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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?

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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 $\phi = \psi$.

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  • $\begingroup$ Great, thanks. I was definitely missing the first row in your answer. Thanks again $\endgroup$ – popoolmica Oct 2 '15 at 14:36

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