Schrödinger equation, Hamiltonian I

Recall from Particle Energy that

∣ψ(0)⟩=∑kck∣Ek⟩\begin{gather*} \left\lvert \psi(0) \right\rangle=\sum_kc_k\left\lvert E_k \right\rangle \end{gather*}

Because we learned this thing called Particle Phase, we can say that particle. oscillate at different phases. This is shown using phases

∣ψ(t)⟩=∑kcke−iωkt∣Ek⟩=∑kcke−iEkt/ℏ∣Ek⟩\begin{gather*} \left\lvert \psi(t) \right\rangle=\sum_kc_ke^{-i\omega_kt}\left\lvert E_k \right\rangle\\ =\sum_kc_ke^{-iE_kt/\hbar}\left\lvert E_k \right\rangle\\ \end{gather*}

If we take the derivative from both sides

ddt∣ψ(t)⟩=ddt[∑kcke−iEkt/ℏ∣Ek⟩]=∑kck⋅(−iEkℏ)e−iEkt/ℏ∣Ek⟩=−iℏ∑kck⋅Eke−iEkt/ℏ∣Ek⟩\begin{gather*} \frac{d}{dt}\left\lvert \psi(t) \right\rangle=\frac{d}{dt}\left[\sum_kc_ke^{-iE_kt/\hbar}\left\lvert E_k \right\rangle\right]\\ =\sum_kc_k\cdot \left(-i\frac{E_k}{\hbar}\right)e^{-iE_kt/\hbar}\left\lvert E_k \right\rangle\\ =-\frac{i}{\hbar}\sum_kc_k\cdot E_ke^{-iE_kt/\hbar}\left\lvert E_k \right\rangle \end{gather*}

Let the Hamiltonian HH be an operator where this is true

H∣Ek⟩=Ek∣Ek⟩\begin{gather*} H\left\lvert E_k \right\rangle=E_k\left\lvert E_k \right\rangle \end{gather*}

so

=−Hiℏ∑kck⋅e−iEkt/ℏ∣Ek⟩=−iℏH∣ψ(t)⟩\begin{gather*} =-H\frac{i}{\hbar}\sum_kc_k\cdot e^{-iE_kt/\hbar}\left\lvert E_k \right\rangle\\ =-\frac{i}{\hbar}H\left\lvert \psi(t) \right\rangle \end{gather*}

which gives us Schrödinger’s equation (in state-vector form)

iℏddt∣ψ(t)⟩=H∣ψ(t)⟩\begin{gather*} i\hbar\frac{d}{dt}\left\lvert \psi(t) \right\rangle=H\left\lvert \psi(t) \right\rangle \end{gather*}