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Let $(x,t) \in \mathbb{R}^2$ and consider the following partial differential equation for the real-valued function $U(x,t)$: \begin{equation} \frac{\partial^2 U}{\partial t^2} = - \frac{\hbar^2}{4m^2} \frac{\partial^4 U}{\partial x^4}, \end{equation} where $m$ and $\hbar$ are positive constants. In the following we shall be quite sloppy, and we shall assume that given (smooth enough) initial conditions $U(x,0)$ and $\frac{\partial U}{\partial t}(x,0)$ (lying in some space) there exists a unique (smooth enough) solution $U$ (lying in some space) to this partial differential equation. Let us call the set of solutions $\mathcal{E}$.

Let us define for ease of notation $D_{x}^k F=\left( F, \frac{\partial F}{\partial x}, \dots, \frac{\partial^k F}{\partial x^k} \right)$ for every non-negative integer $k$ and for every smooth enough function $F(x,t)$. I ask whether there exist some non non-negative integer $k$ and some (smooth enough) real-valued functions $p \geq 0$ and $j$, with $p$ non-constant, such that, by setting \begin{equation} P(x,t)=p \left((D_{x}^k U)(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right), \\ J(x,t)=j \left((D_{x}^k U)(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right), \end{equation} the following properties hold:

(i) for every $U \in \mathcal{E}$ the following continuity equation holds \begin{equation} \frac{\partial P}{\partial t} + \frac{\partial J}{\partial x}=0; \end{equation}

(ii) for the special solution $U(x,t)=\cos\left(\sqrt{\frac{2m \omega}{\hbar}}x-\omega t \right)$, we have that $P(x,t)$ is independent of $\omega > 0$.

The answer should be negative, but I have no idea of a possible proof. Obviously, since we have not formulated the problem in a rigorous way, we do not expect to get a rigorous proof, but only some heuristic, convincing argument in this direction.

NOTE (1) Let us explicitly note a trivial consequence of property (ii). Since for the special solution \begin{equation} U(x,t)=\cos\left(\sqrt{\frac{2m \omega}{\hbar}}x-\omega t \right), \end{equation} $P(x,t)$ is independent of $\omega > 0$ for every $(x,t) \in \mathbb{R}^2$, we have that $(x,t) \mapsto P(x,t)$ is a constant function. Indeed, for this special solution, we have that $P(x,t)=F(x-vt)$ for some function $F$, where $v=\sqrt{\frac{\hbar \omega}{2m}}$. If we had $F'(\xi)\neq 0$ for some $\xi \in \mathbb{R}$, we would get for $x-vt=\xi$: \begin{equation} \left. \frac{\partial P}{\partial t}(x,t) \middle/ \frac{\partial P}{\partial x}(x,t) \right. = -v = \sqrt{\frac{\hbar \omega}{2m}}, \end{equation} a contradiction.

NOTE (2) This problem, as the notation shows, has a physical background, and the mathematical formulation of the problem that I give here is my personal interpretation of a physical exposition given by the great XXth century physicist David Bohm in his wonderful treatise Quantum Theory published in 1951. Bohm explicitly states that the problem has a negative answer, without giving any proof or heuristic argument. For all the physical details about this problem see my post Nonexistence of a Probability for Real Wave Functions.

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    $\begingroup$ If you compute the continuity equation at t=0 you get a relation which depends entirely on the initial conditions for U, which are completely arbitrary. I have the feeling that by playing with suitable choices of U(t=0) and U_t(t=0) you can prove P,J to be constant $\endgroup$ – Piero D'Ancona Nov 6 '18 at 18:13
  • $\begingroup$ Well, if you compute the continuity equation at $t=0$, by using the PDE in order to express $U_{tt}$, you actually get an equation which, given the arbitrariness of $U(t=0)$ and $U_t(t=0)$ and the Borel's Lemma, gives us a PDE for $p$ and $j$. But for sure this PDE has non-constant solutions. Actually, there exist non-constant functions $p \geq 0$ and $j$ satisfying the continuity equation: see the example given by Bohm in my post Nonexistence of a Probability for Real Wave Equations. $\endgroup$ – Maurizio Barbato Nov 6 '18 at 18:42
  • $\begingroup$ We conclude that in any case property (ii) must play some role in the proof of the non-existence result (if it is true). $\endgroup$ – Maurizio Barbato Nov 6 '18 at 18:42
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This is not an answer, but only a remark too long to be posted as a comment.

