In the quantum mechanics, one asked if the complex number was necessary? A typical answer was that it was not, or that it's simple direct product of real numbers.

However, consider rational number to real number, with additional uncountable irrational numbers. One then think, just because rational number to real number added infinite irrational number, was that necessarily for quantum mechanics?

For a simple argument, without the imaginary number, the solution to $x^2=-1$ was unsolvable, but then do we necessarily need the solution to $x^2=2$, since $$\prod_{k=0}^{100} \left(1-\frac{1}{(4k+2)^2} \right)$$ was precipice enough? or even volume computation with $\pi$ with similar procedures.

However, from rational number to real number and from real number to complex number, in each case an uncountable number of elements was added into the set, while rational number was countable, thus there might be some thing there in the view of number theorists.

But after all, quantum mechanics probably had failed at $10^{-100}$ m, the total range of number that quantum mechanics worked was a finite set!

Do we necessarily need real number for quantum?

Math Manifestly Rational Quantum Mechanics:

  1. Derivatives: $\displaystyle y'(t_0)=\lim_{t\rightarrow 0} \frac{y(t_0+t)-y(t)}{t} \Rightarrow \lim_{n\rightarrow \infty,n\in \mathbb{Z}} \frac{y(t_0+t_n)-y(t_n)}{t_n}$ where $t_n\in \mathbb{Q}$ and $\lim_{n\rightarrow \infty} t_n=0$. (In case any question about $\lim_{n\rightarrow \infty} t_n=0$ from analysis, set $\epsilon_n\in \mathbb{Q}$).

  2. If $y'(t_0)=A\in \mathbb{Q}$ case resolved, if $A$ being a rational number, one make the following argument with $-i\frac{\hbar^2}{m} \frac{\partial^2}{\partial x^2} \Psi(x,t)$.

  3. From uncertainty principle, $\Delta x\Delta p\geq\frac{\hbar}{2}$, one adopt the idea from statistical manifold, that the space had fuzziness, that a minimum length was discrete as somewhat ontic. This should be consistent with the current view of physics, for example, pick $\min(x)=l$ such that the $\frac{1}{l}$ was comparable with all the energy in the observable universe, and $\min(p)=m$ to be such that such that a change in $\frac{1}{m}$ would be much larger than the size of the universe. This was achievable since $\mathbb{Q}$ was dense in $[0,1]$.

  4. Now that there's $l,m$, the next was to argue on how to use them.

  5. Let $y'(t_0)=A$ be in real domain. One recognize that, because of the fuzziness in physical space, $A$ was not actually the irrational $A$. Rather, one make the following equivalence class $[x_n]$ where $\alpha\sim x_n$ if $\alpha \in [n\cdot l,(n+1)\cdot l)$. Claim $y'(t_0)=[A]\in \mathbb{Q}$. Just in case if someone worry about the decimal places, change the previous $l$ with $l/10^{82}$, i.e. divided by all the atoms there was in the universe. Also $\hbar^2/m\sim [\hbar^2/m] \in \mathbb{Q}$ by similiar argument, since the rational field was closed.

  6. First, quantum was "scaled back" from $\mathbb{C}$ to $\mathbb{R}^2$ though vectors and an obvious bilinear operation.("enlarged" in another point of view, actually, since a specific bilinear operation was required to capture the analytic(cauchy) condition) Thus, in the component form $-i\frac{\hbar^2}{m} \frac{\partial^2}{\partial x^2} \Psi(x,t)\Rightarrow [\frac{\hbar^2}{m}]\frac{\partial^2}{\partial x^2} \Psi(x,t)$.

  7. Consider $\displaystyle \frac{\partial^2}{\partial x^2} \Psi(x,t)=\lim_{n\rightarrow,n\in \mathbb{Z}} \frac{\Psi(x+t_n,t)-2\Psi(x,t)+\Psi(x-t_n,t) }{t_n^2}=[p_n]^2$

  8. Here's something interesting $[n m,(n+1) m)$ taking second order power was sent to $[n^2m^2,(n^2+2n+1)m)$ the length of the interval changed!! from $[m]$ to $[(2n+1)m^2]$ with $m$ fixed. (Now if one took a transformation to make $m,l$ large number and stare back to the uncertainty principle and think... That's not the point here, the point was $(2n+1)m$ was in comparable with the size of an integer.) That's why I mentioned before to divide the original $l,m$ by a factor of $10^{82}$, so that the $(2n+1)m$ was still much smaller than the original minimal length $\min(x)$.

