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I have this very bad habit of going to the scratch, discarding all the developments of a theory and worldly knowledge, and ask some fundamental (mostly stupid and naive, as some may say) questions as to why why we needed so and so assumption, why we had to consider this way, could we assume $X$ instead of $Y$ and get a different theory and so on and so forth. As a part of this, is the following question :

Way back in the early twentieth century, physicists struggled to explain certain phenomenon like photo electric effect which needed light waves to behave a particles (photons), and the interference effect of electrons (diffraction) which needed them to behave as waves.

So they said, hey, consider a wave (don't ask me know what it is, but just consider it)... ok $$\Psi(x,t) = e^{i(kx-\omega t)},$$ now without asking what is $\Psi$, we can explain the interference of electrons and also the photo electric effect, basically the wave/particle duality, if we make an analogy between wave and particle nature as $p = \hbar k$, $E = \hbar \omega$.

My seemingly blunt question is, if you want to explain wave nature and interference by considering a wave function $\Psi$, why the heck do we need complex numbers, why not just real function? Cant we consider something like $\Psi(x,t) : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ or something like that. Hold on, Water waves are successfully explained by real wave function for example $\Psi(x,t) = \cos(kx-\omega t)$, so why the heck we need complex waves for making an anology for wave-particle duality? What if we just unlearn all the QM and start with a real function for waves, what is going to happen?

Sorry for being a bit presumptuous, but some gentlemen will start talking about probability amplitude, uncertainty principle, and so on and so forth, but Gentlemen, wait, I haven't gone that far! I don't have Born interpretation and I don't have uncertainty principle or for that matter the Schrodinger equation yet, So your logic will lead to circular arguments!

After all our goal is to explain physical phenomenon...what if we venture into this jungle of real functions and come up with a totally different theory which explains physical phenomenon.

((If you ask me to do research in theoretical physics, I'll throw all the QM books in garbage (no disregard though) and start thinking from this point of view...Thats my style of working!)

My expected answer is in this spirit, "Hey if you go in that direction, you are bound to end up in a quick sand, for so and so reason"

Qmechanic
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Rajesh D
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  • Possible duplicates: http://physics.stackexchange.com/q/8062/2451 , http://physics.stackexchange.com/q/32422/2451 and links therein. – Qmechanic Dec 23 '13 at 14:04
  • @JohnRennie : My question is different from them in this sense, they say " Can we have a theory of the same physics without complex numbers? If so, is the theory using complex numbers easier?", I say "can we have a completely different theory but explains existing phenomenon and even more" – Rajesh D Dec 23 '13 at 14:07
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    Complex formulations allow rather more info to be placed into the equation. It's not a duality thing; just a way to get phase and amplitude (and their interactions) into a compact representation. – Carl Witthoft Dec 23 '13 at 14:08
  • @CarlWitthoft : I am not asking about math, I am asking about a new theorybeing possible by going in a completely different direction. – Rajesh D Dec 23 '13 at 14:11
  • Founders of QM did not really had a philosophy to begin with, they did something which could explain certain things, and then they started a philosophy, right? – Rajesh D Dec 23 '13 at 14:12
  • Water waves can be measured, but not the wavefunction. – jinawee Dec 23 '13 at 14:34
  • @jinawee : Is the goal is to come up with something we cannot measure, to begin with. Thas really funny! – Rajesh D Dec 23 '13 at 14:37
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    Personally, I don't necessarily see why the question is being downvoted, although admittedly I don't have an answer as to why a purely real-valued formalism couldn't possibly be cooked up by some wacky means. However, that said, the current formalism used for quantum mechanics essentially encompasses all experimental observations correctly. Thus, if it existed, any "real-valued" formalism would necessarily either be fully isomorphic to the current formalism, or in conflict with reality. – DumpsterDoofus Dec 23 '13 at 16:27
  • rajesh bhai; i don't know about quantum machanics but the fact: $A\cdot \cos(\omega t + \theta) = A \cdot \frac{e^{i(\omega t + \theta)} + e^{-i(\omega t + \theta)}}{2}$ and $A\cdot \cos(\omega t + \theta) = \operatorname{Re} \left{ A\cdot e^{i(\omega t + \theta)}\right} = \operatorname{Re} \left{ A e^{i\theta} \cdot e^{i\omega t}\right}.$ allows to reduce effort in calculations.this fact is common in circuit analysis. Voltage and current are repersntd as phasors,after calculation Real part is considered only. – user31782 Dec 23 '13 at 18:43
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    @RajeshD: If you want to see if there is another way to approach QM theory (or philosophy if you prefer) then try it. Maybe you will find another way. Sending out a missive like this will not do it. – Mike Dunlavey Dec 24 '13 at 01:24
  • Check this book section 1.1 Stern-Gelach experiment. – user5402 Dec 24 '13 at 20:00

4 Answers4

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After all our goal is to explain physical phenomenon...what if we venture into this jungle of real functions and come up with a totally different theory which explains physical phenomenon.

