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The time-symmetry and indeterminacy of elementary processes

  1. Jun 3, 2006 #1
    Is there are conflict between the time-symmetry of some elementary processes and the evolution of the wave functions for those processes? For instance, the emission of a photon from atom A subsequently absorbed by atom B should AFAIK be time symmetric such that it could be viewed as an emission from B followed by an absorption at A going backwards in time.

    But the wave function for the photon emitted (or to be emitted) from A spreads out through space until it collapses when absorbed by B. Similarly, when time-reversed, the wave function of the photon following an emission by B should spread out until it collapses when absorbed by A. These two functions will look different.

    How does this get resolved?

    I have read a view that the wave function for photon emission should not be viewed as the wave function for the photon itself but for the emitter. This makes some sense to me, since the wave function is based on not knowing when the photon will be emitted; that is, it cannot be the wave function for the photon since it probably doesn't exist yet. But this, as it seems to me, would not support the entirely indeterministic picture of quantum transfers. It would be a case of not being able to know, rather than an intrinsic uncertainty of where the photon's wavefunction will collapse following an emission event.

    Sorry, this is a very elementary question. I just haven't read anything that treats it exactly yet.

    Thanks,

    El Hombre.
     
  2. jcsd
  3. Jun 5, 2006 #2
    Wow! 46 views and no replies. I assume this is an even dumber question than I feared, and everyone is being very, very polite.

    So, to change the question into something less repulsive, can anyone explain to me if and why the question is crackers? Guide me down the right path here! Or just post anything! Anything at all! Just to break the ice! Do you...... golf?
     
  4. Jun 6, 2006 #3

    Physics Monkey

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    Hi El Hombre Invisible,

    Your question is not stupid, but it requires some care in answering. The short answer is that there is no conflict: the wavefunction or state vector always evolves continuously. I propose to step through a very simple system to show you how things work.

    The system I have in mind consists of a two level atom which is coupled to a single photon mode. This is a prototype model for a cavity QED experiment where an atom interacts strongly with a particular resonant mode of the cavity. The atomic states are labeled [tex] |g\rangle [/tex] and [tex] |e\rangle [/tex] for ground and excited respectively. The energy spacing between the two levels is [tex] E [/tex]. Take the photon to have frequency E (hbar = 1). This single mode is described by an annihilation operator [tex] a [/tex] (just a simple harmonic oscillator). We can also define a kind of annihilation operator for the atom [tex] c = | g \rangle \langle e | [/tex] which has the effect of removing a particle in the excited state and putting it in the ground state. The Hamiltonian for this system can be written as
    [tex] H = E c^+ c + E a^+ a + \lambda (a c^+ + a^+ c) [/tex]. The first two terms count the energies of the atom and the photons. The third term says that an atom can go from e to g by emitting a photon or vice versa. If you aren't familiar with the formalism then don't worry, I just wanted to provide some background in case you were interested.

    It is possible to exactly solve this simple model, that is to find the energy eigenvectors and eigenvalues. A simple computation will reveal that [tex] | g 1 \rangle \pm |e 0 \rangle [/tex] are eigenstates (the number refers to the number of photons present). The eigenvalues are [tex] E \pm \lambda [/tex]. This immediately tells you that the states of definite energy don't have a definite number of photons. We can now ask how the state [tex] |g 1 \rangle [/tex] would evolve in time. If I name the energy eigenstates [tex] | \pm \rangle = \frac{1}{\sqrt{2}}(|g 1 \rangle \pm | e 1 \rangle ), [/tex] you can easily see that [tex] |g 1 \rangle = \frac{1}{\sqrt{2}}(|+\rangle + | - \rangle )[/tex].

    We are thus able to calculate the time development of our state, and we find [tex] e^{-iEt} \frac{1}{\sqrt{2}}(e^{-i \lambda t} |+\rangle + e^{i \lambda t}| - \rangle ) = e^{-iEt}(\cos{\lambda t} |g 0 \rangle - i \sin{\lambda t} |e 1 \rangle )[/tex]. This is the important result. The thing to notice is that the wavefunction changes smoothly from a state with 1 photon at t = 0 to a state with 0 photons at [tex] \lambda t = \pi/2 [/tex]. The moral of the story is that there is no sudden absorbtion or emission of the photon so long as the system remains undisturbed. These smooth oscillations of the wavefunction are called Rabi oscillations, and they are perfectly reversible. Note that even though I described the Hamiltonian in terms of absorbtion and emission, the actual evolution is continuous.

    The qualifier "undisturbed" deserves some explanation. If we have some apparatus that is able to measure the energy state of the atom, then after the measurement the atom is either in g or e but not in a superposition (interpretations aside). We say that the atom has absorbed a photon if we prepare it in g and measure it in e. Clearly our chance to find e or g in such a measurement depends crucially on when we look. Such jumpy evolution (if indeed it is ultimately jumpy) is always associated with measurement. As I said above, the unitary evolution is always nice and smooth.

    I have no idea what your level is, so I hope I didn't overwhelm you with too much gibberish. Your system of two atoms plus field is no more complicated in principle, but it requires a proliferation of notation since positions, etc are now important. Anyways, I hope this helps.
     
    Last edited: Jun 6, 2006
  5. Jun 6, 2006 #4
    The photon is virtual not real, it doesn't matter where it "is" between the points of emission and absorbtion, only the end-points are relevent. The process of emission and absorbtion at vertices is describable as described by Physics Monkey or by a propogation amplitude and a superposition (of points in space-time where the emission/absorbtion can take place, these rules being contained in the Feynman rules of QED). The intermediate photon is not physical, in the sense that it is off-shell. In reversing time you're not just changing the direction of four-momentum exchange you're exchanging external particles for their anti-particles.
     
    Last edited: Jun 6, 2006
  6. Jun 6, 2006 #5
    Thanks once again, Physics Monkey. Your answer was easy enough to follow. In short, as the wavefunction evolves, you still have a chance to find the would-be emitting atom in the excited state. This will evolve as such until one of the atoms is measured and the [insert interpretative choices such as 'wavefunction collapse' here] occurs, or until it evolves to a ground state for the emitting atom. Excellent!

    Perturbation: your QED-based approach hit me last night when I was reading a paper about sum-over-histories. It basically stated that time reversibility was accounted for in this method. (I was actually about to declare my thread dead because I'd got an answer.) Sounds like it's accounted for in Schrodingers too.
     
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