$\newcommand{\ud}[1]{{#1^{\dagger}}} \newcommand{\bra}[1]{\left\langle #1\right|} \newcommand{\ket}[1]{\left| #1\right\rangle} \newcommand{\braket}[2]{\left\langle #1 | #2\right\rangle} \newcommand{\braketo}[3]{\left\langle #1 \right| #2 \left| #3\right\rangle} \newcommand{\ketbra}[2]{\left| #1\right\rangle\left\langle #2\right|} \newcommand{\angle}[1]{\left\langle #1\right\rangle}$
Toggle Menu

Time Evolution

Schödinger Equation

Postulate 6

The time evolution of a quantum system is determined by the Hamiltonian (total energy operator, $H(t)$) through the Schrödinger Equation: $$i\hbar\frac{d}{dt}\ket{\psi(t)}=H(t)\ket{\psi(t)}$$
The Hamiltonian is an observable, so it is hermitian. Its eigenvalues are the possible outcomes of a total energy measurement (allowed energies, $E_n$) and its eigenstates are the energy eigenstates, $\ket{E_n}$, which form an orthonormal basis. Thus, we have the energy eigenvalue equation as: $$H\ket{E_n}=E_n\ket{E_n}$$ We can write any quantum state in the energy basis: $$\ket{\psi(t)}=\sum_n c_n(t)\ket{E_n}$$ where $c_n(t)$ is the expansion coefficient, where the state's time dependence resides. If we substitute this into the Schrödinger equation and define the initial state as $\ket{\psi(0)}=\sum_n c_n\ket{E_n}$, we get the time evolution of the quantum state as: $$\ket{\psi(t)}=\sum_n c_n e^{-iE_nt/\hbar}\ket{E_n}$$

Stationary States

Consider a state that is initially in an energy eigenstate: $\ket{\psi(0)}=\ket{E_1}$. At a later time t, $\ket{\psi(t)}=e^{-iE_nt/\hbar}\ket{E_1}$. Since the overall phase does not affect any measurements, the state remains in its initial state. This is known as a stationary state.

The Schödinger Equation Recipe

Given a Hamiltonian $H$ and an initial state $\ket{\psi(0)}$, we can calculate the probabilty or expectation value of an outcome $a_i$ of an observable $A$ at time $t$ by applying the following steps:
  1. Diagonalize $H$ to find the eigenvalues $E_n$ and eigenvectors $\ket{E_n}$. Note that this recipe can only be performed with the state written in the energy basis
  2. Write $\ket{\psi(0)}$ in terms of the energy eigenstates: $\ket{\psi(0)}=\sum_n c_n(0)\ket{E_n}$
  3. Multiply each eigenstate coefficient by the time-dependent phase factor, $e^{-iE_nt/\hbar}$ to obtain $\ket{\psi(t)}$.
  4. Calculate the probability: $P_{a_i}=|\braket{a_i}{\psi(t)}|^2$, or the expectation value: $\angle{A}=\braketo{\psi(t)}{A}{\psi(t)}=\sum_i a_i P_{a_i}(t)$
Ex 1. Given $H\doteq\frac{\hbar\omega_0}{2}\left(\begin{matrix}1&0\\0&-1\end{matrix}\right)$, what is the probability that a state is found to be spin-up along the $z$-axis?
  1. Since $H$ is already diagonal, we have the following eigenvalues and eigenvectors: $$E_+=\frac{\hbar\omega_0}{2}\quad\ket{E_+}\doteq\left(\begin{matrix}1\\0\end{matrix}\right),\quad E_-=-\frac{\hbar\omega_0}{2}\quad\ket{E_-}\doteq\left(\begin{matrix}0\\-1\end{matrix}\right)$$ Since the overall phase does not matter and $e^{i\pi}\left(\begin{matrix}0\\-1\end{matrix}\right)=\left(\begin{matrix}0\\1\end{matrix}\right)$, we can let $\ket{E_-}\doteq\left(\begin{matrix}0\\1\end{matrix}\right)$.
  2. We can assume that the initial state is the general state $\ket{+}_n=\cos\frac{\theta}{2}\ket{+}+\sin\frac{\theta}{2}e^{i\phi}\ket{-}$. Thus, $$\ket{\psi}\doteq\cos\frac{\theta}{2}\left(\begin{matrix}1\\0\end{matrix}\right)+\sin\frac{\theta}{2}e^{i\phi}\left(\begin{matrix}0\\1\end{matrix}\right)=\left(\begin{matrix}\cos\frac{\theta}{2}\\\sin\frac{\theta}{2}e^{i\phi}\end{matrix}\right)$$
  3. Multiplying by the time-dependent phase factor, $$\ket{\psi(t)}\doteq\left(\begin{matrix}e^{-iE_+t/\hbar}\cos\frac{\theta}{2}\\e^{-iE_-t/\hbar}\sin\frac{\theta}{2}e^{i\phi}\end{matrix}\right)=\left(\begin{matrix}e^{-i\omega_0t/2}\cos\frac{\theta}{2}\\e^{i\omega_0t/2}\sin\frac{\theta}{2}e^{i\phi}\end{matrix}\right)$$ Since the global phas has no physical significance, we can multiply the state by $e^{i\omega_0t/2}$ to get $$\ket{\psi(t)}\doteq\left(\begin{matrix}\cos\frac{\theta}{2}\\e^{i(\omega_0t+\phi)}\sin\frac{\theta}{2}\end{matrix}\right)$$
  4. Finally, we calculate the probability. $$P_+=|\braket{+}{\psi(t)}|^2=|\cos\frac{\theta}{2}|^2=\cos^2\frac{\theta}{2}$$
Note that the state is time-dependent, while the observable $S_z$ is not. The probability is also time-independent becuase the $S_z$ eigenstates are also energy eigenstates for the problem => $H$ and $S_z$ commute.

