8 ms·
Quantum fluctuations have been shown to affect macroscopic objects
- perl4ever 6y ago"A special case of squeezed light, known as the squeezed vacuum, forms when the average amplitude of the light is zero." I love this. Don't ask me what it means.
- knodi123 6y agoSuch conceptually dense language! Isn't the average amplitude of every wave just zero?
- danbruc 6y agoThe amplitude is the difference between maximum and minimum, not the momentary value of a wave at some point.
- frabert 6y agoThen how can the average amplitude be zero? If at any point the amplitude is nonzero, it must be positive (as it is the absolute value of the difference), thus the average can only be positive
- ben_w 6y agoIIRC the minimum is supposed to be 0.5 photons of any given wavelength. This is what gets called “zero point energy”. Except there are ways to go below that, like a Casimir cavity. I find these things confusing because when people explain Casimir cavities they never bother to say why the justification for zero point energy existing in the first place doesn’t apply within the cavity.
- fsh 6y agoIndeed this is not possible for a classical wave. However, in a quantum description there exists a so-called "vacuum state" for which the expectation value of the electric field operator vanishes at all times, but the expectation value of the square of this operator does not. In the classical approximation of quantum optics, the expectation value of the electric field operator corresponds to the amplitude of the classical wave, whereas the expectation value of the square of the electric field operator corresponds to so-called shot (or quantum) noise on the classical wave. Of course in nature there is no such thing as a "classical wave", so this description has to break down at some point. This is the case for vacuum fluctuations which simply do not have a classical explanation.
- bollu 6y agoDoes this have to do with the fact that the wave is complex-valued?
- raxxorrax 6y agoWould be disappointed at the "wave" at that point.
- stallmanite 6y agoHmm, at least with an audio waveform one can have a DC offset wherein the average is not zero.
- alanbernstein 6y agoThe DC offset does not manifest when the waveform is translated into sound, which is an actual propagating wave, the appropriate analogy to light.
- phonethrowaway 6y agoparticle/wave duality. it means neither fully encapsulates "light".
- sharpneli 6y agoAlmost but not quite. With light the electric field can go negative and the average is generally 0. With sound in air the pressure component has a ”DC” offset of 1bar and it cannot go negative, you will just get nonlinear effects when you start nearing that limit.
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- macawfish 6y agoThe DC offset can be considered as approximating a low frequency component that is so low frequency, relative to the measurement window, that it appears to be constant. I think in the case described here, the zero average amplitude thing has to do with something a little bit different.
- trianglem 6y agoSo just rapidly oscillating light?
- amelius 6y agoIt looks like physicists are building the wrong abstractions. Abstractions are here to help us, not fool us.
- simonh 6y agoThe abstractions are there to help physicists so they don't get fooled. We're just along for the ride, all I try to do is not fall off too often.
- Chris2048 6y agoI found an article with this description: Exotic as the name seems, squeezed states are just Heisenberg’s uncertainty principle in action. In quantum optics, a light field is described by two “quadratures,” one for phase (P) and one for amplitude (X). Per Heisenberg, the minimum uncertainty of those two quadratures in a given measurement is given by the relation ΔXΔP = ħ/2. In other words, for a given measurement, the better you know the phase, the more potential error there is in the amplitude, and vice versa. For a coherent state, such as a laser field, the phase and amplitude uncertainties are equal, giving rise to a circularly symmetric “fuzzball” of potential error between the two quadratures. But using nonlinear-optics techniques, the circle can be “squeezed” into an ellipse. That means that uncertainty in one quadrature—the one that’s relevant to your sensor—can be reduced, while the uncertainty in the other, less relevant quadrature are increased. -- https://www.osa-opn.org/home/articles/volume_30/september_2019/departments/better_sensors_via_squeezed_light/ https://www.osa-opn.org/home/articles/volume_30/september_20...
- artpop 6y agoOn quantum interferometer, I’ve always wondered if you could use it for navigation in space. Say you send a probe to another system. Navigating the uncharted system with precise gravity measurements that also expose its kinematics would be handy.
- foxes 6y agoIn gravitational wave detectors. Heard of the Casimir effect? Two plates can experience a force from vacuum fluctuations.
- kkylin 6y agoSmall but important correction: two conducting plates. https://en.wikipedia.org/wiki/Casimir_effect https://en.wikipedia.org/wiki/Casimir_effect Also works in other geometries, e.g., sphere and plate, two spheres, etc. First experimental verification was a sphere and a plate, I think? https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.78.5 https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.78...
- martinclayton 6y agoOut of context perhaps, but rings true: "...SQL is a direct consequence of the Heisenberg uncertainty principle..."
- The_rationalist 6y agoI don't understand yet I know the uncertainty principle, would you mind to explain?
- tadhgf 6y agoIt's a joke about about SQL databases misbehaving, not physics
- deleted 6y ago[deleted]
- plafl 6y agoPhysicist help needed: From my near null knowledge of quantum mechanics it seems they decrease randomness in phase and amplitude measurement by increasing uncertainty in mirrors position. So, this is interesting because the mirrors are macroscopic objects that interacted with the light in a rather simple way and conditions. On the other hand... Don't they require the mirror position in a well determined position to measure the gravitational waves?
- c1ccccc1 6y agoThey are increasing the precision of the phase while decreasing the precision of the amplitude. Since the amplitude of the light determines how hard it pushes on the mirrors, this has the side-effect of also making the positions of the mirrors more uncertain. To get the most accurate measurement possible, I'd guess that they would squeeze the light until getting any additional precision in the phase was not worth the added noise in mirror position.