We can make the formulation of the problem more explicit in the following way. Let $y=(y_0,y_1,\dots,y_k) \in \mathbb{R}^{k+1}$ and $z=(z_0,z_1,\dots,z_k) \in \mathbb{R}^{k+1}$. Then to say that the functions $(y,z) \mapsto p(y,z)$ and $(y,z) \mapsto j(y,z)$ satisfy property (i) means that for every $U \in \mathcal{E}$ \begin{equation} \nabla_{y} p \cdot D_{x}^{k} \frac{\partial U}{\partial t} + \nabla_{z} p \cdot D_{x}^{k} \frac{\partial^2 U}{\partial t^2} + \nabla_{y} j \cdot D_{x}^{k} \frac{\partial U}{\partial x} + \nabla_{z} j \cdot D_{x}^{k} \frac{\partial^2 U}{\partial x \partial t}=0, \end{equation} which is equivalent to say that \begin{equation} \nabla_{y} p \cdot D_{x}^{k} \frac{\partial U}{\partial t} - \frac{\hbar^2}{4 m^2} \nabla_{z} p \cdot D_{x}^{k} \frac{\partial^4 U}{\partial x^4} + \nabla_{y} j \cdot D_{x}^{k} \frac{\partial U}{\partial x} + \nabla_{z} j \cdot D_{x}^{k} \frac{\partial^2 U}{\partial x \partial t}=0. \end{equation} Now Borel Theorem tells us that for every real sequence $(a_{n})_{n=0}^{\infty}$ and every fixed $x_0 \in \mathbb{R}$ there exists a smooth function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that \begin{equation} \frac{d^m f}{d x^m}(x_0)=a_{n} \quad (n=0,1,2,\dots), \end{equation} so that from the arbitrariness of the initial data $x \mapsto U(x,0)$ and $x \mapsto \frac{\partial U}{\partial t}(x,0)$, by replacing $\frac{\partial^m U}{\partial x^m}$ by $y_m \in \mathbb{R}$ and $\frac{\partial^{m+1} U}{\partial^m x \partial t}$ by $z_m \in \mathbb{R}$ for every non-negative integer $m$, we conclude that $p$ and $j$ satisfy the following equation:

\begin{equation} \sum_{m=0}^{k}\frac{\partial p}{\partial y_m} z_{m} - \frac{\hbar^2}{4 m^2}\sum_{m=0}^{k} \frac{\partial p}{\partial z_m} y_{m+4} + \sum_{m=0}^{k} \frac{\partial j}{\partial y_m} y_{m+1} + \sum_{m=0}^{k} \frac{\partial j}{\partial z_m} z_{m+1} =0 \qquad (I). \end{equation} From this equation we immediately deduce that $p$ does not depend on $z_{k-2}, z_{k-1}, z_k$ and that $j$ does not depend on $z_k$. Moreoever, we also get that \begin{equation} - \frac{\hbar^2}{4m^2} \frac{\partial p}{\partial z_{k-3}}+ \frac{\partial j}{\partial y_k} = 0, \end{equation} and since $p$ and $j$ do not depend on $z_k$ it also holds \begin{equation} \frac{\partial p}{\partial y_k} + \frac{\partial j}{\partial z_{k-1}}=0. \end{equation}

There are no other simple consequences of Equation (I). Now, the issue in the post amounts to see whether there exist solutions $p \geq 0$ and $j$ of Equation (I), with $p$ non-constant and satisfying property (ii). This latter request makes all the matter quite involved.

Let us explicitly note there are simple non-constant functions $p(y,z) \geq 0$ which satisfy (ii). Consider e.g. $p(y,z)=(y_1 z_1 - z_0 y_2)^2$. It is easy to see that in this case there is no $k$ and no function $j$ such that $p$ and $j$ satisfy equation (I). Indeed, equation (I) would imply in our case that $j$ depends only on $y_0,\dots, y_4, z_0,z_1$. Moreover it would also imply that

\begin{equation} \frac{\partial j}{\partial y_4}= \frac{\hbar^2}{2 m^2}(y_1 z_1 - z_0 y_2) y_1, \\ \frac{\partial j}{\partial z_1}= 2(y_1 z_1 - z_0 y_2) z_0. \end{equation}

From the last equation we would get for some function $g$: \begin{equation} j=2(y_1 z_1 - z_0 y_2) z_0 z_1 + g(y_0,\dots,y_4,z_0), \end{equation} so that we should have \begin{equation} \frac{\partial g}{\partial y_4} = \frac{\partial j}{\partial y_4} = \frac{\hbar^2}{2 m^2}(y_1 z_1 - z_0 y_2) y_1, \end{equation}

which is a contradiction since $g$ does not depend on $z_1$.

The difficulty of the problem is to show in general that conditions (i) and (ii) are not compatible, and I have no idea for now of how to tackle the problem.

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