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Manfred13y00

Why do you think I would say that the time-dependent Schrodinger equation doesn't do anything if the universe is in an energy eigenstate?

DanielLC13y00

I guess I misread that. I still don't understand it.

The time-independent equation is how the world works. The time-dependent one also applies, since it's a more general case. The fact that it applies shows that the universe still looks like you'd expect it to, and it all adds up to normality.

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Timeless Physics Question

by DanielLC

28th Apr 2012

1 min read

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Timeless physics is what you end up with if you take MWI, assume the universe is a standing wave, and remove the extraneous variables. From what I understand, for the most part you can take a standing wave and add a time-reversed version, you end up with a standing wave that only uses real numbers. The problem with this is that the universe isn't quite time symmetric.

If I ignore that complex numbers ever were used in quantum physics, it seems unlikely that complex numbers is the correct solution. Is there another one? Should I be reversing charge and parity as well as time when I make the standing real-only wave?

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[-]Manfred13y40

assume the universe is a standing wave,

A standing wave is a 3d thing. The universe is 4d. There is no "second time" through which it can be a standing wave. It's just an object. Probably.

[-]DanielLC13y-20

Maybe "the multiverse" would be a better term. I mean the configuration space for the whole universe. It's at least 10^80d, and many suspect it to be infinity d.

A simple way to make a standing wave would be to take one possible configuration space, and add versions of itself that are off by time t (with tricks to make it come out finite). Thus, you end up with the universe being a standing wave using the time we're all used to, and all the predictions being pretty much the same.

[-]Manfred13y20

This is not possible because different particles oscillate with different frequencies, depending on their energy. And though it would be possible to make a "standing wave universe" work by simply adding a universe to its complex conjugate, writing it as a sum is probably the shortest way to write this universe, so it's not like we actually got rid of complex numbers in any equations.

[-]DanielLC13y00

This is not possible because different particles oscillate with different frequencies, depending on their energy.

The universe oscillates with exactly one frequency as it moves through configuration space.

As for why it looks like different particles have different frequencies, I'm not sure. Apparently, I don't understand entanglement as well as I thought I did.

And though it would be possible to make a "standing wave universe" work by simply adding a universe to its complex conjugate, writing it as a sum is probably the shortest way to write this universe, so it's not like we actually got rid of complex numbers in any equations.

Wouldn't just writing the real parts of the boundary conditions be simpler?

[-]Manfred13y20

The universe oscillates with exactly one frequency as it moves through configuration space.

As for why it looks like different particles have different frequencies, I'm not sure. Apparently, I don't understand entanglement as well as I thought I did.

Our universe is not an energy eigenstate, that's why.

[-]DanielLC13y00

I said that wrong.

The universe has exactly one amplitude. It does not have all the amplitudes of the constituent particles. It does not move in all the ways the constituent particles do.

I figured out why it looks like different particles have different frequencies, even if the universe only has one. I don't think I could explain it well though.

Our universe is not an energy eigenstate

How do you know? Each individual particle is not in an energy eigenstate, but that doesn't mean that the system isn't. If you add the waveform of a system where particle a has energy 1 and particle b has energy 2 to a system where particle a has energy 2 and particle b has energy 1, you end up with a system with an energy eigenstate of 3, but each particle is not in an energy eigenstate.

You could, in theory, check whether or not the universe you're in is where you'd expect a node to be, but if I understand this right, the nodes are all within tiny fractions of a Planck length of each other. You'd have to know the position of every particle in the universe with a root mean square error smaller than that.

[-]Manfred13y00

How do you know?

The short answer is that energy eigenstates don't change over time, while the universe does.

If you add the waveform of a system where particle a has energy 1 and particle b has energy 2 to a system where particle a has energy 2 and particle b has energy 1, you end up with a system with an energy eigenstate of 3, but each particle is not in an energy eigenstate.

This is a good point. What I said didn't mean what I thought it meant. But this system seems like an example of the power of entanglement. If the particles were unentangled, there would be change over time. But they are entangled, and there isn't any change over time. A computer living in this system would not actually move any electrons around to do any computation.

[-]DanielLC13y00

while the universe does.

How do you know? You can only see what's happening now.

