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David Bohm in Section (4.5) of his wonderful monograph Quantum Theory gives an argument to show that in order to build a physically meaningful theory of quantum phenomena, the wave function $\psi$ must be a complex function.

His argument goes as follows. Consider by simplicity the one-dimensional case, and let $$\psi=U + i V,$$ with $U$ and $V$ real. From Schrödinger's equation you easily see that $U$ and $V$ satisfy the following uncoupled second-order equations:

\begin{equation} \frac{\partial^2 U}{\partial t^2} = - \frac{\hbar^2}{4m^2} \frac{\partial^4 U}{\partial x^4},\\ \frac{\partial^2 V}{\partial t^2} = - \frac{\hbar^2}{4m^2} \frac{\partial^4 V}{\partial x^4}. \end{equation} We could think then to replace Schrödinger's equation for the complex-valued function $\psi$ with e.g. the above equation for $U$, so to build all quantum mechanics on $U$. In order to do so, we should be able to define a positive conserved probability $P$, defined as a function of $U$, and the partial derivatives of $U$ of order at most 1 with respect to $t$ (since the initial data for our second-order equation are $U(x,0)$ and $\frac{\partial U}{\partial t}(x,0)$, we require $P(x,t)$ to be a function of the state $U(x,t)$ and $\frac{\partial U}{\partial t}(x,t)$ of the system at time $t$ and their spatial derivatives). Try e.g. \begin{equation} P=\frac{1}{2} \left( \frac{\partial U}{\partial t} \right)^2 + \frac{\hbar^2}{8m^2} \left( \frac{\partial^2 U}{\partial x^2} \right)^2. \end{equation} One can easily see that the following continuity equation holds \begin{equation} \frac{\partial P}{\partial t} + \frac{\partial J}{\partial x}=0, \end{equation} with $J$ defined by \begin{equation} J= \frac{\hbar^2}{4m^2} \left( \frac{\partial U}{\partial t} \frac{\partial^3 U}{\partial x^3} - \frac{\partial^2 U}{\partial x^2} \frac{\partial^2 U}{\partial x \partial t} \right). \end{equation} So $P$ is positive and conserved. Nonetheless, this $P$ is not physically acceptable, since if we consider the solution $U(x,t)=\cos(kx-\omega t)$, where $\omega=\frac{\hbar k^2}{2m}$, then we get \begin{equation} P=\frac{\omega^2}{2}=\frac{E^2}{2 \hbar^2}. \end{equation} In a nonrelativistic theory it should be possible to choose the zero of the energy arbitrarily, and still obtain an equivalent theory. But, in our case, we could choose the zero of energy so to obtain $P=0$, and we must conclude that our definition of probability is not physically acceptable.

Now Bohm states that this conclusion holds not only for the specific probability function we have defined above, but for any probability we could define involving $U$ and the partial derivatives of $U$ of order at most 1 with respect to $t$.

Does someone have any idea why it should be generally so?

NOTE (1). Bohm does not put his statement in a rigorous mathematical form, since he does not define precisely what he means by an "acceptable probability function". He only gives the following vague definition.

Let $P(x,t)$ be a function only of $U(x,t)$, and the partial derivatives of $U$ of order at most 1 with respect to $t$ all computed in $(x,t)$, that is assume there exists a non-constant function $p$ such that $P(x,t)=p \left((D_{x}^d U)(x,t), \left(D_{x}^{d} \frac{\partial U}{\partial t}\right)(x,t) \right)$, where $D_{x}^d F$ is the set of all partial derivatives of $F$ with respect to $x$ from order $0$ (that is the function itself) to order $d$.

Then, we say that $P$ is an acceptable probability function if it satifies the following properties:

  1. P is real and never negative;

  2. the quantity $\int P(x,t) dx$ is conserved over time for every solution $U(x,t)$ of the wave equation above, so that after having normalized $P$ we can get $\int P(x,t) dx=1$ for all $t$;

  3. the significance of $P$ does not depend in a critical way on any quantity which is known on general physical grounds to be irrelevant: in particular this implies (since we are dealing with a nonrelativistic theory) that $P$ must not depend on where the zero of energy is chosen.

These properties, with the exception of the first, are not precisely formulated, so they need to be put in a little more precise mathematical form. In particular, as for property (2), Bohm seems to interpret it in the sense that there exists a function function $j$, such that, if we put $J(x,t)=j \left((D_{x}^d U)(x,t), \left(D_{x}^{d} \frac{\partial U}{\partial t}\right)(x,t) \right)$, the following continuity equation holds \begin{equation} \frac{\partial P}{\partial t} + \frac{\partial J}{\partial x} = 0. \end{equation} As for Property (3), by following Bohm, we could interpret it as requiring that for the particular solution $U(x,t)=\cos\left(\sqrt{\frac{2m \omega}{\hbar}}x-\omega t \right)$ we must get a probability function $P(x,t)$ which is independent of $\omega$.

