Conservation of momentum for nonlinear Schrödinger equation I am having trouble proving the following momentum conservation law.
Given smooth compactly supported solution $u(x,t)\in \mathbb{C}$ where $(x,t)\in \mathbb{R}^n\times \mathbb{R}$ of $iu_t+\Delta u= |u|^{p-1}u$, and
$$ \vec{p}(t)= \text{Im} \int_{\mathbb{R}^n} \overline{u}\, \nabla{u}\, dx,$$
how do I show that $\partial_t \vec{p}(t)=0$, i.e. the above quantity is conserved ?
For simplicity, take $p$ to be odd integer.
I tried to make use of the equation and did integration by parts but it becomes a mess, and I think I must have missed something. Thank you.
 A: Here's a *formal* way of proving this conservation law by using the Hamiltonian structure of the equation (for more details, see these notes by Holmer & Zworski, first section). 
EDIT. See the bottom of this post for another proof based on integration by parts.
Preliminaries.
Introducing the symplectic form 
$$
\omega(f, g)=\Im \int_{\mathbb{R}^n} f(x)\overline{g(x)}\, dx, $$
and the Hamiltonian function 
$$
H(u)=\int_{\mathbb{R}^n}\frac{\lvert \nabla u\rvert^2}2 +\frac{\lvert u\rvert^{p+1}}{p+1}\, dx,$$
the nonlinear Schrödinger's equation can be formally written as 
$$\tag{NLS}
u_t=\nabla_\omega H(u), $$
where the "symplectic gradient" $\nabla_\omega H(u)$ denotes the unique function with the following property:
$$\omega(v, \nabla_\omega H(u))=\left.\frac{\partial}{\partial \epsilon}H(u+\epsilon v)\right|_{\epsilon =0}\qquad \forall v.$$ 
Remark: 
The symplectic form is anti-symmetric: $\omega(f, g)=-\omega(g, f)$
The formulation (NLS) allows the manipulation of the equation from an abstract point of view. Specifically, we consider the action of  differentiable transformations $S_\epsilon=S_\epsilon(u)$, with the property that $S_0(u)=u$. Denote 
$$
S'(u)=\left.\frac\partial{\partial \epsilon}S_\epsilon(u)\right|_{\epsilon=0}.$$
The chain rule gives:
$$
\begin{split}
\left.\frac\partial{\partial \epsilon} H(S_\epsilon(u))\right|_{\epsilon=0}&=\omega\left(S'(u), \nabla_\omega H(u)\right) \\
[\text{using (NLS)}]&= \omega(S'(u), u_t).
\end{split}
$$
This last identity is the key ingredient in the following recipe.

A recipe for finding conserved quantities.
Assume that $S'$ commutes with $\partial_t$ and that it is
  anti-symmetric (AS) with respect to $\omega$, that is $$\tag{AS}\omega(u, S'(v))=\omega(v, S'(u)), \qquad \forall u, v.$$ Then one has
  $$\begin{split}\frac\partial{\partial t}\omega(S'(u), u)&=
 \omega(S'(u), u_t) + \omega(S'(u_t), u) \\
 &=2\left.\frac\partial{\partial \epsilon}
 H(S_\epsilon(u))\right|_{\epsilon=0}.\end{split}$$ In particular, if the Hamiltonian is invariant (IN)
  $$\tag{IN}H(S_\epsilon(u))=H(u), \qquad \text{for all $\epsilon$ in a
 neighborhood of $0$}$$ then the quantity $\omega(S'(u), u)$ is
  preserved by the (NLS) flow.

The conservation law the OP asks about comes from the transformations $$S^{j}_\epsilon(u)(x, t)=u(x+\epsilon e_j, t), $$
(where $e_j=(0\ldots 1 \ldots 0)$, the $1$ being in the $j$-th position). It is manifest that the Hamiltonian is invariant under these transformations. Since $$(S^{j})' (u)=\frac{\partial u}{\partial x_j}, $$
the anti-symmetric property (AS) is easily checked by integration by parts. This completes the proof.
Remark. More generally, any family of differentiable transformations with the following property (symplectomorphism):
$$\omega(S_\epsilon(u), S_\epsilon(v))=\omega(u, v)$$
satisfies property (AS), as it is seen by differentiating with respect to $\epsilon$ and then using the anti-symmetry of the symplectic form $\omega$.
EDIT (A more direct approach). In practice, one expects from the previous discussion (or, in the OP case, from the assignment) that the conserved quantity must be of the form 
$$
\Im \int S’(u) \overline u.$$ 
where $S’(u)=u_{x_j}$. One can therefore multiply the equation termwise by $u_{x_j}$, take the imaginary part and integrate, yielding 
$$\tag{*}
\Im \int u_t \overline{u_{x_j}} - \Im i\int \Delta u \overline{u_{x_j}} + \Im i\int |u|^{p-1}u\overline{u_{x_j}}=0.$$ 
We observe that
$$i\int \Delta u \overline{u_{x_j}} \in \mathbb R, $$ 
by integration by parts (in Dirac’s notation, this integral coincides with the sandwich $\langle u | \Delta \cdot (-i\partial_{x_j})| u\rangle$, which is real by self-adjointness). So the  second summand in (*) vanishes. The third summand also vanishes because of the identity 
$$
|u|^{p-1}u \overline{u_{x_j}} = \partial_{x_j} \left( \frac{|u|^{p+1}}{p+1} \right),$$
so $\int |u|^{p-1}u\overline{u_{x_j}} = \int \partial_{x_j}(\mathrm{something}) = 0$, again by integration by parts. We are left with 
$$
\Im \int u_t \overline{u_{x_j}} = 0, $$ 
which is almost what we need. We use the identity 
$$
\partial_t ( u \overline{u_{x_j}} ) = u_t \overline{u_{x_j}} + u \overline{u_{t\, x_j} }$$ 
to obtain 
$$
\partial_t \Im \int u \overline{u_{x_j}} + \Im \int u \overline{ u_{t\, x_j}} =0.$$ 
It remains to show that the second summand vanishes. We use the equation to obtain 
$$
u_{t\, x_j}= i\Delta u_{x_j} +i\left( |u|^{p-1} u\right)_{x_j} $$ 
and so 
$$
\Im \int u \overline{u_{t\,x_j}} = \Im \int u \overline{\Delta i\partial_{x_j} u} + \Im \int u\overline{ i(|u|^{p-1} u)_{x_j}}.$$ 
Both summands here vanish: the first is the imaginary part of a purely real number and the second is the integral of a derivative. We conclude that 
$$
\partial_t \Im \int u \overline{u_{x_j}} =0.$$ 
Conclusion This is probably not the most efficient way to carry out this computation. These things tend to be quite messy, and that’s why the abstract framework provided above can be useful to at least know where one is going.
