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 \begin{equation} \rho:=|\psi|^2=\sum_{k=1}^{4}|\psi_{k}|^2 \end{equation} 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 \begin{equation} |\phi|^2=\langle \phi, \phi \rangle=\langle U\psi, U\psi\rangle \end{equation} 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!


1 Answer 1


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 $$|\phi|=|\psi| ~~~\longleftrightarrow ~~~ \|\psi\|_{L^2(\mathbb{R} , \mathbb{C} )}=\|\phi\|_{L^2(\mathbb{R} , \mathbb{C} )} $$?

  • $\begingroup$ Thank you very much for your answer. Correct me if I am wrong, but this only holds assuming that one has a complete set of states for the theory. I am guessing nothing can be said without such an assumption? $\endgroup$ Nov 8, 2021 at 19:15
  • $\begingroup$ Yes, this is a cornerstone of the formulation. $\endgroup$ Nov 8, 2021 at 19:24
  • $\begingroup$ 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. $\endgroup$ Nov 8, 2021 at 19:29
  • $\begingroup$ ? 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. $\endgroup$ Nov 8, 2021 at 19:32
  • $\begingroup$ 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. $\endgroup$ Nov 8, 2021 at 19:53

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