A rotating frame of reference (since you rotate your BEC to generate these), where the energy is giv

Spencer Lutz

Spencer Lutz

Answered question

2022-05-13

A rotating frame of reference (since you rotate your BEC to generate these), where the energy is given by:
E ~ [ Ψ ] = E [ Ψ ] L [ Ψ ] Ω ,
Where Ω is the rotational velocity which you apply to the BEC.
Now the argument that the term L [ Ψ ] Ω should be substracted comes from the fact that in a rotating frame your system loses that fraction of rotational energy. Now I was wondering if anyone knew where the form of L [ Ψ ] Ω came from?
I know that in a rotating frame of reference you have that v = v r + Ω × r . If you fill this in into the kinetic energy and use some basic definitions, you get that the extra effect of rotation is given by:
1 2 I Ω 2 = 1 2 J Ω .
This yields half of the value that is used in the book, is there something that I'm missing or not seeing right ?

Answer & Explanation

Athena Blanchard

Athena Blanchard

Beginner2022-05-14Added 13 answers

The L Ω term comes directly from the change of frame of reference, especially from the transformation from the static frame of reference to the rotating frame of reference.
Let R ( x , y , z ) the initial static frame, and R ~ ( x , y , z ) the rotating frame at a constant velocity Ω = Ω z ^ . We study the mouvement of a free particle which hamiltonian reads :
H ^ = p ^ 2 2 m expressed in the R frame.
The particule state | ψ ( t ) follows the Schrodinger equation:
H ^ | ψ ( t ) = i t | ψ ( t ) in the R frame.
If we change from R to R ~ frame, there exists an unitary transformation U ^ ( t ) so that the particule state is now | ψ ~ ( t ) = U ^ ( t ) | ψ ( t ) . To derive the corresponding Schrodinger equation on | ψ ~ ( t ) , one can start to calculate :
i t | ψ ~ ( t ) = i [ d U ^ d t | ψ ( t ) + U ^ t | ψ ( t ) ]
Then, by using the Schrodinger equation in the R frame and the fact that | ψ ( t ) = U ^ | ψ ~ ( t ) , we have :
H ~ ^ | ψ ~ ( t ) = i t | ψ ~ ( t ) in the R ~ frame
with H ~ ^ = U ^ H ^ U ^ + i d U ^ d t U ^
By simple assumptions on momentum conservation, one can take :
U ^ ( t ) = exp ( i Ω t L ^ z )
where L ^ z = x ^ p ^ y y ^ p ^ x = i φ momemtum operator along z* L ^ z commutes with H ^ given the fact the kinetic energy is invariant by rotation transformation.
After some calculations, one can show that:
i d U ^ d t U ^ = Ω L ^ z = Ω L ^

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