The general formulation of quantum mechanics is done by describing quantum mechanical states by vectors $|\psi_t(x)\rangle$ in some Hilbert space $\mathcal{H}$ and describes their time evolution by the Schrödinger equation $$i\hbar\frac{\partial}{\partial t}|\psi_t\rangle = H|\psi_t\rangle$$ where $H$ is the Hamilton operator (for the free particle we have $H=-\frac{\hbar^2}{2m}\Delta$).

Now I have often seen used spaces like $\mathcal{H}=L^2(\mathbb{R}^3)$ (in the case of a single particle), but I was wondering whether this is correct or not. In fact shouldn't we require to be able to derivate $\left|\psi_t\right>$ twice in $x$ and thus choose something like $\mathcal{H} = H^2(\mathbb{R}^3)$?

If we treat directly $\psi(t,x) := \psi_t(x)$, shouldn't we require them to be in something like $H^1(\mathbb{R};H^2(\mathbb{R}^3))$? i.e., functions in $H^1(\mathbb{R})$ with values in $H^2(\mathbb{R}^3)$, e.g. the function $t\mapsto\psi_t$.

  • $\begingroup$ Quantum mechanics rests on the basic fact that wave functions are square integrable and the conservation of probability. My question is, are the spaces you mentioned subspaces of the Hilbert space? $\endgroup$ Commented Jul 26, 2013 at 20:51
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    $\begingroup$ @TorstenHĕrculĕCärlemän The space $L^2(X)$ is the space of square integrable functions on $X$, and you can look at the space $H^k(X)$ as the space of all functions on $X$ which are square integrable and such that all their derivatives up to order $k$ are also square integrable. Look up Sobolev spaces if you want more informations on the subject. $\endgroup$ Commented Jul 26, 2013 at 20:57
  • $\begingroup$ One more question, are functions in $L^2(X)$ not twice differentiable as in the Sobolev space $H^2(X)$? $\endgroup$ Commented Jul 26, 2013 at 21:07
  • $\begingroup$ @TorstenHĕrculĕCärlemän No. The only requirement you have on functions of $L^2(X)$ is to be square integrable. You could for example take $X = [0,1]$ and $f(x) = \chi_{\mathbb{Q}\cap[0,1]}(x)$ (the characteristic function of $\mathbb{Q}\cap[0,1]$) would be in $L^2([0,1])$. However $f$ is nowhere differentiable. $\endgroup$ Commented Jul 26, 2013 at 21:10
  • $\begingroup$ @TorstenHĕrculĕCärlemän A note: I've cheated a bit with my previous example. In fact the elements of $L^2(X)$ are equivalence classes of functions, where two functions are equivalent if their difference squared integrates to zero. In fact you can always find a representative which is at least piecewise continuous. But I assure you there are functions in $L^2$ which are nowhere differentiable. $\endgroup$ Commented Jul 26, 2013 at 21:15

4 Answers 4


Unfortunately, taking $\mathcal{H}=H^2(\mathbb{R}^3)$ won't work. If $f\in H^2(\mathbb{R}^3)$ then, in general, it is not true that $\Delta f\in H^2(\mathbb{R}^3)$. This is a problem, because the Hamiltonian must be an operator from $\mathcal{H}$ to $\mathcal{H}$.

The only mathematically sound way out is taking $\mathcal{H}=L^2(\mathbb{R}^3)$ and considering the Laplacian as an unbounded operator, meaning that it is not defined on the whole $L^2(\mathbb{R}^3)$ space but only on a dense subspace. This idea dates back to von Neumann.

  • $\begingroup$ Nice answer, uses much higher tech than mine, and addresses the point more directly, I was gonna say something about the function space involved in the eigenvalue problem $L u = \lambda u $ is pretty different from the elliptic problem $Lu = f$, then lost it in writing the examples. +1 $\endgroup$
    – Shuhao Cao
    Commented Jul 26, 2013 at 23:50

For the linear Schrödinger equation on $\mathbb{R}^n$ $$ \begin{cases} \partial_t u = i\Delta u \\ u(x,0)=u_0(x) \end{cases} $$ the solution propagator $e^{it\Delta}u_0=(e^{-4\pi it|\xi|^2}\hat{u_0})^{\vee}$ forms a unitary group on $H^s$ for every $s \in \mathbb{R}$ as it's a Fourier multiplier of modulus 1. Thus we wouldn't have a solution in $H^1(\mathbb{R}:H^s(\mathbb{R}^n))$ (no decay in time of $H^s$-norm), but $C(\mathbb{R}:H^s(\mathbb{R}^n))$.

