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Eigenstates: What would happen if you had a laser aimed at a mirror 45 degrees.

  1. Mar 19, 2008 #1
    Eigenstates: What would happen if you had a laser aimed at a mirror 45 degrees.......

    I know this post might fit under some of the other post in this forum but i would like to ask it as a new topic because I am just getting into a lot of Quantum information ever since I noticed this part of the forum, and now i'm hooked!


    You have a laser aimed at a mirror that is offset by 45 degrees of the lasers original path. In classical physics the laser should reflect off this point perpendicular to it's original path correct? Now when the laser beam gets to the first mirror it's Eigenstates are infinite, it could reflect back at any point of degrees or does reflect back at every point. It's not until the observer looks that it reflects perpendicular. [from what i understand]

    Now what would happen if you had a bunch of mirrors and just made the beam bounce around in a zig zag type pattern, until it hits a wall at the end. If you isolate the system so that you can only see the wall at the end, then turn the laser on, what would happen? (i think i know what would actually happen)

    If you "weren't there" to see the laser's first mirror point, would it still reflect perpendicular to it's original path and eventually shine on the wall?

    Man this stuff is confusing the heck out of me and taking up too much of my free time trying to figure it out!
  2. jcsd
  3. Mar 19, 2008 #2
    I would think the probability of reflection at some angle other than 45 degrees decreases exponentially for the 'quantum' laser beam (Gaussian distribution centered at 45 degrees).
  4. Mar 19, 2008 #3
    The angle shouldn't really matter though should it? The point of the senario is that it is reflected back and fourth between multiple mirrors, but if you aren't there to observe the first reflection, will you see the last?
  5. Mar 20, 2008 #4


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    Not at all. The mirror acts as an interaction with the laser beam, and this will influence the time evolution of the state vector of the beam. Everything will reflect. The reflection will simply be part of the time evolution operator (or, if you like, the Schroedinger equation).
  6. Mar 20, 2008 #5

    Ken G

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    It might help you to understand the concepts of "phase" and "interference". The behavior you seek is determined by wave mechanics, which means you could set up water waves to do precisely the same thing as that laser. If you give the wave a very short wavelength, you can focus it tightly into a beam, and it will reflect just like the laser. The reason is that the wave follows the path where its components constructively interfere, and does not go where its components destructively interfere. The "short wavelength" constraint is needed to get highly efficient interference-- otherwise, the interference is too weak and the wave tends to spread out ("diffract"). The wave doesn't need to know its "history" or who is looking at it-- even though it appears to follow a beam, a snapshot of the wave's current amplitude everywhere suffices to determine its future behavior, because it allows you to predict how it will interfere.

    The laser beam only needs to be treated quantum mechanically in applications where you are interested in the behavior of the individual "quanta", or "photons". Then the picture I'm describing only allows you to predict probabilities, as mentioned above. The probabilities are "actualized" when you make a measurement at the wall-- but the behavior of the "wave function" you use to make that prediction is still just like the water wave. In fact, if you "look" at the photon before it gets to the end, you interfere with that wave function and can get something different from the reflecting beam (like the way putting your finger in the water to feel the wave go by will send out ripples of its own).
    Last edited: Mar 20, 2008
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