# How does a Hamiltonian 'generate' a unitary?

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' ?

• A hamiltonian doesn't generate a unitarity, that operator has the property of being a unitary operator in that the sum of the probabilities of all outcomes is $1$. – Triatticus Feb 12 '19 at 18:34
• I don't understand this question - your eq. (1) is the definition of what it means for $H$ to generate a unitary transformation. – ACuriousMind Feb 12 '19 at 18:36
• I see but the word 'generate' is frequently used in this context. Something is generating that operator I am just not sure what exactly it is, hence my post. I'd like to understand this whole process a bit better. For example read upcommons.upc.edu/bitstream/handle/2117/107050/… page 14. – CatoMaths Feb 12 '19 at 18:38
• @ACuriousMind My question is which $H$ exactly generates the unitary given in eq (2). I want to plug it in and then calculate it myself to see how $e^{iHt} \rightarrow U$ – CatoMaths Feb 12 '19 at 18:40
• As long as you don't fix a value for $t$, that question is unanswerable. Once you fix it, it seems straightforward to take the logarithm of the matrix and divide by $\mathrm{i}t$ to get a corresponding $H$, is there a problem with that? – ACuriousMind Feb 12 '19 at 18:42

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.

• Ah I see, thank you so much ! I have 2 minor follow-up questions: 1. If we start at $t_1=0$, and for example end at $t_2=\pi/2$ does $t_2$ then denote the duration of the process? 2. I see. The way I understand it now is that given a specific $U$ such as the one I gave in eq. (2) there are more than 1 ways to find a $H$ and corresponding $t$ to 'generate' the given $U$. Is that correct? – CatoMaths Feb 12 '19 at 21:23
• @CatoMaths 1. Yes. 2. I added some stuff regarding the lack of uniqueness. – ZeroTheHero Feb 12 '19 at 22:40
• Perfect that was very good. What this means to me is that even for a fixed process duration $t= \pi/2$ a given $U$ can be generated by an arbitrary amount of Hamiltonians $H$. This was very good, thank you – CatoMaths Feb 13 '19 at 15:45

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.