Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 32A / d

Table of contents
- Solution d

Solution

Anti-Hermiticity makes

\overset{z}{ˉ}_{j} A_{jk} z_{k}

purely imaginary, so

I_{A}

is real. Its complex derivatives are

\frac{\partial I _{A}}{\partial z _{l}} = \frac{1}{2 i} \overset{z}{ˉ}_{j} A_{j l}, \frac{\partial I _{A}}{\partial z ˉ _{l}} = \frac{1}{2 i} A_{l k} z_{k} .

The complex-coordinate bracket from part (c) therefore gives

{I_{A}, I_{B}} = \frac{i}{2} \overset{z}{ˉ}_{j} (A_{j l} B_{l k} - B_{j l} A_{l k}) z_{k} = \frac{i}{2} \overset{z}{ˉ} [A, B] z = - \frac{1}{2 i} \overset{z}{ˉ} [A, B] z .

Hence the anti-Hermitian quadratic Hamiltonian identity is

{I_{A}, I_{B}} = - I_{[A, B]} .

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version