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--- the_schroedinger_equation [2021/03/16 05:06] – [3.iii.2 Argument 2: Conservation of Probability] admin
+++ the_schroedinger_equation [2021/03/17 21:25] (current) – [3.iii.3 The Continuity Equation] admin
@@ Line 60: / Line 60: @@
 The Born rule says that the probability of getting the outcome $a_n$ when measuring $\hat{A}$ at time $t_0$ is
 \[p(a_n,t_0) = \QProb{a_n}{\psi(t_0)} = \braket{\psi(t_0)}{a_n}\braket{a_n}{\psi(t_0)}.\]
-This implies, the probability of obtaining any outcome at all, which should be $=1$, is
+This implies, the probability of getting //some// outcome at $t_0$, which should be $=1$, is
 \[\sum_{n}p(a_n,t_0) = \sum_n \braket{\psi(t_0)}{a_n}\braket{a_n}{\psi(t_0)} = \braket{\psi(t_0)}{\psi(t_0)} = 1,\]
 where we have used the Dirac Notaty to remove the identity $\hat{I} = \sum_n \proj{a_n}$.
@@ Line 66: / Line 66: @@
 Now consider the situation at time $t$.  The probability of getting the outcome $a_n$ when $\hat{A}$ is measured at time $t$ is
 \begin{align*}
-p(a_n,t) & = \Qprob{a_n}{\psi(t)}) \\
+p(a_n,t) & = \QProb{a_n}{\psi(t)}) \\
-& \Abs{\sand{a_n}{\hat{U}(t,t_0)}{\psi(t_0)}}^2 \\
+& = \Abs{\sand{a_n}{\hat{U}(t,t_0)}{\psi(t_0)}}^2 \\
-& = \sand{\psi(t_0)}{\hat{U}(t,t_0)}{a_n}\sand{a_n}{\hat{U}(t,t_0)}{\psi(t_0)},
+& = \sand{\psi(t_0)}{\hat{U}(t,t_0)^{\dagger}}{a_n}\sand{a_n}{\hat{U}(t,t_0)}{\psi(t_0)},
 \end{align*}
 so, applying the Dirac Notaty again, the probability of obtaining any outcome at all is
-\[sum_n p(a_n,t) = \sand{\psi(t_0)}{\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0)}{\psi(t_0)}.\]
+\[\sum_n p(a_n,t) = \sand{\psi(t_0)}{\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0)}{\psi(t_0)}.\]
-We want this to equal $1$ as well, so, in particular, comparing with $\sum_{n}p(a_n,t_0)$ gives
+We want this to equal $1$ as well, so, in particular, we should have
-\[\sand{\psi(t_0)}{\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0)}{\psi(t_0)} = braket{\psi(t_0)}{\psi(t_0)},\]
+\[\sand{\psi(t_0)}{\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0)}{\psi(t_0)} = \braket{\psi(t_0)}{\psi(t_0)},\]
 which we can also write as
-\[\sand{\psi(t_0)}{\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0)}{\psi(t_0)} = sand{\psi(t_0)}{\hat{I}}{\psi(t_0)}.\]
+\[\sand{\psi(t_0)}{\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0)}{\psi(t_0)} = \sand{\psi(t_0)}{\hat{I}}{\psi(t_0)}.\]
-Now, if $\ket{\psi(t_0)}$ can be any vector in the Hilbert space then we have that
+Now, suppose that the initial state $\ket{\psi(t_0)}$ can be any vector in the Hilbert space. Then, this implies that
-\[\sand{\psi}{\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0)}{\psi(t_0)} = sand{\psi}{\hat{I}}{\psi},\]
+\[\sand{\psi}{\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0)}{\psi} = \sand{\psi}{\hat{I}}{\psi},\]
 for all vectors $\ket{\psi}$.  In an in class activity you will show that this implies that $\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0) = \hat{I}$, i.e. the operator $\hat{U}(t,t_0)$ must be unitary.
