# B Finite Universe vs. Uncertainty Principle

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1. Apr 5, 2017

### Helios

If I have this right, when we have exact certainty of a particle's momentum, the bounds of this particle's location cannot be determined. Now there are some who believe in a universe of finite volume and so this particle has to be within this volume. So there seems to be a contradiction. Does the Uncertainty Principle contradict the finite universe?

2. Apr 5, 2017

### Comeback City

The Uncertainty Principle states that the more accurately we measure a particle's momentum, the less accurately we will be able to measure its position. Your claim is like saying that 4 to the power of 3.4581 is somewhere between 1 and 100,000,000. The universe is rather huge (as far as we know).

3. Apr 5, 2017

### rootone

Quantum cosmology. is difficult.
So different that it barely makes sense at all.
Then again bus time tables can be like that.
Clue: it's about probability, but nobody knows why,

Last edited: Apr 5, 2017
4. Apr 5, 2017

### Staff: Mentor

This is true, but it's irrelevant, because we can never have exact certainty of a particle's momentum (nor can we have exact certainty of its position). States of a particle that have an exact momentum, or an exact position, are not physically realizable.

No. See above.

5. Apr 6, 2017

### PeroK

The Heisenberg UP doesn't say that at all.

6. Apr 6, 2017

### Comeback City

It's the most basic way of explaining it I could think of. If it's plain out wrong, feel free to explain as I would like to know my misunderstanding...

7. Apr 6, 2017

### PeroK

Formally, the HUP is about the state of a particle. See:

The critical point is that it is not about the accuracy of any measurements, but the statistical range of measurements you would expect. For example, take a large ensemble of particles all in the same state and for half measure momentum and the other half measure position.

If the momentum measurements are (expected to be) in a small range, then the position measurements are (expected to be) in a large range and vice versa. Note that when it comes to measurements, the HUP is a statistical law. So, it doesn't say anything about any particular measuremets but about the variance of a large number of measurements on identically prepared particles.

8. Apr 6, 2017

### Comeback City

According to your explanation, saying that it "doesn't say that at all" about my claim seems as a bit of an overstatement, but I do see what you're saying with the range of values.

9. Apr 6, 2017

### PeroK

The HUP says absolutely nothing about the "accuracy" of measurements.

10. Apr 6, 2017

### Comeback City

In that case I'm using certain words/phrases interchangeably where I should't be.

11. Apr 6, 2017

### PeroK

"Accuracy" refers to how good your measurement is. For example, you could measure the position of a particle to the nearest $nm$. That would be a certain accuracy. If the measurements you got for an ensemble of identically prepared particles was:

$100nm, 105nm, 90nm, 98nm, 107nm$

Then, that set of data would have an expected value of $100nm$ and a standard deviation of about $6nm$.

The HUP relates to the standard deviation, which is a measure of the spread of data from several measurements. It does not relate to the accuracy with which the measurements are made.

12. Apr 7, 2017

### Chalnoth

The HUP places a fundamental lower bound on the measurement error, dependent upon other specific measurements.

I don't think that the distinction between the Heisenberg Uncertainty Principle relating to the state of the wavefunction rather than the accuracy of measurements is useful in this context. The end result is largely the same: if you measure a particle's position to an accuracy described by a certain variance, then the variance of subsequent momentum measurements will be at least as large as that determined by the Heisenberg Uncertainty Principle.

13. Apr 8, 2017

### PeroK

This is not correct at all. The variance in the HUP is not related to accuracy. For example, if you roll a die. The expected value is 3.5 and the variance is whatever. But, if you roll a 3, you roll a 3 exactly. There is no error or uncertainty after the measurement. All the uncertainty is prior to the measurement.

Moreover, in a general statistical context, variance is very different from accuracy of measurement. The HUP explicitly relates to variance (not to specific measurement errors).

PS Any measuring equipment has an expected error, which itself will have a variance, but that is a different variance from the variance in the particle's position, which is what the HUP deals with. In particular, the HUP puts no lower bound on the accuracy of any particular measuring equipment.

14. Apr 8, 2017

### Chalnoth

As I said, I don't think this distinction has any relevance at all. Regardless of whether we're talking about the fact that the wavefunction of the particle is spread out over position (or momentum, or some other variable), the fact remains that subsequent measurements will have a certain variance, and no matter how precise your experiment is, that variance will not drop below what the HUP demands.

15. Apr 8, 2017

### PeroK

The variance in the HUP has to do with the expected position of the particle. Explicitly, the HUP says:

$\sigma_x \sigma_p \ge \frac{\hbar}{2}$

There is nothing in that that relates the accuracy of any measuring equipment. The variance in the HUP is entirely to do with the variance in the particle's position. Errors in measurement may add to this, but those errors are not implied or demanded by the HUP. Whereas, the variance in the position of the particle is demanded by the HUP.

16. Apr 8, 2017

### hilbert2

What about a system where there's a particle of definite momentum on a "3-sphere", i.e. a space where you get back to the starting point by going far enough to some direction? Would't the momentum observable be quantized then? I'm just trying to point out that the HUP doesn't necessarily work in a non-euclidean geometry...