Let us explicitly note that Bohm, in his previous discussion about the definition of probability for the Schrödinger equation, requires a further property, which in our context would read:

  1. the probability $P$ is large when $|U|$ is large and small when $|U|$ is small.

Anyway, we cannot require this further property here. Indeed, if we took seriously property (4), by interpreting it in the sense that $P_{U_1}(x,t) \geq P_{U_2}(x,t)$ for every two solutions $U_1$ and $U_2$ with $|U_1(x,t)| \geq |U_2(x,t)|$, then we would get that $P(x,t)=p(|U(x,t)|)$, where $p$ is a non-decreasing function (for a proof see the Lemma in this answer). But then, by considering the particular solution $U(x,t)=U_0 \cos\left(\sqrt{\frac{2m \omega}{\hbar}}x-\omega t \right)$, where $U_0$ is a constant, we see that the independence of $P(x,t)$ from $\omega$ would imply that $p$ is a constant function.

For a detailed mathematical formulation of Bohm's statement see my post on MathOverflow Conserved Positive Charge for a PDE.

NOTE (2). A different and maybe more natural interpretation of Property (3) can be given in the following way. If we start from the Schrödinger equation for a particle in the case of a potential $W(x)$

\begin{equation} i \hbar \frac{\partial \psi}{\partial t}(x,t) = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\psi(x,t) + W(x) \psi(x,t), \end{equation} then we easily get that the real part $U$ of $\psi$ satisfies the following second-order equation \begin{equation} \frac{\partial^2 U}{\partial t^2} = - \frac{\hbar^2}{4 m^2} \frac{\partial^4 U}{\partial x^4}+ \frac{W}{m} \frac{\partial^2 U}{\partial x^2} +\frac{W’}{m} \frac{\partial U}{\partial x} + \left( \frac{W’’}{2m} - \frac{W^2}{\hbar^2} \right) U . \end{equation} Now property (3) can be given the following interpretation: if $U$ is the solution corresponding to the potential $W(x)$ and given initial conditions, and $\tilde{U}$ is that corresponding to the same initial conditions and to the potential $W(x)+W_0$, where $W_0$ is a constant, then $P(x,t)$ should be the same function when computed for $U$ and $\tilde{U}$. I doubt that such a probability function $p$ exists, and maybe this is the core of Bohm's statement. By using this idea, I've given an alternative mathematical interpretation of Bohm's statement in the post Conserved Current for a PDE.

NOTE (3). See also my related post Nonexistence of a Probability for the Klein-Gordon Equation, in which essentially the same issue arises in a relativistic setting in relation to the well-known Klein-Gordon Equation. In the other post of mine Uniqueness of the Probability Function for the Schrödinger Equation an analogous problem is discussed with respect to the Schrödinger equation. Presumably, Bohm had in mind the same mathematical tool to deal with all these three issues.

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  • $\begingroup$ Well what have you tried to do so far? $\endgroup$ Oct 10, 2018 at 9:55
  • $\begingroup$ Dear Aaron, I have no idea of a possible argument, also because my intuition suggests that the statement is false. I posted the issue here because, having been Bohm's book very popular, I thought someone had studied these issues before. $\endgroup$ Oct 10, 2018 at 14:00
  • $\begingroup$ Unfortunately this is not my area of expertise, so this is why I asked. I don't know why you would think it is false, but I would understand if you said that the statement seems to have no proof in general. I guess what you would have to do to prove this is write out a general function of $U$ and its partial derivatives, and then show that we get nonsensical results using this general function. Perhaps this could be a good question on the mathematics stack exchange site if you framed it in the right way? $\endgroup$ Oct 10, 2018 at 14:52

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Looks like Bohm's proof deals with a free particle, which is not very realistic. However, if you consider the Schrödinger or Klein-Gordon equation, say, in electromagnetic field, you can make the wave function real by a gauge transform (at least locally), as noticed by Schrödinger in Nature (1952), v.169, p.538. Just one real function can also be sufficient for the case of the Dirac equation in electromagnetic field, a shown in my article in J. Math. Phys.

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After a long search in the literature I must conclude that the issue raised by Bohm has attracted no interest, maybe because no one has ever entertained some doubt about the impossibility of building quantum mechanics with a real wave function.

I tried to give two precise mathematical formulations of Bohm's nonexistence statement in my post Conserved Current for a PDE, but, as I argued in my answer there, neither of the two seems to be a faithful mathematical translation of Bohm's physical statements.

