# Classic Oscillator

1. Mar 1, 2014

### chimay

Hi guys.
Recentely I'm approaching Quantum Mechanics starting from the mathematical basics.
In order to understand the benefit of representing a certain matrix in its eigenvectors basis my book makes the example I attached ( Principles of Quantum Mechanics by Shankar ).
Using matrix form it can be easily shown we can write this:
$$\begin{bmatrix} x_{1}'' \\ x_{2}'' \end{bmatrix}= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix}$$
a, b, c, d being proper coefficient.

The author says that both the matrix and the vectors refers to the canonic basis; how can we be so confident about this?
I mean that analysing the physical problem does not require any reference to the basis we will refer when we use the matrix form; yet representing an operator in matrix form requires to have specified what is the basis...

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2. Mar 1, 2014

### UltrafastPED

Shankar is positing that condition. You must always use the same basis for all objects (vectors, matrices), except when you are converting from one basis to another.

So in this case you would write the Hamiltonian for the blocks and springs problem. If they are the coordinates for a valid Hamiltonian, then they are canonical.

If you instead write down any old coordinates and write the Lagrangian - well, your dynamics will be correct, but the coordinates will not be canonical until you carry out the transform to go from (p,q) of the Lagrangian to the (P,Q) of the Hamiltonian.

3. Mar 1, 2014

### MisterX

Aren't the coordinates in Lagrangian mechanics $q_i, \dot{q}_i$ ?

Anyway I think maybe what was meant by "canonic basis" w.r.t. the OP is that they are the positions and accelerations of the individual masses - as opposed to normal coordinates.

4. Mar 3, 2014