How does a Hamiltonian 'generate' a unitary?

Question

I know that the unitary (propagator) is given by

$$U=e^{iHt}\tag{1}.$$

But I actually never saw how a Hamiltonian translates into a unitary. For example when I consider a two-level rotation in a qubit such as

$$U = \Bigg{(}\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}\Bigg{)}\tag{2}$$

how would the corresponding $H$ look like?

Bonus: A unitary like in eq.(2) is able to switch the probabilities of the system to be in the corresponding state. For example

$$\mid \psi\rangle = \frac{1}{\sqrt{4}}\lvert 0 \rangle + \frac{\sqrt{3}}{\sqrt{4}}\lvert 1\rangle$$

$$U\lvert \psi\rangle =\frac{\sqrt{3}}{\sqrt{4}}\lvert 0 \rangle + \frac{1}{\sqrt{4}}\lvert 1\rangle \, .$$

But in some papers I've read (using the same type of 2-level rotations) the authors write

Let us consider a rotation between the energy levels $m$ and $n$ with probabilities $P_n$ and $P_m$ by an angle $\theta$.

This sounds like they would rotate the energy levels but in fact they are just rotating the probabilities corresponding to these energy levels.

Why this weird 'notation' ?

Answer 1

Let $$ H=\left( \begin{array}{cc} -1 & 1 \\ 1 & -1 \\ \end{array} \right) \tag{1} $$ Then the continuous transformation $$ U(t)=e^{i t H}= \left( \begin{array}{cc} \frac{1}{2} \left(1+e^{-2 i t}\right) & \frac{1}{2}-\frac{1}{2} e^{-2 i t} \\ \frac{1}{2}-\frac{1}{2} e^{-2 i t} & \frac{1}{2} \left(1+e^{-2 i t}\right) \\ \end{array} \right) $$ gives $$ U(\pi/2)=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right)\, . \tag{2} $$ $H$ "generates" a one-parameter group of unitaries $U(t)$ because $U(t)$ is constructed using powers $\hat H$, in the same way that a finite group is generated by taking powers of its generating elements (although here you're also adding powers of $H$, something you can't do for elements of a finite group). Here, $U(t)$ happens to coincide with your transformation at $t=\pi/2$.

Since the elements of the unitary $U(t)$ can be complex and do not sum to $1$ along lines or columns, it does not really the probabilities directly; the probabilities are actually moduli squared (thus real) of projections: these moduli squared, when summed along a line or a column, do sum to $+1$.

In the reverse direction, given a unitary $U$ is it not directly possible to obtain $H$ without some tinkering. Whereas $$ \frac{dU}{dt}\vert_{t=0}=iH $$ would allow you to recover $H$, you can't really take the derivative of Eq.(2) no more than it makes sense to take the derivative of a function evaluated at one point. One way to find Eq.(1) is to write the most general $2\times 2$ Hermitian matrix (which would depend on $4$ real parameters), take its exponential and select the parameters to give you (2). This isn't so bad for $2\times 2$ matrices but it can become quite complicated for larger matrices.

A more systematic way starts with $$ U=\left(\begin{array}{cc} 0 & 1 \\ 1 &0\end{array}\right) $$ and write $$ U=\left( \begin{array}{cc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{array} \right)\,\left(\begin{array}{cc} -1&0\\ 0&1\end{array}\right) \left( \begin{array}{cc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{array} \right) $$ and find $$ u(t)=\left(\begin{array}{cc} e^{-i k_1 t}&0\\ 0&e^{ik_2t}\end{array}\right) $$ so that $$ u(t_0)=\left(\begin{array}{cc} -1&0\\ 0&1\end{array}\right) $$ Thus for instance, take $t_0=\pi/2 $ (as before) to get $k_1=2, k_2=4$ and then $$ H=\left( \begin{array}{cc} 3 & 1 \\ 1 & 3 \\ \end{array} \right)\, . \tag{3} $$ You can then verify that $e^{i pi H/2}=U$. If anything, this show that $H$ is not unique: in fact this $H$ of Eq(3) different from $H$ of Eq(1) by a multiple of the unit matrix. (The choice $k_2=0$ gives $H$ in Eq.(1).) Moreover, choosing different $t_0$ will produce different $H$'s.

Answer 2

From the time-dependent Schrodinger equation

$$i\hbar\frac{d}{dt}|\Psi(t)\rangle=\hat{H}|\Psi(t)\rangle,$$

one sees that the Hamiltonian operator $\hat{H}$ is the "infinitesimal generator" of time translation:

$$\begin{align} |\Psi(t+dt)\rangle&=|\Psi(t)\rangle+dt\frac{d}{dt}|\Psi(t)\rangle\\ &=\left(1-\frac{i}{\hbar}dt\,\hat{H}\right)|\Psi(t)\rangle.\end{align}$$

By dividing a finite time interval from $0$ to $t$ into $n$ smaller intervals, and letting $n$ to go infinity so that the smaller intervals become infinitesimal, we “generate” the following finite time translation:

$$|\Psi(t)\rangle=\lim_{n\rightarrow\infty}\left(1-\frac{i}{\hbar}\frac{t}{n}\hat{H}\right)^n|\Psi(0)\rangle=e^{-i\hat{H}t/\hbar}|\Psi(0)\rangle$$

Here we've used the mathematical identity

$$\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n=e^x.$$

The word "generator" comes from Lie algebra and is suggestive of the relationship between a Lie algebra and a Lie group: the infinitesimal transformations can be used to "generate" the finite transformations by composing them an infinite number of times.