In quantum mechanics, Ehrenfest's theorem states that ⟨px⟩=mddt⟨x⟩. My question is, does there exist a similar relationship between ⟨Lz⟩, the expectation value of the z-component of the orbital angular momentum operator, and the time derivative of ⟨θ⟩, the expectation value of the position operator ˆθ in spherical coordinates?
If not, is there any way to relate ⟨Lz⟩ to the time derivatives of expectation values of one or more operators? For instance if you knew what ⟨x⟩, ⟨y⟩, ⟨z⟩, ⟨px⟩, ⟨py⟩, and ⟨pz⟩ are as a function of time, would that give you enough information to determine ⟨Lz⟩?
And are the answers to these questions affected at all by whether the particle has spin or not? By the way, this question was inspired by the comment section of this answer.
Answer
Thanks to @udrv, I found the answer in this journal paper. Let's work in cylindrical coordinates (r,θ,z). Let cosˆθ and sinˆθ be defined by Taylor series, and let ˆLz=mˆr2ˆωz. Then we can write the result in two forms:
ddt⟨cosˆθ⟩=⟨−12(ˆωzsinˆθ+sinˆθˆωz))⟩
ddt⟨sinˆθ⟩=⟨12(ˆωzcosˆθ+cosˆθˆωz))⟩
The reason why why we can't simply use the operator ˆθ is that ˆLz is only a Hermitian operator if its domain is restricted to periodic functions, and ˆθ maps periodic functions to non-periodic functions. So if we want to keep things within the domain of Lz we need to work with an operator f(ˆθ) where f is a periodic function. And the simplest periodic functions which make f(ˆθ) a Hermitian operator are sine and cosine. (It needs to be Hermitian if we want our Ehrenfest result to be between observable quantities.)
EDIT: The paper also provides a more general result for arbitrary periodic functions f with period 2π:
ddt⟨f(ˆθ)⟩=⟨12(ˆωzf′(ˆθ)+f′(ˆθ)ˆωz)⟩
where again f(ˆθ) and f′(ˆθ) are defined via Taylor series.
Note that while this formula is true for all such functions f, in order it to be a result between observable quantities f(ˆθ) needs to be a Hermitian operator. I posted a question here to find out what functions f make f(ˆθ) Hermitian.
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