Global phase of wave function in quantum mechanics and Fubini-Study metric

Question

I have a basic question about projective representations in quantum mechanics. In projective representation we identify the class of normalized states in Hilbert space as the same physical state as follows :

\begin{equation} \psi_1\sim \psi_2~~~iff~~~\psi_1=e^{i\lambda}\psi_2,~~~\lambda\in\mathbb{R} \end{equation}

to make sure that both $\psi_1$ and $\psi_2$ satisfy Schrodinger equation, $\lambda$ have to be just a number and not a real function. This is how I understood the projective representation.

On the other hand, there is a natural metric over the finite dimensional Hilbert states which is Fubini-Study metric as follows:

\begin{equation} ds^2=\frac{(\bar\psi.\psi)d\psi.d\bar\psi-(\bar\psi.d\psi)(\psi.d\bar\psi)}{(\bar\psi.\psi)^2} \end{equation}

where $\bar\psi.\psi=\bar\psi_i\psi_i$ is the usual inner product. We can check directly that the curve $$\psi(t)=e^{if(t)}\psi(t=0)$$ for some function $f(t)$ and normalized states ($\bar\psi(t=0).\psi(t=0)=1$) satisfy the null distance condition : \begin{equation} ds^2=0 \end{equation}

and the distance between $\psi(t)$ and $\psi(t=0)$ is zero and we should conclude that these states are the same physical state and indeed the same point in the Fubini-Study metric.

My question is as follows: how are the states $\psi$ and $e^{if(t)}\psi$ the same physical state when they are not both the solution of Schrodinger equation simultaneously unless $f(t)$ would be a constant function?

Answer 1

Let's first get some terminology straight:

Yes, by the postulates of quantum mechanics, the vectors $\psi$ and $\mathrm{e}^{\mathrm{i}\lambda}\psi$ for any $\lambda\in\mathbb{R}$ in Hilbert space $H$ represent the same state. This leads us to consider the projective Hilbert space $P(H) := H/\sim$ where $$ \psi \sim \psi ' \iff \exists c\in\mathbb{C} \psi' = c\psi.$$ This is not "the projective representation", it's just a projective space. If we would now consider how groups (the physical groups of symmetries or transformations of this space of states) act on this, we would be led to the idea of projective representations (see also this Q&A of mine), but there are no representations in the question as written.

There is, however, the Fubini-Study metric $d_\text{FS}$. This is a metric on $P(H)$ in the proper sense, i.e. $$d_\text{FS}(\psi,\psi') = 0 \iff \psi = \psi'.$$ Now, the question seems concerned about the "trajectory" $\psi(t) = \mathrm{e}^{\mathrm{i}f(t)}\psi(0)$ being a the same state at all times even when $f(t) \neq -Et$, i.e. $\psi(t)$ is not a solution to the Schrödinger equation.

But - regardless of whether we phrase it in terms of projective spaces and the FS metric or not - this is just what you should expect from the postulate: $\psi(0)$ and $\psi(t)$ are, by definition, the same state when they are related by multiplication by a complex number, i.e. when they lie in the same "ray" in Hilbert space, and $\mathrm{e}^{\mathrm{i}f(t)}$ acts as multiplication by a complex number at all times.

So this "trajectory" is, in fact, the constant trajectory that remains the same physical state at all times, regardless of whether it is a solution to the equations of motion or not.