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Uhm...forgive me for asking, but a continuous quantum jump is...what exactly? How is this in any way consistent with the rest of quantum mechanics? Edit: I do

Uhm...forgive me for asking, but a continuous quantum jump is...what exactly? How is this in any way consistent with the rest of quantum mechanics? Edit: I don't mean to sound snide, I am genuinely confused about what this experiment means.

Anon84 7y ago
This is a common misconception. Quantum jump are jumps between two stable levels... there's nothing preventing "jumps" to unstable states, the particle just doesn't stay long in that specific state (although it can stay there long enough to be useful as in the case of two-photon absorption [0] and microscopy [1]). If I understand it correctly, what this seems to imply is that it's not so much a (discontinuous) jump but rather a continuous(-ish) transition from one state to another. [0] https://en.wikipedia.org/wiki/Two-photon_absorption https://en.wikipedia.org/wiki/Two-photon_absorption [1] https://en.wikipedia.org/wiki/Two-photon_excitation_microscopy https://en.wikipedia.org/wiki/Two-photon_excitation_microsco...
- joycian 7y ago
  But in a quantum harmonic oscillator, the energy levels can only be certain values. As in the Hamiltonian is: H = omega*planckbar(N + 0.5) where N is the state counter (number operator). As a direct reply to your links: So did the fact that two photon excitation works then not make this entire paper not that interesting or novel?
  - pas 7y ago
    The interesting bit is the experiment here as far as I know, but that is completely consistent with QM (basically validates it). https://www.reddit.com/r/slatestarcodex/comments/bwlhh9/anyone_who_understands_quantum_physics_how/ https://www.reddit.com/r/slatestarcodex/comments/bwlhh9/anyo...
trevyn 7y ago
It’s a process of transition between two distinct states. Say a bit flips from 0 to 1; there is a short time period during which that bit is physically “in transition”, but it would be incorrect to say that the bit has a value of, say, 0.5 during the transition. Also remember that quantum mechanics is a way to make statistical predictions of outcomes of certain experiments; it does not claim to explain what is actually happening underneath.
- joycian 7y ago
  With your second paragraph firmly in the back of my mind: In the sister comment thread, two photon excitation seems to indicate that this intermediate state (0.5 in your coin example) is not invalid. Can you help me understand better? Edit: Also, I think your coin "in transition" implicitly assumes that the system has more degrees of freedom than 1 bit (You can only flip a coin in 3 dimensions).
- gaze 7y ago
  "it would be incorrect to say that the bit has a value of, say, 0.5 during the transition." In what sense? It has a <z>=0, meaning there's a statistically 50%/50% chance of measuring 1 or 0 but the system is in principle coherent. This means that half way through the jump, if you rotated it 90 degrees, you'd measure a _definite_ z value of -1 or 1, or you'd measure a bit value of 0 or 1, depending on whether you twisted it clockwise or counterclockwise.
  - wnoise 7y ago
    In the sense that it ignores that coherence. Any state on the disc of the Bloch sphere's equator has <z> = 0, and 50:50 chance of measuring 1 or 0. If you want to assign a real value of 0.5 to a bit, then the only one that makes sense is the center, completely decoherent mixture.
- DiogenesKynikos 7y ago
  I strongly disagree with your second paragraph. Quantum theories do, we think, correspond to what is actually happening underneath. They might not be perfect correspondences, in that our theories are incomplete or approximate, in the same way that Newtonian gravity is an approximation of General Relativity in the weak-gravity regime. Bell's inequality places a strong constraint on what sort of physical theories can explain quantum phenomena. If you believe in locality (the universe has no global variables, and information propagates outwards through local interactions), then the wavefunction is a real thing, and the actual state of the universe.
gaze 7y ago
A continuous jump means that the system travels coherently through the superposition of |0>+|1> through the jump, and other such a|0>+b|1> states for a^2+b^2=1. It isn't in 0, and then suddenly 1.

Anon84 7y ago

This is a common misconception. Quantum jump are jumps between two stable levels... there's nothing preventing "jumps" to unstable states, the particle just doesn't stay long in that specific state (although it can stay there long enough to be useful as in the case of two-photon absorption [0] and microscopy [1]). If I understand it correctly, what this seems to imply is that it's not so much a (discontinuous) jump but rather a continuous(-ish) transition from one state to another. [0] https://en.wikipedia.org/wiki/Two-photon_absorption https://en.wikipedia.org/wiki/Two-photon_absorption [1] https://en.wikipedia.org/wiki/Two-photon_excitation_microscopy https://en.wikipedia.org/wiki/Two-photon_excitation_microsco...

joycian 7y ago
But in a quantum harmonic oscillator, the energy levels can only be certain values. As in the Hamiltonian is: H = omega*planckbar(N + 0.5) where N is the state counter (number operator). As a direct reply to your links: So did the fact that two photon excitation works then not make this entire paper not that interesting or novel?
- pas 7y ago
  The interesting bit is the experiment here as far as I know, but that is completely consistent with QM (basically validates it). https://www.reddit.com/r/slatestarcodex/comments/bwlhh9/anyone_who_understands_quantum_physics_how/ https://www.reddit.com/r/slatestarcodex/comments/bwlhh9/anyo...

trevyn 7y ago

It’s a process of transition between two distinct states. Say a bit flips from 0 to 1; there is a short time period during which that bit is physically “in transition”, but it would be incorrect to say that the bit has a value of, say, 0.5 during the transition. Also remember that quantum mechanics is a way to make statistical predictions of outcomes of certain experiments; it does not claim to explain what is actually happening underneath.

joycian 7y ago
With your second paragraph firmly in the back of my mind: In the sister comment thread, two photon excitation seems to indicate that this intermediate state (0.5 in your coin example) is not invalid. Can you help me understand better? Edit: Also, I think your coin "in transition" implicitly assumes that the system has more degrees of freedom than 1 bit (You can only flip a coin in 3 dimensions).
gaze 7y ago
"it would be incorrect to say that the bit has a value of, say, 0.5 during the transition." In what sense? It has a <z>=0, meaning there's a statistically 50%/50% chance of measuring 1 or 0 but the system is in principle coherent. This means that half way through the jump, if you rotated it 90 degrees, you'd measure a _definite_ z value of -1 or 1, or you'd measure a bit value of 0 or 1, depending on whether you twisted it clockwise or counterclockwise.
- wnoise 7y ago
  In the sense that it ignores that coherence. Any state on the disc of the Bloch sphere's equator has <z> = 0, and 50:50 chance of measuring 1 or 0. If you want to assign a real value of 0.5 to a bit, then the only one that makes sense is the center, completely decoherent mixture.
DiogenesKynikos 7y ago
I strongly disagree with your second paragraph. Quantum theories do, we think, correspond to what is actually happening underneath. They might not be perfect correspondences, in that our theories are incomplete or approximate, in the same way that Newtonian gravity is an approximation of General Relativity in the weak-gravity regime. Bell's inequality places a strong constraint on what sort of physical theories can explain quantum phenomena. If you believe in locality (the universe has no global variables, and information propagates outwards through local interactions), then the wavefunction is a real thing, and the actual state of the universe.

gaze 7y ago

A continuous jump means that the system travels coherently through the superposition of |0>+|1> through the jump, and other such a|0>+b|1> states for a^2+b^2=1. It isn't in 0, and then suddenly 1.