Time-dependent Hamiltonian

If we are given a time-dependent Hamiltonian, we must find a way to remove its time-dependence to apply it to the recipe. We do this by substituting the state into the Schrödinger equation.
Ex 1. Given $H\doteq\frac{\hbar}{2}\left(\begin{matrix}\omega_0&\omega_1e^{-\omega t}\\\omega_1e^{\omega t}&-\omega_0\end{matrix}\right)$, what is the probability that a state is found to be spin-up along the $z$-axis?
Since we aren't using the recipe, we are free to choose our own basis. For convenience, we choose the basis that $H$ is written in. The state vector becomes $\ket{\psi(t)}=c_+(t)\ket{+}+c_-(t)\ket{-}$.
Substituting into the Schrödinger equation, $$i\hbar\frac{d}{dt}\left(\begin{matrix}c_+(t)\\c_-(t)\end{matrix}\right)=\frac{\hbar}{2}\left(\begin{matrix}\omega_0&\omega_1e^{-\omega t}\\\omega_1e^{\omega t}&-\omega_0\end{matrix}\right) \left(\begin{matrix}c_+(t)\\c_-(t)\end{matrix}\right)$$ This gives us the following differential equations: $$i\hbar\dot{c_+}=\frac{\hbar\omega_0}{2}c_+(t)+\frac{\hbar\omega_1}{2}e^{-i\omega t}c_-(t)\\ i\hbar\dot{c_-}=\frac{\hbar\omega_1}{2}e^{i\omega t}c_+(t)-\frac{\hbar\omega_0}{2}c_-(t)$$ To solve these equations, we consider the problem from the rotating frame. The state vector after a frame transformation becomes $$\ket{\tilde{\psi}(t)}=c_+(t)e^{i\omega t/2}\ket{+}+c_-(t)e^{-i\omega t/2}\ket{-}$$ If we let $c_+(t)=e^{-i\omega t/2}a_+(t)$ and $c_-(t)=e^{i\omega t/2}a_-(t)$, we can rewrite the state vector in the non-rotating frame as $$\ket{\psi(t)}=a_+(t)e^{-i\omega t/2}\ket{+}+a_-(t)e^{i\omega t/2}\ket{-}$$ Substituting this into the differential equations, we get $$i\hbar\dot{a_+}=-\frac{\hbar\Delta\omega}{2}a_+(t)+\frac{\hbar\omega_1}{2}a_-(t)\\ i\hbar\dot{a_-}=\frac{\hbar\omega_1}{2}a_+(t)+\frac{\hbar\Delta\omega}{2}a_-(t)$$ where $\Delta\omega=\omega-\omega_0$. These differential equations can be thought of as a transformed Schrödinger equation: $$i\hbar\frac{d}{dt}\ket{\tilde{\psi}(t)}=\tilde{H(t)}\ket{\tilde{\psi}(t)}\text{, }\tilde{H}\doteq\frac{\hbar}{2}\left(\begin{matrix}-\Delta\omega&\omega_1\\\omega_1&\Delta\omega\end{matrix}\right)$$ Now that the Hamiltonian is time-independent, we can solve it using the recipe and use the transformation equations to find the solution to the original problem.