What does periodicity of $e^{-iHt/\hbar}$ mean in physical terms?

The unitary time evolution operator $U(t)=e^{-iHt/\hbar}$ has some distinct flavour of periodicity to it because of $e^{x+2\pi i}=e^x$.

Is this periodicity reflected in any way in physical systems? Does it have any physical meaning?

(Sorry for a rookie question. In case this is too simple for this site, tell me where to migrate it.)

• no need to migrate. – jerk_dadt Mar 16 '14 at 5:56

Although unitarity is necessary for conservation of the norm, indeed there is some periodicity ingraved in its form. For time-invariant bounded systems (the Hamiltonian does not depend on time and has a discrete, or at least non-dense number of eigenstates) there is always a recurrence time $\tau_r$ after which the state of the system is again the initial state (or as close as you can measure it). The simpler quantum system that you can imagine, a two-level system, can be represented as a point on the surface of a sphere, moving at constant speed along a circle (for time-invariant Hamiltonians). It will obviously revisit the inital state periodically. More complex systems (but bounded and time-invariant) described by n-level Hamiltonians also revisit the initial state following richer dynamics. You can google for information under the name of revival (and partial revival) times.

If you think carefully, there is nothing surprising about it. Similar phenomena is also found in Classical Physics under the name of Poincaré recurrence time. Stable systems tend to be periodic or quasi-periodic. This has helped their study and the birth of Physics since the early days. However, as the complexity of the system increases, or in other words, as the number of eigenstates that participate in the dynamics is larger, so it will be the recurrence time, which eventually may be larger than the time related to the inverse of the precision of the Hamiltonian eigenstates, or in other words, larger than the time disposed to follow or measured the system. In fact for most physical systems the recurrence time is clearly unphysical. This is also reflected in the fact that for increasingly complex systems it becomes harder to guarantee that they are discrete and time-invariant, or closed, or isolated.

• This answer is misleading in that it severely understates the effect of quasi-periodicity in recurrence times, as well as the unimaginably long timescales involved in Poincaré recurrences. It takes as little as three states with irrationally related energy differences to preclude exact recurrences and to require long timescales for any $\varepsilon$-recurrence. Having more eigenstates does make this (qualitatively) longer, but a lot of finer argumentation is required to make this precise. Note also that Poincaré recurrences typically occur at times much longer than the age of the universe. – Emilio Pisanty Mar 18 '14 at 11:29
• @EmilioPisanty Very interesting point. If I understand correctly, Poincaré recurrences pertain to classical physics. Are recurrence times as long in the quantum setting? – მამუკა ჯიბლაძე Mar 18 '14 at 11:47
• As for quasiperiodicity, see my comment below to @WetSavannaAnimalakaRodVance – მამუკა ჯიბლაძე Mar 18 '14 at 11:49
• @Pisanty- I don't understand your criticism. Firstly I related the periodicity to the recurrence, which is what the question was about. Secondly I pointed to some physical observations where recurrence or quasi-recurrence shows up (revivals and partial revivals). The exactness is out of the world of experiments, of course. And finally I did say that in most systems that are not cooked up, the recurrence time cannot be observed because the required time is unphysical. Misleading? – perplexity Mar 18 '14 at 15:46
• @perplexity The main point is that, as WetSavannaAnimal points out, the apparent periodicity is simply an illusion, because only for very few hamiltonians does $e^{-iHt}$ share the periodicity properties of $e^{-iEt}$, and quasiperiodicity is hardly a substitute unless you're very specific about it, and you clearly define for which systems this happens and for which ones it doesn't. Saying "there is always a recurrence time" without mentioning the time may be unphysical is, I think misleading. I also note the conspicuous absence of hamiltonians with continuous spectra. – Emilio Pisanty Mar 18 '14 at 20:21

The operator, as you say, seems to evoke periodicity, but this is in general illusory, aside from when the quantum system has energy eigenstates whose energies are rational number multiples of one another. The quantum harmonic oscillator is a simple example of this, indeed one of the very few possible examples when we talk about countably infinite dimensional (separable) quantum state spaces. I discuss the special nature of the quantum harmonic oscillator in my answer here.

