What is quantum field theory and why was it developed?

[slyboy]

I have an alternative point of view, which is that we need to find some physically meaningful way to jump straight to the correct quantum theory without first writing down a classical field theory and then trying to quantize it. Of course, no-one really has a clue how to do such a thing, so we are stuck with the ambiguities involved in quantization.

 

[selfAdjoint]

Here's a thought that has troubled me for years. Planck's quantum is an indivisible chunk of action. This means that a Lagrangean, an expression for the action of some system, must always be an integer multiple of h. So the proper branch of mathematics to study quantized Lagrangeans is number theory - say partition theory for example. No need to bring calculus into it at all, unless you want to study the well-defined analytical number theory. Is this why the Riemann zeta function keeps coming up mysteriously in advanced physics (google on zeta function regularization or renormalization)?

 

[lethe]

Originally posted by selfAdjoint
Here's a thought that has troubled me for years. Planck's quantum is an indivisible chunk of action.

i have seen you mention this before, but i think perhaps you didn't see my response to this issue.

basically, it sounds like you are describing Bohr-Sommerfeld quantization. one of the results of that kind of quantization is that the action is an integral multiple of Plank's constant.

however, Bohr-Sommerfeld quantization is just wrong. it doesn't work except for some simple cases.

i have never heard anyone say that action takes integral values in a modern theory. does modern quantum mechanics predict integral values for the action for the SHO, for example? i would be surprised.

This means that a Lagrangean, an expression for the action of some system, must always be an integer multiple of h. So the proper branch of mathematics to study quantized Lagrangeans is number theory - say partition theory for example. No need to bring calculus into it at all

well, as i am sure you know, the relationship between the action and the Lagrangian is an integration over a spacetime, so there is definitely still calculus involved.

unless you want to study the well-defined analytical number theory. Is this why the Riemann zeta function keeps coming up mysteriously in advanced physics (google on zeta function regularization or renormalization)?

i know the Riemann zeta function has a lot to do with number theory, but i think of zeta regularization as being a complex analysis result, not to germaine to number theory. but then, i don't really know much number theory, so i could be way off base there.

 

[selfAdjoint]

Yes, Bohr-Sommerfeld quantization did equate the action to nh, but they did nothing except the obvious with the idea. Also your statement that the integral of the action requires calculus is worng; the integral of an integer valued function between finite limits is a finite sum.

 

[Mike2]

Originally posted by selfAdjoint
Yes, Bohr-Sommerfeld quantization did equate the action to nh, but they did nothing except the obvious with the idea. Also your statement that the integral of the action requires calculus is worng; the integral of an integer valued function between finite limits is a finite sum.

Correct me if I'm wrong. But just because the integral equates to an integer does not mean the integrand is an integer. The Gauss-Bonnett theorem of the curvature on a surface also equates to an integer, but the curvature itself is a continuous function, right?

 

[selfAdjoint]

Mike you've got it backward. I didn't say the integrand is an integer because the integral is, I said the integral is a sum because the integrand is always integer valued.

 

[lethe]

Originally posted by selfAdjoint
Yes, Bohr-Sommerfeld quantization did equate the action to nh, but they did nothing except the obvious with the idea.

right. Bohr-Sommerfeld quantization was dead by 1915. so my question is, why are you still talking about it?

modern quantum mechanics, as far as i know, does not stipulate that the action is integer valued.

Also your statement that the integral of the action requires calculus is worng; the integral of an integer valued function between finite limits is a finite sum.

huh?

who said anything about the integral of an integer valued function? the Lagrangian is a real valued fuction, whose integral is the action. so if you are talking about the action being an integer valued function (which i object to), then you have a real valued function whose integral is an integer valued function.

you do not have the integral of an integer valued function.

and what in the world are you saying about calculus and finite sums? since when do you not need calculus to evaluate integrals?

