# What if h=0?

1. Feb 4, 2005

### DrChinese

Suppose that h had a different value in the uncertainty relations. For the sake of argument, suppose:

h=0

Would we then live in a fully deterministic universe? Does this question even make sense?

2. Feb 4, 2005

### mathman

The question makes logical sense, but what is the point? We know that h has a specific value and quantum theory works.

One possible problem with h=0 is the original subject that Planck was addressing when he discovered h - specifically, the blackbody spectrum. With h=0, you end up with an infinite amount of energy out of the black body.

3. Feb 4, 2005

### Edgardo

Hi Dr. Chinese,

I once heard a professor saying: $\hbar=0$
means classical physics. I am not sure, but I think
it's because of $E= \hbar \omega$ for the energy of
a photon. So if $\hbar=0$ there are no photons.

That's what I think.
......

http://www.lns.cornell.edu/spr/2004-04/msg0060332.html

Last edited: Feb 4, 2005
4. Feb 4, 2005

### kirovman

Well everyone knows $\hbar=1$

In natural units.

But if $\hbar$ were any other value, the universe would be totally different to what it is today.

You could also discuss what if the unit of electronic charge were slightly different?

What if protons were twice their mass?

This belongs to philosphy discussion I believe.

5. Feb 4, 2005

### vanesch

Staff Emeritus
Well, if h were "any other non-zero value" with all the rest the same, we wouldn't notice it ! As you point out, in natural units, hbar = 1, and if you would then carefully derive our unit system (you know, the speed of light is ... and a second is so many oscillations of a cesium atom transition etc...) you would find that h = 6.6256 10^(-34) in THAT unit system !

So we only have physics with h given that value, or with h = 0 (which would indeed give a lot of troubles).

Exactly the same can be said about the "value of lightspeed".

The only thing that really makes sense are dimensionless numbers.
All the other constants just define the unit system.

cheers,
Patrick.

6. Feb 5, 2005

### Galileo

From the de Broglie relation:

$$\lambda=\frac{h}{p}$$

if h=0, the wave-nature of particles does not exists.
Likewise, in optics, when we ignore the wavelike nature of light, we get geometrical optics (no diffraction effects etc.)
So in that sense, geometrical optics is to light, what classical mechanics is to quantum mechanics.

7. Feb 5, 2005

### vanesch

Staff Emeritus
In fact, classical EM wouldn't exist either. Otherwise you'd hit the UV catastrophy of blackbody radiation. Gravity is a bit more difficult, but I'd guess you would have a Newtonian universe with point particles and contact interactions, and no fields.

cheers,
Patrick.

8. Feb 5, 2005

### DrChinese

Thanks everyone. I guess the lesson is that h must be greater than 0.

I have previously wondered as follows: can the future influence the past? Could the value of h represent the extent that the future influences the past? I.e. a relationship between the past and the future in an otherwise mostly time-symmetric universe? If the future influenced the past, that would certainly explain a lot about where the source of randomness might be.

Just wondering... i.e. perhaps much about the specification of a system might be due to its past. The remainder would be filled in by effects propagating from the future. I know this is totally speculative, just wondered if anyone else had any knowledge or thoughts about it.

P.S. Edgardo, thanks for the ref.

9. Feb 5, 2005

### ahrkron

Staff Emeritus
I don't agree with this. If there was a sudden change in the value of h, while all other constants remained the same, many dimensionless numbers would change and many physical systems would show it.

... unless all physical parameters were ultimately dependent on the value of h, but I see that as extremely unlikely.

10. Feb 5, 2005

### ahrkron

Staff Emeritus
It might, but I find it interesting to think about how nature would solve the "ultraviolet catastrophe" in a classical universe.

Either black bodies would show an extremely bizarre behavior (what would it be? would they stay at almost absolute zero temperature? would they emit all their energy in a flash the minute they became black?), or some more basic laws (like Maxwell's) would need to be different (... but then it is hard to see where to stop, or what principles to keep).

Actually, understanding this better may shed some light on the implications and, abusing language, the logical necessity of quantization.

11. Feb 5, 2005

### Haelfix

Your proffessor was correct, the limit hbar --> 0 is the classical limit.

Its quite easy to see, the poisson bracket of two operators in quantum mechanics is

[Q(t),P(t')] = i hbar. Sending hbar --> 0 implies the operators commute for all t,t' spanning the phase space. Ergo by definition the system is deterministic.