 Now, we also want that $\hat{U}(t_0,t_0) = \hat{I}$, i.e. evolving for no time does nothing, and we also want the time evolution to be continuous, so that the state stays close to the initial state if the period of time evolution is small $\Delta t = t - t_0 \ll 1$. Therefore, we will assume that, to first order in $\Delta t$,
-\[\hat{U}(t,t_0) = \hat{I} + \delta t \hat{A}.\]
+\[\hat{U}(t,t_0) = \hat{I} + \Delta t \hat{A}.\]
 In an in class activity, you will show that the unitarity of $\hat{U}(t,t_0)$ implies that $\hat{A}^{\dagger} = -\hat{A}$, i.e. $\hat{A}$ is anti-Hermitian.  Therefore, we have
@@ Line 97: / Line 97: @@
 \[i\hbar\frac{\partial \ket{\psi}}{\partial t} = \hat{H}\ket{\psi}.\]
-This shows that the Schrödinger equation must be correct for //some// Hermitian operator $\hat{H}$, but not that $\hat{H}$ must be the Hamiltonian (the operator representing the energy).  More sophisticated arguments can show that the generator of time translations, i.e. $\hat{H}$ must be the Hamiltonain, and you will learn about this if you go to grad school for physics.
+This shows that the Schrödinger equation must be correct for //some// Hermitian operator $\hat{H}$, but not that $\hat{H}$ must be the Hamiltonian (the operator representing the energy).
 ====== 3.iii.3 The Continuity Equation ======
 The previous argument shows that the total probability is conserved in time by Schrödinger evolution, i.e.
-\[\braket{\psi(t)}{\psi(t)} = \braket{\psi(0)}{\psi(0)},\qquad\text{so }\frac{\D}{\D t}\left ( \braket{\psi}{\psi}\right ) = 0.\]
+\[\braket{\psi(t)}{\psi(t)} = \braket{\psi(t_0)}{\psi(t_0)},\qquad\text{so }\frac{\D}{\D t}\left ( \braket{\psi}{\psi}\right ) = 0.\]
 In fact, we can derive a more detailed result describing the //local// conservation of probability.  This is normally done for the probability density of a position measurement, which is what we will do here, but note that you could derive a similar equation for the conservation of probability for any observable.
-The basic idea should be familiar if you have studied continuity equations in fluid dynamics or electromagnetism.  In a fluid, consider the mass of fluid contained in a small volume.  The increase in mass in that region over short period of time is the net current density flux into the region over that time.  Similarly, in electromagnetism the increase in electric charge in a small region is the net electric current flux into that region over that time.  The general form of a continuity equation is
+The basic idea should be familiar if you have studied continuity equations in fluid dynamics or electromagnetism.  In a fluid, consider the number of particles contained in a small volume.  The increase in particle number in that region over short period of time is the net particle flux into the region over that time.  Similarly, in electromagnetism the increase in electric charge in a small region is the net charge flux into that region over that time.  The general form of a continuity equation is
 \[\frac{\partial \rho(\vec{r},t)}{\partial t} + \vec{\nabla}\cdot \vec{J}(\vec{r},t),\]
 where $\rho(\vec{r},t)$ is the //density// and $\vec{J}(\vec{r},t)$ is the //current//.
@@ Line 112: / Line 112: @@
 In quantum mechanics, we are going to derive an equation like this for the //probability// density $\rho(\vec{r},t) = \Abs{\psi(\vec{r},t)}^2$.  This means that $\vec{J}(\vec{r},t)$ will be a //**probability current**//.
-It is worth stopping to consider the meaning of this before we proceed.  We normally think about continuity equations being about the flow of density of actual particles.  For example, a fluid is made out of a large number of small particles.  The density describes the amount of particle mass within a small volume and the current describes the rate of flow of those particles.  In quantum mechanics, we are dealing with the wavefunction of a //single particle//.  It is the probability density of this particle being found in a small volume that we are describing in the continuity equation and $\vec{J}(\vec{r},t)$ represents the rate of flow of this probability density.  No physical particles are necessarily flowing into or out of the region, just the likelihood of finding the particle there if we make a measurement.
