How can there be fluctuations in the inflaton field value?

  • #1
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http://www.counterbalance.org/cq-turok/etern-body.html
However, the field φ, like every other field, is subject to quantum mechanical fluctuations. As it rolls down the hill, in some regions φ fluctuates downwards and in others it fluctuates upwards.
https://ned.ipac.caltech.edu/level5/Kolb/Kolb3_1.html)
During inflation there are quantum fluctuations in the inflaton field.
http://www.astronomy.ohio-state.edu/~dhw/A873/notes7.pdf)
The field φ experiences quantum fluctuations, as the uncertainty principle tells us it must.

According to QFT, if you make repeated measurements of some property of the field then you will in general measure a fluctuating value. But the fluctuation is in the measurement of the field not in the field itself. Quantum fields do not fluctuate.

So what causes the inflaton field to fluctuate to a higher value in some regions and to a lower value in some others?
Fluctuations from what?
 
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  • #2
kimbyd
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http://www.counterbalance.org/cq-turok/etern-body.html


https://ned.ipac.caltech.edu/level5/Kolb/Kolb3_1.html)


http://www.astronomy.ohio-state.edu/~dhw/A873/notes7.pdf)



According to QFT, if you make repeated measurements of some property of the field then you will in general measure a fluctuating value. But the fluctuation is in the measurement of the field not in the field itself. Quantum fields do not fluctuate.

So what causes the inflaton field to fluctuate to a higher value in some regions and to a lower value in some others?
I don't think this is an accurate understanding of QFT. My understanding is as follows:
1) If the field is in a steady state, then it does not fluctuate. Typically this means an eigenstate of minimum energy. Measurement doesn't impact this much, but repeated measurements over a short period of time can "freeze" a field to prevent it fluctuating.
2) If the field is not in a steady state, then it will experience fluctuations. The inflaton field is not in a steady state (in typical models, the field is approaching a local potential minimum during inflation). And so it experiences fluctuations.
 
  • #3
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I don't think this is an accurate understanding of QFT. My understanding is as follows:
Thanks Kim. It’s what I understand from reading these answers about QFT in the links below. I’m a layman so please feel free to correct me if I misunderstood something.

There’s a discussion about fluctuations of a field.
The quantum field does not fluctuate.
https://physics.stackexchange.com/q...ns-in-the-value-of-a-quantum-field-at-a-point
In quantum field theory, fields undergo quantum fluctuations.
— Nope. You won't find this assertion in any book on QFT.
https://physics.stackexchange.com/a/257700
 
  • #4
kimbyd
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A really simple way to think of quantum fluctuations is using the Simple Harmonic Oscillator. In the SHO system, if it is in an energy eigenstate, the system is static, in that any observables you might try to measure about the system are completely unchanged no matter when you measure the system. If, however, the system is in a superposition of states, e.g. a superposition of energy levels 1 and 2, then the wave function of the particle changes over time: its position seems to sort of slosh back and forth.

This system is over-simplified, because it is an isolated system and the particle has no electric charge. If you add an electric charge, then only the lowest-energy eigenstate is stable: all others decay as the particle has a probability of emitting a photon at any time. So in the real world, you get oscillations in most any situation. The exception is when the system really is in its minimum-energy state, as could be the case with a scalar field in the late universe (e.g. dark energy). But inflation wasn't at all like that. Inflation was highly dynamic. And so you have the "sloshing" represented by the simple harmonic oscillator mentioned above. So, yeah, you'll get zero-point fluctuations in the inflaton field because it is a dynamical field that changes its field value over time. And those fluctuations will result in some regions having greater density than others.

Note: this is a pretty over-simplified description, but I think it's reasonably-accurate.
 
  • #5
PeterDonis
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In the SHO system, if it is in an energy eigenstate, the system is static, in that any observables you might try to measure about the system are completely unchanged no matter when you measure the system.
Isn't this too strong as you state it? The expectation values of all observables are constant in time; but I don't think you are guaranteed to get the same values on measuring all observables no matter when you measure them. For example, if you measure the position ##\hat{x}## of the oscillator, you won't always get the same value.
 
  • #6
kimbyd
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Isn't this too strong as you state it? The expectation values of all observables are constant in time; but I don't think you are guaranteed to get the same values on measuring all observables no matter when you measure them. For example, if you measure the position ##\hat{x}## of the oscillator, you won't always get the same value.
To state it more precisely, if you set up the SHO system in an energy eigenstate, the probability distribution of any measurements you might make are completely independent of your choice of how long to wait to make those measurements.

Different measurements can result in different results, and measuring once can impact the next measurement made (since measurements will modify the system). But if you imagine setting up thousands of SHO's, and measuring them at random times, the distribution of those measurements will be independent of your choice of time for each measurement.

Edited to fix pedantic error: the system is only time-independent if it's in an energy eigenstate. If it's in a mixed state, as described above, it changes over time.
 
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  • #7
PeterDonis
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To state it more precisely, if you set up the SHO system in a particular state, the probability distribution of any measurements you might make are completely independent of your choice of how long to wait to make those measurements.
Yes, this I agree with.
 

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