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In this question, the answer says that unitary time evolution means that probability is conserved. Is this the same as saying that a system obeys the Schrödinger equation?

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    $\begingroup$ One could imagine a system which was unitary, but obeyed something else besides the Schrodinger equation. $\endgroup$
    – Jbag1212
    Aug 9, 2023 at 12:34
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    $\begingroup$ The content of the Schrodinger equation is that there exists a Hermitian operator $H$ on the vector space of states such that $-iH$ (setting $\hbar = 1$) is the generator of time evolution. This is in principle slightly less general than postulating a unitary time evolution operator $U(t_0, t)$, but actually entails no loss of generality in almost all cases of interest (we usually have $U(t_0, t) = e^{-iH(t-t_0)}$). In fact, I don't know of a situation we care about where there is a loss of generality. Perhaps another user can provide an example. $\endgroup$ Aug 10, 2023 at 0:45

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As other answers have pointed out, assuming unitary time evolution does get you part of the way there — but not all the way. In particular, you could call it a postulate that the Hermitian operator which governs time evolution is the "same" Hamiltonian (after the canonical substitution) that is used in classical mechanics. See Sakurai, Modern Quantum Mechanics, chapter 1 and chapter 2, section 2.1.

Assumptions of unitary time-evolution are enough to guarantee that $$i \hbar \frac{d}{dt} |\Psi(t) \rangle = \hat{H} | \Psi(t)\rangle$$ for some Hermitian operator $\hat{H}.$ If you want to stop here and call this the Schrödinger equation, then you are done. Generally though, the Schrödinger equation requires that $\hat{H}$ must be the Hamiltonian of some physical system. The Hamiltonian can be defined as corresponding to the total energy of the system — or, in the case of time-varying potentials, the Legendre transformation of the Lagrangian of the system.

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Unitary evolution means evolution is given by $\psi(t) = U(t,t_0)\psi(t_0)$, where $U(t,t_0)$ is a unitary operator with $U(t_0,t_0) = \mathbb{1}$. It happens that this is enough to guarantee that $$i \hbar \frac{\mathrm{d} \psi}{\mathrm{d} t} = H \psi$$ for some Hermitian operator $H$, and vice-versa.

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    $\begingroup$ So the answer is ‘yes’? $\endgroup$
    – Riemann
    Aug 9, 2023 at 13:03
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    $\begingroup$ See also Stone's theorem. You need some more mathematical conditions. $\endgroup$ Aug 9, 2023 at 13:10
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    $\begingroup$ This does not show that the $H$ here is the same $H$ used in the Schrodinger equation. $\endgroup$
    – Jbag1212
    Aug 9, 2023 at 14:16
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    $\begingroup$ @Riemann it depends on whether you call this the "Schrödinger equation" or not. This argument holds for all quantum systems with unitary evolution (up to the mathematical details mentioned by Tobias), but some people only give the name "Schrödinger equation" for the motion of a non-relativistic particle $\endgroup$ Aug 9, 2023 at 16:12
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    $\begingroup$ If you want $H = - \frac{\hbar^2}{2m}\nabla^2 + V$, then unitarity is not enough. This Hamiltonian leads to unitary evolution, but it is not the only Hamiltonian that does so. $\endgroup$ Aug 9, 2023 at 16:13
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The answer of course is yes as Níckolas has pointed out, but I want to give a brief and popular explanation for why this is the case.
So let's say we have a unitary time evolution operator $U$, which acts on the state vector like so $$U(t)\psi(t')=\psi(t'+t)$$ This means of course that $U(0)$ is the identity on the state space. Now assume this operator is smooth as a function of time; this allows us to write $$U(\varepsilon)\approx\mathbb{I}+\varepsilon dU$$ for some operator $dU$ evaluated at time $0$. Seeing as $U$ is unitary, and also, up to first order, $U^\dagger=\mathbb{I}+\varepsilon dU^\dagger$ we have \begin{equation}(\mathbb{I}+\varepsilon dU)(\mathbb{I}+\varepsilon dU^\dagger)\approx \mathbb{I} \iff dU=-dU^\dagger \end{equation} where we only kept terms up to first order. So what we learned is that if the time evolution is unitary, its derivative is anti-Hermitian; this sucks because our observables have to be Hermitian operators. But there's a simple way to construct a Hermitian operator from an anti-Hermitian one: just set $H=i\, dU$. We can clearly see then that $H^\dagger=-i \,dU^\dagger=i\, dU=H$, so $H$ is indeed a Hermitian operator.
Now the Schrodinger equation falls out of these considerations, seeing as $$(U(\varepsilon)-U(0)\psi(t)=-i\, \varepsilon H \psi(t) \iff i\frac{\partial}{\partial t}\psi(t)= H\psi(t)$$ dividing by epsilon and taking the limit. Of course this is all in natural units.
So, to conclude, the answer is yes; but let me note that requiring unitarity is not necessary to arrive at the general form of Schrodinger's equation. If $U$ had not been unitary, you could still have written $$\frac{\partial}{\partial t}\psi(t)= dU\psi(t)$$ The Schrodinger eq is not special in that regard. The regard in which it is special is, although, that of course had $U$ not been unitary the expression above would have been meaningless, for $dU$ might not be Hermitian in general, and the time evolution of a state would not have been an observable. So the crux of the question, it seems to me, of "does unitarity imply Schrodinger's eq" is not that yes it indeed does, but that unitarity ensures that time evolution is an observable, giving meaning to the question of "how does a state change in time".

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    $\begingroup$ Yes, but this answer does not show that the $H$ you have must be the Hamiltonian as it is normally defined. $\endgroup$
    – Jbag1212
    Aug 9, 2023 at 14:15
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    $\begingroup$ @Jbag1212 The Hamiltonian here described is general, it might be the identity, it might be proportional to Pauli matrices, and it might be of course the Hamiltonian for a massive particle that you usually are first introduced to. The identification of $H$ with energy, and that particular energy which I think you might be referring to, has of course to do with it being the generator of time translations, and so being the natural generalization of the classical Hamiltonian which was related to the mechanical energy, but that was out of the scope of the question. $\endgroup$ Aug 9, 2023 at 14:23
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    $\begingroup$ The scope of the question is about the Schrodinger equation. The Schrodinger equation very much uses the Hamiltonian. $\endgroup$
    – Jbag1212
    Aug 9, 2023 at 14:26
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    $\begingroup$ @Jbag1212 but what I call the Schrodinger equation is not $\partial _t \psi=(\partial_x ^2 +V)\psi$. OP wanted to know if unitarity yielded the SE, so I showed how (what I call the SE is what I wrote about). As you pointed out in your answer and your comment, of course unitarity doesn't choose the Hamiltonian for you; that must be done by you depending on the system. $\endgroup$ Aug 9, 2023 at 14:32
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    $\begingroup$ Point taken, but I still think that to be called a Schrodinger equation you need the $H$ to be a Hamiltonian, and you can't just call any Hermitian operator a Hamiltonian. That is, the definition of a Hamiltonian corresponds to the total energy of the system - or, it must correspond to a Legendre transformation of the Lagrangian. $\endgroup$
    – Jbag1212
    Aug 9, 2023 at 14:52

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