[-]Manfred13y00

If particles "move around," but the state doesn't change at all (because it's an energy eigenstate), then no robot made of particles will ever write anything new to its hard drive. The particles don't remember that they moved around - that's one of the whole points of quantum mechanics.

[-]DanielLC13y00

The particles are in that position. They are in that position because the boundary conditions of the universe are such that them being in that position has a relatively high amplitude.

If it's an energy eigenstate, it still has to work as a universe. It still has to have a past for every future and a future for every past. If it has the big bang, it will still have the people who remember existing that must inextricably follow. It's just that it has them all at once.

[-]Manfred13y00

So the time-dependent Schroedinger equation is how the world works, but it doesn't do anything, and by some separate miracle the things that exist look like the time-dependent Schroedinger equation? :D

Sounds neat!

[-]DanielLC13y00

but it doesn't do anything

What do you mean?

and by some separate miracle

It's only a miracle if it's false. It would be surprising for there to be a simpler explanation than the true one, but it's only expected for there to be a more complex one.

[-]Manfred13y00

Why do you think I would say that the time-dependent Schrodinger equation doesn't do anything if the universe is in an energy eigenstate?

[-]DanielLC13y00

I guess I misread that. I still don't understand it.

[-]Shmi13y40

Please note that, while an appealing idea, timeless physics is not a physical theory but only a hope for one. As far as I know, it has no solid mathematical foundations separate from the mainstream and makes no new testable predictions.

With this caveat, why do you want to ignore the complex numbers, given that they are just a set of real commuting 2x2 matrices? Or do you want to get rid of all matrices (i.e. of all linear algebra) in physics? If so, what would be your motivation?

[-]DanielLC13y-30

Please note that, while an appealing idea, timeless physics is not a physical theory but only a hope for one.

Really? Isn't it just the idea that Schroedinger's time-independant equation is the correct one? Is there not a time-independent version of the correct one (as opposed to the non-relativistic approximation I've seen)?

With this caveat, why do you want to ignore the complex numbers, given that they are just a set of real commuting 2x2 matrices?

If you can remove complex numbers in this manner, you could use R, or R^2 in the same manner as you'd use C, but you could also use R^3, R^4, R^5, etc. It must be more likely that any of those is correct than that R^2 is correct in particular.

[-]Shmi13y40

Really? Isn't it just the idea that Schroedinger's time-independant equation is the correct one? Is there not a time-independent version of the correct one (as opposed to the non-relativistic approximation I've seen)?

The best of what people have come up with so far is the Wheeler-de Witt equation. Unfortunately it is not good enough.

If you can remove complex numbers in this manner, you could use R, or R^2 in the same manner as you'd use C, but you could also use R^3, R^4, R^5, etc. It must be more likely that any of those is correct than that R^2 is correct in particular.

Commuting 2x2 real matrices are the smallest real representation of C.

[-]DanielLC13y-20

The best of what people have come up with so far is the Wheeler-de Witt equation. Unfortunately it is not good enough.

From what I can find, that looks like some attempt at quantum gravity. We can't do that with MWI either, as far as I know. Am I mistaken about this?

Commuting 2x2 real matrices are the smallest real representation of C.

I guess I'll take your word for it.

Do you really need C for quantum physics though? You can't multiply two amplitudes together. The only thing I've seen is rotating by 90 degrees in Schrodinger's time-dependent equation. If I accept timeless physics, it doesn't even do that.

[-]Mitchell_Porter13y20

You can't multiply two amplitudes together.

You do this whenever you calculate the amplitude contributed by a single history within a sum over histories. The amplitude for an event is exp(i.action) and action is additive, so the amplitude for two events forming a single history is the product of their individual amplitudes, exp(i.action1+i.action2). In this respect it's just like ordinary probability theory, where you multiply probabilities for conjunction of events and add them for disjunction.

I don't understand the motivation or the assumptions for what you are doing. Quantum cosmology is such a guessing game that even unusual formal generalizations might lead somewhere, and I would like to offer you useful feedback, but I'm wondering if there's some basic misconception about QM motivating you.

[-]DanielLC13y-20

the amplitude for two events forming a single history is the product of their individual amplitudes

I'm not sure I understand. What is an "event"?

I've noticed that the amplitude of a system is equal to the product of the amplitudes of the component particles, but that's just mathematical shorthand. Individual particles don't have their own amplitude. Only the universe does.

