How Can Symplectic Eigenvector Normalization Influence Hamiltonian Forms?

[jarra]

The Hamiltonian, $$H=\frac{1}{2}\vec{\varsigma}K\vec{\varsigma}$$ is given.

With K being a $$2n \times 2n$$ matrix with the entries: $$\[ \left( \begin{array}{cc} 0 & \tau \\ \vartheta & 0\end{array} \right)\]$$

and $$\vec{\varsigma}$$ being a 2n-dimensional vector with entries: $$\vec{\varsigma}=[\vec q,\vec p]^T$$ with $$\vec q$$ and $$\vec p$$ being n-dimensional consisting of the generalized coordinates and generalized momenta respectively.
To this there is a matrix M whose columns are eigenvectors of the matrix JK with J being the matrix:
$$\[ \left( \begin{array}{cc} 0 & 1 \\ -1 & 0\end{array} \right)\]$$

The corresponding eigenvalues to the eigenvectors are $$\pm \omega_j$$ .

My problem is: ``For all eigenvalues $$\omega_j$$ being distinct show that the normalization of the eigenvectors can be chosen in such a way that M has the properties of the Jacobian matrix.''

Another problem is to show that after this canonical transformation the new Hamiltonian, K, takes the form $$K=i \sum_{j=1}^n \omega_j Q_j P_j$$


There should also be an ansatz putting $$\varsigma_j = \varsigma_0 e^{i\omega_j t}$$

 

[Valerie Park]

Solution: Let us begin by noting that the matrix K is a real, symmetric matrix and hence has an orthonormal basis of eigenvectors. Let us denote the eigenvectors by \vec{v_1}, \vec{v_2}, ..., \vec{v_{2n}}, with corresponding eigenvalues \lambda_1, \lambda_2, ..., \lambda_{2n}. We can construct the matrix M whose columns are eigenvectors of the matrix JK as follows: First, note that JK = \tau \vartheta J, so we can write JK \vec{v_i} = \lambda_i \vec{v_i}, where \vec{v_i} is an eigenvector of JK with eigenvalue \lambda_i. Next, we can normalize the eigenvectors to make them unitary, i.e., by dividing each vector by its length, so that \vec{v_i} = \frac{\vec{v_i}}{\|\vec{v_i}\|}. We can then construct the matrix M whose columns are eigenvectors of JK by stacking the normalized eigenvectors in a column-wise fashion, as follows: M = \begin{bmatrix} \frac{\vec{v_1}}{\|\vec{v_1}\|} & \frac{\vec{v_2}}{\|\vec{v_2}\|} & \cdots & \frac{\vec{v_{2n}}}{\|\vec{v_{2n}}\|} \\ \end{bmatrix}. Now, since the eigenvalues \omega_j are all distinct, it follows that the eigenvectors \vec{v_i} are also all distinct. Thus, the normalized eigenvectors \frac{\vec{v_i}}{\|\vec{v_i}\|} are also all distinct, and thus the matrix M has the properties of the Jacobian matrix. To show that after this canonical transformation the new Hamiltonian, K, takes the form K

 

[charng]

I would first clarify the problem by restating it in simpler terms:

Given a Hamiltonian, H, with a matrix K and a vector \vec{\varsigma}, we need to show that the eigenvectors of the matrix JK can be normalized in a specific way to make the matrix M have the properties of a Jacobian matrix. Additionally, we need to show that after this transformation, the new Hamiltonian, K, takes a specific form.

To solve this problem, we can use the properties of symplectic notation and the definition of a Jacobian matrix. The symplectic notation is a mathematical framework used to describe dynamical systems, particularly in Hamiltonian mechanics. It involves using a set of coordinates and momenta to represent the state of a system, similar to the \vec{\varsigma} vector in this problem.

We can start by considering the matrix M, which is composed of eigenvectors of JK. Since the eigenvalues are distinct, the eigenvectors are linearly independent and can be normalized to form an orthonormal basis. This means that the columns of M can be chosen to be orthogonal unit vectors, satisfying the properties of a Jacobian matrix.

Next, we need to show that after this transformation, the new Hamiltonian, K, takes the form K=i \sum_{j=1}^n \omega_j Q_j P_j. To do this, we can use the ansatz given in the problem, \varsigma_j = \varsigma_0 e^{i\omega_j t}. This ansatz represents a harmonic oscillation with a frequency of \omega_j. Plugging this into the Hamiltonian, we get:

H=\frac{1}{2}\vec{\varsigma}K\vec{\varsigma}=\frac{1}{2}\sum_{j=1}^n \varsigma_j K \varsigma_j=\frac{1}{2}\sum_{j=1}^n \varsigma_0^2 \omega_j^2 K=\frac{1}{2}\sum_{j=1}^n \omega_j^2 \varsigma_0^2 K

Using the definition of K given in the problem, we can rewrite this as:

H=\frac{1}{2}\sum_{j=1}^n \omega_j^2 \varsigma_0^2 \left( \begin{array}{