Exchange symmetry question I am trying to learn Quantum mechanics using 1.
I have a question regarding exchange symmetry Sec 10.1 , page 214 (Image added below).
My question is why is $\delta$ in equation 10.3 not a function of $\mathbf{a},\mathbf{b},t$ ?
To give a crude example,  if one assumes
$$\psi(\mathbf{a},\mathbf{b},t) = \left(\left\lVert\mathbf{a}\right\rVert_2+\left\lVert\mathbf{b}\right\rVert_2\right)e^{j\left\lVert\mathbf{a}\right\rVert_2*t}  $$
Then equation 10.2 is satisfied but neither 10.5 nor 10.6 is satisfied.
I am guessing we are using some property from Schrödinger's equation to argue that $\delta$ is independent of $\mathbf{a},\mathbf{b},t$.
Can someone please explain what I am missing?
1 Philips A.C., Introduction to Quantum Mechanics

 A: Let's first take time out of the picture, because it complicates things. The wavefunction is only defined up to an overall phase anyway; so even if the wavefunction has an overall phase $e^{i \delta(a,b)}$, we can get rid of this overall phase by redefining the wavefunction.
The time dependence of the phase of the wavefunction is determined by the Schrodinger equation. For a system of identical particles, the Hamiltonian must be invariant under interchanges of these particles. Therefore if we start in a state where the wavefunction only changes up to a sign, the time evolution of the Schrodinger equation can never introduce a phase which breaks the interchange symmetry. More mathematically, you can introduce an exchange operator $P$ which exchanges $a$ and $b$. This operator commutes with the Hamiltonian, $[H,P]=0$, so a state that starts as an eigenstate of $P$ will remain under time evolution.
In terms of your example, a few points.

*

*First, a sort of trivial point for this question, but important in general, is that the wavefunction you wrote is not normalizable. This is easy to fix if you replace $|a|^2+|b|^2$ by something like $e^{-|a|^2 - |b|^2}$.

*Second, your example relies on an overall phase $e^{i |a|^2 t}$. At $t=0$, this phase vanishes, and your wavefunction is symmetric under $a\leftrightarrow b$. (Alternatively, we could pick an arbitrary reference time $t_0$, and multiply the wavefunction by an overall phase $e^{-i |a|^2 t_0}$, in accordance with paragraph 1, to get a symmetric wavefunction). Evolving the state from $t$ (or $t_0$, multiplying by $e^{-|a|^2 t_0}$) can never break this symmetry, because the Hamiltonian by assumption is invariant under the interchange of $a$ and $b$.

*I suspect Philips took this to be understood when he said "When this condition for an acceptable wavefunction is satisfied" (emphasis mine), but I agree this is a subtlety that's worth making explicit (I had never thought of it in this way before).

