Why must the stationary state vectors rotate?

Question

In quantum mechanics, pure states are the one-dimensional orthogonal projections or from another perspective, the one-dimensional subspaces, or yet from another, the so-called rays (the "up to modulus 1 complex factor" equivalence classes of unit vectors ) of a Hilbert space. Symmetries are transformations of the ray space (or of the projective Hilbert space) that preserve the so-called "transition probabilities" defined as $$\mathbf a\cdot \mathbf b:= |\langle a,b\rangle| \text{ where } a\in \mathbf a, b\in\mathbf b, |a|=|b|=1 \tag{1}$$ In quantum mechanics, elements of the ray space (e.g. $\mathbf a$ and $\mathbf b$ above) are called pure states while elements of the Hilbert space (e.g. $a$ and $b$ above) state vectors.

By Wigner's theorem, for every symmetry, there is a unitary or anti-unitary transformation of the Hilbert space resulting in this symmetry when applying it to one-dimensional subspaces (or to rays) instead of individual vectors.

Due to Stone's theorem, the possible processes (i.e. the change of the states in time) of a quantum mechanical system are described by Schrödinger's equation, which describes the possible evolutions of the state vectors:

$$-i\hbar\dot\psi= H\psi\tag{2}$$

In the special case when $\psi$ is an eigenvector of $H$ with eigenvalue $\lambda$, the solution of (2) is $$\psi(t) = \psi(0)e^{-i\lambda t/\hbar}\tag {3}$$ that is, the state vector "rotates" with constant angular velocity $\lambda/\hbar$, but the state (the one-dimensional subspace spanned by $\psi(t)$) is constant. There isn't any solution of (2) where $\psi$ itself is also constant in spite of that a constant state of course can be represented also with a constant state vector. I don't see but I'm curious that is this a consequence of Wigner's theorem (i.e there aren't unitary or anti-unitary transformations having fixed points in any subspace that is a fixed point of a symmetry) or of Stone's theorem (which requires strongly continuous one-parameter unitary groups).

Answer 1

In contrast to Qiaochu Yuan, I think that Wigner is responsible for the rotation. More precisely, the shift from the one-dimensional subspaces (i.e. from the rays) to their representatives.

According to Wigner, for every transition probability keeping transformation $T$ of the rays, there is a unitary or anti-unitary transformation $U$ that $[Ux] = T[x]$ for every vector $x$ where $[x]$ denotes the ray which contains $x$. The situation is the clearest when the eigenvectors $\psi_1,\psi_2,\dots$ of $H$ (belonging to the eigenvalues $\lambda_1,\lambda_2,\dots$) form a basis of the Hilbert space. In this case, the time evolution (which is a probability-keeping transformation $T_t$ for every $t\in\mathbb R$) leaves $[\psi_i]$ unchanged. If we want to realize $T_t$ (different from the identity) with a transformation $U$ of the Hilbert space so that unchanged rays are represented by unchanged vectors under $U$ then clearly, this transformation can't be linear (nor anti-linear) because a linear or anti-linear transformation which leaves a basis unchanged can only be identity. So, we lose the possibility to shift from the rays to their representatives so that unchanged rays are represented by unchanged vectors, when we chose the transformation to be linear.

Answer 2

There isn't any solution of (2) where $\psi$ itself is also constant

This is false. The "standing wave" solutions $\psi(t) = \psi(0) e^{-\frac{i}{\hbar} \lambda t}$ are constant iff $\lambda = 0$, so $\psi$ itself can be constant iff it corresponds to an energy eigenstate with energy zero. This has nothing to do with Wigner's theorem; the connection to Stone's theorem is that it forces you to write down the Schrodinger equation (although it doesn't tell you anything about what $H$ actually looks like a priori; that requires some discussion of the classical limit) but it sounds like you're asking something else and I don't really know what.

It's worth noting that 1) it's impossible to physically tell whether $\psi$ itself is constant or not, since we never observe $\psi$ but only at best the ray it defines, and 2) this is mirrored by the fact that $H$ is only well-defined up to an additive constant; the observable physics does not change if we replace $H$ with $H + c$ for any constant $c$, which has the effect of shifting all the energy eigenvalues and hence changing the rate at which any individual energy eigenstate "rotates," but does not change any transition probabilities. Only the relative energy of different eigenstates is physically meaningful.