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So suppose we have the wave function $\phi(x,t)$ in the context of quantum mechanics, that satisfies the Schrodinger's equation. We want to see that if we normalize this function at $t=0$ then it will normalized for the rest of time so, i have been reading this for Griffiths book on quantum mechanics, he goes like this $\frac{d}{dt} \int_{-\infty}^{+\infty}\text dx|\phi(x,t)|^2$ and then he switches the derivative with the integral, now im not seeing why we could do this, by this i mean i tried to use the Lebesgue dominated convergence theorem but i didnt get anywhere,i dont know if this has to do with the fact that the function is a solution to the equation and that tell us something, so any help is aprecciated, sorry for the more mathematical question.

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    $\begingroup$ We often get questions about how physics textbooks don't seem mathematically rigorous. They are not. There are Hamiltonians where these "hand waving" arguments fail. See for example The Inverse Cube Force Law - John Baez. $\endgroup$ Mar 15 '20 at 5:03
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Standard treatments of quantum mechanics start by assuming the Schrodinger equation, but it is better to do it the other way round. There are much deeper mathematical arguments starting from the probability interpretation, which unfortunately are usually omitted from text books. Starting with the probability interpretation we naturally normalise the wave function. We then show that preserving the probability interpretation requires unitarity. Then, by way of Stone's theorem, we can prove the Schrodinger equation, showing that it is actually a theorem not a postulate. I have given the following derivation in my book The Mathematics of Gravity and Quanta. It applies to fundamental behaviours in relativistic quantum mechanics, rather than specific systems.

If at time $t_0$ the ket is $|f(t_0) \rangle$, then the ket at time $t$ is given by the evolution operator, $U(t,t_0):\mathbb{H} \rightarrow \mathbb{H}$ (this is not specific to one particle Hilbert space), such that $$|f(t) \rangle = U(t,t_0)|f(t_0) \rangle.$$ The evolution of kets is expected to be continuous because kets are not physical states of matter, but are probabilistic statements about what might happen in measurement given current information. Probabilities apply to our ideas concerning the likelihood of events (this is a fundamental principle of a modern Bayesian interpretation of probability theory). Whether or not reality is fundamentally discrete, probability is properly described on a mathematical continuum (in the sense that a continuum means that measurements of position are sufficiently precise that discreteness has no practical impact on predictions). A discrete interaction will not lead to a discrete change in probability because we do not have exact information on when the interaction takes place. This being so, $U$ is expected to be a continuous function of time.

If the ket at time $t_0$ was either $|f(t_0) \rangle$ or $|g(t_0) \rangle$, then it will evolve into either $|f(t) \rangle$ or $|g(t) \rangle$ at time $t$. In the absence of further information, any weighting in OR will be preserved (superposition is properly interpreted as OR in a many valued logic). So, $U$ is linear, $$U(t,t_0)[a|f(t_0)\rangle + b|f(t_0)\rangle]= aU(t,t_0)|f(t_0)\rangle + bU(t,t_0)|f(t_0)\rangle . $$ Since local laws of physics are always the same, and $U$ does not depend on the ket on which it acts, the form of the evolution operator for a time span $t$, $$U(t) = U(t+t_0,t_0)$$ does not depend on $t_0$. We require that the evolution in a span $t_1 + t_2$ is the same as the evolution in $t_1$ followed by the evolution in $t_2$, and is also equal to the evolution in $t_2$ followed by the evolution in $t_1$, $$ U(t_2)U(t_1) = U(t_2 + t_1) = U(t_1)U(t_2) .$$ Using negative $t$ reverses time evolution (put $t=t_1=-t_2$) $$U(-t) = U(t)^{-1} . $$ In zero time span, there is no evolution, $$U(0)=1 .$$

The result of the calculation of probability of a measurement result $g$ at time $t_2$ given an initial condition $f$ at time $t_1$ is not affected by the time at which it is calculated (parameter time for Hilbert space). Since kets can be chosen to be normalised we may require that $U$ conserves the norm, i.e., for all $|g\rangle$, $$\langle g |U^\dagger U |g\rangle = \langle g|g \rangle .$$ Applying this to $|g\rangle + |f\rangle$,$$ ( \langle g| + \langle f | )U^\dagger U (|g\rangle + |f\rangle)= ( \langle g| + \langle f | )(|g\rangle + |f\rangle) $$

By linearity of $U$, $$ ( \langle g|U^\dagger + \langle f |U^\dagger ) (U|g\rangle + U|f\rangle)= ( \langle g| + \langle f | )(|g\rangle + |f\rangle) .$$ By linearity of the inner product, we can multiply out the brackets. After cancelling equal terms $$ \langle g| U^\dagger U |f\rangle + \langle f| U^\dagger U |g\rangle = \langle g |f\rangle + \langle f| g \rangle . $$ Similarly, applying the argument to $|g\rangle + i|f\rangle> $ $$ \langle g| U^\dagger U |f\rangle - \langle f| U^\dagger U |g\rangle = \langle g |f\rangle - \langle f| g \rangle . $$

Adding shows that for all $|g\rangle, |f\rangle$ $$ \langle g| U^\dagger U |f\rangle = \langle g| f \rangle .$$ I.e. $U$ is unitary.

This establishes the conditions for Stone's theorem. In the book I give a proof of Stone's theorem and find the Schrodinger equation as a simple corollary.

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  • $\begingroup$ Alright thanks alot !! Yeah thats my problem with physics textbooks , cause im a mathematics student and sometimes they are not very rigorous and i just cant ignore these crucial steps in the proofs. $\endgroup$
    – Something
    Mar 15 '20 at 8:58
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    $\begingroup$ Me too. Actually I think all the conceptual problems in quantum mechanics, as well as the divergence problems in QED have been due to a lack of mathematical rigour in physics treatments. This is why I wrote the books. $\endgroup$ Mar 15 '20 at 9:11

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