+It is worth stopping to consider the meaning of this before we proceed.  We normally think about continuity equations being about the flow of actual particles.  For example, a fluid is made out of a large number of small particles.  The density describes the amount of particle mass within a small volume and the current describes the rate of flow of those particles.  In quantum mechanics, we are dealing with the wavefunction of a //single particle//.  It is the probability density of this particle being found in a small volume that we are describing in the continuity equation and $\vec{J}(\vec{r},t)$ represents the rate of flow of this probability density.  No physical particles are necessarily flowing into or out of the region, just the likelihood of finding the particle there if we make a measurement.
 Having understood this slightly weird concept, we know that $\rho(\vec{r},t) = \Abs{\psi(\vec{r},t)}^2$, so our aim is to use the Schrödinger equation to find a probability current $\vec{J}(\vec{r},t)$ that satisfies the continuity equation.
+To do this, we start with the Schrödinger equation written out in the position representation.
+\[i\hbar \frac{\partial \psi(\vec{r},t)}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2 \psi(\vec{r},t) + V(\vec{r},t)\psi(\vec{r},t).\]
+We now multiply this equation by the complex conjugate of the wavefunction $\psi^*(\vec{r},t)$ to obtain
+\begin{equation}
+\label{eq1}
+i\hbar \psi^*(\vec{r},t)\frac{\partial \psi(\vec{r},t)}{\partial t} = -\frac{\hbar^2}{2m}\psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t) + V(\vec{r},t)\psi^*(\vec{r},t)\psi(\vec{r},t).
+\end{equation}
+Next, we take the complex conjugate of the Schrödinger equation, which is
+\[-i\hbar \frac{\partial \psi^*(\vec{r},t)}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2 \psi^*(\vec{r},t) + V(\vec{r},t)\psi^*(\vec{r},t),\]
+and we multiply this equation by the wavefunction $\psi(\vec{r},t)$ to obtain
+\begin{equation}
+\label{eq2}
+-i\hbar \frac{\partial \psi^*(\vec{r},t)}{\partial t}\psi(\vec{r},t) = -\frac{\hbar^2}{2m}\psi(\vec{r},t)\nabla^2 \psi^*(\vec{r},t) + V(\vec{r},t)\psi^*(\vec{r},t)\psi(\vec{r},t).
+\end{equation}
+The next step is to subtract equation \eqref{eq2} from equation \eqref{eq1}, which gives
+\[i\hbar \left [ \psi^*(\vec{r},t)\frac{\partial \psi(\vec{r},t)}{\partial t} + \frac{\partial \psi^*(\vec{r},t)}{\partial t}\psi(\vec{r},t) \right ] = -\frac{\hbar^2}{2m} \left [ \psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t) - \psi(\vec{r},t)\nabla^2 \psi^*(\vec{r},t) \right ].\]
+Before doing anything else, let's just divide both sides of this equation by $i\hbar$, which gives
+\begin{equation}
+\label{eq3}
+\psi^*(\vec{r},t)\frac{\partial \psi(\vec{r},t)}{\partial t} + \frac{\partial \psi^*(\vec{r},t)}{\partial t}\psi(\vec{r},t) = \frac{i\hbar}{2m} \left [ \psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t) - \psi(\vec{r},t)\nabla^2 \psi^*(\vec{r},t) \right ].
+\end{equation}
+Using the product rule for derivatives, we have
+\[\frac{\partial \rho(\vec{r},t)}{\partial t} = \frac{\partial \psi^*(\vec{r},t)\psi(\vec{r},t)}{\partial t} = psi^*(\vec{r},t)\frac{\partial \psi(\vec{r},t)}{\partial t} + frac{\partial \psi^*(\vec{r},t)}{\partial t}\psi(\vec{r},t),\]
+so the left hand side of equation (\ref{eq3}) is just $\frac{\partial \rho(\vec{r},t)}{\partial t}$.
