Symmetries of a 1 qubit universe I am studying an isolated 1 qubit system trying to find its symmetries. By that I mean see what I can change without changing the physics. But I am having problems understanding what is the physics of such an isolated system.
Here is what I know.
$$|\psi(0)> = \alpha|0>+\beta|1>$$
$$H = 
\begin{pmatrix}
E_0 & 0 \\ 0 & E_1
\end{pmatrix}$$
Then unitary evolution is
$$|\psi(t)> = e^{-it\hat{H}}|\psi> = \alpha e^{-itE_0}|0>+\beta e^{-itE_1}|1>$$
I am considering a 1 qubit isolated system because it acts as a toy model for a quantum universe. I do not consider measurements in the sense of the many world. ie, There is apparent collapse of the wavefunction in our universe, but this 1 qubit universe is too small to split into system and environment.
The 4 numbers that define this simple "universe", are $(\alpha,\beta,E_0,E_1)$.
My question is. What transformation can I apply to theses numbers without changing the physics?
Here is as far as I got
The quantities I look at are, expectation values $<\hat{A}>$, eigenvalues and the trajectory of the system in Hilbert space.
Can I rescale the Hamiltonian. $\hat{H} \rightarrow \hat{H}-E_0\hat{I}$. I get that the expectation value $<\hat{H}> = |\alpha|^2E_0 + |\beta|^2E_1\rightarrow |\beta|^2(E_1-E_0)$ changes. I don't think it is a physical change, since the expectation value is the average over infinitely repeated experiments, and does not make sense here as we do not measure our system.
What about dynamics?
Under $\hat{H} \rightarrow \hat{H}-E_0\hat{I}$, we get,
$$|\psi(t)> = e^{-it\hat{H}}|\psi> = \alpha |0>+\beta e^{-itE_1}|1>$$
So there is no longer any movement in the dimension of $|0>$.
Does that count as changing the physics? I am not sure
In the case of the expectation value, $\hat{H} \rightarrow \lambda\hat{H}$ looks like the same ambiguity.
For dynamics we get,
$$|\psi(t)> = e^{-it\hat{H}}|\psi> = \alpha e^{-it\lambda E_0}|0>+\beta e^{-it\lambda E_1}|1>$$
Again, unsure if this changes the physics, since redefining $t'=\lambda t$, means that $|\psi(t')>$ has the same trajectory as |\psi(t)>.
Added question: What other transformations to my Hamiltonian (or wavefunction) leave my physics unchanged?
Thanks
Edited for clarity and added a question I asked in a comment.
 A: First of all you have to be a bit careful what you mean by symmetry here: An easy definition would be that all expectation values of observables are invariant. A second definition (which implies the first) would be that all measurements yield the same results i.e. that all eigenspaces are left invariant. The second is normally more useful to look at. Now you mentioned trajectories in Hilbert space which are valid to consider. However you have to be careful as Hilbertspace has considerable structure that makes some operations that look like changes in the trajectory at first just identity operations. The reason is that the Hilberspace is actually not the space of all states $\alpha |1\rangle + \beta |0\rangle$ but rather is this space together with an equivalence relation relating states along all complex lines. What this means is that in the Hilbert space two states are considered identical if
$|\psi\rangle = c |\tilde{\psi}\rangle \quad \forall c \in \mathbb{C}$.
you might know this condition reformulated by saying we only consider normalized states and that global phases do not matter which together amounts to the condition set above.
Now why do we want this? Essentially this just amounts to picking a different representative in some Eigenspace:
Take $|lambda\rangle$ as an eigenvector to some Observable $\hat{O}$ such that
$\hat{O} |\lambda\rangle = \lambda |\lambda\rangle$. Then trivially this whole equation still holds if we multiply both sides by $c$ such that $|c\lambda \rangle \equiv  c|\lambda \rangle $ is still an eigenvector with the same Eigenvalue.
Now as we saw before one definition of symmetry is that all measurements yield the same result. However measurement just produce eigenvalues (as they are by definition normalized) and as such scaling a state by some complex number does not change any measurement.
We can now apply this to your problem:
Take $\hat{H} - E_0 \hat{I}$ so you find
$e^{it(\hat{H} - E_0 \hat{I})} |\psi\rangle$ = $e^{-itE_0\hat{I}} e^{it\hat{H}} |\psi\rangle = \hat{I} e^{-itE_0}e^{it\hat{H}}|\psi\rangle = e^{itE_0} |\psi (t)\rangle$.
Where $|\psi(t)\rangle$ is defined to be the time evolution under $\hat{H}$. Now we find that this transformation amounts to a global phase shift in your evolution. However we found that such states are identified in the Hilbert state such that this is essentially (applied globally) the identity transformation if you look at measurements.
For the second transformation, so scaling the Hamiltonian your reasoning is sound. It is exactly the same as changing the timescale. So rescaling is essentially changing the energyscale you are looking at. However to some degree, this is just looking at the system in different 'units'. E.g instead of seconds we now measure in $\lambda$seconds. Now as the units for energy and time are related and as such rescaling one of them changes both:
Thus rescaling energies has the effect that processes will complete faster e.g. where before we would have measured some outcome $A$ at time $t$ we will now find the identical outcome at time $\lambda t$. So does this change the physics? It depends!
If you only have your one cubit system and nothing else then there is nothing to compare it to so the scale you picked at the start of your setup was arbitrary anyhow, only ratios of energies are really unique as we can just rescale units to change absolute values. Now if you do however have a second system as a reference (e.g. in a lab you have the lab itself with its clocks) then this is a relative change between the lab system and the qubit system and as such changes the physics (though it does so in a very simple and predictable way).
The reason for this is that your system does not have any intrinsic energy scale apart from the one you give it, i.e. by defining the scale of $\hat{H}$.
Edit answering comment:
The real question here I guess is what you mean by 'gauge transformation'. Normally one defines gauge invariance as a nonphysical property anyhow such that a gauge transformation just describes a redundancy of your description of the physical system. However, defining symmetries generally, requires some quantity that is invariant under such a transformation (both gauge and normal symmetry), and if it is not measurement values what is it in your system? Well the obvious answer is that 'the physics' should be invariant. But that's just rewording the question to asking what 'the physics' actually includes. My answer would be all measurable quantities. However, there are definitely other (maybe) sensible answers that I'm not willing to go into as I don't have a foundations background.  Sticking with this definition then yes what you talk about could be considered 'gauge transformations' i.e. symmetries that arise due to the redundancy in the description of a state (exactly the redundancy that states in the Hilberspace are actually Rays in a vectorspace such that considering points leaves you with a gauge redundancy).
