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- Thread starter Damon Turney
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Nugatory

Mentor

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You said "mixed state", but what you're describing sounds like a superposition not a mixture. Could you clarify which you're asking about?If that nucleus is in a "mixed state" at time zero, such that it's wavefunction is 50% up and 50% down at time zero....

- #3

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My question was motivated by this NMR example. Basically, I'm confused as to how the nuclei can only absorb energy in quanta (energy of up state - energy of down state) yet the nuclei are usually NOT in the down state.

- #4

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Let me first assume that by "mixed" state you mean superposition. Let me denote the down and up state by ##|0\rangle## and ##|1\rangle##, respectively, and the quantum of EM energy by ##|1_{\rm EM}\rangle##. Before the interaction the full state is

$$ \frac{1}{\sqrt{2}}|0\rangle|1_{\rm EM}\rangle + \frac{1}{\sqrt{2}} |1\rangle|1_{\rm EM}\rangle $$

while after the interaction it becomes

$$ \frac{1}{\sqrt{2}}|1\rangle|0_{\rm EM}\rangle + \frac{1}{\sqrt{2}} |1\rangle|1_{\rm EM}\rangle $$

where ##|0_{\rm EM}\rangle## is the EM vacuum. You can easily check that

The case when "mixed" really means

- #5

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Your answer is very illuminating (no pun intended)! Thank you!

- #6

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Let me first assume that by "mixed" state you mean superposition. Let me denote the down and up state by ##|0\rangle## and ##|1\rangle##, respectively, and the quantum of EM energy by ##|1_{\rm EM}\rangle##. Before the interaction the full state is

$$ \frac{1}{\sqrt{2}}|0\rangle|1_{\rm EM}\rangle + \frac{1}{\sqrt{2}} |1\rangle|1_{\rm EM}\rangle $$

while after the interaction it becomes

$$ \frac{1}{\sqrt{2}}|1\rangle|0_{\rm EM}\rangle + \frac{1}{\sqrt{2}} |1\rangle|1_{\rm EM}\rangle $$

where ##|0_{\rm EM}\rangle## is the EM vacuum. You can easily check thataverageenergy before the interaction is the same asaverageenergy after the interaction. Note, however, that the state after the interaction is in a superposition. Even though the nucleus is entirely in the ##|1\rangle## state, the EM state is neither entirely ##|0_{\rm EM}\rangle## nor entirely ##|1_{\rm EM}\rangle##. In your analysis you tacitly assumed that the EM quantum is entirely absorbed after the interaction (i.e. that its state after the interaction is entirely ##|0_{\rm EM}\rangle##), but that assumption was wrong.

The case when "mixed" really meansmixedis almost the same. Loosely speaking, the only important difference is that instead of ##+## in the equations above you have to put a logicalor.

Demystifier, after researching this issue a couple weeks, I have one final question. So it seems that the NMR process is similar to your answer. But I'm confused. The incoming photon in NMR is carefully selected to have "resonant frequency" with the nuclei spin precession, meaning it has exactly enough energy to INSTANTLY flip the nuclei spin from anti-parallel to parallel with the magnetic field, but the NMR textbooks all say that the incoming photons rotate the nuclei spins monotonously and smoothly in time (i.e., they do not TOTALLY flip a nuclei spin instantly), which makes me think that a given photon leaves the collision in a superposition or mixed state like in your answer. So the photons would all leave the collision in a mixed state and would only transfer a little bit of energy? It's confusing because the photon perspective suggests the transfer of energy occurs in increments (quanta) and could TOTALLY flip a nuclei, but the typical NMR textbook (like Slichter) makes it sound like energy transfer from the photons is smoothly monotonously and each photon-nuclei collision only rotates the nuclei a small amount. Which is it? And can you recommend a textbook that explains this?

- #7

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I'm not an expert in NMR, so I don't know the details. However, it is possible that a nucleus absorbs photon of energy E and soon later emits another photon with energy E'<E. This is a two-step process which effectively looks as if the nucleus absorbed only a part of the photon's energy. Perhaps something like that happens in NMR.Demystifier, after researching this issue a couple weeks, I have one final question. So it seems that the NMR process is similar to your answer. But I'm confused. The incoming photon in NMR is carefully selected to have "resonant frequency" with the nuclei spin precession, meaning it has exactly enough energy to INSTANTLY flip the nuclei spin from anti-parallel to parallel with the magnetic field, but the NMR textbooks all say that the incoming photons rotate the nuclei spins monotonously and smoothly in time (i.e., they do not TOTALLY flip a nuclei spin instantly), which makes me think that a given photon leaves the collision in a superposition or mixed state like in your answer. So the photons would all leave the collision in a mixed state and would only transfer a little bit of energy? It's confusing because the photon perspective suggests the transfer of energy occurs in increments (quanta) and could TOTALLY flip a nuclei, but the typical NMR textbook (like Slichter) makes it sound like energy transfer from the photons is smoothly monotonously and each photon-nuclei collision only rotates the nuclei a small amount. Which is it? And can you recommend a textbook that explains this?

- #8

A. Neumaier

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These are two different ways of modeling the same situation. It is like with switching on the light in your home. Generally you think of it as being instantaneous but when paying careful attention you know of course that it happens (on a finer time scale) in a continuous way.INSTANTLY flip [...] rotate the nuclei spins monotonously and smoothly in time

In the quantum case you can model an absorption process as a quantum jump process (instantaneous jumps) or as a quantum diffusion process (continuous jumps). A useful related paper is ''Interpretation of quantum jump and diffusion processes illustrated on the Bloch sphere'' by Wiseman and Milburn.

Note that both quantum jump and quantum diffusion processes are dissipative, not unitary. In unitary quantum physics, there are neither jumps nor definite absorption.

- #9

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Thank you very much -- that PRA paper is outstanding.

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