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.

  • $\begingroup$ What kind of solutions do you want to consider? $\endgroup$
    – MaoWao
    Feb 18, 2019 at 16:39
  • $\begingroup$ As far as the meaning of the derivatives of the PDE is concerned? I have no bias towards weak/strong solutions at this point. The remainder of my biases is explained in the post. $\endgroup$
    – Vergilius
    Feb 18, 2019 at 20:19

2 Answers 2


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


  1. $\psi(x,0)=0\,$ 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.

  • 2
    $\begingroup$ What exactly are you saying here, and what does it have to do with the original question (asked by yourself 4 days earlier)? $\endgroup$ Oct 6, 2020 at 21:01
  • $\begingroup$ @Vergilius I was reading your answer because of this similar question (math.stackexchange.com/questions/4638435/…). In assumption 1. you interchanged on the left the integral and the derivative with respect to $t$. For that, it seems to me that some assumption on $\psi$ is needed (in addition to 2.), something like $|\psi_t(x,t)|\leq g(x)$ for some $g\in L^1$ ? $\endgroup$
    – user111
    Feb 15 at 11:09
  • $\begingroup$ @user111: no, it is just a type of weak formulation of the Schrödinger equation. It is legitimate to start out with such a formulation from the beginning in stead of trying to derive it from a strong formulation. $\endgroup$
    – Vergilius
    Feb 15 at 16:23
  • $\begingroup$ But, then, your analysis doesn't apply to strong solutions ? (sorry, I am not an expert on weak formulations...). $\endgroup$
    – user111
    Feb 15 at 17:40
  • $\begingroup$ I think it does. Since $\partial_t \psi$ is continuous if $\psi$ is $C^1$ , it is locally integrable and this local integrability suffices since the integrand contains a multiplication with a smooth compactly supported function. $\endgroup$
    – Vergilius
    Feb 15 at 17:44

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 $H$
  1. 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:

  1. 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:

  1. 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:

  1. 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)

  1. 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.

  • $\begingroup$ I never mentioned "quantum mechanics" in my question, so your paragraph titled "PoW of theor. physics" is irrelevant. Let me mention however that 1) the Laplacian is not the quantization of position but kinetic energy. 2) The whole self-adjoint/Stone theorem/time-translation invariance paradigm does involve taking self-adjoint extensions of symm. operators, which is in a sense a non-local enterprise, so I don't accept your criticism of certain of my paragraphs at face value. I can just stand with my counterexample given in my question and say that you ignored it or did not address it. $\endgroup$
    – Vergilius
    Jun 27, 2021 at 14:44
  • $\begingroup$ That aside, I will look at the paper that you referenced to (What are the subscripts on the $u$'s? Does the inclusion into $W^{1,2}(...)$ have to hold for all $t$ or did you write a typo and should it have been $W^{1,2}(e^{a|x|^{\alpha}dxdt})$?). $\endgroup$
    – Vergilius
    Jun 27, 2021 at 14:47
  • $\begingroup$ @ThibautDemaerel you are right about the quantization of position of course. I'll edit the answer. Concerning the rest, you certainly mentioned Born's rule, which is definitely within QM. Also, being the free Schrödinger equation a central subject of QM, there is hardly the need to mention it. My point was simply that Born's rule (whether assumption or outcome) is desirable in QM, and not an obstacle. About 2) I have no clue what you mean by saying that taking self-adjoint extensions is a "non-local enterprise". I am happy to discuss this point but it is clear that we have a disagreement... $\endgroup$
    – AlephBeth
    Jun 27, 2021 at 16:59
  • $\begingroup$ as to what local means. Coming to the counterexample, I have commented in the answer that outside the Tikhonov class you cannot expect uniqueness of solutions (the Tikhonov class being defined by the conditions in "6. Theorem"). The subscript just denotes evaluation in time, i.e. $u_0:=u(\cdot, 0)$ and similarly for $u_1$, so there is no typo. $\endgroup$
    – AlephBeth
    Jun 27, 2021 at 17:02
  • $\begingroup$ But it seems I do have a result in my answer which does not require global integrability of the (weak) derivative of the wave-function (something that you seem to require in the function classes that you mention). Regarding the term "non-local", for me global is synonymous to non-local in the context of this question, i.e. not the non-local of Bell or something like that. A global integrability condition = a "non-local" integrability condition. $\endgroup$
    – Vergilius
    Jun 27, 2021 at 17:31

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