Now, the configuration space was labeled by integers, so does all the physical meaningful quantities such as energy, momentum, etc. Not only this was a rational quantum mechanics, it's practically an integer quantum mechanics. The precision should match whatever the calculation one had in mind with the quantum. If not, divide $l,m$ by extra $10^{1000...}$ power. This was just one of such attempt to directly scale back the usual quantum mechanics to rational or even integer domain.

It's interesting to see what was gained or what was lost during the procedure.

  • 1
    $\begingroup$ How is this specific to quantum mechanics? Classical mechanics uses the real number system as well. $\endgroup$ – Emilio Pisanty Mar 25 at 15:29
  • $\begingroup$ @EmilioPisanty But in classical mechanics, rarely one uses complex number. All configurations spaces took the real number. In fact, anything one do with classical mechanics was rational because of the computer, for engineering, $\pi=3$ etc. The solution doesn't even need irrational number, as most undergraduate did with speed and velocity. i.e. in certain degree, classical mechanics was using rational number already. $\endgroup$ – ShoutOutAndCalculate Mar 25 at 15:34
  • $\begingroup$ @EmilioPisanty On another thought, one of the arguments for this question was that quantum mechanics had a lower bound from $\lim_{L\rightarrow \epsilon}1/L$ but it did not have an upper bound. Thus, the theory itself was of infinite order. But that would work only if the universe had infinite size. However, for now, it's questionable how quantum would work beyond the size of the universe. $\endgroup$ – ShoutOutAndCalculate Mar 25 at 15:50
  • 1
    $\begingroup$ So you are basically asking if we can model space discretely? $\endgroup$ – BioPhysicist Mar 25 at 17:25
  • 3
    $\begingroup$ Potential duplicate of Can physics get rid of the continuum? $\endgroup$ – Emilio Pisanty Mar 25 at 17:43

There is nothing special about quantum mechanics here. Identical arguments hold for basically every aspect of physics, which uses the real number system throughout.

To give a simple answer: physics as we know it would not work without the real number system.

The reason for this is that the real numbers are complete, and this is what allows limits to work. Without completeness, there are very few circumstances in which you can guarantee that a limit exists. In turn, this makes it impossible to define derivatives and integrals $-$ and much less to write down differential equations of motion, which form the bedrock of just about every field of physics.

Now, is it possible to produce an equivalent formalism using only approximate derivatives (or, to make things work, infinite sequences of approximate derivatives) based only on the rational numbers? Sure. After all, you can construct the reals out of the rationals using quite a wide array of methods. Would doing that be helpful (let alone make anything simpler)? No.

  • $\begingroup$ That's the thing, points in spaces doesn't need to "not having gap", as long as there was "fuzzy". The model and math one built was still consistent. $\endgroup$ – ShoutOutAndCalculate Mar 25 at 20:11
  • 2
    $\begingroup$ @ShoutOutAndCalculate Your last comment (along with most of the question, and all of your previous comments) is word salad. I'm providing a clear answer just so that there is one on the record, but I'm utterly uninterested in back-and-forth on this topic. $\endgroup$ – Emilio Pisanty Mar 25 at 20:14
  • $\begingroup$ Thank you for the help. It clarified lots of things but one, quote my instructor:"Statistical manifold have properties of the continuum and of discrete space". Those was even weaker condition than demanded and properly constructed. There by the "equivalent formalism" you mentioned and doesn't necessarily based on real number as for the configuration space. (But real numbers were nice so one might use it anyway. I'm just trying to get some inside on what went on with QM with a view on the fields they operate on.) $\endgroup$ – ShoutOutAndCalculate Mar 25 at 20:23

I don't know how to take your question, but I will treat it seriously.

@EmilioPisanty's link and answer covers pretty much everything, but some points.

  1. QM in its current form needs real numbers since the square root of many integers/rationals are irrational, and the formalism of QM involves square roots. QM also needs complex numbers, realistically. They crop up because of the need to solve specific algebraic equations. Dirac matrices can be seen as a quaternion solution to the KG equation. Octonions anyone?

  2. These is no evidence that Quantum Mechanics fails at $10^{-100}$ m. Something goes wrong but it is not clear that it is the continuum. Even if something like spin-networks underlie reality, I doubt attempting to purge real numbers from QM formalism is the way forward.

  3. Real numbers are a strict total ordering. If 2 measurements are made, the results can always be compared. The completeness of real numbers means that no measurements will fall into any "theoretical holes" E.g. Calculations show E(x) = $\frac{1}{\sqrt{2}}$ but irrational numbers are not allowed. Oops.


Not the answer you're looking for? Browse other questions tagged or ask your own question.