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In some textbooks about quantum mechanics, the position-momentum uncertainty principle is treated as being valid for an individual "particle", with $\Delta x\cdot\Delta p\geq\hbar/2$ referring to the theoretical limit on the uncertainty in the position and momentum of one particle at the same time.

Other sources state that the position-momentum uncertainty principle only applies to the standard deviations of $x$ and $p$ in a large number of identically prepared particles.

Which, if any, of these presentations is "correct" or at least acceptable at a pre-university level, and why?

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  • $\begingroup$ What exactly do these sources mean by "$\Delta x$ being exactly zero"? To me, "standard deviation" is the definition of what $\Delta x$ means, but if you define it to be e.g. the difference between the expectation value and what you actually measure, then it may well be zero. Therefore, we can't decide whether the second statement is true, false or meaningless without further context. $\endgroup$ – ACuriousMind Oct 9 '16 at 22:17
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    $\begingroup$ From this source, I understand $\Delta x=0$ in my second paragraph to mean that the position of a specific (individual) particle can be known exactly. Can we apply the uncertainty principle to an individual particle or only to a large number of identically prepared particles? $\endgroup$ – gamma1954 Oct 9 '16 at 22:32
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It is best to first understand the uncertainty relation as a mathematical property of the wavefunction of a particle. So what it is saying is that is mathematically impossible to construct a wavefunction which does not satisfy the uncertainty relation.

Then we can move on to think about what physical meaning this relationship may have. To avoid getting bogged down with interpretational questions when first learning QM it is best to stick to experimental (operational) procedures where we can test the uncertainty principle. (This is just deferring deeper questions - not saying that these deeper questions are of no interest or meaning).

Now to relate the theoretical predictions of QM (which have a probabilistic nature) to experimental results we inevitably will need to do multiple experiments with many particles all prepared in the same way so they have the same initial wavefunction. If we do a large set of measurements of position we can calculate a standard deviation of position; if we do another large set of measurements of momentum (on a different set of particles but prepared the same way) we can calculate a standard deviation of momentum. These standard deviations will necessarily satisfy the uncertainty relation whatever method we used for preparing the particles.

I'm not saying that is the end of the story - just a good way to start thinking about it.

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Both of the statements are correct, and you have a good and fundamental question of what it is to be a wave function. Suppose that a particle is in a certain state $\psi$, and say that the wave function tells you the $exact$ position of the particle (in the lingo we call this a state of definite position or position eigenstate; if that doesn't make sense, don't worry). Then, you cannot know anything about what momentum it has (this is the statement of the uncertainty principle). Now, this is not just true for $this$ particle. Rather, this relationship is true for $any$ particle in this state (i.e. for any particle in a state of definite position). This is what they mean by an "identically prepared system": two systems such that they are in the exact same state $\psi$. Now we can think of taking, not just two, but $N$ systems, all of which are in $\psi$, and what we will find is that after we measure, say, the momentum of each system and calculate the standard deviation $\Delta p$ of all of those measurements that it must be greater than or equal to $\frac{\hbar}{2\Delta x}$.

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    $\begingroup$ "Then, you cannot know anything about what momentum it has" - this might be interpreted as the particle (in a position eigenstate) has a definite momentum but one cannot know anything about that momentum. $\endgroup$ – Alfred Centauri Oct 9 '16 at 23:52
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    $\begingroup$ Which would not be right as it would imply a hidden deterministic variable. $\endgroup$ – Bob Bee Oct 10 '16 at 6:05

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