Exploiting Schrödinger equation and its conjugate we can show that $$ \dot{\Psi} = \frac{i \hbar}{2m} \nabla^2 \Psi - \frac{i}{\hbar} U \Psi $$ $$ \dot{\Psi}^* = -\frac{i \hbar}{2m} \nabla^2 \Psi^* + \frac{i}{\hbar} U \Psi^* $$ So $$ \frac{\partial |\Psi|^2}{\partial t} = \frac{i \hbar}{2m} (\Psi^* \nabla^2 \Psi - \Psi \nabla^2 \Psi^*) $$ Exploiting the vector identity $f \nabla^2 g = \nabla \cdot (f \nabla g) - \nabla g \cdot \nabla f$ (with $f$ and $g$ scalar fields) and the divergence theorem (we integrate over $V$), we can rearrange in this way $$ \frac{d}{dt} \int_V |\Psi|^2 dV = \frac{i \hbar}{2m} \oint_S [\Psi^* \nabla \Psi - (\nabla \Psi^*) \Psi] \cdot d \mathbf{S} $$ You can find this expression in Bransden & Joachain's "Physics of atoms and molecules" (page 69 of 2nd ed). Now, to show that $\Psi$ remain normalized as time goes, we should show that $$ \oint_S [\Psi^* \nabla \Psi - (\nabla \Psi^*) \Psi] \cdot d \mathbf{S} \to 0 $$ when the surface $S$ embraces, so to speak, all the 3-dimensional space (and if $\int_V |\Psi|^2 dV$ is finite). In chapter one, Griffiths shows this in the one dimensional case (and not in a completely satisfactory way), but what about the more realistic three-dimensional case?


For all practical purpose in physics, the existence of $\int_V |\Psi|^2dV$ comes with $\Psi(x,t)$ going to 0 as $\|x\|$ goes to $+\infty$ whereas the spatial derivatives stay bounded. Thus your last integral goes to 0.

As Valter Moretti pointed out in the comments, it is easy to build a $\Psi$ such that $\int_{\mathbb{R}^3} |\Psi|^2dV$ is finite but $\Psi$ goes to infinity as $\|x\|$ goes to $+\infty$. It is already possible to do that in 1D. See this question on Mathematics Q&A and the handful of answers to get the idea, then blend that with this trick to get smooth functions. Shake and despair at Mathematics… Fortunately, it turns out we physicists have managed to safely ignore such issues without significant backlash.

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    $\begingroup$ Well, actually the fact that if $|f|^2$ has finite integral then $f$ vanishes at large argument is false even considering smooth functions. I think that the proof of conservation of the norm cannot be given this way as it strictly depends on the fact that the Hamiltonian operator is self-adjoint (not simply Hermitian). Self-adjoint operators are not defined on spaces of smooth functions. At most this way could give some suggestions... $\endgroup$ Oct 10 '17 at 19:45
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    $\begingroup$ Indeed I was just referring to Stone's theorem. The Hamiltonian $H$ of the S. equation viewed a PDE is usually only essentially self-adjoint on a space of smooth functions. To pass to the unitary temporal evolutor one has to enlarge the domain to the closure $\overline{H}$ of $H$. The operator prserving the norm is $e^{-it \overline{H}}$. $\endgroup$ Oct 10 '17 at 20:01
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    $\begingroup$ @ Luc J. Bourhis Ideed it is physics. But you were referring to some mathematical property. My view is the following. There are two possibility. Or one proves assertions with mathematics or one assumes plausible physical hypotheses. The OP's statement is physically plausible ($\psi$ vanishes at infinity) and one could stop there. Trying to fix delicate details requires delicate mathematics. $\endgroup$ Oct 10 '17 at 20:03
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    $\begingroup$ @Fausto Vezzano The point is that the problem is not well-posed from scratch. Schroedinger equation is not a standard PDE because the $t$-derivative is computed with respect to the topology of the Hilbert space whereas the other derivatives are standard only in a suitable domain. Conservation of the $L^2$-norm is guaranteed within this formulation by Stone theorem. Attempts to obtain the same result in a pure PDE interpretation are very difficult. $\endgroup$ Oct 13 '17 at 15:19
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    $\begingroup$ First of all one should prove that the solutions associated to a certain class of initial conditions are rapidly vanishing in space (including spatial derivatives) at every fixed time and the existence of this class strongly depends on the nature of $U$. If simply $U=0$ the afore-mentioned class is Schwartz space: initial condition therein produces solutions therein. It is plausible that localized smooth functions $U$ have analogous properties concerning solutions with initial data in Schwartz space, but I do not know results of this type: I interpret Schroedinger equation in $L^2$ sense. $\endgroup$ Oct 13 '17 at 15:25

@LucJ.Bourhis's answer supplies the needed information for why $\oint_S \rightarrow 0$ as the size of $S\rightarrow \infty$. Put more concretely, examine the normalization integral in spherical coordinates: $$N\equiv\int |\Psi|^2 r^2\sin\theta \operatorname{d} r \operatorname{d}\theta \operatorname{d}\phi.$$ In order for $N$ to be finite, $\lim_{r\rightarrow\infty} r^{3+\epsilon} |\Psi|^2 = 0$ for some $\epsilon > 0$ (i.e. the argument of the radial integral has to vanish faster than $1/r$). If you examine the components of the gradient of $|\Psi|^2$ (equal to the argument of $\oint_S$ by the product rule) you'll find that it vanishes like at least the derivative of the upper bound on $|\Psi|^2$. With the upper bound on $|\Psi|^2$ being $\propto r^{-3-\epsilon}$, then the upper bound on $\left|\partial_r |\Psi|^2\right|$ is $\propto r^{-4-\epsilon}$. This question on math.stackexchange is relevant to whether this proof is sound.

