Is unitary time evolution the same as obeying the Schrödinger equation?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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".