# Experimental perspective in understanding the Heisenberg Uncertainty Principle

I need to confirm whether or not I understand Heisenberg Uncertainty Principle. So the crucial thing is that you need an "ensemble" of measurements:

$$\delta x \delta p \ge \frac{\hbar}{2}.$$

If I were to conduct an experiment trying to validate this with particles, then I would first measure, say, the position in the $X$ direction of an "ensemble" of particles right? Because when you measure the position of a particle, you get a single number.

Thus, "had I tried to measure the momentum in the $X$ direction at the same time measuring position in the $X$ direction of a particle", I would also get a single number for the momentum corresponding to that particle.

It is until I combine all of my measurements of $x$ and $p$ together in and plot them in a graph that you would literally "see" the uncertainty relation. This is my understanding.

So you would calculate $\delta x$, which is the standard deviation of $x$ measurements and $\delta p$ as the standard deviation of $p$ measurements and multiply them together. You calculate the standard deviation by $\sqrt{\sum(x-x_{avg})^2/n}$, where $n$ is the number of particles and $x_{avg}$ is the average of all the position measurements.

I do the same thing with the $p$ values for momentum. Am I correct?

Furthermore, I want to solidify my understanding further... So can I think of the nice probability density curve that I typically see for the position of a particle is the ideal plot for an ensemble of a zillion measurements right? Its usually a Gaussian. The distribution depends on the potential that you plug into the wavefunction. Nevertheless, the calculated distribution is "as if" you've done a zillion measurements on an ensemble of identically prepared systems. Am I right here?

So, this is why when you have "precise $\delta x$" meaning a very small value for $\delta x$", the spread of the position measurements is very narrow. Following the principle, you MUST end up with a very fat or wide curve for the momentum function because its standard deviation must be large to preserve the principle.

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Note that this site supports MathJax for equation rendering. I've edited this post for you. Have a look at the FAQ or search the web for more info on how it works. –  Michael Brown Mar 25 '13 at 6:27
You may see that the distribution is fuzzy - the phase space cells are diluted to the area of at least $\hbar$ - only if you combine many measurements. On the other hand, you should appreciate that quantum mechanics yields (probabilistic) predictions for every experiment, and each measurement in the ensemble has the same probabilities, so the uncertainty is a fact about the probability distributions that hold for one single measurement, too. From this viewpoint, your addition of many experiments is just a way to "visualize" what the word "probability" means, and isn't essential. –  Luboš Motl Mar 25 '13 at 6:28
When you say, "so the probability distributions hold for a single measurement too." Do you mean, if we had calculated the probability distribution through solving the wavefunction with the appropriate potential, then a single measurement would be equivalent to a random pick from that calculated distribution? –  QEntanglement Mar 25 '13 at 6:33
Yes, your description is the most general description/definition of the term "probability" combined with the QM's prescription how these probabilities are calculated. That's what quantum mechanics is all about. The probabilities of different outcomes exist - are calculable - even before a single measurement, and it's these probabilities that are constrained by the uncertainty relationship. –  Luboš Motl Mar 25 '13 at 6:43

If I were to conduct an experiment trying to validate this with particles, then I would first measure, say, the position in the X direction of an "ensemble" of particles right? Because when you measure the position of a particle, you get a single number.

When you measure the position of a particle experimentally you get a single number with its instrumental measurement error, i.e. a number anyplace within this measurement error, which usually follows a gaussian probability curve.

Thus, "had I tried to measure the momentum in the X direction at the same time measuring position in the X direction of a particle", I would also get a single number for the momentum corresponding to that particle.

In the micro regime, where the HUP is significant, it tells you that if you try to measure the error for that single particle in the p_x momentum, the error will be constrained by the HUP relationship. Even though your instruments might be able to measure momentum to much smaller accuracies, the HUP relationship constrains the uncertainty in p_x, for each individual simultaneous measurement .

That is the reason why deltas are used in the HUP instead of sigmas ( the usual error symbol). One of the pair is quantum mechanically constrained to be uncertain as a function of the measurement error of the other.

Furthermore, I want to solidify my understanding further... So can I think of the nice probability density curve that I typically see for the position of a particle is the ideal plot for an ensemble of a zillion measurements right? Its usually a Gaussian. The distribution depends on the potential that you plug into the wavefunction. Nevertheless, the calculated distribution is "as if" you've done a zillion measurements on an ensemble of identically prepared systems. Am I right here?

You can get a good statistical probability gaussian for the x measurement, as long as that is your only measurement. If, for each particle, you measure x with a specific sigma error in your measurement and at the same time you measure p_x, you will not be able to constrain the statistical dispersion of the p_x measurement within the possible accuracy of your instruments. The dispersion will be controlled by HUP.

So, this is why when you have "precise δx" meaning a very small value for δx", the spread of the position measurements is very narrow. Following the principle, you MUST end up with a very fat or wide curve for the momentum function because its standard deviation must be large to preserve the principle.

"Standard deviation" has a precise statistical meaning. What you will be seeing is the dispersion due to the HUP. If you do the opposite, require great instrument accuracy for p_x, the x position will be uncertain within HUP and will show dispersion.

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