+Now, we define the //**probability current density**// to be
+\[\boxed{\vec{J}(\vec{r},t) = \frac{i\hbar}{2m} \left [ \psi(\vec{r},t)\vec{\nabla}\psi^*(\vec{r},t) - \psi^*(\vec{r},t)\vec{\nabla}\psi(\vec{r},t)\right ]}.\]
+Again, by the product rule for derivatives, we have
+\begin{align*}
+\vec{\nabla} \cdot \vec{J}(\vec{r},t) & = \frac{i\hbar}{2m} \left [ \psi(\vec{r},t) \nabla^2 \psi^*(\vec{r},t) + \vec{\nabla}\psi(\vec{r},t) \cdot \vec{\nabla} \psi^*(\vec{r},t) - \vec{\nabla}\psi^*(\vec{r},t) \cdot \vec{\nabla} \psi(\vec{r},t) - \psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t)\right ] \\
+& = \frac{i\hbar}{2m} \left [ \psi(\vec{r},t) \nabla^2 \psi^*(\vec{r},t) +  - \psi^*(\vec{r},t)\nabla^2 \psi(\vec{r},t)\right ].
+\end{align*}
+In other words, the right hand side of equation \eqref{eq3} is $-\vec{\nabla}\cdot \vec{J}(\vec{r},t)$.  Therefore, equation \eqref{eq3} can be rewritten as a continuity equation
+\[\frac{\partial \rho(\vec{r},t)}{\partial t} + \vec{\nabla}\cdot \vec{J}(\vec{r},t),\]
+for the probability density $\rho(\vec{r},t) = \Abs{\psi(\vec{r},t)}^2$ with the probability current density
+\[\vec{J}(\vec{r},t) = \frac{i\hbar}{2m} \left [ \psi(\vec{r},t)\vec{\nabla}\psi^*(\vec{r},t) - \psi^*(\vec{r},t)\vec{\nabla}\psi(\vec{r},t)\right ].\]
+If you think about it, what we have just done is pretty weird.  Usually, when we derive a continuity equation we assume that there is some density of particles $\rho(\vec{r},t)$ and some flow of particles described by a current density $\vec{J}(\vec{r},t)$.  We then //impose// the continuity equation to ensure that particles do not jump discretely from one location to another, but rather, if they go from point $A$ to point $B$ then they have to do so by traversing a continuous path consisting of points between $A$ and $B$.  That is really all that a continuity equation says.
+Instead, what we have done here is start by assuming that there is some continuity equation for $\rho(\vec{r},t)$ and defined a current $\vec{J}(\vec{r},t)$ that satisfies it.  But, in quantum mechanics, is it really true that probability "flows" in space according to the current density $\vec{J}(\vec{r},t)$ as we have defined it?  You cannot observe this flow of probability in quantum mechanics, but only measure the position of the particle at various times.  There are in fact, other possible current densities that would satisfy the same continuity equation.  For example, if we define
+\[\vec{J}'(\vec{r},t) = \vec{J}(\vec{r},t) + \vec{F}(\vec{r},t),\]
+where $\vec{F}(\vec{r},t)$ is any divergence-free vector field, i.e. $\vec{\nabla}\cdot \vec{F}(\vec{r},t)$, then $\vec{J}'(\vec{r},t)$ would still satisfy the continuity equation.
+Nonetheless, $\vec{J}(\vec{r},t)$ as we have defined it is the simplest form for a probability current density that satisfies the continuity equation, and it is a useful picture to imagine that probability density flows according to $\vec{J}(\vec{r},t)$, even if we cannot confirm this directly.
+====== 3.iii.4 Formal Solution of the Schrödinger Equation ======
+We saw in section 3.iii.2 that the time-evolution of a quantum system from time $t_0$ to time $t > t_0$ is described by a unitary operator $\hat{U}(t,t_0)$, i.e.
