3
$\begingroup$

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' ?

$\endgroup$
6
  • $\begingroup$ 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 $1$. $\endgroup$ – Triatticus Feb 12 '19 at 18:34
  • 2
    $\begingroup$ I don't understand this question - your eq. (1) is the definition of what it means for $H$ to generate a unitary transformation. $\endgroup$ – ACuriousMind Feb 12 '19 at 18:36
  • $\begingroup$ 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. $\endgroup$ – CatoMaths Feb 12 '19 at 18:38
  • $\begingroup$ @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 $e^{iHt} \rightarrow U$ $\endgroup$ – CatoMaths Feb 12 '19 at 18:40
  • $\begingroup$ 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? $\endgroup$ – ACuriousMind Feb 12 '19 at 18:42
4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ – CatoMaths Feb 12 '19 at 21:23
  • $\begingroup$ @CatoMaths 1. Yes. 2. I added some stuff regarding the lack of uniqueness. $\endgroup$ – ZeroTheHero Feb 12 '19 at 22:40
  • $\begingroup$ 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 $\endgroup$ – CatoMaths Feb 13 '19 at 15:45
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.