Maybe, we will never know what physical and mathematical argument Bohm had in his mind not whether he had actually one. He could have maybe quoted without hesitation in his book some remark made by R.J. Oppenheimer (whose lectures at the University of California at Berkeley inspired a large part of Bohm's treatise) or he simply stated something he considered intuitively evident without worrying about a possible proof. We cannot know how the mind of a genius works ... and David Bohm was an absolute genius!

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  • $\begingroup$ I'm guessing it hasn't gained much interest because there isn't any pay off to proving or disproving this. Complex wave functions work very well, and it's not like we got any issues using them where having only real wavefunctions is needed. $\endgroup$ Oct 23, 2018 at 10:21
  • $\begingroup$ Well, maybe you're right ... but you must admit that it is at least a strange fact that you have to handle complex quantities in quantum mechanics, a situation completely different from that of all earlier physical theories, where complex quantities were introduced only as mathematical tricks to solve some specific problems or to have more compact notations. I think this kind of feeling pushed Bohm to explore the possibility of discarding complex wave functions in favour of real ones. $\endgroup$ Oct 23, 2018 at 11:13
  • $\begingroup$ Yeah... QM is weird :) haha $\endgroup$ Oct 23, 2018 at 11:14
  • $\begingroup$ I would agree with your statement @MaurizioBarbato, in that one often hears "QM is necessarily complex, it's not just useful" and I can't remember a good reason from the lecture for it to be so. $\endgroup$ Oct 23, 2018 at 11:33
  • $\begingroup$ @AtmosphericPrisonEscape Yes, this foundational issue has been completely overlooked as many others. I think this is the effect of the abstract way of thinking which became more and more prevalent in the physics of the XXth century (and this trend continues unrestrained today). It gives QM and subsequent theories a completely different flavour with respect to classical theories as the Mechanics of Galilei and Newton or the Electromagnetic theory of Faraday and Maxwell. $\endgroup$ Oct 23, 2018 at 12:37
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After having thought about this question for a long time, I have arrived at the conclusion that what Bohm really meant must be far simpler than I had conjectured in my previous speculations.

I already remarked in my NOTE (2) to the post that the impossibility result stated by Bohm is trivial if we assume Property (4) in the form given there. Actually, I have now realized that even though we assume a kind of "weak form" of that property we can still prove Bohm's impossibility "theorem". More precisely, I will show in this answer that under the assumption that $p(X_1,\dots,X_{2d+2})$ is a polynomial such that $p(0,\dots,0)=0$, if $P(x,y)=p \left((D_{x}^d U)(x,t), \left(D_{x}^{d} \frac{\partial U}{\partial t}\right)(x,t) \right)$ is independent of $\omega$ for the special solution \begin{equation} U(x,t)= \cos\left(\sqrt{\frac{2m \omega}{\hbar}}x-\omega t \right), \end{equation} then we must have that $(x,t) \mapsto P(x,t)$ is identically equal to zero. Now since $P$ must have the meaning of a probability density, an everywhere zero probability density for a non-zero wave function is not physically acceptable, which proves Bohm's impossibility result. Note that the assumption that $p(0,\dots,0)=0$ is very natural since it simply says that for a locally zero wave function the probability density must be zero. Note also that the assumption that $p(X_1,\dots,X_{2d+2})$ is a polynomial is also quite natural since we are seeking conserved quantities with a "simple" expression.

Now the proof. For ease of convenience we shall write $\omega$ as a function of $k$, instead of making the converse as we did in the post. So we write our special solution as \begin{equation} U(x,t)= \cos\left(kx- \frac{\hbar k^2}{2m} t \right), \end{equation} We first show that, for this solution, under our assumptions the map $(x,t) \mapsto P(x,t)$ must be a constant. Property (3) is equivalent so say that $P(x,t)$ is independent of $k > 0$ for every $(x,t) \in \mathbb{R}^2$. Now, for the special solution $U(x,t)$ considered, we have that $P(x,t)=F(x-vt)$ for some function $F$, where $v=\frac{\hbar k}{2m}$. 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 = \frac{\hbar k}{2m}, \end{equation} a contradiction.