+\[\ket{\psi(t)} = \hat{U}(t,t_0)\ket{\psi(t_0)},\]
+and such that $\hat{U}(t_0,t_0) = \hat{I}$.
+Using the Schrödinger equation, we can find an equation for $\hat{U}(t,t_0)$ as follows.  The Schrödinger equation is
+\[i\hbar \partial{\ket{\psi(t)}}{\partial t} = \hat{H} \ket{\psi(t)},\]
+but this is equivalent to
+\[i\hbar \partial{\hat{U}(t,t_0)\ket{\psi(t_0)}}{\partial t} = \hat{H} \hat{U}(t,t_0)\ket{\psi(t_0)}.\]
+Now, $\ket{\psi(t_0)}$ does not depend on $t$ and this equation has to be true for all possible initial states $\ket{\psi(t_0)}$, so we have
+\[i\hbar \partial{\hat{U}(t,t_0)}{\partial t} = \hat{H} \hat{U}(t,t_0).\]
+Dividing both sides by $i\hbar$ gives
+\begin{equation}
+\label{eq4}
+\partial{\hat{U}(t,t_0)}{\partial t} = -\frac{i}{\hbar}\hat{H} \hat{U}(t,t_0).
+\end{equation}
+The unitary operator $\hat{U}(t,t_0)$ must always satisfy this equation, but it has a particularly simple solution when the Hamiltonian $\hat{H}$
+is independent of time, i.e. since the kinetic energy term is independent of time, this happens when the potential $V(\vec{r},t)$ does not depend on time.  In this case, the solution is
+\[\boxed{\hat{U}(t,t_0) = e^{-i(t-t_0)\hat{H}/\hbar}}.\]
+This is known as the //**propagator**// in quantum mechanics and it represents the finte-time dynamics of the system.  Since it only depends on $t-t_0$, we will often denote it $\hat{U}(t-t_0)$ rather than $\hat{U}(t,t_0)$
+To understand why $\hat{U}(t,t_0) = e^{-i(t-t_0)\hat{H}/\hbar}$ is the solution to equation \eqref{eq4}, recall that a function of an operator is defined by its power series so, in particular, if
+\[\hat{B}(t) = e^{\alpha\hat{A}t},\]
+where $\alpha$ is a constant and $\hat{A}$ is independent of time, then
+\[\hat{B}(t) = \sum_{n=0}^{\infty} \frac{\alpha^n \hat{A}^n}{n!}t^n,\]
+and hence
+\begin{align*}
+\frac{\partial \hat{B}}\partial{t} & = \sum_{n=0}^{\infty} \frac{n\alpha^n \hat{A}^n}{n!}t^{n-1}.
+\end{align*}
+Due to the factor of $n$, the $n=0$ term is zero, so we have
+\begin{align*}
+\frac{\partial \hat{B}}\partial{t} & = \sum_{n=1}^{\infty} \frac{n\alpha^n \hat{A}^n}{n!}t^{n-1} \\
+& = \sum_{n=1}^{\infty} \frac{\alpha^n \hat{A}^n}{(n-1)!}t^{n-1} \\
+& = \alpha \hat{A}^n \left ( \sum_{n=1}^{\infty} \frac{\alpha^{n-1} \hat{A}^{n-1}}{(n-1)!}t^{n-1}\right ) \\
+& = \alpha \hat{A} \left ( \sum_{n=0}^{\infty} \frac{\alpha^{n} \hat{A}^{n}}{n!}t^{n}\right ) \\
+& = \alpha \hat{A} e^{a\hat{A}t}.
+\end{align*}
+Using this result, it is easily checked that $\hat{U}(t-t_0) = \hat{C}e^{-i(t-t_0)\hat{H}/\hbar}$ is a solution to equation \eqref{eq4}, where $\hat{C}$ is an arbitrary time-independent operator and, in fact, this must be the general solution because it is a first order differential equation and the solution has one arbitrary constant.  However, we also have the boundary condition $\hat{U}(t_0-t_0) = \hat{U}(0) = \hat{I}$ and this implies that
+\[\hat{I} = \hat{C}e^{0} = \hat{C}\hat{I} = \hat{C},\]
+so we must have $\hat{C}=\hat{I}$ and so $\hat{U}(t-t_0) = e^{-i(t-t_0)\hat{H}/\hbar}$ is the only solution satisfying $\hat{U}(0) = \hat{I}$.