12. Feb 6, 2005

### vanesch

Staff Emeritus
Planck's constant is, if I'm not mistaking, just a derived number of the definition of the SI unit system.
If you start out in Planck units, all numbers are dimensionless. Let us take the standard model: you can express all its fundamental parameters as dimensionless numbers. Assuming we take all these numbers to remain the same (because otherwise we're not changing only Planck's constant) ALL numbers we calculate in Planck units (and which are of course dimensionless) remain the same. In order to map them onto SI units, you have to work out the definitions of the units as a function of things you calculate in the standard model.
Now let us suppose we have a dimensionless number which represents a certain action A(Planck). In order to have it in SI units, we have to multiply it with h of course. So A(SI) = h(SI)/(2pi) .A(Planck)
Can we choose h(SI) ?
Let us find out.
h(SI) can be found by looking at the wavelength of, say, a photon.
We have that the wavelength is h/p = hc /E, so from this, we find:

h = E lambda/c

In Planck units this just tells you that lambda(Planck) = 1/E(Planck), and in SI units we have:

h(SI) = E(SI) lambda(SI) / c(SI)

c(SI) is a fixed number in the SI unit system, by definition.
We now have to work on lambda(SI) and on E(SI).

We can calculate lambdaCs(Planck) for the Cesium transition, and we have a fixed SI number N(meterSI) of these transitions defining the meter in SI.

lambda(SI) = Lambda(Planck) / { lambdaCs(Planck).N(meterSI) }

Finally, the energy in SI units. There is a problem with the definition of the Kg in SI units, because it has not yet been defined by a fundamental process, but just a sample in Paris somewhere. But this will change. Let us take for sake that we already have such a definition, namely one which fixes the electron mass (that's not the most accurate technique, but hey). I could do this in detail but it comes down to saying that 1 kg is N(kgSI) times the mass of an electron.

m(SI) = m(Planck) / {m_electron(Planck) N(kgSI)}

This then fixes our unit of energy:
E(SI) = E(Planck) / { c(SI)^2 m_electron(Planck) N(kgSI)}

And now we're done:
h(SI) = E(Planck) / { c(SI)^2 m_electron(Planck) N(kgSI)} Lambda(Planck) / { lambdaCs(Planck).N(meterSI) } / c(SI)

h(SI) = E(Planck).Lambda(Planck) / {c(SI)^3 m_electron(Planck) N(kgSI) lambdaCs(Planck) N(meterSI)}

We know that for a photon, E(Planck) = 1/Lambda(Planck) so this drops and we find:

h(SI) = 1 / {c(SI)^3 m_electron(Planck) N(kgSI) lambdaCs(Planck) N(meterSI)}

So the number for h(SI) is fully determined by conventional quantities for the SI unit system such as c(SI), N(kgSI) and N(meterSI), and by dimensionless constants which are fundamental (m_electron(Planck)) or derived (lambdaCs(Planck)) in the standard model (of which we assumed that they remained constant of course ; otherwise it wouldn't be a change of Planck's constant, but a change of the mass of the electron or so we would be talking about).

cheers,
Patrick.

EDIT:

It is even more fun: note that the energy levels of atoms are proportional to the reduced mass, which is essentially the effective mass. So it even seems that m_electron(Planck) will be eliminated by lambdaCs(Planck) of which only remains some dimensionless numbers of combinations of properties of the angular functions Y and so on.

Last edited: Feb 6, 2005
13. Feb 6, 2005

### WORLD-HEN

What if the value of h increases with time. Maybe the universe started classically with h=0 and becomes a quantum universe. crazy idea.

14. Feb 6, 2005

### Edgardo

Hey, thanks for pointing that out :shy: .

15. Feb 7, 2005

### vanesch

Staff Emeritus
Yes, but things aren't that easy. It also implies that Q and P, as any other observable, commute with H. So there is no time evolution anymore, either. Every state is a stationary state.

cheers,
Patrick.

16. Feb 7, 2005

### Chronos

'h' cannot be zero. A universe such as ours cannot exist if h=0.

17. Feb 7, 2005

### vanesch

Staff Emeritus
But maybe another universe does

cheers,
patrick.

18. Feb 7, 2005

### Haelfix

Err the Lie bracket of the hamiltonian operator and say momentum is zero in the classical limit.

You know, i hbar Dp/Dt = [p,H] --> 0. That just says you can simultaneously measure energy and momentum observables classically. It does not imply the poisson bracket is zero.

You are correct in the sense that its not the full story. Really the classical limit is a bit nefarious, you really want to send the Fock particle count and temperature counts high as well.

The more troubling bit is the Dirac correspondance principle breaks down under general conditions. In general for arbitrary dynamic variables that are higher than quadratic order, you have corrections in the commutator algebra that may or may not depend on powers of hbar. Physicists call this operator ordering ambiguities. I guess the sensible thing to do is to say that it makes no sense to take a classical limit of a nonsensible quantum theory, even if you prequantized it to begin with.

Last edited: Feb 7, 2005
19. Feb 8, 2005

### Chronos

'h' cannot be zero. The speed of light would then become infinite.

20. Feb 8, 2005

### Haelfix

What? That doesn't make much sense. In classical physics, before we discovered the ultraviolet catastrophe, there was no notion of h at all. However we still had Maxwell's equations, which gives c a constant value. Special relativity and General relativity are semi classical theories, no notion of quantum mechanics is there either.

I suspect you are looking at some equation with derived properties, not fundamental first principles.