The Heisenberg Uncertainty Principle and velocity

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Hello all

I have a question concerning The Heisenberg Uncertainty Principle. The principle mathematically looks like this-

$$\Delta x\Delta p \geq \hbar/2$$

The principle states that you can not measure more than two quantities simultaneously. If you know a particle's position very precisely, then you won't know its momentum very precisely and vise versa. I want to know how it is used, so let's set up an example.

I have a caliper that can measure to an accuracy of ±.05mm. I confine an electron which has a mass of 9.11*10^-31 kilograms in a space of 10mm. Predict the error of velocity. So let's start with.

$$\Delta x\Delta p \geq \hbar/2$$
Lets solve for
$$\Delta v$$

$$\Delta x\Delta p \geq \hbar/2$$

$$\Delta x\Delta m\Delta v \geq \hbar/2$$

$$\Delta v \geq \hbar/(2 \Delta x\Delta m)$$

Ok now I just plug in values. If I do something wrong or stupid, don't be afraid to say something. I want to see how this works.

$$\Delta v \geq \hbar/(2*.05*9.11*10^-31)$$
For an answer I get 1.15759767 m / s. Now I have a question about this answer. Does it mean that the electron will have a velocity of v±1.15759767? I don't know what the final answer is suppose to represent.

Thanks for your help!

 

[diazona]

The uncertainty principle is, strictly speaking, a statement about distributions - the inherent statistical uncertainty in measurements of two quantities. The fact that your calipers can measure to an accuracy of 0.5mm isn't really that important. (Yes, the principle is often stated in a misleading way)

What I mean about statistical uncertainty is this: let's say that by some unspecified method, you are able to get a very very large number of electrons all in the same quantum state. (For example, the ground state of a free hydrogen atom.) Suppose you go through these electrons one by one, and for each one, randomly choose to measure either the position or the momentum. Not all your position measurements will give the same result; instead, you'll get a distribution of positions, which you can compute the mean and standard deviation of. Same for the momentum measurements: they won't all be the same, but you'll get a distribution, which you can compute the mean and standard deviation of.

Now, what the uncertainty principle says is that if you multiply the standard deviation of the positions by the standard deviation of the momenta, the product will be no less than $$\hbar/2$$. And the reason for this is not because your measurements were not good enough; it's because of the intrinsic properties of the quantum state. No matter how carefully you perform this experiment, you will never wind up with a product of standard deviations that is less than $$\hbar/2$$.

For what it's worth, you could sort of loosely apply the uncertainty principle to the situation you described, with the calipers - if you confine the position of the electron to within 0.5mm, it does have a minimum allowed momentum, so if you measure its speed, you should get at least some particular value which might be 1.16 m/s. Or if you confine the electron to within 10mm (your box) and measure its speed, you should get at least some value which might be 5.8 mm/s. But I'd be a little wary of making statements like that without doing the math to back them up (it's not hard but I can't do it off the top of my head at 1:00 in the morning ;-) The real, precise meaning of the position-momentum uncertainty principle (at least, my understanding) is just what I described in the previous two paragraphs about the distribution of measurements.