1. Apr 20, 2006

### JustinLevy

If I have two positive definite Hermitian NxN matrices A and B, if I adiabatically change the components of A to B (constraining any intermediate matrices to be Hermitian as well, but not necessarily positive definite) while \"following\" the eigenvalues ... will the mapping of the eigenvalues of A to the eigenvalues of B be independent of the adiabatic path?

Obviously the answer is yes for N=1.
And I have convinced myself the answer is yes for N=2 (given that the path doesn\'t take you through the zero matrix).

But I am not sure how to approach this in a general enough manner to figure this out.

Can anyone here offer some advice?

I eventually want to find the mapping of the eigenvalues of A to B. So if this is a solved problem, by all means point me to the solution so that I may read up on it.

Last edited: Apr 20, 2006
2. Apr 21, 2006

### Rach3

This probably belongs in the Quantum Physics subforum.

You seem to be thinking about quantum adiabatic evolution? As it is your question makes little sense - what does "adiabatic" mean in a mathematical context? What do you mean by "follow the eigenvalues"? I think you're misinterpreting some physical concept - that of a ground-state quantum system evolving adiabatically, "following" the lowest eigenvalue of the Hamiltonian?

My question - what are you reading, and how did this question come up?

3. Apr 23, 2006

### JustinLevy

This is not a quantum mechanics problem.
Yes, this question came up because of some physics, but I don\'t really want to get into that as it is extraneous info. This is meant to be a math question. I appologize if I did not specify my terms mathematically well enough. Let me try again.

Let us parameterize the path of changing from A-> B with the parameter s.

Let s=0 correspond to A, and s=1 correspond to B.

By \"adiabaticly changing\" from A to B, I mean that the components should be a smooth function of s (continuous and all derivatives are continuous).

Similarly, the eigenvalues will be a smooth function of s.

Therefore, by \"follow the eigenvalues\" I mean finding these functions (or more specifically, evalutating the functions at s=0 and s=1 to see which eigenvalue of A corresponds to which eigenvalue of B according to the chosen adiabatic path). There is only an ambiguity in these functions if the eigenvalues cross (become degenerate) at some point along the path, but I believe the requirement that the functions be smooth eliminates this ambiguity and makes \"follow the eigenvalues\" well defined.

Here is an example of the simplest path (meeting the requirements), which is just linear:

F(s) = A + s*(B-A)

The eigenvalues of the matrix F(s) can be solved to obtain how the eigenvalues evolve along this path.

But that is just an example, as my question in more general than this. Will the mapping of the eigenvalues of A to the eigenvalues of B be independent of the adiabatic path chosen?

Last edited: Apr 23, 2006
4. Apr 23, 2006

### matt grime

None of the terms you have used is widely (if at all) known in mathematics. Adiabatic for instance appears to have *no* mathematical meaning at all.

You are asking for a matrix valued smooth function s(t) from [0,1] to the space of hermitian matrices (of some fixed dimension).

Now, how do you even begin to pick out one particular eigenvalue of s(0)? And what does it mean for this to correspond to any eigenvalue of s(1)? The path does not map an eigenvalue to another eigenvalue. The map s does not act on eigenvalues, it doens't even act on matrices, it acts on the interval [0,1].

Perhaps, we should think of this instead by looking at the characteristic poly det(x-s(t)), and looking at the locus of zeroes as t varies. Is that better? Probably that is how I should have read it. It makes more sense. Now, what, in this context, is your question?

Last edited: Apr 23, 2006
5. Apr 23, 2006

### matt grime

I thihnk I begin to see what you're thinking. If we plot the graph of the eigen values against time, so we have N strands, if you will, you wish to know if the start and end points are independent of the path s(t), got it. Right.

The first thing that springs to mind is that smoothness, or otherwise of s doesn't have any bearing on whether the strands can cross or not (become degenerate). It is rathe clear that it is possible to go from the matrix diag(1,2) to diag(2,1) smoothly via the path diag(1+s,2-s) and that the eigenvalues 'cross' at t=1/2.

6. Apr 24, 2006

### JustinLevy

I appologize. I am trying my best to explain mathematically what I mean. Even if people can not answer the question, if I can get it worded in a completely mathematical way ... maybe I could then approach some math professors for help.

