Grand canonical hamiltonian and thermodynamic ensembles

Question

In many-body theory (and quantum field theory I suppose) we often work in the grand canonical ensemble, where the number of particles in the system is only fixed on average. The density operator used to compute expectation values is

$$ \rho = \frac{e^{-\beta (H - \mu N)}}{Z_{G}} = \frac{e^{-\beta K}}{Z_{G}} $$

where $K$ is the so-called grand canonical hamiltonian.

My problem is when we substitute $H$ for $K$ in the expression of the evolution operator $ U(t) = e^{-iHt} \rightarrow e^{-iKt} $, which is done most of the time because it simplifies calculations. It seems equivalent to saying that Schrodinger equation is invariant under the change $H \rightarrow K$.

The justifications I've seen so far, are based on the fact that since the original hamiltonian conserves the number of particles and thus commutes with the operator $N$, this replacement is just a displacement in energy and does not essentially changes the dynamics of the system. I'm not really convinced by this because the replacement with $K$ is not equivalent with adding a simple constant to the hamiltonian.

Additionally, it seems that if this argument is true, then in general one would be justified to construct a new hamiltonian $H' = H + \mathcal{O}$ to describe the dynamics of a system, as long as $[H, \mathcal{O}]=0$.

Given that, my questions are as follow :

1. Is the above statement that you can replace a hamiltonian $H' = H + \mathcal{O}$ when $[H, \mathcal{O}]=0$ true? If it'sn't, is there anything special with the case $\mathcal{O} = \mu N$, or are there some caveats?
2. Thermodynamics ensembles are often defined by density operators on the form $\rho=e^{-\beta S}$ with $S$ that is a linear combination of $H$ and various conserved operators $c_j \mathcal{O}_j$. Could we get rid of those ensembles by just including the operators $c_j \mathcal{O}_j$ in the hamiltonian in the first place and finding $c_j$ such that $\langle \mathcal{O_j} \rangle$ is the value we want? $c_j$ would have the physical interpretation of a classical field that couples with $\mathcal{O}_j$.

Answer 1

The Schrödinger equation is not invariant under the change $H\rightarrow K$. However, in the grand canonical ensemble with some values of inverse temperature $\beta$ and chemical potential $\mu$, the fluctuation of the number of particles $N$ is of the order of $\sqrt{\langle N \rangle }$, where $\langle N \rangle $ is the average number of particles in this ensemble. This fluctuation is much less than $\langle N\rangle$, so the states of this ensemble most probably have a number of particles that is very close to $ \langle N \rangle $, and $K=H-\mu N$ gives pretty much the same evolution as $K^\prime=H-\mu \langle N\rangle$ for this specific grand canonical ensemble. And $K^\prime$ is indeed just $H$ with a constant shift.

Answer 2

Actually, replacing $H \to H - \mu N$ does affect the dynamics substantially, but it does so in a rather trivial way. This is precisely because $H$ and $N$ commute. If two operators $A$ and $B$ commute, and only then, we can factor an exponential as $e^{A + B} = e^{A} e^{B}$. This allows us to conclude that $e^{-iHt} = e^{iN\mu t} e^{-iKt}$. That is, we are free to evolve with $K$ as long as we remember that at the end of the day we have to apply $e^{iN\mu t}$ to recover the "true" time evolution. This is generally pretty easy since $N$ is a very simple operator. Another way to think about it is that we are working in a "rotating frame" -- we study the evolution not of the actual state of the system $|\psi(t)\rangle$, but of the "rotated" state $e^{iN\mu t} |\psi(t)\rangle$.

The answer above, which claimed that the differences are insignificant because $\sqrt{ \langle (N - \bar{N})^2 \rangle }$ is small, is not correct. In the thermodynamic limit, this quantity still goes to infinity. Its ratio with $\bar{N}$ goes to zero, but it would have to go to zero in absolute value to not affect the dynamics, and this is not the case. In fact, one can verify that, for example, the expectation value (in a bosonic system) of a boson creation operator $\langle a^{\dagger} \rangle$ rotates at a different frequency under $K$ than under $H$ (specifically, the difference in frequency is $\mu$). This is precisely accounted for by the extra rotation by $e^{iN \mu t}$ discussed above.