Questions about normalizing wavefunctions learning QM and just have a few questions regarding normalizing wavefunctions.

*

*Thus far, every initial wavefunction that we've normalized has had an undefined constant explicitly put out front (i.e., something like $\Psi (x,0) = A(a - x^{2})$, as an example) I was recently asked to normalize a wavefunction that didn't have any undetermined constant out front of the function involving x. Now it's okay for me to multiply the undefined constant, say $A$, to the wavefunction? Since, of course, the purpose of normalizing the wavefunction is finding a form such that the probability is 1 over all space.


*If I'm given a complete wavefunction (i.e., a solution to the time dependent Schrodinger equation - $\Psi(x,t) = \psi(x)\phi(t)$ I can just perform the normalization at $t=0$ right? In other words, normalize the initial wavefunction $\Psi(x,0) = \psi(x)\phi(0)$. Since once a wavefunction is normalized, it remains normalized for all time, correct? Or do I need to perform the normalization at arbitrary time $t$ if given the complete wavefunction?


*If I've just normalized a wavefunction and am determining the constant $A$ but have $|A|^{2}$ =  $K$ where $K$ is just some constant then I necessarily have $A$ = $\pm\sqrt{K}$ but the problem makes no mention of the nature of the constant (i.e., doesn't say anything like "where $A,a,b,$ etc... are all positive real constants) then how am I to determine the sign of $A$? Does it even matter? I imagine it wouldn't in the verification of the normalization but it would change the sign of the wavefunction, so it must have some importance?


*If I get a wavefunction such that $|\Psi(x,0)|^{2} = 1$ would that tell me that this wavefunction is non-normalizable? Since the integral doesn't converge.
 A: *

*Yes. Normalizing the wavefunction means multiplying it by whatever constant you need to in order to make its norm equal to $1$.

*Yes, because time evolution in quantum mechanics is unitary - in other words, you can write $\psi(t) = \hat U(t) \psi(0)$ for a unitary operator $\hat U(t)$ called the propagator, and $\Vert \hat U \psi\Vert = \Vert \psi\Vert $ for all unitary operators $\hat U$. Caveat: this is no longer true if you perform a measurement, in which case your wavefunction will undergo (non-unitary) projective evolution and you may need to renormalize afterward.

*If $|A|^2 = K$, then $A=e^{i\theta} \sqrt{K}$ for any arbitrary phase angle $\theta\in[0,2\pi)$.  You may make any choice you want, and it will not impact any physically meaningful predictions of the theory. Perhaps you could check to see whether multiplying the wave function by some $e^{i\theta}$ changes e.g. the probabilities of any measurement outcomes.

A: *

*yes

*yes

*the constant is determined up to an arbitrary phase. Remember that this is also the case for the wave function.

A: Since people had already answered correctly questions 1 and 2, I'd like to answer/point some details as for 3 and 4:


*It is a postulate in quantum mechanics that the states of a (quantum) dynamical system are in one-to-one correspondence with some complex projective Hilbert space, then yes, the sign of your normalization doesn't matter, nor a complex phase (a unit modulus complex number), not even you need to normalize the wavefunction at first place (this would mean you have to take the adequate care when computing things such as operator's expectation values, projected states after measurement/collapse etc.). The reason why one computes normalizations is because it makes further calculations "cleaner".

*The complex projective Hilbert space after which we usually model our (quantum) dynamical system is the projective space of $L^2(V)$, where $V \subseteq \mathbb{R}$ (in order to allow every momenta value to be a possible eigenstate of the Hamiltonian). So, if for any reason $V$ happens to be compact, then your function is square-integrable.

