A quantum particle at rest viewed from a merry-go-round.

In summary, the conversation discusses the transformation of the wavefunction of a particle at rest in 3D space when it is observed from different frames of reference. It also mentions the effect of rotation on the wavefunction and its relation to angular momentum. The concept of boosts and rotations forming a group is also mentioned.
  • #1
Spinnor
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Feynman's Lectures on Physics has an interesting graphic in volume 3, page 7_3, "Fig. 7-1. Relativistic transformation of the amplitude of a particle at rest in the x-t systems.", see scan below. Say ψ is the wavefunction of a particle at rest in 3D space, ψ = exp[-iEt], hbar = 1.

If I now move with some velocity v the particle at rest now has some momentum so ψ --> ψ' = exp[-i(-mv.r' - Et')]?

If on the other hand, instead of going some velocity v, I instead rotate say on a merry-go-round how will ψ transform? Will I say the particle now has angular momentum?

Related question? What does a large field of simultaneous clocks look like from the reference frame of someone riding a merry-go round?

From the three points of view,at rest, moving, and rotating, what do the hyperplanes ψ = constant look like? Feynman's graphic gives clue for the first two? Does rotation cause the hyperplanes to curve or are they "flat"?

Does this problem have anything to do with the "magic" of boosts and rotations forming a group? Boosting around some point results in some rotation?

Thanks for any hints or help!
 

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  • #2
If on the other hand, instead of going some velocity v, I instead rotate say on a merry-go-round how will ψ transform? Will I say the particle now has angular momentum?
You mean, if the observer is in a non-inertial reference frame?

Hint: What normally happens in this case?
 
  • #3
Simon Bridge said:
You mean, if the observer is in a non-inertial reference frame?

Hint: What normally happens in this case?

If I'm on a spinning merry-go-round with no roof and it's raining vertically I would say from my frame of reference the rain has angular momentum? The further from the center the greater the momentum? Relativity adds an additional "twist" because of Thomas Precession?
 
  • #4
Hint: What normally happens in this case?
I give up -- what?
 
  • #5
Bill_K said:
I give up -- what?
Bill_K the Science Advisor ... is trying to trip me up :)
It's a pedagogical gambit Bill - admittedly I've been neglecting this thread... but since you are here, perhaps you'd like to try answering the question?
 

FAQ: A quantum particle at rest viewed from a merry-go-round.

What is a quantum particle at rest?

A quantum particle at rest refers to a particle that is not moving and has a defined position. In the field of quantum mechanics, particles can exist in a state of rest, where they have a specific location and their momentum is zero.

What is the perspective of viewing a quantum particle at rest from a merry-go-round?

Viewing a quantum particle at rest from a merry-go-round adds a rotational component to the particle's motion. This means that the particle's position and momentum are constantly changing due to the rotation of the merry-go-round.

How does the observer's motion affect the observation of a quantum particle at rest on a merry-go-round?

The observer's motion on the merry-go-round will affect the observation of the quantum particle at rest. This is because the observer's frame of reference is constantly changing, causing the particle's position and momentum to also change.

What is the concept of frame of reference in relation to a quantum particle at rest on a merry-go-round?

Frame of reference refers to the perspective or point of view from which an object's motion is observed. In the case of a quantum particle at rest on a merry-go-round, the frame of reference of the observer is constantly changing due to the rotation, affecting the observation of the particle's position and momentum.

Can a quantum particle at rest on a merry-go-round be accurately observed?

No, it is not possible to accurately observe a quantum particle at rest on a merry-go-round. This is due to the uncertainty principle in quantum mechanics, which states that the more precisely we measure a particle's position, the less precisely we can measure its momentum, and vice versa. The changing frame of reference from the merry-go-round adds to this uncertainty, making it impossible to accurately observe the particle's state.

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