As the notation correctly implies, the time variation of each energy eigenstate is the time harmonic function $\exp\left(-i\,\frac{E}{\hbar}\,t\right)$. But if there are two or more of these eigenfunctions present, a linear superposition of them can only be harmonic if all the energies are rational number multiples of one another. To understand this statement, think of the two eigenstate superposition: $A\,exp(-i\alpha\,t) + B\,\exp(-i\beta\,t)$. This comes back to its beginning value only if $exp(-i\alpha\,t) = \exp(-i\beta\,t) = 1$ which can only be so if $\beta\,t = 2 b \pi;\,\alpha\,t = 2 a \pi$ for $a,\,b\in\mathbb{N}$. Otherwise put, $\alpha / \beta = a/b$, so the ratio must be rational.

A wonderful way to picture this is to think of the quantum state space as the torus: the Cartesian product of two circles. Trajectories through time for a superposition of the two eigenstates are helices that wind themselves around the torus like a wire on a toroidal inductor. If the eigenfrequencies are rationally related, the trajectory meets up with itself and a periodic cycle follows. If not, the thread winding around the torus never gets back to its beginning point and indeed the trajectory is dense on the torus! As the ratio becomes "less rational" i.e. $a/b$ in its standard form (with all cancelations done) becomes the ratio of bigger and bigger numbers, the period becomes longer and longer.

Now if you add into the mix many eigenfrequencies, all of which can be irrationally related, you can see that pretty quickly the superposition is going to become highly complicated and all likeness of periodicity will vanish, aside from for special cases like the QHO.

• This is the only correct answer so far. +1. – Emilio Pisanty Mar 16 '14 at 14:47
• Well fapp any quasi-periodic dynamics of the kind you describe is as good as periodic in the sense that from time to time it goes through states which are arbitrarily close to the initial state. More precisely, I think for any $\varepsilon$ there is a $T$ such that time intervals between moments when the system is in a state differing from the initial one "by less than $\varepsilon$" are less than $T$. – მამუკა ჯიბლაძე Mar 18 '14 at 5:51
• @მამუკაჯიბლაძე True, but check the topics in the answer you accepted, particularly the Poincaré recurrence theorem. I'm pretty sure you'll find that $T$ is exponentially related to the number of eigenstates, so pretty quickly becomes unimaginably big. – Selene Routley Mar 18 '14 at 11:16
• @WetSavannaAnimalakaRodVance Yes I believe this is true. Still you see I am not a physicist, my interest was more of "conceptual" nature. It follows that no closed system can "escape its beginning" in any essential way as eventually it will always return to states arbitrarily similar to the initial one infinitely many times. This is both disappointing and striking, is not it? :) – მამუკა ჯიბლაძე Mar 18 '14 at 11:33
• @მამუკაჯიბლაძე Also, Poincaré recurrence doesn't apply to (countably) infinite quantum systems. As you your last sentence, I wouldn't at all mind doing this life thing again! BTW what language is that stunningly beautiful script writing? – Selene Routley Mar 19 '14 at 2:46

A nice example is the harmonic oscillator, where we see that the periodicity of the operator just corresponds to the periodicity of the motion.

For a harmonic oscillator the hamiltonian is

$$\hat H = \frac{\hat{\mathbf p}^2}{2m} + \frac{1}{2} m \omega^2 \hat{\mathbf x}^2.$$ and the stationary solutions can for example be written $\left| n \right>$ with eigenvalues $$E_n = \hbar \omega \left( \frac{1}{2} + n \right) .$$ Let us now look at the time evolution of e.g. the state $\left| 0 \right> = \left| n = 0 \right>$. In this case we get $$\left| \psi_0 (t) \right> = e^{i \hat H t / \hbar} \left| 0 \right> = e^{i \omega t / 2} \left| 0 \right> ,$$ where we see that the periodicity of the operator gives rise to the a periodic motion of the oscillator (or, perhaps more correctly, the phase of the wave function of the oscillator).

