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So measure an electron, take down it's position $p$. Then measure the electron a second time and take down it's new position $p'$. Note the time between measurements, $t$. What does physics say about the average speed $$v=\frac{d(p,p')}{t}$$ between the two measurements? Is it possible that $v$ must have been larger than the speed of light.

Bohr radius (radius of the hydrogen atom) is about $5\cdot 10^{-11}$ m, so take twice that, $10^{-10}$ m as the distance between opposite "sides" of an atom. Letting $v=c$, we solve for $t$, $$t=\frac{10^{-10}}{3\cdot 10^8}= 3.33\cdot 10^{-19} \text{ sec}.$$ That's more than Planck time!

So my question: Is the probability of the wave function somehow limited by the speed of light in quantum mechanics?

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  • $\begingroup$ Why is it relevant that the time you found is greater than the Planck time? $\endgroup$ – user4552 Jul 26 '14 at 21:30
  • $\begingroup$ I also don't understand the relevance of the Bohr-model calculation. $\endgroup$ – user4552 Jul 26 '14 at 21:44
  • $\begingroup$ related: physics.stackexchange.com/q/107261 $\endgroup$ – user4552 Jul 26 '14 at 22:14
  • $\begingroup$ How can you be certain it's the same electron? It could have moved in from an adjacent atom and the electron you first observed is on another atom. $\endgroup$ – LDC3 Jul 26 '14 at 23:07
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You seem to be assuming that there is a measurement process that can be applied to the electron on a single, isolated hydrogen atom twice within somewhere under 10–19 seconds, and that because an electron in a hydrogen atom's ground state (or any $s$ state) is equally likely to be found along any direction from the nucleus, there's some chance that the electron will "jump" from one side of the nucleus to other.

But you are forgetting about the uncertainty principle, which governs the precision with which you may make a measurement. If you want to measure which side of the atom the electron sits on, you must have a precision better than the Bohr radius, $\Delta x < a_0 \approx \frac12\times10^{-10}\,\mathrm m$. After such a measurement, the electron must have a momentum uncertainty (in energy units) of $$ \Delta p\, c \gtrsim \frac{\hbar c}{2\Delta x} = \mathrm{\frac{0.2\,GeV\cdot fm}{2\cdot \frac12\times10^{5}\,fm}} = \mathrm{2\times10^{-6}\,GeV} = 2\,\mathrm{keV} $$ We can assume that the most likely value for $p$ is zero, since the electron began at rest. But a momentum of $pc = 0\pm 2\,\mathrm{keV}$, normally distributed, implies a typical squared momentum $(pc)^2 \approx (2\,\mathrm{keV})^2$, or a typical energy \begin{align} E = \frac{p^2}{2m} \frac{c^2}{c^2}\approx \mathrm{\frac{(2\,keV)^2}{2\cdot 500\,keV}} \approx 4\,\mathrm{eV} \end{align} Remember that the hydrogen ground state has energy $-13.6\,\mathrm{eV}$, and that the first excited state has energy $-13.6\,\mathrm{eV}/2^2 = -3.4\,\mathrm{eV}$, neary 10 eV away. A position measurement with enough precision to distinguish "this side" from "that side" of a hydrogen atom involves nearly enough energy to promote the electron into a different state.

However it's also not really enough precision to say that you've measured faster-than-light travel for the electron: since the electron's $s$-wave wavefunction is roughly a three-dimensional Gaussian wavepacket whose width is the Bohr radius, you're mostly likely to find successive measurements of electron position separated by $1\sigma$. If you wanted to typically find electron positions separated by $3\sigma$ ("$3\sigma$ is a measurement, $4\sigma$ is a discovery," as one of my mentors likes to say, but $3^2=10$ is a nice round number), you'd have to improve your precision by a factor of three; this takes the energy associated with the measurement to 40 eV and means that your first position measurement will definitely ionize the atom, and your assumption that both measurements take place in the atom's ground state is broken.

Or alternatively, you could prepare your atom in an excited state (with radius $na_0$ for principal quantum number $n$) and keep the precision of your measurement the same. However, in that case also your first measurement would have enough energy to change the atom's state, or ionize it completely.