For conjugate exponents $p,p'$ with $p' \in [1,2]$ we have that the propagator $e^{it\Delta} : L^{p'}(\mathbb{R}^n) \rightarrow L^p(\mathbb{R}^n)$ is continuous with $$ \|e^{it\Delta}f\|_p \leq c|t|^{-n/2(1/p'-1/p)}\|f\|_{p'}. $$ This can be proved by interpolating the $L^2$ isometry result along with the case $p=1,p'=\infty$ (which is handled by Young's inequality). This leads to the so called global smoothing effects or Strichartz estimates. For example, the above result along with Hardy-Littlewood-Sobolev inequality implies $$ \left(\int_\mathbb{R} \|e^{it\Delta}f\|_p^q \; dt \right)^{1/q} \leq c\|f\|_2 $$ given some relations on $n,p$ and $q$.

Now you can attack nonlinear problems like the semilinear Schrödinger equation $$ \begin{cases} i\partial_t u = -\Delta u - \lambda |u|^{\alpha-1}u \\ u(x,0)=u_0(x) \end{cases} $$ for $\alpha>1$. One approach is to use Duhamel's principle and solve the (weak) integral formulation $$ u(t) = e^{it\Delta}u_0 + i\lambda \int_0^t e^{i(t-t')\Delta}(|u|^{\alpha-1}u)(t') \; dt' $$ via the contraction mapping principle on an appropriate space. The Strichartz estimates play a key role in deriving the contraction property. For a result about local existence for $H^1$ data, see this paper of Kato in which he produces a solution in $C(I:H^1) \cap C^1(I:H^{-1})$ for some interval $I$ given assumptions on the nonlinear term. More references may be found on the Dispersive Wiki. For more mathematical details on Strichartz estimates and the contraction argument, check out Cazenave's Semilinear Schrödinger Equations.

  • $\begingroup$ This answer is meant to provide mathematical background to results you may encounter in physics literature. $\endgroup$
    – dls
    Commented Jul 26, 2013 at 23:19

My first attempt at a weak formulation for your problem would be: \begin{equation} \left(i\partial_t \psi,\phi\right) - \tfrac{\hbar}{2m}\left(\nabla \psi,\nabla\phi\right)=0, \end{equation} where $(.,.)$ is the $L^2$-inner product. This would (analogous to parabolic equations) require a space like $$ W = \left\{v\in L^2(I;V)\Big| i\partial_t v \in L^2(I;V')\right\} $$ Here, $I$ is a suitable time interval, $V=H^1_0(R^3)$, and $V'$ its dual.

  • $\begingroup$ Yes, your space $V$ is basically what I mean by $H^1(\mathbb{R};H^2(\mathbb{R}^3))$. Notice that you don't need $i\partial_tv\in L^2(I,V')$ (in the sense that you don't need the dual $V'$) because the space is Hilbert. $\endgroup$ Commented Jul 26, 2013 at 20:53
  • $\begingroup$ The space $W$ does not involve second derivatives, since due to integration by parts, only a single derivative is needed $\endgroup$ Commented Jul 26, 2013 at 21:01
  • $\begingroup$ Yes, if you look for weak solutions. Also, shouldn't those be in fact $H^1$ instead of $L^2$ in your definition of $W$? $\endgroup$ Commented Jul 26, 2013 at 21:03
  • $\begingroup$ The distinction of $V$ and $V'$ is important in the definiton of $W$. These two spaces are isomorphic by the Riesz representation theorem, but not equal, see [math.stackexchange.com/a/26291/87363]. $\endgroup$ Commented Jul 26, 2013 at 21:34
  • $\begingroup$ Ok, yeah, you're right. I just never got the hang of the difference... Anyway, this makes sense from the point of view of PDEs, but do you know by any chance some references or similar stuff treating QM in full formality with all the correct spaces? I would be interested to see how it is really done. $\endgroup$ Commented Jul 26, 2013 at 21:38

Short answer: it is technically okay because smooth functions are just subset of the $L^2(\mathbb{R}^3)$.