Yes, I believe you understand the question now.
But just to make sure: all the start points are defined by the eigenvalues of A, and all the end points are defined by the eigenvalues of B. So these are constant. If I define a path evolving A to B (parameterized by s), and therefore obtain N \"strands\", I can make a one to one map of As eigenvalues to Bs eigenvalues (by looking at the endpoints of each individual \"strand\").

I want to know if this map is independent of how I define the path. (And ignoring the trivial cases which go through the zero matrix.)

If they did not become degenerate, this would be a boring problem (as then of course the map of As eigenvalues to Bs will remain unchanged). It is the very fact that they can become degenerate that makes this interesting.

When two \"strands\" cross, I believe that the restriction that the strands are a smooth function of s lifts any ambiguity of relating the strands that enter the crossing with the strands that exit it.

The only cases where degeneracy should be an issue is if A or B are degenerate, as the mapping of As eigenvalues to Bs is not necessarily uniquely 1 to 1. So let us ignore the cases where A or B are degenerate (but again, definitely the path may take us through matrices that are indeed degenerate).

Also, please note again that A and B are positive definite (I do not know if that is helpful but thought I would throw in a reminder), but the only restriction for the path is that the matrix is hermitian.

Last edited: Apr 24, 2006
7. Apr 24, 2006

### Rach3

I have no idea why you think this is important. If you ignore all the degeneracies (defeating the physics!), your "mapping of eigenvalues" depents critically on the path. You can define any mapping you want, merely by choosing the path! For instance, a linear interpolation (like matt grime's example) maps the |1> to |2> and the |2> to |1>. You can easily define a map which takes |1> to |1> and |2> to |2>, just using rotation matrices. You could do this in two steps; in s <= 0.5, continuosly map diag(1,2) to diag(1,-2), and in the remaining time s>0.5, perform a pi/2 rotation diag(1,-2) to (2,1). In this process, the 1 eigenvalue always stays at 1, while the 2 eigenvalue goes down to -2 and back up to +2. This is general - you can "map" any eigenvalue to any other eigenvalue in this general way, using a sequence of properly defined rotations.

It would help us greatly if you told us what you are currently reading (obviously it's in the physics literature), and what physical problem is causing you grief. Replacing words like "adiabatic evolution" with words like "smooth function" does nothing good, just makes nonsense.

8. Apr 24, 2006

### JustinLevy

Why do you think I am ignoring the degeneracies? I said that degeneracies along the path are allowed.

Your example is a piece wise definition, and thus the components, nor the eigenvalues will be a smooth function of s. So I do not consider that a valid path.

I would prefer to leave the physics out of this, but if you insist.

I have a collection of masses connected with springs. With the harmonic approximation I obtain the K matrix (force constant matrix in mass weighted cartesian coordinates (or some other basis expressed as a linear combination of such)). The eigenvalues of K are the squared frequencies of the normal modes.

Now let us \"adjust\" the springs. We get another K matrix. How can the normal modes of the two arrangements be related to each other? In essence, chemists and physicists do this all the time when looking at vibrational spectra between related molecules. It is apparent that if you specify a chemical path connecting the two molecules, the relation of the modes can be specified. I want to know if this relation is actually independent of the chemical path ... because if not, then the relation is actually pointless.

I have done much literature searching, and they never approach this from a mathematical point of view (even when a chemical path is not specified, they just arbitrarily chose a \"linear\" path connecting the matrix elements, and assume the result is indeed unique). I am in physics, not math, and was hoping that I could rephrase this into a mathematical question.

Last edited: Apr 24, 2006
9. Apr 24, 2006

### Rach3

Okay then, a smooth example, all matrices Hermitian, in which the eigenvalues are always constant!:

$$f(s)=\mathbf{R}(-s \pi/2)\cdot A\cdot\mathbf{R}(s\pi/2)$$

where
$$R(t)=\left(\begin{array}{cc}\cos t & \sin t\\ -\sin t & \cos t\end{array}\right)$$

so that
$$f(0)=A=\left(\begin{array}{cc}1&0\\0&2\end{array}\right)$$
and
$$f(1)=B=\left(\begin{array}{cc}2&0\\0&1\end{array}\right)$$

It's obvious what this does if you think in terms of the eigenvectors of the matrix.

Last edited by a moderator: Apr 24, 2006