Explaining why the angular components are irrelevant is left as an exercise.

  • $\begingroup$ why I have to average? $\frac{\partial |\Psi|^2}{\partial t} = \dot{\Psi}^* \Psi + \Psi^* \dot{\Psi}$, where is the 2 factor? $\endgroup$ Oct 10 '17 at 21:53
  • $\begingroup$ @FaustoVezzaro Removed criticism - it is incorrect. $\endgroup$ Oct 10 '17 at 22:13

For a system with Hamiltonian $\hat H$, the time-evolution of an arbitrary state $|\Psi\left(\textbf{r},t\right)\rangle$ is governed by $$i\hbar\frac{\partial}{\partial t}|\Psi\left(\textbf{r},t\right)\rangle = \hat H |\Psi\left(\textbf{r},t\right)\rangle \tag{1}\label{SE}.$$

We can take the adjoint of Eq.\eqref{SE} and it reads $$-i\hbar\frac{\partial}{\partial t}\langle\Psi\left(\textbf{r},t\right)| = \langle\Psi\left(\textbf{r},t\right)| \hat H \tag{2}\label{adjSE}.$$

Note that we assume the Hamiltonian to be Hermitian, therefore in the right-hand-side of Eq. \eqref{adjSE}, $\hat H^ \dagger$ was replaced by $\hat H$.

Now we get to the original question namely, the time evolution of the normalization $ \mathcal{N}(t) := \langle\Psi\left(\textbf{r},t\right)|\Psi\left(\textbf{r},t\right)\rangle$ of the wavefunction. Therefore we would like to evaluate the following quantity -- $$\mathcal{\dot N}(t) = \frac{\partial}{\partial t}\langle\Psi\left(\textbf{r},t\right)|\Psi\left(\textbf{r},t\right)\rangle.$$

We proceed as \begin{align} \mathcal{\dot N}(t) & = \frac{\partial}{\partial t}\langle\Psi\left(\textbf{r},t\right)|\Psi\left(\textbf{r},t\right)\rangle, \\ \\ & = \left(\frac{\partial}{\partial t} \langle\Psi\left(\textbf{r},t\right)|\right)|\Psi\left(\textbf{r},t\right)\rangle + \langle\Psi\left(\textbf{r},t\right)|\left(\frac{\partial}{\partial t} |\Psi\left(\textbf{r},t\right)\rangle\right), \\ \\ & = \frac{1}{-i\hbar}\left(-i\hbar\frac{\partial}{\partial t} \langle\Psi\left(\textbf{r},t\right)|\right)|\Psi\left(\textbf{r},t\right)\rangle + \langle\Psi\left(\textbf{r},t\right)|~\frac{1}{i\hbar}~\left(i\hbar\frac{\partial}{\partial t} |\Psi\left(\textbf{r},t\right)\rangle\right), \\ \\ & = \frac{1}{-i\hbar}~\left(\langle\Psi\left(\textbf{r},t\right)|\hat H\right)|\Psi\left(\textbf{r},t\right)\rangle + \langle\Psi\left(\textbf{r},t\right)|~\frac{1}{i\hbar}~\left(\hat H|\Psi\left(\textbf{r},t\right)\rangle\right), \mathrm{Using~Eqs. \eqref{SE} \& \eqref{adjSE}} \\ \\ & = \frac{1}{-i\hbar}~\langle\Psi\left(\textbf{r},t\right)|\hat H|\Psi\left(\textbf{r},t\right)\rangle + \frac{1}{i\hbar}~\langle\Psi\left(\textbf{r},t\right)|\hat H|\Psi\left(\textbf{r},t\right)\rangle, \\ \\ & = 0. \end{align}

Since the time-derivative of the normalization is zero, therefore the normalization remains constant in time.

Things to note:

  • For the normalization to be time-independent, one requires unitary time-evolution of the states $|\Psi\left(\textbf{r},t\right)\rangle$, namely $$|\Psi\left(\textbf{r},t\right)\rangle = \mathcal{U}\left(t,t'\right)|\Psi\left(\textbf{r},t'\right)\rangle.$$
  • For $\mathcal{U}\left(t,t'\right)$ to be unitary operator of the form $$\mathcal{U}\left(t,t'\right) = \exp\left(-i\hat H (t-t')/\hbar\right),$$ one requires $\hat H$ to be Hermitian.
  • The fundamental reason for the states to stay normalized in a Schrodinger-type time-evolution, is the Hermiticity of the Hamiltonian. Although, there are situations where the states can stay normalized under time-evolution for Hamiltonian which are not Hermitian. However, the time-independence of the normalization is guaranted for a unitary time-evolution, namely for a Hermitian Hamiltonian, under the Schrodinger-type equation.

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