Minimal conditions to ensure uniqueness of solutions of Schrödinger equation IVP

Question

Let us consider the free non-relativistic Schrödinger equation $$i\partial_t \psi =-\frac{1}{2}\partial_x^2 \psi=:H\psi.$$ Adapting Fritz John's pathological solution to the heat equation, I find that the non-zero smooth function $$\varphi:\mathbb{R}^2 \to \mathbb{C}:(x,t) \mapsto \sum_{n=0}^\infty f^{(n)}(t)\frac{x^{2n}(-2i)^n}{(2n)!}, \qquad f(t)\equiv e^{-1/t^2}$$ solves the free Schrödinger equation while reducing identically to zero as $t\to 0$. This establishes that the Schrödinger equation, regarded as a PDE at face value, never offers a unique solution to an initial value problem.

Traditionally, we add in the constraint that the solution of the Schrödinger equation ought to maneuver inside $L^2(\mathbb{R})$ in order to make the Born rule operable. However, usual treatments also add in strong-continuity-like ingredients so that we can finally handle the Schrödinger equation with a cosy and standard functional-analytic framework. However, the physical interpretation and requirement of these continuity ingredients is a bit obscure to me, certainly so since they are to a certain degree non-local (e.g. in the semigroup context, it is demanded that there is a dense core of "classical solutions" $t \mapsto \psi(t)$ characterized by $\forall t \in \mathbb{R}: H\psi(t) \in L^2(\mathbb{R})$, which indeed has the flavour of a non-local weight condition).

Q: Does Born's integrability condition $\psi(t)\in L^2(\mathbb{R})$ suffice to select unique solutions for the Schrödinger-equation-related IVP (or do we really need the additional $\partial_x^2\psi(t)\in L^2(\mathbb{R})$ or similar strong-continuity requirements)?

EDIT(18/02/19): One is of course tempted to use $\psi \in L^2(\mathbb{R})$ to our advantage by allowing us to use the Fourier transform in the direction of $x$: the Schrödinger equation then reads $i\partial_t \hat{\psi} = p^2 \hat{\psi}$ from where uniqueness seems easy to obtain. I'm unsure though what to say about the necessary "differentiations under the integral sign" and partial differentiations that are required along this line of thinking.

Answer 1

Let us make the following three assumptions:

1. The Schrödinger equation is interpreted in the following weak sense: $\forall v \in C_0^{\infty}(\mathbb{R})$ $$i\frac{d}{dt}\int_{-\infty}^{\infty}v(x)\psi(x,t)dx = -\frac{1}{2}\int_{-\infty}^{+\infty} v''(x)\psi(x,t)dx$$ $(\psi(.,t)$ Borel measurable for all $t \in \mathbb{R})$

2. $\forall t_1,t_2 \in \mathbb{R}:\,\psi(.,t_1)\in L^2(\mathbb{R})$ and

$$\int_{t_1}^{t_2}dt\,\|\psi(.,t)\|_2<+\infty.$$

3. $\psi(x,0)=0\,$ for almost all $x\in \mathbb{R}$

(the motivation for the 3rd condition comes from the linearity of the problem at hand: If $\psi_1$ and $\psi_2$ both obey conditions 1 and 2 and agree $\psi(.,0)=\psi_2(.,0)$ a.e., then $\psi:=\psi_1-\psi_2$ obeys condition 1, 2 and 3.)

Then $\forall t \in \mathbb{R}:\,\psi(x,t) = 0$ for almost all $x\in \mathbb{R}$.

To show this, we consider the family $v_{R,p}(x):=\eta(x/R)e^{ipx}$ of compactly supported functions where $\eta$ is a standard bump function: $\eta \in C_0^\infty$ and $\eta(0)=1$. Inserting $v= v_{R,p}$ in the Schrödinger equation above and abbreviating $\hat{\psi}_R(p,t):=\int_{-\infty}^{\infty}v_{R,p}(x)\psi(x,t)dx$, we get $$(i\partial_t-\frac{1}{2}p^2)\hat{\psi}_R(p,t)= -\frac{1}{2R}\int_{-\infty}^{\infty}\underbrace{\left(2ip\eta'(x/R) + \eta''(x/R)/R\right)e^{ipx}}_{=:f_R(x)}\psi(x,t)dx=:\Delta_R(p,t)$$ Now, $$\partial_t\left|\hat{\psi}_R(p,t)\right|^2=2\text{ Im}\left[-i\overline{\hat{\psi}_R(p,t)}\Delta_R(p,t)\right]\leq |\hat{\psi}_R(p,t)||\Delta_R(p,t)|\leq |\hat{\psi}_R(p,t)|\frac{\|f_R\|_2\|\psi(t)\|_2}{2R}.\qquad(*)$$ Dividing (*) by $(1+\left|\hat{\psi}_R(p,t)\right|^2)$ and using the identity $\{\forall \xi>0:\,\frac{\xi}{1+\xi^2}\leq 1\}$, we get $$\partial_t\log\left(1+\left|\hat{\psi}_R(p,t)\right|^2\right)\leq \frac{\|f_R\|_2\|\psi(t)\|_2}{2R}$$ Since $f_R$ has a support with diameter of order $R$ and amplitude of order 1, we can find $M>0$ s.t. $\forall R>0:\,\|f_R\|_2\leq MR^{1/2}$. By integrating the previous inequality, this yields $$\int_0^T dt \,\partial_t\log\left(1+\left|\hat{\psi}_R(p,t)\right|^2\right)\leq \frac{M}{2R^{1/2}}\int_0^Tdt\,\|\psi(t)\|_2.\qquad(**)$$ To deal with the left hand side of this inequality, we note that assumption 1 enables repeated differentiation of $\hat{\psi}_R(p,.)$ (differentiation under the integral) so that this function is $C^\infty$ so that the integrand of the LHS of (**) is $C^\infty$ and therefore subject to the simplest version of the fundamental theorem of calculus, which implies $$\log\left(1+\left|\hat{\psi}_R(p,T)\right|^2\right)\leq \frac{M}{2R^{1/2}}\int_0^Tdt\,\|\psi(t)\|_2.$$ Since $\psi(.,t)\in L^2(\mathbb{R})$ and $\eta(./R)\psi(.) \overset{R \to \infty}{\to} \psi$ in $L^2(\mathbb{R})$, Plancherel's theorem (unitarity of Fourier transform) asserts that $\hat{\psi}_R(.,t)\overset{R \to \infty}{\to}({\cal F}\psi)(.,t)\overset{\text{abbrev.}}{=}\hat{\psi}(.,t)$ in $L^2(\mathbb{R})$. So, for almost all $p \in \mathbb{R}$, $$\left|\hat{\psi}(p,t)\right|=\lim_{R\to \infty}\left|\hat{\psi}_R(p,t)\right|\leq \lim_{R\to \infty}\left(\exp\left(\frac{M}{2R^{1/2}}\int_0^T dt\,\|\psi(t)\|\right)-1\right)=0$$ by virtue of assumption 2. Again invoking Plancherel's theorem, this means that $\psi(.,t)=0$ in the sense of $L^2(\mathbb{R})$, i.e. $\psi(.,t)$ is zero a.e.