Good luck to you. First of all you are wrong that classical physics did not use imaginary functions. The solutions of Maxwell's equations expressed as imaginary functions are more general and universal than sines and cosines.

The simplest set of solutions to the wave equation result from assuming sinusoidal waveforms of a single frequency in separable form

$$\mathbf{E}(\mathbf{r}, t)=\mathrm{Re}\{\mathbf{E}(\mathbf{r} e^{i\omega t}\}$$

Imaginary functions are a useful tool in integrations and descriptions of real data.

((If you ask me to do research in theoretical physics, I'll throw all the QM books in garbage (no disregard though) and start thinking from this point of view...Thats my style of working!)

With such blind spots I am sure nobody will ask you to do research in theoretical physics.

The difference between the classical use of imaginary functions from the solutions of the wave equations and the quantum mechanical one is the postulate the posits that the square of the wavefunction is real and gives the probability of an elementary particle (or nuclear) interaction to be observed. When in the microcosm quantum mechanics reigns. There one cannot take a ruler, mark it and measure, it was found that the theories and data agreed when the probability postulate was imposed. One has to make many measurements and get the probability distribution for a particularly desired value.

The above link discusses the postulates of quantum mechanics which were not imposed out of a freak imagination, but were necessary to be able to calculate and fit known observations, like the hydrogen atom, and predict the outcome of experiments and observations.

EDIT to address the last part of the question:

((If you ask me to do research in theoretical physics, I'll throw all the QM books in garbage (no disregard though) and start thinking from this point of view...Thats my style of working!)

That works for art, art is much less dependent on data bases of observations and the tools that can be used.

The fact that for two thousand years people have been creating models of physical observations, and particularly the last 300 a data base of mathematical tools too, constrains creativity in science. The mathematical tools have been used to model all observations up to now. These models are in a way a shorthand description of nature that could be used in many ways instead of going back to the data itself. There exists a frontier of experimental research where the models have not been validated , and that is where new thinking can come in.

My expected answer is in this spirit, "Hey if you go in that direction, you are bound to end up in a quick sand, for so and so reason"

If you go into the direction of throwing everything away you will end up with vague models like the Democritus atomic model, or the phlogiston theory, in your own words. The mathematical models used now are validated, some of them to great accuracy. New mathematical tools to model the already modeled data would only be worth the attention if something new and unexpected is predicted and found in the experiments.

There are people working off the beaten track theories, trying to explain quantum mechanics by underlying deterministic theories. These people have a thorough knowledge of existing mathematical tools and the physics models that have been validated. They just want to work at the frontier by ignoring that mainstream physics considers their effort contradictory or impossible/prohibited by the postulates of quantum mechanics and special relativity. An example is the current research interests of G.'t Hooft who has also participated here a while ago .

So if you go in that direction you will end in quick sand surely if you do not have a thorough knowledge of the data and mathematical tools used by physics up to now. If you make the effort to acquire them, then of course you are free to prove mainstream physics "wrong" , as long as your new theory can accommodate the data shorthand of the models up to now . All new theories as they appeared in physics joined smoothly with the old ones, as limiting cases.

anna v
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  • My question is not about use of complex numbers in physics in general (classical/modern or whatever). My question is very specific, why do we start with a complex function for waves to explain wave-particle nature. In spirit, my question to speak is directed to Debroglie or who ever started this?.......on other note, I am a signal processing engineer/researcher who deals with complex numbers all the time without any problem. – Rajesh D Dec 23 '13 at 14:33
  • Nobody asked Einstein to invent relativity theory, he did it when he was a patent clerk, during is off hours. – Rajesh D Dec 23 '13 at 14:44
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    Lorenz transformations existed in the Maxwell equations, that is why they are called Lorenz and not Einstein. His out of the box thinking, consistent with the data they were gathering at the time, was to use it for massive particles . He did not create out of the blue,but as all new physics, used previous steps to go further in explaining observations. – anna v Dec 23 '13 at 15:17
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I believe the key words here are time shift invariance, flow, state, linearity (or homogeneity), continuity and unitarity and the answer to your question is quite independent of whether or not we are doing quantum mechanics.

Suppose we are groping around in the dark trying to describe some new phenomenon as the early (actually 1920s) physicists were. In the tradition of Laplace, you hope that a deterministic system description will work (and the part of QM that uses complex numbers, namely unitary state evolution, is altogether and utterly deterministic). So we are going to begin with a state: some information - an array of numbers $\psi$ - real ones for now if you like - that will wholly determine the system's future (and past) if the system is sundered from the rest of the Universe.

