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How do you build a wave packet and inital conditions

  1. Jan 21, 2014 #1
    Hi, I'm trying to get my head around Schrödinger's equation and quantum wave theory. I'll try to shortly state how I understand it so you may see where I'm wrong and better answer the question.

    In classical mechanics if you solve a linear differential equation, the sum of the solutions is also a solution. This happens in quantum mechanics too obviously, but there's a difference: in classical mechanics you must choose one solution based on initial conditions, i.e. the speed and shape of a string, or the speed of a particle inside a gravitational or electromagnetic field when t=0. Then you get a single solution.

    In quantum mechanics you solve the equation, but you can't possibly know the initial conditions because of the uncertainty principle. So instead we add up all the solutions to form a localized wave packet, which represents both the uncertainty in position and momentum, through the Fourier transform. I understand the maths behind this process, what I don't understand is the physics. How do you determine the weigh of each possible solution? For a free particle, for instance, why would you build a gaussian packet instead of another one? On a similar note, with bound states (particle in a box for example) how do you weigh each possible state and in which way do these things model and reflect a particular problem and initial conditions? Is there even such a thing as initial conditions in QM?

    Those are the main questions, but I'll sneak a last one here. With bound states, you get a wave function which is the weighed sum of all the solutions of the SE. Each of these has a determined energy so we call them energy states. We say that the particle exists in a superposition of all these states until it's observed, where it must collapse to one and only one of the states: it can only have one energy value and it must be that of one of the possible energy states. Does this imply that each state is linearly independent from every other state? I can verify it in the particle in a box experiment, but does it always happen? Can it be proved from the SE? I could elaborate on the math to say why I think they should be LI but I'm away from home now and I need my notes for this. If it is unclear, I'll try and explain it better or post another question when I get back home.

    Thanks a lot in advance!
     
  2. jcsd
  3. Jan 21, 2014 #2
    In QM, just as in classical mechanics, it is possible to prepare a system--say, a single particle--in a definite, known initial state. But in QM, the state of a particle is not given by its position and velocity. The state of a particle is given by its *wave function*. The wave function *is* the state. It is possible to construct a definite initial state--i.e., a definite initial wave function. This process is called "state preparation."

    The exact initial wave function will depend on how the particle is prepared. From then on the evolution of the wave function is given by the Schrodinger equation.

    This is very analogous to classical mechanics. There we are given an initial state for a particle--its position and velocity--and the evolution of the particle's position and velocity is thereafter determined by Newton's second law.

    A Gaussian wave packet is simply a convenient analytically tractable example and not necessarily a wave function that will actually occur in any particular experiment.

    You can prepare various bound state wave functions, and superpositions thereof, by appropriate experimental procedures. Take a hydrogen atom. You can prepare a hydrogen atom in the ground state just by letting it sit for a while; it will spontaneously relax to the ground state. Perhaps you instead want to prepare a hydrogen atom in a certain definite excited state. To do this you can expose the ground-state atom to an electromagnetic wave of a certain frequency and duration.

    Your statement is too general. The wave function will only collapse to an energy eigenstate if we measure the energy of the particle. If we measure, say, the momentum of the particle, the wave function will not collapse to an energy eigenstate but a momentum eigenstate.

    Yes. This fact is independent of the Schrodinger equation. For any Hermitian operator, including the Hamiltonian, all its eigenstates are linearly independent and this fact is not too hard to prove. In fact it ought to be proven in any class or textbook on QM.
     
  4. Jan 21, 2014 #3
    Thanks for the fast and great answer, it is much clear now. I'm actually majoring in engieneering so I only took an introductory class which was way too basic, sorry if the questions are too basic as well :P

    There's one more thing I'd like to ask. I understand you can prepare a certain state in a lab, but when you don't have that kind of control, how do you determine the inital state? Maybe you just don't because you don't have enough information. What I'm getting at is that if you have, for example, a free electron emitted in a cathode ray tube, how do you choose an inital condition? In classical mechanics you would say it just starts with a certain velocity in a certain direction. How do you model stuff you can't prepare in a lab? Maybe this is too general and each case requires distinct treatment. All this comes from me trying to run a simulation of the double slit experiment and not knowing how to choose an inital condition :P

    Thanks again!
     
  5. Jan 21, 2014 #4
    To simulate the double-slit experiment, you'll probably want to start off with a plane wave or a broad wave packet (Gaussian if you like). Take the width of the wave packet to be much larger than the wavelength; this corresponds to small fractional uncertainty in the momentum. A plane wave is probably simplest and corresponds to an infinitely broad wave packet and perfectly definite momentum. Then let this plane wave or wave packet impinge on the double slit and look at the waves that come out the other side. The details of the shape of the initial wave packet won't matter too much as long as the wave packet is rather broader than the slits and the slit separation.

    The momentum of cathode ray electrons is fairly definite. It is determined by the voltage through which the electrons are accelerated. So the wave functions of cathode ray electrons are some kind of broad wave packets (at least, broad compared to their wavelength).
     
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