 

[selfAdjoint]

huh?

who said anything about the integral of an integer valued function? the Lagrangian is a real valued fuction, whose integral is the action. so if you are talking about the action being an integer valued function (which i object to), then you have a real valued function whose integral is an integer valued function.

you do not have the integral of an integer valued function.

and what in the world are you saying about calculus and finite sums? since when do you not need calculus to evaluate integrals?

Right, the action is the integral of the energy X dt, and both of these are continuous (at least in current models). But the integral only takes on integer values of h! Because action is not a continuous quantity, it's quantized. This is not something I assumed, it's the basic definition of the quantum. It isn't something that just went away when the incorrect model of Bohr and Sommerfeld was replaced.

And the Lebesge-Stieltjes integral becomes a sum when the integrand in integral, which happens not in the initial step, but later.

 

[lethe]

Originally posted by selfAdjoint

Right, the action is the integral of the energy X dt, and both of these are continuous (at least in current models).

a minor nit: action is the integral of the lagrangian, not the energy

But the integral only takes on integer values of h! Because action is not a continuous quantity, it's quantized. This is not something I assumed, it's the basic definition of the quantum. It isn't something that just went away when the incorrect model of Bohr and Sommerfeld was replaced.

you keep saying this, but i have never heard this in my life. can you please provide a reference?


the action for a nonrelativistic free particle is
$$S=\int dt \frac{1}{2}m\dot{x}^2$$
in a momentum eigenstate, this action can take on any value. it does not appear to me to be restricted to integer values. can you provide some evidence for your claim?

And the Lebesge-Stieltjes integral becomes a sum when the integrand in integral, which happens not in the initial step, but later.

the integrand is the Lagrangian. are you also claiming that the Lagrangian is integer valued?

 

[jeff]

Hi selfAdjoint,

Can we not write down actions whose absolute values are less than Planck's constant? Really, Planck's constant just gives the scale at which quantum effects become important.

 

[lethe]

suppose that selfAdjoint were correct, in quantum physics, the action is an integral multiple of h.

then let's see what this tells us about the path integral

$$S[\phi]=nh$$

$$Z=\int\mathcal{D}\phi\ e^{iS[\phi]/\hbar}=\int\mathcal{D}\phi\ e^{2\pi ni}=\int\mathcal{D}\phi$$

so if what selfAdjoint is telling us is true, then all dynamics is trivial, the partition function of any theory has no functional dependence on $\phi$, and so all correlation functions are zero.

i don't even want to talk about selfAdjoint's other statements concerning integrals turning into sums, because i am having a lot of trouble even making heads or tales of that what might be supposed to mean.

 

[lethe]

well, selfAdjoint, are you not going to respond?

 

[selfAdjoint]

Not a constant n, silly. n(t) varies but only takes integer values. Probably you need an appropriate measure for your integral.

 

[lethe]

Originally posted by selfAdjoint
Not a constant n, silly. n(t) varies but only takes integer values. Probably you need an appropriate measure for your integral.

regardless of whether n is constant or not, so long as it only takes integer, $e^{i2\pi n}=1$

so, your rebuttal is ridiculous.

now can you please provide a reference for your position that the action (and also the Lagrangian?) of a quantum system takes integer values?

perhaps you can point out why the action of a free particle must be integer valued?

or you know, if you are going to make completely wild and baseless assertions, call me incorrect and silly, without ever providing any evidence, then i will conclude that you are a rambling crackpot, and get on with my life, and not read your posts anymore.

 

[selfAdjoint]

Lethe, consider that you have won. You know I just wondered, and you dragged me into this argument. But how would you explain that action is certainly expressed as a multiple of h? For that matter how can people do quantum mechanics over a Galois field?

 

[Nereid]

Guys, please (I think you're both guys), this isn't about 'winning' or 'losing'

I'd love to be shown to be wrong, but most readers of this thread can barely keep up with the concepts, let alone appreciate how much math is behind them.

It's an exciting time to be here; there's lots of really interesting ideas being proposed, examined, discussed, debated, ... even arguments There is an unprecedented access to experimental data, and the promise of vastly more to follow.

Let's have more discussions, without 'winners' or 'losers'.