So for some constant $c \in \mathbb{R}$ we have $P(x,t)=c$ for all $k > 0$ and all $(x,y) \in \mathbb{R^2}$. Put by ease of notation $y=kx- \frac{\hbar k^2}{2m} t$. Note that $P(x,t)$ is a polynomial in $\sin y$, $\cos y$ and $k$. By replacing $\cos^2 y$ by $1-\sin^2 y$, $\cos^4 y$ by $(1-\sin^2 y)^2$, ... and $\cos^3 y$ by $(1-\sin^2 y)\cos y $, $\cos^5 y$ by $(1-\sin^2 y)^2 \cos y $, ..., we get that \begin{equation} P(x,t)= \sum_{l=0}^{L} a_l(k) \sin^l(y) + \cos(y) \sum_{n=0}^{N} b_n(k) \sin^n(y), \end{equation} where $a_l(k)$ and $b_n(k)$ are polynomials in $k$. Fix a value of $k$. Since $P(x,t)$ is constant for any $(x,t)$, we get for any value of $y$: \begin{equation} \sum_{l=0}^{L} a_l(k) \sin^l(y) + \cos(y) \sum_{n=0}^{N} b_n(k) \sin^n(y) = \\ = \sum_{l=0}^{L} a_l(k) \sin^l(\pi-y) + \cos(\pi-y) \sum_{n=0}^{N} b_n(k) \sin^n(\pi-y), \end{equation} so that \begin{equation} \cos(y) \sum_{n=0}^{N} b_n(k) \sin^n(y)=0, \end{equation} which implies that for any $y \neq \frac{\pi}{2} + r \pi$, with $r$ integer, we have \begin{equation} \sum_{n=0}^{N} b_n(k) \sin^n(y)=0, \end{equation} so that the polynomial $Q(z)= \sum_{n=0}^{N} b_n(k) z^n=0$ admits infinitely many zeroes. We conclude that $b_0(k)=\dots=b_N(k)=0$. But then \begin{equation} c=P(x,t)=\sum_{l=0}^{L} a_l(k) \sin^l(y), \end{equation} and so now the polynomial $R(z)=- c + \sum_{l=0}^{L} a_l(k) z^l$ has infinitely many zeroes, which implies $a_1(k)=\dots=a_L(k)=0$ and $a_0(k)=c$. But we have $a_0(k)=p(0,\dots,0)=0$, so that we conclude that $c=0$. QED

Now that I have ended the proof I realize that a much simpler proof can be given, which actually can be used to obtain a much more general result.

Theorem Let $p:\mathbb{R}^{2d+2} \rightarrow \mathbb{R}$ be a continuous function such that $p(\mathbf{0})=0$. Let \begin{equation} U(x,t)= \cos\left(kx- \frac{\hbar k^2}{2m} t \right), \end{equation} and define the quantity \begin{equation} P(x,y)=p \left((D_{x}^d U)(x,t), \left(D_{x}^{d} \frac{\partial U}{\partial t}\right)(x,t) \right). \end{equation} Assume that for each $(x,t) \in \mathbb{R}^2$ $P(x,t)$ is independent of $k > 0$. Then we have $P(x,t)=0$ for all $(x,t) \in \mathbb{R}^2$.

Proof First of all, let us note that under our assumptions $(x,t) \mapsto P(x,t)$ is a constant map. Indeed, given the form of $U(x,t)$, we have $P(x,t)=F(x-vt)$ for some continuous function $F:\mathbb{R} \rightarrow \mathbb{R}$, where $v=\frac{\hbar k}{2m}$ as above. Now, for any given $t > 0$, the value of $P(0,t)=F(-vt)$ does not depend on $k > 0$, so that we get that$F$ is constant on $[0,\infty)$. By the same argument with now $t < 0$, we deduce that $F$ is constant on $(-\infty, 0]$. Hence $F$ is constant on $\mathbb{R}$, so that for some $c \in \mathbb{R}$, we have $P(x,t)= c$ for all $(x,t) \in \mathbb{R}^2$. Now, for any fixed value of $k > 0$, by picking values of $(x,t)$ such that $kx- \frac{\hbar k^2}{2m} t = \frac{\pi}{2}$, we get \begin{equation} p\left(0,-k,0,k^3,\dots,0,(-1)^r k^{2r-1},\frac{\hbar k^2}{2m},0,-\frac{\hbar k^4}{2m},0,\dots,(-1)^{r-1}\frac{\hbar k^{2r}}{2m},0\right)=c \end{equation} when $d=2r-1$ is odd, and \begin{equation} p\left(0,-k,0,k^3,\dots,0,(-1)^r k^{2r-1},0,\frac{\hbar k^2}{2m},0,-\frac{\hbar k^4}{2m},0,\dots,(-1)^{r-1}\frac{\hbar k^{2r}}{2m},0,(-1)^{r}\frac{\hbar k^{2r+2}}{2m}\right)=c \end{equation} whene $d=2r$ is even. By taking the limit of the above expressions as $k \rightarrow 0$, we get in both cases $c=0$. QED

I think this very simple result is what Bohm should have in his mind when he made his statement. It actually belongs to the kind of heuristic considerations that guide physicists in their search of the right physical law of a new phenomenon when no clear theory is at hand, and this was exactly the case at the beginning of the twentieth century when quantum mechanics was born. Bohm in his fantastic book tried exactly to replicate the process of the discovery of quantum laws, because he was absolutely convinced that only by speculating about the possible paths one can take in developing quantum mechanics, we can develop a true familiarity with this strange and unintuitive theory and we can really try to understand the physical meaning of the concepts hidden behind its abstract mathematical formalisms.

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