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Moving electron - finding the wavefunction

  1. Aug 7, 2013 #1
    1. The problem statement, all variables and given/known data
    On our modern physics class e did a problem:
    At first i said: "Oh i know this!" and solved the case like this.

    2. Relevant equations
    Lorentz invariant: ##E=\sqrt{{E_0}^2 + p^2c^2}##
    Schrodinger equation where ##V(x)=0##.

    3. The attempt at a solution
    The energy ##100eV## must be the kinetic energy of the electron. So i said ok this kinetic energy is very small compared to the rest energy and i can say that ##pc \ll E_0## which means i have a classical limit where:

    \begin{align}
    E=\sqrt{{E_0}^2 + p^2c^2}\\
    E=\sqrt{{E_0}^2 + 0}\\
    \boxed{E=E_0}
    \end{align}

    So now i can write the general wavefunction for a free right-mooving particle like this:

    $$\psi=Ae^{iLx}\quad L=\sqrt{\tfrac{2mE}{\hbar^2}}$$

    So if i want to get the speciffic solution i need to calculate the constant ##L## and then normalise the ##\psi##. Because ##E=E_0## i calculated the constant ##L## like this:

    $$L=\sqrt{\frac{2mE_0}{\hbar^2}}$$

    while my professor states that i should do it like this:

    $$L=\sqrt{\frac{2mE_k}{\hbar^2}}$$

    where ##E_k## is the kinetic energy of the electron. Who is wrong? I mean whaaaaat? The constant ##L## is afterall defined using the full energy and not kinetic energy...
     
  2. jcsd
  3. Aug 7, 2013 #2

    vela

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    You're wrong. Both your expression for L and the Schrodinger equation are non-relativistic. It's not appropriate to use the relativistic expression for the energy of the electron here.
     
  4. Aug 7, 2013 #3
    Thank you.

    But can you explain to me why don't we generaly use the rest energy when dealing with a classical approximations. In books everyone explains the rest energy but what it realy is? I mean i know how to calculate it and whatsoever but when does it appear and why do we have to take it into the calculation?

    I am confused only because of that Lorentz invariance which says that if a particle is mooving slowly most of its energy is the rest energy. On the other hand we just neglect it like it isn't there... It seems to me like a contradiction...
     
  5. Aug 7, 2013 #4

    vela

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    It's just the way you're doing the classical approximation isn't correct. A plane wave is given by $$\psi = e^{i(kx-\omega t)}$$ where ##p= \hbar k## and ##E = \hbar \omega##. For the spatial dependence, you need to find the classical approximation for the momentum p:
    \begin{align*}
    pc &= \sqrt{E^2 - (mc^2)^2} \\
    &= \sqrt{(mc^2+K)^2 - (mc^2)^2} \\
    &= \sqrt{2mc^2 K + K^2} \\
    &\cong \sqrt{2mc^2 K}
    \end{align*} where K is the kinetic energy. The rest energy term cancels out.

    The other way to look at it is that ##k = \sqrt{\frac{2mE}{\hbar^2}}## is equivalent to the relationship ##E = \frac{(\hbar k)^2}{2m} = \frac{p^2}{2m}##. This latter expression is clearly the classical quantity for the kinetic energy of a particle of mass m, not the total energy.
     
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