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The Simplest Eigenfunction for Schrödinger's Wave EQ?

  1. Jun 11, 2014 #1
    I am just getting into quantum after a long absence of working on modern physics. I am having a go at "Introduction to Modern Physics" by Griffiths

    What is the simplest equation that satisfies Schrödinger's Wave Equation. It looks like $$Ae^{x}$$ would do the trick, but it would not solve the boundary conditions of being zero at +/- ∞. So would $$Ae^{-|x|}$$ be the simplest? If it is, does it correspond to anything in reality?

    Sorry for the possibly lame question,

    Chris
     
  2. jcsd
  3. Jun 11, 2014 #2

    Matterwave

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    That function is not differentiable at x=0, there's a kink, so that particle would have undefined momentum at x=0.

    Also, that function is NOT an eigenfunction of the time independent Schrodinger equation for any common potentials V(x). Which potential are you trying to solve?
     
  4. Jun 11, 2014 #3
    Yes, this is true (the discontinuity). I was thinking that might be a problem. I was musing on what possible different solutions would it have for a free particle, so V=0. Does the function always have to be complex? I was also thinking of the time independent version of the equation.

    Edit: I would imagine that if the answer is "it has to be complex", it does make sense to me as far as where the "wavyness" comes from (specifically, the function goes to zero on both extremes).

    Chris
     
    Last edited: Jun 11, 2014
  5. Jun 12, 2014 #4

    Matterwave

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    The time independent Schrodinger equation for a particle in potential V(x)=0 is:

    $$\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial^2 x}\psi=E\psi$$

    Solving this eigenvalue equation, you should get that the eigen-functions are:

    $$\psi(x)=Ae^{ix\sqrt{2mE}/\hbar}+Be^{-ix\sqrt{2mE}/\hbar}$$

    These are sinusoidal. Making the identification ##p=\sqrt{2mE}## gives the usual:

    $$\psi(x)=Ae^{ip\cdot x/\hbar}+Be^{-ip\cdot x/\hbar}$$


    Notice however, that these eigenfunctions are not square-integrable.
     
  6. Jun 12, 2014 #5
    Do you mean that it is not $$|\Psi|^2$$ (mod squared) friendly? Could not one, just pick the first or the second term?

    Chris
     
  7. Jun 12, 2014 #6

    Matterwave

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    Let's just take the positive solution:

    $$|\psi|^2=|A|^2$$

    This is just a number, so when integrated over all ##x## I will get infinity:

    $$\int_{-\infty}^{\infty}|\psi|^2 dx=\int_{-\infty}^{\infty}|A|^2 dx=\infty$$

    This is a purely oscillatory function, it does not go to 0 as x goes to plus/minus infinity.
     
  8. Jun 12, 2014 #7
    Very true! Learning I am :D

    Chris
     
  9. Jun 12, 2014 #8

    Matterwave

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    The point I was trying to make is that the eigenfunctions of the free Hamiltonian are not normalizable. As such, they are not, strictly speaking, part of the Hilbert space.
     
  10. Jun 12, 2014 #9
    Is that because the particle is unbound, so that it in essence can be anywhere? That would mean the one could know the momentum precisely. No?

    Chris
     
  11. Jun 12, 2014 #10

    Matterwave

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    This means that one should always construct wave-packets.

    Given an unbound particle, it's energy will take on a continuum of values. Knowing the energy exactly is impossible as there is always uncertainty involved with any experiment. So particles are never physically in an energy eigenstate. But we can use the eigenstates to construct wave packets that DO represent physical particles. In some applications we might use an energy eigenfunction for simplicity, but we should be aware of the simplification we are making.
     
  12. Jun 12, 2014 #11
    Thank, you. This little write up explains the concept nicely. I have a vague familiarity with the Fourier transform mathematically, but a fairly solid understanding of it conceptually.

    Here is the article:
    http://www.tcm.phy.cam.ac.uk/~bds10/aqp/handout_1d.pdf

    Thanks,
    Chris
     
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