Answer 2

PoW of theoretical physics. In the mathematical take to quantum mechanics, $\psi_t:=\psi(t)$ being in $L^2$ is no additional constraint. It comes built-in in the standard axiomatization:

1. Axiom (Dirac - von Neumann) Quantum states are modelled as elements of the projective pace $\mathbb P H$ for some separable complex Hilbert space $H$

2. Theorem (vv) The free evolution of a quantum state is determined by the free Schrödinger equation

From this we conclude that $\psi_t\in L^2$. Whereas a priori (from the pow of intuitionist QM) this is related to Born's rule as you write, a posteriori (from the pow of axiomatic QM) it can be derived from other axioms:

3. Theorem. (Gleason) Every non-contextual measurement of a quantum system satisfies Born's rule.

In other words, Born's rule can be proven from the other axioms of QM, so we can dispense from assuming it.

Secondly, this

they are to a certain degree non-local (e.g. in the semigroup context, it is demanded that there is a dense core of "classical solutions" $t \mapsto \psi(t)$ characterized by $\forall t \in \mathbb{R}: H\psi(t) \in L^2(\mathbb{R})$, which indeed has the flavour of a non-local weight condition).

is not correct, neither from the pow of physics, nor from the pow of mathematics. I'll come back to the mathematics below, but let me just stress that there is nothing non-local here (neither in space nor in time) and that the existence of a core is a theorem, not a 'demand'. Back to the physics:

4. Axiom (Dirac - von Neumann) Observables are (possibly unbounded) self-adjoint operators on $H$

The observable kinetic energy is the free particle Hamiltonian $(\Delta, W^{2,2})$, i.e. the closed $L^2$-Laplacian (not any Laplacian on some arbitrary space of functions). This fact is derived from the quantization of the corresponding Newtonian description and can be found in every textbook on the subject. Now:

5. Theorem (Stone) $(A,\mathrm{dom}(A))$ is a self-adjoint operator on $H$ $\iff$ $e^{it A}$ is a strongly continuous one-parameter unitary group

No 'demands' here. We start from $(\Delta, W^{2,2})$ because we choose to (axiom), and we get a strongly continuous unitary group by proving so (theorem).

At this point:

uniqueness is an obvious consequence of unitarity of $e^{it\Delta}$

since $\psi_t=e^{it\Delta}\psi_0$, i.e. we have written the solution as a function of the initial datum.

PoW of mathematics. While from the pow of the standard axiomatization of quantum mechanics, solutions to the free Schrödinger equation must be elements of $L^2$, from the point of view of mathematics it might make sense to ask about uniqueness of solutions to the free Schrödinger equation when $\Delta$ is some Laplacian on some function space different from $L^2$.

If we choose to discuss $W^{1,2}$-weak solutions, then we immediately fall back to $(\Delta, W^{2,2})$, because the $W^{1,2}$-norm is the energy form of $(\Delta, W^{2,2})$, which motivates why your answer post, although maybe correct, is unnecessary complicated.

Standard alternatives are $W^{1,p}$-weak solutions for $p\neq 2$, or strong solutions (at least $\mathcal C^0$). $W^{1,p}$-solutions are strongly related to $W^{1,2}$-solutions since there is no potential involved in the equation that can prescribe more stringent summability conditions.

Coming to strong solutions, we have the following uniqueness result (which I adapt to the case of the free Schrödinger, but the authors discuss the case of quite general potentials)

6. Theorem (L. Escauriaza , C. E. Kenig , G. Ponce & L. Vega, 2006, Thm. 1.1) If $u\in \mathcal C([0,T], W^{2,2})$ is a strong solution to the free Schrödinger equation, and there exist $\alpha>2$ and $a>0$ such that $$u_0, u_1\in W^{1,2}(e^{a|x|^\alpha}\mathrm{d}x)$$ then $u\equiv 0$.

For the free equation, this can be understood as a consequence of Hardy's uncertainty principle. Outside this class there exist non-trivial solutions, as noted in the op. This is the same as for the heat equation, where one cannot expect unique solutions outside the Tikhonov or Åsplund classes.