+====== 3.iii.5 From the Formal Solution to Useful Solutions ======
+At this point, we have shown that the solution to the Schrödinger equation is
+\[\ket{\psi(t)} = e^{-i(t-t_0)\hat{H}/\hbar}\ket{\psi(t_0)}.\]
+However, I do not recommend trying to solve practical problems by writing down $e^{-i(t-t_0)\hat{H}/\hbar}$ as a power series in the Hamiltonian.  We can do something a bit more clever.
+Recall that, since $\hat{H}$ is a Hermitian operator, its eigenvectors form a complete orthonormal basis for the Hilbert space.  Also, since $\hat{H}$ represents the energy of the system, the possible values we can obtain in a measurement of energy are the eigenvalues of $\hat{H}$ (see section 3.iii on postulate 3).  Denote the energy eigenvalues $E_1,E_2,\cdots$.  If $\hat{H}$ is nondegenerate then the eigenvalue equation $\hat{H}\ket{E_n} = E_n\ket{E_n}$ has unique solutions for the normalized energy eigenstates $\ket{E_n}$ (up to a global phase).  If it is degenerate, we can find an orthonormal basis $\ket{E_n,1},\ket{E_n,2},\cdots$ for the eigenspace corresponding to $E_n$ and then the vectors $\kat{E_n,m}$ will form a complete orthonormal basis for the Hilbert space.
+Now, recall that if $\ket{a}$ is an eigenvector of $\hat{A}$ with eigenvalue $a$ then $f(\hat{A})\ket{a} = f(a)\ket{a}$, i.e. $\ket{a}$ is also an eigenvector of $f(\hat{A})$ with eigenvalue $f(a)$ for any function $f$.  Therefore, we have
+\[e^{i(t - t_0)\hat{H}/\hbar}\ket{E_n,m} = e^{-i(t-t_0)E_n/\hbar}\ket{E_n,m}.\]
+This equation says that if the system starts in an energy eigenstate $\ket{E_n,m}$ at time $t_0$ then it will remain in the same energy eigenstate at all times.  It just gets multiplied by an overall global phase $e^{-i(t-t_0)E_n/\hbar}$, which, as we know, does not change the physical state of the system.
+The system need not start out in an energy eigenstate, but, since the states $\ket{E_n,m}$ form an orthonormal basis, any initial state can be written as
+\[\ket{\psi(t_0)} = \sum_{n,m} c_{n,m}(t_0) \ket{E_n,m},\]
+where the components $c_{n,m}(t_0) = \braket{E_n,m}{\psi(t_0)}$ are defined by the initial conditions.  At some later time $t > t_0$ it will be
+\[\ket{\psi(t)} = \sum_{n,m} c_{n,m}(t) \ket{E_n,m},\]
+where the new components $c_{n,m}(t) = \braket{E_n,m}{\psi(t)}$ are generally different.  However, by linearity, we also have
+\begin{align*}
+   \ket{\psi(t)} & = e^{-i(t-t_0)\hat{H}/\hbar} \ket{\psi(t_0)} \\
+   & = e^{-i(t-t_0)\hat{H}/\hbar} \left ( \sum_{n,m} c_{n,m}(t_0) \ket{E_n,m} \right )  \\
+   & = \sum_{n,m} c_{n,m}(t_0) e^{-i(t-t_0)\hat{H}/\hbar}\ket{E_n,m} \\
+   & = \sum_{n,m} c_{n,m}(t_0) e^{-i(t-t_0)E_n/\hbar}\ket{E_n,m},
+\end{align*}
+from which we deduce that
+\[\boxed{c_{n,m}(t) = c_{n,m}(t_0) e^{-i(t-t_0)E_n/\hbar}}.\]
+In other words, to solve the Schrödinger equation for an arbitrary initial state, you have to do the following.
