unitary transformation affecting the inner product of wave-functions as vectors

I am having this quandary over unitary transformation and probability density functions in quantum mechanics. I thought this was the adequate place to post it, but feel free to tell me if you think otherwise.

As I am sure you are all aware, in quantum mechanics it is quite common to have unitary operators acting on wave-functions (these being elements of the Hilbert space $$L^2(\mathbb{R}^3, \mathbb{C}^4)$$, for concreteness). These wave-functions play a crucial role because $$$$\rho:=|\psi|^2=\sum_{k=1}^{4}|\psi_{k}|^2$$$$ where $$\psi_{k}$$ denote the components of $$\psi$$, can be interpreted as probability density functions. Note $$\rho$$ is the norm of the wave-function $$\psi$$ as a vector (and not as an element of $$L^2(\mathbb{R}^3, \mathbb{C}^4)$$). Now, under a unitary transformation $$U$$ (i.e. a map from $$L^2(\mathbb{R}^3, \mathbb{C}^4)$$ to itself that is self-adjoint), $$\psi$$ is turned into $$\phi:=U\psi$$ and therefore $$$$|\phi|^2=\langle \phi, \phi \rangle=\langle U\psi, U\psi\rangle$$$$ where $$\langle\cdot,\cdot \rangle$$ is the dot product of two vectors, and not the dot product in $$L^2(\mathbb{R}^3, \mathbb{C}^4)$$. My question is: are these two quantities related in some way? Obviously since $$U$$ is unitary the $$L^2(\mathbb{R}^3, \mathbb{C}^4)$$ inner product is preserved and $$\|\psi\|_{L^2(\mathbb{R}^3, \mathbb{C}^4)}=\|\phi\|_{L^2(\mathbb{R}^3, \mathbb{C}^4)}$$ but that is very far from implying that $$|\phi|=|\psi|$$.

To be a bit more specific (I think otherwise there's no hope of saying anything about $$|\phi|$$ and $$|\psi|$$, but correct me if I am wrong), the unitary transformation $$U$$ has the form of a Fourier multiplier i.e. $$U=\mathcal{F}^{-1}u\mathcal{F}$$ where $$\mathcal{F}$$ is the Fourier transform (acting component-wise) and $$u$$ is a unitary matrix (I am not if that makes it a unitary map from $$L^2(\mathbb{R}^3, \mathbb{C}^4)$$ to itself). Thus $$|\mathcal{F}\phi|=|\mathcal{F}\psi|$$ (and $$\|\mathcal{F}\phi\|_{L^2(\mathbb{R}^3, \mathbb{C}^4)}=\|\mathcal{F}\psi\|_{L^2(\mathbb{R}^3, \mathbb{C}^4)}$$ but that we already knew by Plancherel’s theorem). Again does not imply that $$|\phi|=|\psi|$$. I am bit confused.

Any help will be much appreciated. thanks, and stay safe!

I cannot claim I understand what is troubling you, but it may be confusion about the complete formal equivalence of wavefunctions to vectors, presumably covered in the Dirac bra-ket notation in the first month of mainstream QM courses. W.l.o.g., go from 3 to 1 dimensions in coordinate space, so $$\langle x|\psi\rangle = \psi(x) ~~~\longleftrightarrow ~~~ |\psi\rangle= \int\!\! dx~~\psi(x) ~|x\rangle,~~\leadsto \\ \langle \psi|\psi\rangle= \int\!\! dx~~\psi^*(x) \psi(x) = \rho,$$ where, of course, $$\langle x'|x\rangle = \delta(x'-x)$$ and $$\int\!\! dx~~ |x\rangle \langle x|=\mathbb{I}$$.
The action of a unitary operator on a vector, then, is equivalent to the action of its coordinate representation $$\langle x|\hat U|x'\rangle=U(x,x')$$, on the wavefunction,
$$|\phi\rangle= \hat U |\psi\rangle ~~~\longleftrightarrow ~~~\phi(x) = \int\!\! dx'~~U(x,x')\psi(x') .$$
Can you now see how $$\hat U ^\dagger \hat U= \mathbb{I} ~~~\longleftrightarrow ~~~ \int\!\! dx'~~U^*(x,x') ~ U(x',x'')=\delta(x-x'')$$ implies $$|\phi|=|\psi| ~~~\longleftrightarrow ~~~ \|\psi\|_{L^2(\mathbb{R} , \mathbb{C} )}=\|\phi\|_{L^2(\mathbb{R} , \mathbb{C} )}$$?
• I see, thank you. And, if I understand it correctly, it also assumes that $\psi$ has been promoted to an operator (i.e. after second quantisation). For a bit of context, I was trying to understand what would hold if $\psi$ was only assumed to be a certain solution to a PDE instead of formulating everything in QFT. Nov 8, 2021 at 19:29
• ? There is no mention of second quantization in the question or the answer! I am only considering states $|\psi\rangle$. Second quantization is a functor, and there should never be an ambiguity for anything, but you must then go on and ask a different question. Nov 8, 2021 at 19:32
• Oh yes, I apologise for this lapsus of mine. Still, I believe my question has been answered: provided one has a complete set of states, one can prove $|\psi|=|\phi|$ if and only $||\psi||_{L^2}=||\phi||_{L^2}$ by the argument in your answer. Nov 8, 2021 at 19:53