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 $P(x,t)=p \left(U(x,t),(D_{x}^k U)(x,t), \frac{\partial U}{\partial t}(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right)$, where $D_{x}^k F$ is the set of all partial derivatives of $F$ with respect to $x$ from order 1 to order $k$.

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 $J(x,t)=j \left(U(x,t),(D_{x}^k U)(x,t), \frac{\partial U}{\partial t}(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right)$, 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.

  • $\begingroup$ Well what have you tried to do so far? $\endgroup$ – Aaron Stevens Oct 10 '18 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. $\endgroup$ – Maurizio Barbato Oct 10 '18 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? $\endgroup$ – Aaron Stevens Oct 10 '18 at 14:52

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!

  • $\begingroup$ 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. $\endgroup$ – Aaron Stevens Oct 23 '18 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. $\endgroup$ – Maurizio Barbato Oct 23 '18 at 11:13
  • $\begingroup$ Yeah... QM is weird :) haha $\endgroup$ – Aaron Stevens Oct 23 '18 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. $\endgroup$ – AtmosphericPrisonEscape Oct 23 '18 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. $\endgroup$ – Maurizio Barbato Oct 23 '18 at 12:37

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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