The first thing that comes to mind is the fact that energy eigenstates have a periodic phase factor, $e^{iEt/\hbar}$. In general we usually say that states are defined up to a phase factor, so for a single particle in an energy eigenstate we can ignore it, but this phase factor can manifest physically when you have a state which is a superposition of energy eigenstates, each with its own oscillating phase. A simple example of this which actually results in periodic behavior would be a spin-$1/2$ particle in a magnetic field. Such a system will have a Hamiltonian proportional to the Pauli spin matrix $\sigma_z$. So our time-evolution operator is $U(t) = e^{i\omega \sigma_z t}$. Recall that the Pauli matrices are idempotent, so when we Taylor expand this we can group our terms together and we will find

$$U(t) = \cos\left(\omega t\right)+ i\sigma_z \sin\left(\omega t\right)$$

For eigenstates of $\sigma_z$ this isn't particularly interesting. The $\sigma_z$ will give us just a + for $\vert \uparrow \rangle$ or a - for $\vert \downarrow \rangle$ and of course the modulus of the state won't change but it will have a time-oscillating phase factor. Instead let's think about a superposition of such eigenstates, like $\vert \psi \rangle \equiv\frac{1}{2}\left(\vert \uparrow \rangle + \vert \downarrow \rangle\right)$.

$$\begin{split} U(t) \vert \psi \rangle = \frac{1}{2}\cos\left(\omega t\right)\vert \left(\vert \uparrow \rangle + \vert \downarrow \rangle\right) + \frac{i}{2} \sin\left(\omega t\right) \left(\vert \uparrow \rangle - \vert \downarrow \rangle\right) \\ = \frac{1}{2} e^{i\omega t} \vert \uparrow \rangle + \frac{1}{2} e^{-i \omega t} \vert \downarrow \rangle \end{split}$$

Not incidentally, this state was chosen to have a spin of $+\hbar/2$ along the $x$ direction, but we can see in the above equation that when we evolve it in time we reach a point where it will have spin $-\hbar/2$ along the $x$ direction, when $\cos(\omega t) = 0$ and $\sin(\omega t) = 1$, so the expectation value of the $x$ spin of this particle is periodic.

Because the Hamiltonian operator is Hermitian by assumption, it can be diagonalized and the operator exponential can be rewritten as a manageable Taylor series, but this nice periodic form only works because the square of our Hamiltonian is proportional to the identity matrix, so we could group terms in the Taylor expansion to give us a cosine multiplied by the identity and a sine multiplied by the $\sigma_z$ operator. In general this procedure does not work out quite as nicely, but you can always think about the interference of the phase factors I mentioned in the first paragraph.

You will probably find it is connected to photons that could excite it, similar to the photo-electric effect.

But in the main, the solution to a particular problem is a wave equation as well as a particle. In the case of electron microscopes, electrons behave as waves, based on a term $\exp(imc^2/hf)$, which allows us to see very small detail.

In essence, from what i read on google, it is connected to Schrodinger's equation. This is the wave-model of a particle trapped in a potential well, such as an electron or whatever.

One of the references i found connects the time operator with spin. But the $H$ operator is a kind of mathematical beast that is beyond by ken.

• Could you explain in more detail? I am not a physicist, so... How does unitary time evolution relate to the term you describe? I understand in general that actually most states in QM are represented by wave functions, but I could not figure out how does this relate to the $U(t)$ above – მამუკა ჯიბლაძე Mar 16 '14 at 8:27
• This answer is incomplete. It does not address the OP question fully and it talks about the resolution of electron microscopes, which the OP was not asking for. – mpv Mar 16 '14 at 9:42
• The unitary time thing defines the frequency. All frequencies not equal to that special time are quickly dampened, as in music-theory. OP's question is about a real-world application of this, and electron microscopes use evident particles to produce a wave-like focusing etc. – wendy.krieger Aug 11 '14 at 10:44