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The wave functions of matter waves give the probability density of the particle being at a certain location. Does this arise because as an outside observer, we have incomplete information about the particle, or is there really some uncertainty in the motion of the particle? What would an uncertainty in the particle's motion or position mean?

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  • $\begingroup$ There are no particles here, at all, at least not in a naive understanding of the word "particle" as "something small". What we are measuring are all properties of quantum fields, which, like normal fields, can have plane wave solutions or somewhat localized wave packets. These fields can only interact by exchanging quanta, which are the smallest units of angular momentum, charge etc.. Quanta, all by themselves, do not have particle properties and they are only giving us statistical information about the state of the fields. That uncertainty is inherent in nature and nothing can remove it. $\endgroup$ – CuriousOne Feb 19 '16 at 0:41
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This is a commonly misunderstood aspect of quantum mechanics, particularly when introductory teachings describe matter as behaving like a "matter and a wave at the same time". "Particles" are, in fact, simply regions of high density of a wavefunction. The particles do not behave as "particles" until you observe them, resulting in a collapse of the wavefunction to a single point. (Also, note that it is actually the "square" of a wavefunction - that is, the wavefunction times its complex conjugate, $\psi^*\cdot \psi$ that gives the probability density of the "particle".) The "particle" representation of it is simply the observed value of position.

But, this is not just the case for position! A wavefunction encodes all quantum information about a particle in it: the wavefunction is the particle ("particle" here meaning more generally, a "thing", such as an electron, group of photons, etc.). The expected outcome of measuring any observable $\Omega$ (any value you can observe, such as momentum, position, energy, spin, etc.) can be calculated with the rather elegant (and general) "bra-ket notataion":

$$\left<\Omega\right> = \left<\psi\right|\Omega\left|\psi\right>.$$

If you've never seen this notation before, it can seem a little off-putting. This is mainly mathematical formulism and isn't important to understand quantum mechanics on a conceptual level, but bra-ket notation generalizes the concept of a wavefunction to a "ket" vector: $\left|\psi\right>$. Multiplying "bras" by "kets" is equivalent to taking the inner product of the bra and ket:

$$\left<\psi|\psi\right>=\int_\mathbb{R} \psi^* \psi \cdot d \Omega=1$$

All this means is that the wavefunction normalizes to 1. Note that $d\Omega$ is used because a wavefunction doesn't have to be expressed in terms of position; it can expressed in terms of a momentum basis $\psi(p)$, an energy basis $\psi(E)$, or the basis of an arbitrary observable $\Omega$.

The representation of this "wavefunction" over any given observable is $\psi(\Omega)=\Omega\left|\psi\right>$. This is read as the "operator $\Omega$ acting upon the state $\left|\psi\right>$". So the "wavefunction" with respect to position that you're probably familiar with is $\psi(x)=x\left|\psi\right>$. So the expression for the expected value of position (where the "particle" is "centered") is:

$$\left<x\right> = \left<\psi\right|x\left|\psi\right> = \int_{-\infty}^\infty \psi^*(x) \cdot x \cdot \psi(x) \cdot d x.$$

There is no "incomplete information" about the particle: if we have the full quantum state $\left|\psi\right>$, we have all the information we could ever (mathematically) obtain about the particle. There is actually a very famous result about this called Bell's theorem, which (paraphrased, simplified, and applied to your specific question) says that quantum mechanics cannot be created with a "local hidden variable" theory. That is, there cannot simply exist a deterministic, but arbitrarily complex, mechanism for selecting the point in space the wavefunction collapses to - it is "truly" random.

Similarly fundamental is the uncertainty between two non-commuting observables (more on this in a second), such as position ($x$) and momentum ($p$). You've probably seen the famous Heisenberg uncertainty principle, summarized as:

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

This is a fundamental mathematical result: it is physically and mathematically impossible for you to know both $x$ and $p$ to arbitrarily high precision. The reason for this is that $x$ and $p$ do not commute. That is, $X(P\left|\psi\right>)\ne P(X\left|\psi\right>)$. You're probably a little confused by this statement, so here is a less mathematical and more intuitive explanation to this.

Consider a "wavefunction" in the position basis. That is, the normal $\psi(x)$ you're likely familiar with. Then the position of the "particle" is related to the "position" of the wavefunction (that is, $x$ is related to the magnitude of the wavefunction at each point in space), while the momentum of the "particle" is related to the "frequency" of the wavefunction. If you have a wavefunction with a perfectly known momentum, such as $\psi(x)=e^{ix}$, which has an exact "frequency" of 1 (since $e^{ix}=\cos x + i \sin x$, which both have frequencies of 1), then you are guaranteed that when you observe the quantum system, your momentum will always be measured to be the same value of $p_0=\hbar$. However, $\psi(x)=e^{ix}$ says exactly nothing about the position of the particle:

$$\psi^*(x) \cdot \psi(x) = (\cos x - i \sin x) \cdot (\cos x + i \sin x) = 1 \text{ at every point in space}.$$

Note that this wavefunction isn't "normalized", but the important conclusion here is that you know the momentum perfectly, while the "particle" is equally likely to be anywhere in space,s o you know nothing about the position.

In case you're interested in a more in-depth explanation of this, the more technical explanation for this is that $x$ and $p$ are non-commuting, and, more specifically, are "canonically conjugate variables". That is, they are "Fourier duals" of each other, meaning that they are related to each other through a Fourier transform, which, in this case, takes a function in position basis and represents it in terms of its frequencies. To reference the example I gave above, the Fourier transform of $\psi(x)=e^{ix}$ would be a spike at $f=1$ ($f=$frequency). So "all of $\psi(p)$ is contained at $p=\hbar$", and nowhere else.

Note that this is just the momentum-position uncertainty principle. In general, any two noncommuting ("incompatible") observables will have an uncertainty relation between them: you cannot know both to arbitrary precision. This is not a lack of knowledge, this is a fundamental feature of quantum mechanics.

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