# Energy of Photon Pulses

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1. Jun 19, 2015

### Julian Blair

I'm confused because the James-Cummings model of an EM field interacting with a 2 state atom uses fields of "n" photons of frequency "omega" But, a pulse of time , t = pi/(2*coupling) creates an equally superimposed state, |superimposed states> = 1/(sqrt2) (|1> - i*|2>) .
My questions are:
(1) what is the energy of the pulse creating this state?
(2) Is this pulse considered a "piece of a photon of frequency omega" and therefore it has less energy than a "whole photon?"

2. Jun 19, 2015

### blue_leaf77

$E=\hbar \omega$, what else could be the energy of a photon?
What leads you to think this way? Actually when you wrote down $|\psi (T) \rangle \propto |1\rangle + i|2\rangle$, you have neglected the quantum properties of the light because that expression only contains the level notation of an electron, otherwise any state must also include that that belongs to the photon. In other words, it comes down from the semi-classical picture of light-atom interaction. Consequently, the talk of photon energy is irrelevant in this case since the interacting light is treated as a classical field.

3. Jun 20, 2015

### Julian Blair

If the quantum state of an atom of two energy levels should not be written as |ψ(T)⟩ = 1/ sqrt(2) * (|1⟩-i|2⟩), then I an certainly in good company, because that expression is widely used in discussions of light interacting with atoms. Are you familiar with the James-Cummings Model?
Also, I'm NOT asking the energy of the photon, but rather the ENERGY OF THE PULSE. Since the pulse does not seem to transfer the entire E=ℏω to the atom, then what part of the photon WAS ACTUALLY ABSORBED?

4. Jun 20, 2015

### blue_leaf77

Actually it is "Jaynes-Cummings".
That's the way the state of the system written when the interacting light is treated as a classical field, while only the atom is treated as a quantum object. There are of course also other possible ways, which are actually more formal to treat the light-atom interaction problem, in which both light and the atom are treated quantum mechanically. In such cases, you also have to indicate that the state's notation also includes the states of the light quanta.
I'm guessing by "pulse", you mean that $\pi$-pulse thing or whatever. I guess the terming pulse here doesn't really refer to the light pulse in the usual definition which is an oscillation of EM wave of finite duration in time and space, which is of course polychromatic. It's a mere naming. In contrast, in the simplest form of Jaynes-Cummings model, the (classical) light is supposed to be quasi-monochromatic.
The energy of a hypothetical monochromatic light is infinite. If this light is in the form of a quasi monochromatic (or not monochromatic at all, i.e. polychromatic centered around $\omega$). However be cautious, that if the bandwidth of the light is broad enough, the semi-classical treatment may greatly differ from that which results in the presence of sinusoidal temporal behavior of the coefficients $c_1(t)$ and $c_2(t)$ which eventually leads to our current discussion of the $pi$-pulse now, then $E = \int P(t) dt$, where $P(t)$ is the radiated power.

Last edited: Jun 20, 2015
5. Jun 20, 2015

### Julian Blair

If I remember correctly, the Jaynes-Cummings model treats the EM field as being made of "n" photons of frequency ω. Since photons are a QM concept, I don't know why you refer to their field treatment as "Classical." I'm not at all versed in QED, so, no doubt, there are deeper quantum treatments of the field.
Actually by 'pulse" I do mean oscillation of EM wave of finite duration in time and space, in fact of time, t = π/(2*Ω), where Ω is the field coupling. This pulse is NOT polychromatic, but only of frequency = ω, the resonance frequency of the atom's energy gap.
If you are correct that the energy transfered to the atom by this pulse is indeed, E=∫P(t)dt, where P(t) is the radiated power, then can you tell me how to calculate P(t) for a single photon of E=ℏω?
If would make sense to me if the energy transfered equaled the expected energy gained by the superimposed atom, E[atom] = 1/2 {E[Excited] - E[Ground]}

thanks!

6. Jun 20, 2015

### blue_leaf77

I think I should have corrected your understanding from the beginning. Jaynes-Cummings model does treat the light as photons, as quantum object as well as you said. If we assume the atom is initially in an upper state $|2\rangle$ and the light in a number state $|n\rangle$, then the state of the system of the interacting 2-level atom and the light (photons) at time t is written as $|\psi(t) \rangle = c_a(t) |2\rangle |n\rangle + c_b(t) |1\rangle |n+1\rangle$. You see that the state of photon also appear in the expression, this is clearly the consequence of treating the light quantum mechanically. On the other hand, equation like $|psi \rangle \propto |1\rangle + i|2\rangle$ only describes the quantum state of the atom because the light has been treated as an environment, in contrast to the Jaynes-Cummings model where both the atom and light act as the "actors" in the play. Therefore, I should have corrected when you said that $|psi \rangle \propto |1\rangle + i|2\rangle$ comes from the Jaynes-Cummings model, in fact the semi-classical model is sometimes called the Rabi model.

No that is not what I meant. I mean E=∫P(t)dt is the energy contained in an arbitrary pulse whose power is P(t), not the energy transferred from the pulse to an atom. When an atom absorbs a single photon of frequency $\omega$, the transferred energy is of course $\hbar \omega$.

Why is there 1/2 if the atom only absorbs one photon?

Last edited: Jun 20, 2015
7. Jun 20, 2015

### Julian Blair

OK, I stand corrected. I see that in the Jaynes -Cummings model, one has Dressed States, which involve Both the Atom and The Field. and that the associated energies are NOT those of the bare states |e> & |g>, but are rather superpositions of these bare states: |e,n> & |g,n+1>.

So, now I'm confused as to how one can create an atomic qubit with constant equal probabilities for the two bare states. The descriptions that I have seen have used the Rabi model, which makes it all seem rather simple, i.e. ... just hit the atom with a pulse at the resonance frequency, but for a length of time. t = π /(2*Ω).
Can you describe to me how to use the more sophisticated quantum model to create this qubit state?

8. Jun 20, 2015

### blue_leaf77

First of all I'm not well acquainted with quantum computing, so if your aim is to seek connection with quantum computing stuff, I cannot help.
Creation of superposition state... Well Jayness-Cummings model is basically the fully QM treatment of Rabi's semiclassical model. So I guess just by shining a near resonance frequency photon, the levels in resonance in the atom are coupled. But probably also there have been a couple of ways to create superposition states of practical use.