Lecture 6: Time Evolution and the Schrödinger Equation

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Recommends Alasdair · 2014-07 · Physics Stack Exchange · on “Solving the Schrödinger equation where the initial wave function is an energy eigenfunction”

I was watching Allan Adams' lecture on energy eigenfunctions [this video] , and there's one part (around 43 minutes into the lecture) that confuses me. Suppose we have the initial wave function $\Psi (x,0)$ such that $\hat{E}\,\Psi (x,0)=E \,\Psi (x,0)$ for some constant $E$. Then, plugging this into the Schrödinger equation, we'd get: \begin{align} i \hbar \frac{\partial}{\partial t} \Psi (x, 0) &= E \, \Psi (x,0) \\ \frac{\partial}{\partial t} \Psi (x, 0) &= \frac{E}{i \hbar} \, \Psi (x,0) \tag{1}\\ \therefore \Psi (x, t) &= \exp\left({-i \frac{E\,t}{\hbar}}\right) \Psi(x,0) \tag{2} \end{align} I'm a bit confused about how to go from $(1)$ to $(2)$. Now if we make the additional assumption that

Recommends Matrix23 · 2021-05 · Physics Stack Exchange · on “Solving the Schroedinger equation with the initial condition as an energy eigenstate”

I was studying quantum mechanics by watching a video lecture series. In the lecture [this video] , the professor tries to solve the Schroedinger equation with the initial condition of being in an energy eigen state. $$\hat{E} \psi(x,0) = E \psi(x,0) \ \ \ \ \ \ \ eqn (1)$$ $$i\hbar\partial_t\psi(x,0) = \hat{E} \psi(x,0) = E \psi(x,0)\ \ \ \ \ \ \ eqn (2)$$ $$\partial_t\psi = \frac{-iE}{\hbar}\psi\ \ \ \ \ \ \ eqn (3)$$ $$\psi(x,t) = e^{-i\frac{Et}{\hbar}}\psi(x,0)\ \ \ \ \ \ \ eqn (4)$$ The above were the series of equations the professor wrote on the blackboard. I don't understand how you get from equation (2) to equation (3). Eqn (2) is valid only at $t=0$ . How do you then get equation (3) whi

Cites Our · 2018-11 · Physics Stack Exchange · on “Does any basis belong to a set of eigenfunctions of some observable?”

In the Quantum Mechanics courses in ocw.mit.edu [this video] it is said that any wave function can be expanded as a superposition of the eigenfunctions of any observable, i.e the eigenfunctions form a eigenbasis for our space. However, if we take an arbitrary basis for our space, is that space an eigenbasis corresponding to an observable? In other words, is any basis a set of eigenfunctions of some observable?