2
$\begingroup$

I'm studing quantum field theory. Especifically the procedure called second quantization for the complex scalar field. I noticed that I can derive the Klein Gordon equation from the Heisemberg equation (the names doesn't mind) by mean of using a sort of generalization of the gauss divergence theorem for $\phi(x)$.

The background is this, I have a map $\phi:\mathbb{R}^{3}\to \mathcal{B}(\mathcal{H})$, where $\mathcal{H}$ is a complex Hilbert space.

I would to know if the following two assestemnts are true and under what hipotesis.

1) $ div(\phi^{\dagger}(x)\nabla\phi(x))=\nabla\phi^{\dagger}(x)\cdot\nabla\phi(x)+\phi^{\dagger}(x)\nabla^{2}\phi(x)$ which for smooth functions $C^{\infty}(\mathbb{R})$ for example is obvious.

2) The expression $\int_{\mathbb{R}^{3}}{div(\phi^{\dagger}(x)\nabla\phi(x))}dx=0$ under certain hipotesis over $\phi$.

I apreciate any help.

$\endgroup$

1 Answer 1

0
$\begingroup$

Derivatives are well-defined also when taking values in $B(H)$. For a reference you may look in e.g. Serge Lang, Real and Functional Analysis, Chapter XIII, notably section 3, properties of the derivative, where he talks about the Leibnitz rule for 'products' in the context of Banach spaces.

The divergence (acting on vector fields) only 'sees' the vector coming from the gradient. This is perhaps more clear if you write your expression as $$\sum_i \partial_i (\phi^\dagger(x) \partial_i \phi(x)).$$ Here, $\partial_i$ acts on the product in the usual Leibnitz way. Integrating with values in $B(H)$ may be defined as usual. And you may integrate the $i$'th term first in the $i$'th direction to get a vanishing integral in your example (provided, of course, that your fields have compact support). The above holds e.g. if fields are $C^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.