 

[outandbeyond2004]

We need the truth to be a winner, never a loser. (This has more than one meaning).

 

[lethe]

Originally posted by selfAdjoint
Lethe, consider that you have won.

what are you saying? you have really confused me now. do you concede that action is not a multiple of h?

You know I just wondered, and you dragged me into this argument. But how would you explain that action is certainly expressed as a multiple of h?

i would explain it very simply: action is not an integral multiple of h, and anyone who says it is, is mistaken.

you told me outright that i was wrong, without providing any evidence. i wasn't trying to drag you into an argument you didn't want to have, i just want to know whether there is any reason why you think these things you think about action. it is very frustrating when you simply respond "you are wrong" or "you are silly", without any explanation, or even worse, explanations that contain arithmetic errors of a high school nature. if you were just wondering aloud about an idea you had, then you shouldn't put it forth as fact.

For that matter how can people do quantum mechanics over a Galois field?

what the **** does that have to do with anything?

 

[Janitor]

I skimmed through Selfadjoint's link to see if it looked like crankery. I found this statement in it:

Suppose we wish to verify experimentally whether addition is commutative, i.e. whether (a + b) = (b + a) is always satisfied. If our Universe is finite and contains not more than N elementary particles then we shall not be able to do this if a + b > N. In particular, if the Universe is finite then it is impossible in principle to build a computer operating with any large number of bits.

Whaaaa?

Is he saying what I think he is saying? Let us suppose the universe contains 10^80 particles. (I just pulled that number out of thin air. Supply your own number if you don't like mine.) Is he saying that an ordinary desktop computer cannot be programmed to add (for example) a pair of base-10 numbers each of which is 81 digits long, first in one order, then in the other order, and check that the results are the same? Surely he isn't that stupid? [?] Am I being stupid?

 

[selfAdjoint]

Well since all the registers in your computer (where you would store the intermediate and final results) could by assumption have less than 10^80 bits (at a minimum 1 electron per bit) it seems to follow. And if you choose to do the math with pencil and paper , consider how many electrons are in a sheet of paper. Even if you reused paper in your calculation, you would have to write down the final answer.

 

[Mike2]

Originally posted by selfAdjoint
Well since all the registers in your computer (where you would store the intermediate and final results) could by assumption have less than 10^80 bits (at a minimum 1 electron per bit) it seems to follow. And if you choose to do the math with pencil and paper , consider how many electrons are in a sheet of paper. Even if you reused paper in your calculation, you would have to write down the final answer.

So are you suggesting that the universe must be infinite? It does not take 10 billion particles to symbolize 10 billion particles. But you are right, we could never "experimentally confirm" by measuring that addition is commutative at high enough values if the universe is finit? What the point? I seem to have missed it.

 

[selfAdjoint]

I think the point of the paper was to see how far you could go in this direction. Notice that it gave them some nice things, but also some things that disagreed with experiment. There are a lot of papers like this on the arxiv, and every now and then one of them makes it into a reviewed journal.

 

[kurious]

If their is a sea of virtual particles interacting with electrons and protons, for example,then what is the temperature of that sea and does it have entropy and exchange heat with normal matter?

 

[selfAdjoint]

No it doesn't. It is possible, if not necessary, to think of virtual particles as just mathematical terms in a power series, with no physical reality at all. Even the Casimir effect, which seems to show their reality, can be explained as an interaction with the plates.

 

[rtharbaugh1]

I just read this thread, with interest but not satisfaction. I thought it was an interesting question, about whether action must occur in integer values of Planck's constant.

Is it possible that this argument reduces to the theory of Pythagorus regarding the sum of squares? For example, if an action involves two steps left and three steps up, the resultant action is the hypoteneus, sqrt 13, not an integer. This seems trivial. Have I missed the argument entirely?

 

[lethe]

No it doesn't. It is possible, if not necessary, to think of virtual particles as just mathematical terms in a power series, with no physical reality at all. Even the Casimir effect, which seems to show their reality, can be explained as an interaction with the plates.

i wonder what on Earth you were talking about before, silly