# Physical reason for diagonal matrix

• Brewer
In summary, a diagonalized matrix with the eigenvalues of the system as the values is an important concept in physics, particularly in quantum mechanics. It is used to find the possible outcomes of a measurement of an observable, as well as to simplify and solve systems of coupled equations. In classical mechanics, it can be used to find clever generalized coordinates for easier problem solving. In general, diagonalization allows for a change of basis and can simplify mathematical calculations.
Brewer
Whats the physical reason for a diagonalized matrix with the eigenvalues of the system as the values?

Reading from wiki, it seems that its something to do with the Schrodinger Eqn, but I don't follow that. If someone could explain the point of it to me in plain English (or as close as it gets!) then I'd be very grateful to them.

Can you be a little more specific? Are you talking about quantum mechanics, where a measurement of a Hermitian operator will always give an eigenvalue of the operator?

Sorry, yes I am. I didn't realize there was a difference.

If you have Griffith's book on QM, it is explained quite clearly in chapter 3; to be more precise in section 3.2. The argument is roughly as follows:
• Recall that the expectation value of an observable A, for a system described by wavefunction $\Psi$ is
$$\langle A \rangle = \int \Psi^* \hat A \Psi \, \mathrm dx$$
• We want the outcome of a measurement to be real, so in particular we want the expectation value (~ average) to be real:
$$\langle A \rangle = \langle A \rangle^*$$
• It is now easy to see that this must lead to the property of A being Hermitian.
• If the system is in a determinate state (a state in which a measurement of A yields a well-defined value; for example stationary states are determinate states of the Hamiltonian) then you want the standard deviation
$$\langle \hat A - \langle A \rangle )^2 \rangle$$
to vanish. This can be shown to lead to
$$\hat A \Psi = \lambda \Psi$$
for some number lambda. In other words, determinate states are eigenfunctions of $\hat A$.
• The eigenfunctions of a Hermitian operator can be shown to be orthogonal (or, in case of multiple eigenfunctions for the same eigenvalue, we can choose them to be such). Moreover, the wave-function can be decomposed into the eigenfunctions $f_n$ of $\hat A$, meaning we can write
$$\Psi = \sum_n c_n f_n$$ with $$c_n = \int f_n(x)^* \Psi(x, t) \, \mathrm dx$$
(in the discrete case, replace the sum by an integral in the continuum case)
• Now using all these properties, you can show that
$$\langle A \rangle = \sum_n \lambda_n |c_n|^2$$
where $\lambda_n$ are the eigenvalues $\hat A f_n = \lambda_n f_n$. This is precisely the expectation value for an experiment where the outcome $\lambda_n$ has probability $|c_n|^2$!

To summarize then: the possible outcomes of a measurement of an observable are the eigenvalues of that observable. This means that we have to calculate the eigenvalues, which you know how to do in the finite case from linear algebra (you can write $\hat A$ as a matrix). Usually, if you're going to work with A (dropping the hat from now on) a lot in your particular problem, it might pay off to express everything in eigenstates of A. From linear algebra you know, that this is equivalent to diagonalizing A such that the eigenvalues are explicit in the way you write it down.

Hope that clarifies.

Brewer said:
I didn't realize there was a difference.
You can do the same thing in classical mechanics. The typical example studied is that of a system of coupled oscillators. You can write the Lagrangian in terms of the displacements of the individual oscillators as the generalized coordinates, but you will find that the differential equations can be quite difficult to solve, because they mix the coordinates. You will find that it is much easier to solve the system if you discover some clever generalized coordinates that can be treated independently, so that you obtain a set of differential equations that each involve a single coordinate. This can be done by diagonalizing a matrix, and the characteristic frequencies are the eigenvalues.

In fact, in any part of physics where linear algebra occurs (and since we like to work with linear things because they are so simple, it occurs almost everywhere - either exactly, or in approximation) you can play such tricks. For then you can work with eigenvalues and eigenvectors, which means diagonalizing the matrix, and in the end you can just take linear combinations afterwards to describe the whole system.

Brewer said:
Whats the physical reason for a diagonalized matrix with the eigenvalues of the system as the values?

Reading from wiki, it seems that its something to do with the Schrodinger Eqn, but I don't follow that. If someone could explain the point of it to me in plain English (or as close as it gets!) then I'd be very grateful to them.

There's a short, generally applicable answer too, but not real physical.

The diagonalised matrix is a change of basis, where all vector quantities undergo dilation or contraction.

Edit: I think that was unclear. We cn represent the matrix in a different coordinate system. In the coordinate system where the matrix is in diagonal form all vector quantities operated upon by the diagonalized matix undergo dilation or contraction.

Last edited:
But Brewer should be aware of the profound physical difference in the interpretation of this mathematics in QM vs. CM. In CM, the interpretation is basically as Phrak suggests, and then one must further interpret what is the meaning of such vector stretching. In QM, the interpretation is completely different, because a stretched version of a state vector is interpretted as the same state, and so the eigenvalues have no effect on the state, but rather the possible results of the measurement. In fact, instead of stretching the vector, in QM the vector is rotated but maintains its length.

## 1. What is a diagonal matrix?

A diagonal matrix is a special type of square matrix where all the elements outside the main diagonal are equal to zero. The main diagonal refers to the elements that run from the top left corner to the bottom right corner of the matrix.

## 2. What is the physical significance of a diagonal matrix?

A diagonal matrix has many physical applications, such as representing linear transformations, coordinate systems, and energy states in quantum mechanics. It also simplifies calculations and makes them more efficient.

## 3. How is a diagonal matrix related to eigenvectors and eigenvalues?

A diagonal matrix is closely related to eigenvectors and eigenvalues. The eigenvalues of a diagonal matrix are the elements on the main diagonal, and the eigenvectors are the corresponding columns of the identity matrix. This relationship is useful for solving equations involving diagonal matrices.

## 4. Can a non-square matrix be diagonal?

No, a non-square matrix cannot be diagonal. Diagonal matrices are only defined for square matrices, meaning they have an equal number of rows and columns.

## 5. How can you determine if a matrix is diagonal?

To determine if a matrix is diagonal, you can check if all the elements outside the main diagonal are equal to zero. You can also calculate the determinant of the matrix, which will be equal to the product of the elements on the main diagonal. If the determinant is zero, the matrix is not diagonal.

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