So we have some $\psi\in\mathbb{R}^N = X$ that is our basic description.

Now an isolated system evolves on its own: with the lapsing of time the system changes in some way. Let $\varphi: X\times\mathbb{R}\to X$ be our time evolution operation: for $x\in X$ $\varphi(x,t)\in X$ is what the state evolves to after a time $t$. Moreover, the description cannot be dependent on when we let it happen: it must have a basic time shift invariance. The experiment's results cannot depend on whether I do it now or whether I wait till I've had my cup of tea. So the description of the change in some time interval $\Delta t$ must be the same as that for the change in any other $\Delta t$. If we're not interacting with the system, then there are no privileged time intervals. Therefore, we must have:

$$\varphi(x,t+s) = \varphi(x,s+t) = \varphi(\varphi(x,t),s) = \varphi(\varphi(x,s),t),\;\forall s,t\in\mathbb{R}$$

and

$$\varphi(x,N \Delta t) = \underbrace{\varphi(\varphi(\cdots \varphi(\varphi(}_{N\ \text{iterations}}x,\overbrace{\Delta t),\Delta t)\cdots,\Delta t)}^{N\ \text{iterations}}$$

so, from our Copernican notion of time shift invariance we have our first next big idea: that of a flow defined by the state transition operator $\varphi$. So our state transition operators form a one parameter, continuous group. Here we have brought our fundamental idea of continuity to bear.

This may seem to be saying (since we now have a one-parameter Lie group) that the only systems that fulfill these ideas are linear ones, but this is not quite so: the Lie group only acts on $X$ so there is always the possibility of some local, $X$-dependent nonlinear "stretching" or "shrinking" of the path. So now we make an assumption of system linearity or of some other notion of a homogeneous action (so that, intuitively, $\varphi(-,t):X\to X$ conserves some local "structure" of our state space $X$). Then the only continuous, linear, homogeneous state transition flow operator on $X$ is:

$$\varphi(x, t) = \exp(H\,t) x \stackrel{def}{=} \left({\rm id} + H\,t + H^2\,\frac{t^2}{2!} + H^3\,\frac{t^3}{2!} + \cdots \right) x$$

for some linear operator $H:X\to X$.

So now, to investigate the behaviour of this basic model, we need to think about the eigenvalues of $H$, so that, for example, we can "spectrally factorise" the operator ("diagonalise" it - or at least come as near as we can to diagonalising it - but this always can be done in QM). The natural home for the eigenvalues of an operator are $\mathbb{C}$, not $\mathbb{R}$, because the former is an algebraically closed field and the latter is not. There exist finite dimensional linear operators whose characteristic equations have solutions in $\mathbb{C}-\mathbb{R}$. Moreover, we must consider the behaviours of $e^{\lambda\,t}$ where $\lambda$ are the eigenvalues of $H$ (this is easy in a finite dimensional space - for an infinite dimensional one, see my answer here to the Physics SE question "Are all scattering states un-normalizable?")- how do these square with our physics?

If, for any eigenvalue, $\lambda$ is real and positive, or if complex and its real part is real and positive, our state vector diverges to infinite length as $t\to\infty$. If they are all complex with negative real part, then our state vector swiftly dwindles to the zero vector. Even before we have crystallised the idea of probability amplitude properly, we may have the idea that we "want as much state to stick around for as long as possible*. The system must end up in some nonzero state: our particles don't all collapse to the same state $0\in X$. So all eigenvalues being wholly imaginary so that $e^{\lambda\,t}$ has bounded magnitude that doesn't rush of to infinity or to nought, might be a fair bet. Bringing this into sharper focus: having $\exp(H\,t)$ conserve length might be another reasonable assumption. The easiest "length" in state space is of course the $\mathbf{L}^2$ one. So now we might postulate, even before our ideas about probability amplitude in QM are fully crystallised, that:

The operator $\exp(H\,t)$ is unitary

This equivalently means that:

$H$ is skew-Hermitian

or that our operator is $\exp(i\,\hat{H}\,t)$ where:

$\hat{H}$ is Hermitian

Hermitian operators, given very mild assumptions, are always diagonalisable, and they always have real eigenvalues. So entities like $\exp(i\,\omega\,t)$ where $\omega \in \mathbb{R}$ are inevitable in our state transition operator expansions.