Then you have the question of how you'll measure the electron's position twice within 10–19 seconds—that's fast! Suppose you bathe the atom in an oscillating electric field with frequency $\nu=10^{19}\,\mathrm{Hz}$. In the quantum-mechanical picture, this field is made of photons, each with energy $E = h\nu \approx 40\,\mathrm{keV}$! It is vanishingly improbable that the atom would remain in its ground state after interacting with such a field. Your experiment would never work.

This second argument is in the spirit of Bohr's arguments during his debates with Einstein about the proper interpretation of quantum mechanics. But to my mind the first argument, based on the observation that bound-state wavefunctions tend to have nearly the minimum uncertainties $\Delta x,\Delta p$ allowed by the Heisenberg principle, is more interesting. In some sense, the different states of the hydrogen atom have the size that they do because of the uncertainty principle. Notice that relativity doesn't enter that argument explicitly, apart from the definition of energy $(E_\text{rest}+E_\text{kinetic})^2 = pc^2 + (mc^2)^2$, which I used in the more familiar low-momentum approximation $E_\text{kinetic} \approx p^2/2m$. The universe conspires in many ways to preserve its fundamental laws.

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As noted in comments, some parts of the question don't make sense to me.

But anyway, I think the answer to your question is that yes, the quantity $v$ you define can be greater than $c$. However, this doesn't necessarily have any consequences for relativity. The only thing that would be problematic for relativity would be if you could transmit information at $v>c$, which is a different thing.

The evidence that $v>c$ is possible in quantum mechanics actually comes from quantum-mechanical tunneling. The experiments are mostly done with photons, not electrons, and the search term you want is "Hartman effect." For example, figure 1 in [Nimtz 2007] describes an experiment similar to, although not identical to, the one you propose. I've listed some other recent papers below.

Another thing that it might be helpful to point out is that in the measurement scheme proposed in the question, the first measurement changes the state of the electron -- you can't just sample the same wavefunction twice. Also keep in mind that the stuff I've referenced, about photons, may not be perfectly analogous with what you have in mind, about electrons. In particular, there is no such thing as a position eigenstate for a photon.

Nimtz and Stahlhofen, Macroscopic violation of special relativity, 2007, http://arxiv.org/abs/0708.0681

Chiao, Tunneling Times and Superluminality: a Tutorial, 1998, http://arxiv.org/abs/quant-ph/9811019

Winful, Do single photons tunnel faster than light?, 2007, http://arxiv.org/abs/0708.3889

Aichmann and Nimtz, The Superluminal Tunneling Story, 2013, http://arxiv.org/abs/1304.3155

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    $\begingroup$ You are addressing the first part of the question and I am addressing the second , given as an example by the OP. Should I make a snide remark about your answer and give you a -1? $\endgroup$ – anna v Jul 27 '14 at 3:44
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The Bohr atom was ( and still is) a useful phenomenological approximation of the underlying quantum mechanical system. It is no use playing with numbers of the Bohr atom as if the electron is a billiard ball. It is not a classical particle. It is a quantum mechanical entity that appears as particle in some interactions, or builds up a probability wave in others, depending on the boundary conditions the observation sets. In the hydrogen atom it builds up probability loci called "orbitals", not orbits.

Actually the experiment of measuring the velocity of a very fast particle , the neutrino, was done . The neutrinos have masses much smaller than the electron. For a while a faster than light velocity hit the physics news, but it was soon found out it was an experimental error, a baddly plugged cable in the complicated measurement and analysis. Electrons travel very close to the speed of light in accelerators, but that is all.

So the answer is, the experimental evidence is that a free particle never goes over the speed of light and a bound electron is in the probability wave manifestation and speed has no meaning in that framework.

Another way at looking in your numbers is to recall the Heisenberg Uncertainty principle, which is inherent in the quantum mechanical phase.

HUP

You chose your electron position accurately, so that means that momentum ( velocity*mass) can be anything, cannot be estimated.

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  • $\begingroup$ The first paragraph doesn't address the question. The second and third paragraphs are an example, but an example doesn't prove a general rule. The final paragraphs don't address the question, since the question asks about the outcome of a measurement process, which is manifestly well defined. $\endgroup$ – user4552 Jul 26 '14 at 22:16

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