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A relativistic electron in a potential box

  1. Nov 6, 2016 #1
    1. The problem statement, all variables and given/known data
    In a potential box (##L = 1.00pm##) an electron moves at a relativistic speed, meaning it's momentum can't be expressed as ##P = \sqrt{2mE}##.

    a) Using the uncertainty principle, show that the speed is indeed relativistic
    b) Derive an expression for the allowed energy states of the electron
    c) By how much (in relative terms) do the relativistic and non-relativistic ground states differ from each other? Answer: The relativistic one is 22% smaller.

    2. Relevant equations
    The uncertainty principle:
    Momentum and position
    \begin{equation}
    \Delta p \Delta x \geq \frac{h}{4\pi}
    \end{equation}

    Energy and time
    \begin{equation}
    \Delta E \Delta t \geq \frac{h}{4\pi}
    \end{equation}

    Time independent Shcrödinger equation:
    \begin{equation}
    \frac{-\hbar^2}{2m} \frac{d^2 \Psi (x)}{dx^2} + V(x)\Psi (x) = E\Psi (x)
    \end{equation}

    3. The attempt at a solution
    a) Using the uncertainty principle (where ##\Delta x = L##):
    \begin{align*}
    \Delta p
    &= m \Delta v \geq \frac{h}{4 \pi \Delta x} &\iff\\
    \Delta v &\geq \frac{h}{4\pi m \Delta x}\\
    &=57 885 643.48 m/s\\
    &\approx 0.19c
    \end{align*}
    Since this is the minimum uncertainty (standard deviation) of a normally distributed speed, the expected value of it must be even larger. Therefore the speed must be relativistic (Not sure my logic holds here).

    b) We are dealing with a potential box, which to me implies that outside of the box, and at it's edges, ##V(x) \rightarrow \infty##, and inside the box's boundaries ##V(x) = 0##.

    Then the time independent Shcrödinger equation becomes:
    [tex]\frac{-\hbar^2}{2m} \frac{d^2 \Psi (x)}{dx^2} = E\Psi (x)[/tex]

    Let's now assume that the equations are of the form: [tex]\Psi (x) = A \sin (kx)[/tex] and [tex]\Psi (x) = B \cos(kx).[/tex] Then by invoking the continuity requirement at ##x = 0##:
    [tex]\Psi (0) = B \cos (k 0) \Rightarrow B = 0.[/tex]

    On the other hand,
    [tex] \Psi (L) = A \sin (kL) = 0 \Rightarrow kL = n \pi \iff k = n \pi L^{-1}.[/tex]

    Finally, since we chose to try the functions above as the solutions, we have in both cases:
    [tex]\Psi ''(x) = -k^2 \Psi (x).[/tex]

    Therefore ##\Psi (x)## is eliminated from the Schrödinger equation, and we have:
    [tex]E = \frac{k^2 \hbar^2}{2m} = \frac{(n \pi L^{-1})^2 \hbar^2}{2m} = n^2\frac{\hbar^2 \pi^2}{2mL^2} = n^2 \frac{h^2}{8mL^2}[/tex]

    This should be our solution. However, isn't this for the classical case? How would I get the relativistic version?

    c) Not sure what to do here, at all. If I had an expression for a relativistic energy, this would be a piece of cake. Just take the ratio [tex]\frac{E_{relativistic}}{E_{non-relativistic}}.[/tex]

    However, I don't have an expression for the numerator that I could use, since I don't know the actual speed of the electron.

    Help would be appreciated.
     
  2. jcsd
  3. Nov 6, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    The Schrodinger equation is non-relativistic. So, you can't use it to determine the relativistic energy.

    The electron is free inside the box, so you are right to treat the electron as having a definite wavelength. Your allowed values of k look good. Can you determine the relativistic energy from k?
     
  4. Nov 6, 2016 #3
    Well, according to de Broglie:
    [tex]\lambda = \frac{h}{p} = \frac{2\pi}{k} \iff p = \frac{hk}{2\pi},[/tex]
    and the total relativistic energy of a particle is
    [tex]E = \sqrt{(pc)^2 + (mc^2)^2}
    = \sqrt{\left(\frac{hk}{2\pi} c\right)^2 + (mc^2)^2}
    = \sqrt{\left(\frac{h\frac{\pi}{L}}{2\pi} c\right)^2 + (mc^2)^2}
    =\sqrt{\left(\frac{h}{2L} c\right)^2 + (mc^2)^2}.
    [/tex]
    I can't believe I didn't remember this before, I just did this a month ago...

    Just to check, the non- relativistic case gives me:
    [tex]
    E = 1^2 \frac{h^2}{8m_e L^2} = 6.024 794 \cdot 10^{-14}J,
    [/tex]
    and the relativistic one is
    [tex]
    E = \sqrt{\left(\frac{h}{2L} c\right)^2 + (m_e c^2)^2} = 1.287 114 \cdot 10^{-13}J.
    [/tex]

    The numbers look a bit off. If I take the ratio ##\frac{E_{relat}}{E_{non-relat}}##, I get something like ##2##, which is way off.
     
  5. Nov 6, 2016 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    The nonrelativistic energy is the kinetic energy of the particle in the box.

    To compare with the relativistic case, you should compare with the relativistic kinetic energy.

    (Your calculations look good.)
     
  6. Nov 6, 2016 #5
    Yeah, that was it.

    [tex]E = \sqrt{\left( \frac{h}{2L} c \right)^2 + m^2c^4}-mc^2.[/tex] This gives the correct answer.

    Thanks.
     
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