# Could a universe with different wavefunctions for elementary particles from those of our universe be self-consistent?

I was thinking of a universe that has the Heisenberg uncertainty principle, and in which there are elementary particles which cannot be subdivided into smaller particles. In this universe the equation for the Compton wavelength of a massive particle and the equation for the wavelength for a massless particle are both the same as they would be in our universe.

However, in this universe the wave functions that elementary particles and systems of elementary particles can have are different from the wave functions that elementary particles can have in the quantum mechanics of our universe. Even the wave functions for free particles that are isolated from the rest of this universe are different from the wave functions for free particles that are isolated from the rest of our universe.

More specifically: I'm talking about a universe in which wavefunctions have only real parts with no imaginary parts, a universe described by something different from the Schrödinger equation for the analog of non-relativistic quantum mechanics, and/or a universe described by something different from the Dirac equation for the analog of relativistic quantum mechanics.

This universe has three dimensions of space and one of time. The postulates of special relativity describe this universe. Also, the only elementary particles are ones that are time-like, and ones that are light-like. In this universe the laws of physics are the same for all positions, directions, and times.

Could this universe be self consistent?

• What exactly do you mean by self-consistent? As it stands, this is one of the big questions of pure physics. What does a theory of everything look like and how many parameters does it have? Parameters here would be something like the masses and charges of elementary particles. But right now, we don’t have a theory of everything and it is quite difficult to be sure that some of these things are constant even within our universe. – Wrzlprmft Jul 4 at 7:38
• The way wave functions are defined and uncertainty principle are very general - in pure mathematics. You would need to stray far from our laws of nature ( adjusting constants will not do the trick) so that the ones we have are no longer valid. It will be quite the exercise to adjust them to still allow for a universe as you describe, but I don't see why not. – bdecaf Jul 4 at 8:32
• What do you mean by "different"? A wavefunction isn't an intrinsic, fundamental property of a particle itself in the same way that charge or spin is; it depends on the surroundings and environment the particle finds itself in. – HDE 226868 Jul 4 at 15:32
• Spacelike elementary particles!? That means they're some form of tachyonic particle. That and lightlike elementary particles. This is a radical departure from physics as we know it. This would take a lot of work even for a theoretical physicist to work out. Is your hypothetical universe self-consistent? Perhaps. – a4android Jul 5 at 1:51
• @a4android Spacelike was a typo. I meant timelike, and so edited my question to correct that mistake. – Anders Gustafson Jul 5 at 1:59

The crux of the question is

I'm talking about a universe in which wave functions have only real parts with no imaginary parts, a universe described by something different from the Schrödinger equation for the analog of non-relativistic quantum mechanics, and/or a universe described by something different from the Dirac equation for the analog of relativistic quantum mechanics.

These ideas are all, I believe, tied together, and at the root of the answer is simply the axiom that probability must be conserved. From this, along with a couple of other assumptions, you can show that wave functions cannot be purely real-valued. From this, the form of the Schrödinger equation becomes apparent.

## Why the wave function must be complex

Let's say that at time $$t=0$$, a particle is in the state $$|\psi(0)\rangle$$ (where I am describing quantum states in bra-ket notation)$$^{\dagger}$$. There must be some operator $$\hat{U}(t)$$, known as the time-evolution operator, that shows how this state evolves in time, i.e. such that for any future state $$|\psi(t)\rangle$$, we can write $$|\psi(t)\rangle=\hat{U}(t)|\psi(0)\rangle$$ Now say that the particle is in a state $$|\psi_1\rangle$$, and we wish to find the probability that it is in $$|\psi_2\rangle$$, which we denote by $$\langle\psi_2|\psi_1\rangle$$. Naturally, if the two states are identical, this probability must be 1: There is a 100% chance of finding the particle in the state it is in. Therefore, we require the following to be true: $$\langle\psi(t)|\psi(t)\rangle=1=\langle\psi(0)|\psi(0)\rangle$$ But we can invoke the time-evolution operator to rewrite $$|\psi(t)\rangle$$ and see that $$\langle\psi(0)|\hat{U}^{\dagger}(t)\hat{U}(t)|\psi(0)\rangle=\langle\psi(0)|\psi(0)\rangle$$ where $$\hat{U}^{\dagger}(t)$$ is known as the adjoint of the operator. For the above equation to be true, we need $$\hat{U}^{\dagger}(t)\hat{U}(t)=1$$, which is the definition of a unitary operator. If this holds, probability is conserved.

This is where complex numbers enter the stage. We can show that any unitary operator can be written in the form of a complex exponential; because $$\hat{U}(t)$$ is unitary, it obeys that line of reasoning, and as such must be complex. In quantum mechanics, it happens to take the form $$\hat{U}(t)=e^{-i\hat{H}t/\hbar}$$ with $$\hat{H}$$ the operator known as the Hamiltonian and $$\hbar$$ is the reduced Planck constant. We see immediately that, in general, $$|\psi(t)\rangle$$ must be complex.$$^{\ddagger}$$

For further reading, see About the complex nature of the wave function? and QM without complex numbers on Physics Stack Exchange. Some of those answers use empirical arguments, but pcr's answer makes the same argument as mine and remains purely theoretical, and is by extension still applicable to your universe.