+  * Solve the eigenvalue equation $\hat{H}\ket{E_n,m} = E_n\ket{E_n,m}$, where $m$ is an extra index labelling an orthonormal basis for degenerate eigenspaces.
+  * Write your initial state as a superposition of energy eigenstates
+  \[\ket{\psi(t_0)} = \sum_{n,m} c_{n,m}(t_0) \ket{E_n,m},\]
+  where $c_{n,m}(t_0) = \braket{E_n,m}{\psi(t_0)}$.
+  * At some later time $t$, this state will evolve to
+  \[\ket{\psi(t)} = \sum_{n,m} c_{n,m}(t_0)e^{-i(t-t_0)E_n/\hbar} \ket{E_n,m}.\]
+Let's illustrate this method with a simple example.  Suppose the Hilbert space of the system is two dimensional and is spanned by the orthonormal basis vectors $\ket{\phi_1}$ and $\ket{\phi_2}$.  Suppose that the Hamiltonian is
+\[\hat{H} = E \left ( \ketbra{\phi_1}{\phi_2} + \ketbra{\phi_2}{\phi_1}\right ),\]
+where $E$ is a constant with the dimensions of energy.  Suppose that the initial state of the system is $\ket{\psi(t_0)} = \ket{\phi_1}$.
+It will be convenient to do the calculation in matrix form in the basis $\ket{\phi_1},\ket{\phi_2}$.  In this basis, the Hamiltonian is represented by the matrix
+\[H = E \left ( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right ),\]
+and the initial state is represented by the column vector
+\[\vec{\psi}(t_0) = \left ( \begin{array}{c} 1 \\ 0 \right ).\]
+The eigenvalues of $H$ are of the form $\lambda E$, where $\lambda$ is an eigenvalue of
+\[\left ( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right ),\]
+so let's find the eigenvalues and eigenvectors of this matrix.
+The characteristic equation is
+\[(-\lambda)^2 - 1 = 0,\]
+so $\lambda = \pm 1$.  The eigenvalues of $\hat{H}$ are therefore $\pm E$.  Since we have two distinct eigenvalues, the Hamiltonian $\hat{H}$ is nondegenerate.
+For $\lambda = 1$, the eigenvector is found by solving
+\[\left ( \begin{array}{cc} -1 & 1 \\ 1 & -1 \end{array}\right ) \left ( \begin{array}{c} x \\ y \end{array} \right ) = \left ( \begin{array}{c} 0 \\ 0 \end{array} \right ),\]
+which has solution $y=x$.  So the eigenvector corresponding to energy $E$ is of the form
+\[\vec{\chi}_1 = C\left ( \begin{array}{c} 1 \\ 1 \end{array} \right ),\]
+for an arbitrary constant $C$.  But, as I keep telling you and someone will nevertheless fail to do on a homework or exam, we should always work with normalized eigenvectors, so we impose $\vec{E}\cdot\vec{E} = 1$ to obtain
+\[\vec{chi}_1 = \frac{1}{\sqrt{2}}\left ( \begin{array}{c} 1 \\ 1 \end{array} \right ).\]
+For For $\lambda = -1$, the eigenvector is found by solving
+\[\left ( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}\right ) \left ( \begin{array}{c} x \\ y \end{array} \right ) = \left ( \begin{array}{c} 0 \\ 0 \end{array} \right ),\]
+which has solution $y=-x$.  So the eigenvector corresponding to energy $-E$ is of the form
+\[\vec{\chi}_2 = \sqrt{1}{\sqrt{2}}\left ( \begin{array}{c} 1 \\ -1 \end{array} \right ),\]
+where I have normalized the vector already.
+Now, we have to express our initial state as a superposition of $\vec{\chi}_1$ and $\vec{\chi_2}$