0
$\begingroup$

I asked this question on math.stack but I got no answer, so I try here.

Let $\phi(t)$ be a solution for the nonlinear Schroedinger equation\begin{equation} i\partial_t\phi(t)=-\Delta\phi(t)+(V*|\phi|^2)\phi(t) \end{equation} inside the Hilbert space $L^2(\mathbb{R}^d)$. I know $\phi(t)$ is also supposed to live in $L^\infty\cap L^2$, however I'm not sure this is really important for the problem. Let's define the projector onto $\phi(t)$ as \begin{equation} p(t):=\left|\phi(t)\right\rangle\left\langle\phi(t)\right|. \end{equation} On the paper http://arxiv.org/abs/0907.4313 I found a formula stating, without proof, that \begin{equation} \|\nabla p(t)\|=\|\nabla\phi(t)\|. \end{equation} I was trying to figure out how to prove this, but I got really confused. I think by definition and integration by part I can write \begin{equation} \|\nabla\phi\|^2=\sum_{i=1}^d\|\nabla^i\phi(t)\|^2=\int dx^d\,\phi_t(x)(-\Delta)\phi_t(x), \end{equation} even though I'm not completely sure I'm using the correct formula for a vector-valued $L^2$ function. But I really can't figure out how to write the norm of the vector of operators $\|\nabla p(t)\|$ to try to compare them. Could anybody help me?

$\endgroup$

1 Answer 1

1
$\begingroup$

I haven't checked the paper, but $\nabla p(t)$ most likely mean the operator $\psi \mapsto \nabla \phi \cdot \langle \phi | \psi\rangle$ and since it looks like by assumption $\phi$ has $L^2$ norm one (since you are using it do define a projector), the equality follows.


To be more detailed, $\|\nabla p(t) \psi\|^2 = \|\nabla\phi\|^2 \langle \phi | \psi\rangle|^2 \leq \|\nabla \phi\|^2 \|\phi\|^2 \|\psi\|^2$ so the operator norm is bounded above by $\|\phi\|^2 \|\nabla\phi\|^2 = \|\nabla\phi\|^2$. To show that this is attained just plug in $\phi = \psi$.

$\endgroup$
1
  • $\begingroup$ Great, thanks. I was definitely missing the first row in your answer. Thanks again $\endgroup$
    – popoolmica
    Oct 2, 2015 at 14:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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