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May 8

Open quantum system tending towards equilibrium

This question uses some of the terminology of quantum mechanics, but it is fundamentally a maths question. If you recognize the physics, I am trying to justify a system tending towards the same unique equilibrium regardless of the initial state.

Let finite sequences of 0s and 1s form the basis of a vector space. For example, $|0010\rangle$ is a element of this basis.

Let $U$ be a unitary operator on this vector space. This operator essentially "appends a 0 or 1", but to a linear combination. One example is:

$U|s0\rangle =\alpha |s00\rangle +\beta |s01\rangle$

$U|s1\rangle =-\beta |s10\rangle +\alpha |s11\rangle$

For any sequence s. Think of U as the next-time-step operator.

Let $f_{x}(v)$ pick all the coefficients of basis vectors that form v that end with x, and takes the sum of their norm-squared. For example,

$f_{0}(c_{1}|00\rangle +c_{2}|01\rangle +c_{3}|10\rangle +c_{4}|11\rangle )=|c_{1}|^{2}+|c_{3}|^{2}$ and $f_{1}(c_{1}|00\rangle +c_{2}|01\rangle +c_{3}|10\rangle +c_{4}|11\rangle )=|c_{2}|^{2}+|c_{4}|^{2}$ .

Now, consider $\lim _{n\rightarrow \infty }f_{x}(U^{n}v)$ for some initial v. It can be shown that $\lim _{n\rightarrow \infty }f_{x}(U^{n}v)={\frac {1}{2}}$ for any initial v and either x.

Furthermore, if you consider sequences of some finite element space with m elements (beyond just 2 elements as in this example) then $\lim _{n\rightarrow \infty }f_{x}(U^{n}v)={\frac {1}{m}}$ . So the equilibrium distribution is uniform over the element space, and is independent of the initial state.

What I am interested in is when the element space becomes infinite. I suspect that the equilibrium distribution is not uniform over the element space anymore in that case. Is the quantity $\lim _{n\rightarrow \infty }f_{x}(U^{n}v)$ still independent of the initial v?

--49.183.55.236 (talk) 05:30, 8 May 2019 (UTC)[reply]

I'm guessing, without getting into details, that the problem you're having is that while the f_x(Uⁿv) all seem to converge to 0 (the equilibrium), their sum remains constant (≠0). Tannery's theorem gives conditions under which limit and infinite sum can be exchanged, and these conditions aren't met in this case. The article is a bit sparse in terms of examples, but it's not hard to come up with simpler cases where limit and sum can't be exchanged; one is a_k(n) = δ_kn. I don't know how this is interpreted with bras and kets, but presumably you need some quantum mechanical version of the conditions in Tannery to hold before you can assume the equilibrium state behaves as expected. --RDBury (talk) 16:26, 8 May 2019 (UTC)[reply]