1$
$\lambda_2, \ v_2$
$\lambda_3, \ v_3, v_4$

So, $A$ has three eigenvalues and the third is two-fold degenerate.

And, for $B$ you might get:

$\mu_1, \ v_1$
$\mu_1, \ v_2$
$\mu_2, \ v_3$
$\mu_3, \ v_4$

And, you have now completely decomposed your (in this case 4-dimensional) vector space using a basis defined by a unique combination of eigenvalues of $A$ and $B$.

Neither $A$ nor $B$ is enough on its own (as I've made $B$ degenerate as well as $A$). But,the combination of $A$ and $B$ is enough.

$v_1$ is defined by $\lambda_1, \mu_1$
$v_2$ is defined by $\lambda_2, \mu_1$
$v_3$ is defined by $\lambda_3, \mu_2$
$v_4$ is defined by $\lambda_3, \mu_3$

Note, however, that no common eigenvector is possible for, say, the combination $\lambda_3, \mu_1$.

8. Jan 11, 2017

### A. Neumaier

Not necessarily; it might be zero.

9. Jan 11, 2017

### goodphy

Am...I'm afraid that my statement gives a possibility of wrong interpretation.

In my statement of "for compatible (commuting) operators {A, B}, we can always find a set of common eigenfunctions of all eigenvalues of both operators", "eigenfunction of all eigenvalues of both operator" does not mean an eigenfunction which corresponds all eigenvalues of both operators. Each common eigenfunction has particular eigenvalues of A and B. I meant there is no missed eigenvalues of A and B by a set of common eigenfunctions of commuting opreators, even if some of eigenvalues are degenerate.

I added this info to remove any confusion, although your comments may not be regarding this at all.

10. Jan 11, 2017

### PeroK

Good point! In that case, $v$ is an eigenvector of $B$ with eigenvalue $\mu = 0$ and $B$ still maps eigenspaces of $A$ into themselves.

11. Jan 11, 2017

### PeroK

Eigenvectors of a Hermitian operator corresponding to distinct eigenvalues are linearly independent, so you can't miss any out.

12. Jan 12, 2017

### goodphy

Hello.

I found and studied a simpler proof of existence (always exist) of a set of common eigenfunctions for commuting operators in degeneracy.

I uploaded this so other people who have a similar question can refer.

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