Now you are right and we could keep everything real, use $\cos$ and $\sin$ instead and keep our state space as $\mathbb{R}^{2 N}$ were we would have $\mathbb{C}^N$ in the conventional description. Whether or not we choose to single out an object like:

$$\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\in U(1), SU(2), SO(3), U(N) \cdots$$

and give it a special symbol $i$ where $i^2=-1$ is a "matter of taste", so in this sense the use of complex numbers is not essential. Nonetheless, we would needfully still meet this object in decomposing our state transition operator and $H$, $\hat{H}$ operators and ones like it and would have to handle statements involving such objects when describing physics - there's no way around this. So, in particular, if we have bounded, but everlasting wave behaviour, we must be encountering pairs of states in our $\mathbb{R}^{2 N}$ representation of the conventional $\mathbb{C}^N$ that evolve with time through linear differential operators with submatrices like the $i$ object above.

So you actually see that complex numbers arise very naturally out of the very classical and indeed Renaissance ideas of people like Laplace, Copernicus, Leibnitz, Galileo and Newton. We simply have a better and more refined mathematical language to talk smoothly about these ideas where these guys were groping in the dark.

And now, if we keep linearity but drop time shift invariance (say we have some "control" of our linear operator $H(t)$: we might have an electron in a classical magnetic field which we can vary to "steer" the electron's state, for example) we still get most of the above ideas. For now we solve:

$${\rm d}_t U(t) = H(t) U(t)$$

which we can think of as having some dials to twiddle to steer our Lie algebra member $H$ in the tangent space at $U$ to the Lie group of state transition operators. $H$ must still be skew-hermitian and $U\in U(N)$ (if our system is finite dimensional). See my answers here to the Physics SE question "Why can the Schrödinger equation be used with a time-dependent Hamiltonian?" as well as, for a practical example, here to the Physics SE question "fixed input qubit state to an arbirary pure state using two variable rotations and one fixed rotation"

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Wave-particle duality is the "prehistory" of Quantum Mechanics. Looking at Quantum Mechanics, it is far more interesting to look at the Heinsenberg representation, where position and momentum are operators depending on time, while the states $|\psi\rangle$ did not.

Quantum constraints are expressed by commutation relations :

$[X(t), P(t')]_{|t=t'} = Constant ~~\mathbb I_d$

Here "Constant" is a real or imaginary value which does not depends on $t$ (at $t=t'$), and $\mathbb I_d$ is the identity operator (which commutes with all operators)

Now, the question is : must we choose this constant as a pure real value, or a pure imaginary value ?

Because $X(t)$ and $P(t')$ are hermitian operators, their commutator is anti-hermitian, so the only mathematical possibility is that the constant is pure imaginary :

$[X(t), P(t')]_{|t=t'} = i \hbar ~~\mathbb I_d \tag{1}$

Now, we may go back to the Schrodinger representation, and to the wavefunction $\psi(x,t) = \langle x|\psi(t)\rangle$

With $\psi(x,t)$, we must find a representation for the operators $X$ and $P$, which apply to $\psi(x,t)$, so now these operarors do not depend on time anymore, but their definition must be compatible with their commutation relations found in $(1)$, so the simplest solution is :

$P = -i\hbar \frac{\partial}{\partial x}$

Now, imagine a wavefunction corresponding to a constant momentum $p$ and energy $E$, then necessarily, it must have the form :

$\psi(x,t) \sim e^{\frac{i}{\hbar}(px-Et)} = e^{i(k x-\omega t)}$

We have $P\psi(x,t) = -i\hbar \frac{\partial}{\partial x} \psi(x,t)= p\psi(x,t)$

[The energy operator $E$ being defined as $E = i\hbar \frac{\partial}{\partial t}$]

So you see that you cannot "escape" to the imaginary representation, it appears totally natural in the Quantum Mechanics context.

Trimok
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Complex numbers are just as real as real numbers. Calling them imaginary is just a naming thing, but they are there. In some applications, they're a great shorthand and a way to avoid tons of goniometric functions and similar "complex" calculations (a great example being applying rotation - instead of calculating sin and cos over and over again, you just multiply by a complex number; very handy in computer graphics). Could you get by without using negative numbers? Or a zero? Sure you could, it would just make everything more complicated. It's just a way to save time and space and make everything easier to express and understand.

If you're treating complex numbers as something special, "not real", you're doing the same mistake ancient people and mathematicians did when they similarly dismissed negative numbers, or zero, or irrational numbers. It's there. It does the job. Why do we use matrices to express a collection of linear equations? We don't have to, it's just convenient. Can you express newtonian physics without vectors? Sure you can, but now you've tripled the number of equations for no reason. If you have a better tool or representation for a job, you use it. That's the only way to keep up with the increasing complexity of the stuff you do. And it doesn't matter if you're talking about mathematics, physics or programming - the guy that keeps using outdated tools gets outdated as well.

All in all, sure you could express QM without complex numbers. But that wouldn't change the underlying principles, you'd just have a different (and actually more complicated) expression of the same thing.

Luaan
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