I don't understand the motivation or the assumptions for what you are doing.

I'm trying to make it simpler. It's not much, but each bit you can shave off of the equations doubles the probability.

[-]Mitchell_Porter13y00

What is an "event"?

An event is a "thing that happens". Relativity made discussion of "events" routine in physics, because one wants to talk about something - the tick of a clock, the emission or absorption of a photon - that is localized in space and time. "Event" is a completely standard term of art in relativity - thus "event horizon". Of course, it is also an elementary everyday word and concept, independent of its use in physics.

In standard probability calculus, P((A and B) or (C and D)) = P(A) x P(B) + P(C) x P(D). You're summing the probabilities of two possibilities: the possibility that A and B occur together, and the possibility that C and D occur together. Feynman's reformulation of quantum mechanics as a "sum over histories" has this schematic form as well, except that it is the complex-valued "probability amplitudes" that we multiply and add. The basic events are "a particle moves from one place to another place" and "a particle is emitted or absorbed", and the amplitude or these events is "e" to the power of "i" times the "action" for this event, action being a concept from classical physics which carries over to quantum theory, and which in fact assumes a fundamental role there.

A standard example of a timeless-looking construction from quantum cosmology is the Hartle-Hawking wave-function of the universe, derived from a "no-boundary condition". This wavefunction assigns an amplitude of three-dimensional configurations of the universe; which sounds like Barbour's "Platonia", But how are these amplitudes calculated? By summing over space-time histories which evolve to the three-dimensional configuration of interest. The amplitude for a configuration X is the sum of the amplitudes of every space-time history which starts from "nothing" (that's why this is the "no-boundary condition") and which evolves to X.

In a timeless framework, you could possibly conceive of an event as the difference between one configuration and its neighbor in configuration space, and the amplitude for the "event" as the weighting for the contribution made by the first configuration to the amplitude of the second configuration, via timeless amplitude "flow". That is, if you have one configuration of particles, and then a neighboring configuration which is the same except that there is now an extra photon on top of one of the electrons, then the "event" corresponding to this configurational difference would be "emission of a photon by that electron", and the usual Feynman amplitude for this event would define the proportional contribution to the amplitude flow entering timelessly into the second configuration's point in configuration space.

It's a standard fact about the Schrodinger and Feynman formulations of quantum mechanics that they are equivalent - the evolution of the Schrodinger wavefunction is equivalent to the cumulative flow of amplitude produced by the converging and diverging Feynman histories - and this should carry over to the timeless case of quantum cosmology... in principle. But in practice, the Feynman formulation seems more relativistic so perhaps it's more fundamental. In any case, you do sometimes multiply amplitudes when you do quantum mechanics as Feynman did it.

[-]DanielLC13y-10

The basic events are "a particle moves from one place to another place" and "a particle is emitted or absorbed", and the amplitude or these events is "e" to the power of "i" times the "action" for this event

Is that like the idea that a particle being in a certain position has an amplitude? It doesn't. The universe does. It's just that if you pretended that a decohered particle was its own universe, you'd get the same results from much simpler calculations.

This does explain why physicists tend to write amplitude as a complex number. I've wondered that for a while.

[-]asr13y00

I'm extremely unclear what you are trying to explain or what problem you are trying to solve. Can you give a better overview here?

[+]DanielLC13y-60

[-]orthonormal13y00

Entropy (in particular, the low-entropy state at t=0) is really important to the timeless physics of our universe; I don't think you'll get the right picture without it.

[-]DanielLC13y00

Is what I said incompatible with that?

Technically, there's no t=0, since there's no t, but it has low entropy and high amplitude at the big bang.

[-]Mitchell_Porter13y00

Are you trying to explain the Born rule, by producing a strictly real-valued superposition of universe-wavefunctions, which therefore looks more like an ordinary probability distribution?

ETA: I have a downvote, I don't know why. My very next observation would be that you would still have to square these real-valued amplitudes in order to get the probabilities, possibly followed by a discussion of the prospects for doing QM over algebraic fields other than R and C, and whether this changes anything.

[-]DanielLC13y-20

Are you trying to explain the Born rule

No. I just figure that real numbers are simpler than complex numbers.

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