## The Schrödinger equation from $$\hat{U}(t)$$

From the time-evolution operator, we can quickly derive a form of the Schrödinger equation by looking at an infinitesimal time translation $$\hat{U}(dt)=1-\frac{i}{\hbar}\hat{H}dt$$ At a time $$t+dt$$, we can find the state of the system from $$\hat{U}(t+dt)$$, which you can convince yourself is just $$\hat{U}(dt)\hat{U}(t)$$: $$\hat{U}(t+dt)=\left(1-\frac{i}{\hbar}\hat{H}dt\right)\hat{U}(t)$$ Rearranging, $$\hat{U}(t+dt)-\hat{U}(t)=\left(-\frac{i}{\hbar}\hat{H}\right)\hat{U}(t)$$ If we divide both sides by $$dt$$, we see that the left just gives us the expression for the time derivative of $$\hat{U}(t)$$. We can then rewrite this as $$i\hbar\frac{d}{dt}\hat{U}=\hat{H}\hat{U}(t)$$ Applying both sides to the initial state $$|\psi(0)\rangle$$ gives us $$i\hbar\frac{d}{dt}|\psi(t)\rangle=\hat{H}|\psi(t)\rangle$$ which is the Schrödinger equation. This is a quick-and-dirty derivation (source: Townsend, A Modern Approach to Quantum Mechanics, second edition, chapter 4).

## The Dirac equation

The Dirac equation is much more complicated. It breaks the wave function into four separate components, and in reality, it's actually four separate coupled first-order linear partial differential equations. I'm not as familiar with the Dirac equation as I am with the Schrödinger equation, so I won't try to do it justice, but I will say that given that it can be thought as arising from taking the square root, so to speak, of the operator $$\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}$$ we can see that perhaps a factor of $$i$$ must creep in somewhere to account for that minus sign.

$$^{\dagger}$$The wave function $$\psi$$ can be found from a quantum state by taking the appropriate inner product. For example, if we want the position space representation of the wave function, we define it by the inner product $$\psi(x)\equiv\langle x|\psi\rangle$$; if we want the momentum space representation, we use the inner product $$\psi(p)\equiv\langle p|\psi\rangle$$. While I've technically focused this answer on quantum states, it's simple to show that the logic by extension also holds for wave functions.

$$^{\ddagger}$$In the case where $$\hat{H}=0$$, we have $$\hat{U}(t)=1$$, and so if $$|\psi(0)\rangle$$ is real, then so is $$|\psi(t)\rangle$$. On the other hand, this is a trivial case that holds only under one particular (and extremely odd) set of circumstances, and in reality, no particle is truly subject to a vanishing Hamiltonian.

• You mentioned that it's possible to derive the form of the Schrödinger Equation. Is it also possible to derive the form of the Dirac Equation? – Anders Gustafson Jul 5 at 21:07
• @AndersGustafson I don't know the Dirac equation well enough to give a good answer there; I can do some reading to try and figure it out. – HDE 226868 Jul 5 at 21:57
• What types of mathematics do I need for deriving the Schrödinger Equation, or the Dirac Equation? – Anders Gustafson Jul 7 at 19:31
• @AndersGustafson I've edited in a fairly basic derivation of the Schrödinger Equation, one you could find in an introductory quantum mechanics textbook (in fact, it's the one I learned). A derivation of the Dirac equation is a bit harder, and I don't know what sort would be the most intuitive to you. Being comfortable with linear algebra and matrix mechanics would be ideal for fully understanding it. – HDE 226868 Jul 7 at 20:38
• Thank You, for the information. – Anders Gustafson Jul 9 at 2:36

Edit: I have misunderstood the question.

An elementary particle IS a wave-function. Your question is just asking could there be different elementary particles. The answer is the same as many of your other similar questions:

There are a bunch of parameters in the laws of physics. For example, the mass of the proton. The equations that govern the proton's behaviour depend on some number $$m_P$$ which is measured experimentally to be about $$1.6726219 × 10^{-27}$$kg.

We have no idea if this number is special. So there is no more reason to believe the current laws are consistent than there is to believe the laws with $$m_P= 2.6726219 × 10^{-27}$$kg are consistent. The second set of laws describe a universe with a heavier proton.

As always a universe with a heavier proton is probably full of only loose energy and is completely uninteresting.

• An elementary particle is not a wavefunction. A wavefunction just describes the state of a particle or group of particles, but different elementary particles of the same type can have different wavefunctions. – StephenG Jul 4 at 9:36
• @StephenG Ah it looks like I have misread the question " in this universe the wavefunctions that elementary particles, and systems of elementary particles can have". – Daron Jul 4 at 9:39
• This looks like the beginnings of a good answer. Please edit it to better match the question. – a4android Jul 5 at 1:42
• @a4android I don't understand the question. I will leave the bad answer up to encourage others to post a better one. – Daron Jul 5 at 11:35
• Fair enough. Always a good idea to encourage others to do better. Keep up the good work. – a4android Jul 6 at 1:49