Long answer: $L^2(\mathbb{R}^3)$ addresses more of the inner product than the differentiability: for normally we want the our eigenfunctions $\{|\psi_j \rangle \}$ are orthogonal to each other, also normalized spacially in $L^2(\mathbb{R}^3)$. $$ \langle \psi_i |\psi_j\rangle = \int_{\mathbb{R}^3} \overline{\psi_i}(t,x)\psi_j(t,x) \,dx = 0, \quad i\neq j. $$ Also normalized: $$ \langle \psi_i |\psi_i\rangle = \int_{\mathbb{R}^3} \overline{\psi_i}(t,x)\psi_i(t,x) \,dx = 1. $$ Such that $$ \frac{d}{dt}\langle \psi_i |\psi_i\rangle = 0 $$ If eigenfunctions satisfy above property, then for any normalized wave function $|\psi\rangle$ spanned using inner product has a diagonal form: $$ |\psi\rangle = \sum^\infty_{i=1} \langle \psi_i |\psi \rangle\, |\psi_i \rangle, $$ translated to mathematical notation: $$ \psi(t,x) = \sum^\infty_{i=1} \alpha_i(t) \,\psi_i(t,x), $$ where $$ \alpha_i(t) = \int_{\mathbb{R}^3} \overline{\psi_i}(t,x)\psi (t,x)dx, $$ so that we can have a physical interpretation, namely, the probability amplitude of the system described by wave function $\psi(t,x)$ being at that state $i$ is $\alpha_i$, a.k.a. $|\alpha_i|^2$ is the probability the system at state $i$ at time $t$. All these benefit from the orthonormality of the basis.

The $\psi_i$'s are found by solving the eigenvalue problem: $$ H\psi_i = E_i \psi_i, $$ where $E_i$ is the eigenvalue for the operator $H$. Mathematical analogy would be the eigenfunctions of minus Laplacian on a square: $$ -\Delta u = k^2u, $$ the eigenfunctions are orthonormal in $L^2$, and orthorgonal in $H^1$.

Differentiability is not the main concern in the quantum world AFAIK.

Firstly, for Bochner space you mentioned, the wave function is normally assumed to be smoothly evolving over time, the obtaining of the wave function for simple models normally follows some Ansatzes, plane wave, decay, etc. The functions we obtain from these Ansatzes are always smooth.

For example, think 1D single particle in a box model: $$ i\hbar\frac{\partial}{\partial t} \psi(t,x) = H\left|\psi \right> = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi(t,x) \tag{$\star$} $$

there is a particle at some state $E$ in $x\in [-1,0]$, then at some time $t_0$ the box length is doubled, $x\in [-1,1]$. What we do is to expand the wave function $\psi(t_0,x)$ at $t_0$ in the old box using the new eigenfunctions $\phi_i$ in the new box $[-1,1]$: $$\psi(t_0,x) = \sum_{i=1}^{\infty} \langle\phi_i(x)|\psi(t_0,x)\rangle \phi_i(x),\tag{1}$$ where the new eigenfunctions $\phi_i$ solves the eigenvalue problem for $(\star)$ in the new box $$ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\phi_i(t,x) = E_i \phi_i(t,x) , \quad \text{ for }x \in[-1,1]. $$ Then $$ \psi(t_0+\Delta t,x) = \sum_{i=1}^{\infty} \langle\phi_i(x)|\psi(t_0,x)\rangle \phi_i(x)e^{ -iE_n \Delta t/\hbar } \tag{2} $$ The question is: Does the wave function evolve smoothly? The answer is yes, well, methodologically, there is no non-smoothness existing. For (1) can be expanded in old eigenfunctions or in new eigenfunctions. An exercise for you is to check what happened when $\Delta t\to 0$ in (2).

Second, the differentiability in space is normally infinite within the domain of interest, and continuous up to the boundary of the domain of interest, zero outside. Just think of that particle in the box model (infinite potential wall): the eigenfunction is set to be zero on boundaries $x=1,-1$.


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