1. Jul 3, 2007

qtheoryquestion

I want to start with my current understanding of QT before i get into my question:

Quantum theory states that any physical system remains in a superposed state of all possibilities until it interacts with the mind of an observer.

Is this essentially correct?

2. Jul 3, 2007

nrqed

No. There is nothing about "mind of an observer". It only has to interact with an external "measuring device" which may be a screen, an electric field, etc. Now, there is some debate about this interaction and how it occurs and whether there is a collapse at all and so on. But there is nothing about "mind of an observer".

3. Jul 3, 2007

qtheoryquestion

I see. I will revise my definition then:

Quantum theory states that any physical system remains in a superposed state of all possibilities until it interacts with an external measuring device.

Okay. I'm going to challenge the theory itself with a revision:

[Revised Theory] states that any unknown physical system remains in a superposed state of no possibility until it interacts with an external measuring device.

It seems to me, that if we are using Quantum theory, we are starting with infinity possibilities, and work towards a limit of 0.

Rather, why wouldn't it be that we start with 0 possibilities, and work towards a limit of infinity, as the [Revised Theory] would state?

Last edited: Jul 3, 2007
4. Jul 3, 2007

nrqed

well, not all possibilities are necessarily present. It depends on the state of the system. I mean, there are situations when the system is in a so-called "eigenstate" of an operator in which case even before making any measurement only one outcome is possible.
I don't understand what you mean. Once you make a measurement, there is only on epossibility: the number you obtained for your measurement. You never observe an inifinite number of measurements, you always measure one value.

EDIT: Do not pay attention, I am just testing something
$$\ddot{a} \framebox{111111112333444543}$$

Last edited: Jul 3, 2007
5. Jul 3, 2007

qtheoryquestion

Not all possibilities are necessarily present, but there is a probability that they are, correct?

If we're using the same definition of QT, we're stating that the system is remaining in a state of "all possibilities" (total possibilities are not known), and when a measurement is made, the value obtained eliminates one possibility out of a total which is not known. No?

Last edited: Jul 3, 2007
6. Jul 3, 2007

nrqed

Well, if you include a probability equal to zero, then yes. My point is that it's possible to have systems which are in well defined states (state of energy or spin or whatever) so that the probability of one possibility is 100% and the probability of all other possibilities is zero. In such a state, we know ahead of time with certainty what the outcome of a measurement will give.
No, the value obtained eliminates all the other possibilities than the one that was measured!

7. Jul 3, 2007

qtheoryquestion

I see where you're coming from. How do we know that it's possible to have a system in a well defined state? Is there such a system already existing?

Thanks for hanging in there with me while I ask these questions. :)

8. Jul 3, 2007

nrqed

Well, it's hard to explain without the maths but the basic idea is this. Let's say you start with a system that you don't know anything about. Then you measure one of its properties (its spin for example). Then, rightafter the measurement is taken we know that the system is still in that state that was measured. Now, under certain conditions the system will remain in the same state. So we know its state with certainty.
You are welcome. It's hard to explain clearly without getting into the technical aspects.

9. Jul 3, 2007

qtheoryquestion

Hold on. We start with a system that you do not know anything about. (i.e. either stable or unstable).

Then, we take a measurement of one of its properties.
Right after, we know the system is still in the same previous state.

If the system remains in the same state, we know with certainty it's state.

In conclusion, we know is that it's not changed it's state. It is still in a stable or unstable state. That's what I gather from what you're saying... am I missing something?

Last edited: Jul 3, 2007
10. Jul 3, 2007

nrqed

That's basically what I am saying, yes. Although I am not sure why you talk about "stable" and "unstable". I am talking about some measurable property, for example the projection of the spin along the z axis of an electron. This is a good example to focus on since a measurement may either yield "spin up" or "spin down" alongthe z axis so in this case, there are only two possible outcomes.

A genreal state is described by two numbers: the probability taht the measurement will yield spin up an dthe probability that it will yield spin down. (of course, the sum of those two numbers must give 1).

Now, if you measure the spin of an electron and get that the result is spin up, you know that right after the measurement it will still be with certainty in a spin up state (or if you want, it is in a state with the probability of being spin up being 100% and the probability of being spin down being 0%). Now, under certain conditions (here it means that the electron must not be subjected to any magnetic field), the spin will remain up for sure. In that case, I know with certainty that I wil get a spin up if I measure again the spin along z let's say 10 minutes later.

11. Jul 3, 2007

qtheoryquestion

I see. Essentially, we're using the first spin measurement to test.

What I meant originally by "stable" and "unstable" (my mind was on another subject), i meant defined. Are we sure that all systems are well defined? I would be assuming so, correct?

Last edited: Jul 3, 2007
12. Jul 3, 2007

nrqed

No. First, before you take your first measurement, you don't really know if it was well defined or not.

But then you may ask: are there any situations when we know for sure that it was not in a defined state before measuring?? his is a deep question and if you have heard about "hidden variable theories" or the "Einstein-Podolsky-Rosen paradox" or "Bell's theorem", it all ties in with those things.

But going back to the spin example, is there anyway to do something that will necessarily put the electron in an undefined combination of spin up and spin down along z?

the answer is yes! How do you do that? It turns out that if you measure the spin along a different axis (let's say the x axis) you will of course either get the measurement to give either spin up or down along x. It turns out that the theory says that an electron in a state of well-defined spin along x is necessarily in a combination of 50% spin up along z and 50% spin down along z! So by measuring the spin along x and therefore having a well-defined spin state along x, you have created an undefined spin state along z.

Hope this makes sense

Patrick

13. Jul 3, 2007

qtheoryquestion

Interesting. That spin example makes perfect sense. It's all very fascinating. Got introduced to quantum mechanics really through reading hawking's brief history of space and time a short while ago and have been thinking about it since.

I want to touch on what you said before, about

Actually, that's not my question! My question is: if it's possibly not a well defined system, does the spin test still hold true in all cases? I would logically think, no it doesn't. My following question would then be, how can taking these measurements determine with certainty if a system is well defined or not?

14. Jul 3, 2007

nrqed

[/quote]
I am not sure what you mean by "spin test". If by that you mean the measurement of the spin, the point is that whenever you measure a quantity, only one precise outcome is obtained (well, for some quantities like position, one only get a range of outcomes but that's a different can of worms! Spin is simpler to discuss).

And just after the measurement, thespin is necessairly in the state you just measured (it can change if some interactions occur. For example, if you letthe electron travel through a magnetic field, even if just afterthe measurement you had got spin up along z you may end up with some nonzero probability of the spin being down )

15. Jul 3, 2007

qtheoryquestion

Hmm... ok. I'm going to go back to where I think I got lost:

My question: