Nonstandard physics/Quantum Hall effect

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This page contains inexplicable physics discussions.
For comparison, see the Wikipedia page on this topic.

This page is about the quantum Hall effect. It builds upon the knowledge of:

It turns out that when a free electron is placed in a magnetic field, its energy levels go from being continuous (by the virtue of being free particles) to being a discrete set. What follows is the calculation of this phenomenom, which is called the Quantum Hall Effect.

Deduction of the quantization of electrons' energy levels[edit]

For a free electrons gas in a magnetic field, the Hamiltonian is:

    H = \frac{1}{2m} \left|-i \hbar \nabla + \frac{e}{c} \textbf{A}\right|^2 

(Note we will be using Gaussian units.) When the magnetic field is constant, \textbf{H} = H \textbf{k}[1], we can choose for our vector potential Landau's gauge:

    \textbf{A} = - H y \textbf{i}

This gauge will simplify the resolution of Schrödinger's equation for this Hamiltonian.

Developping the square in the Hamiltonian (do it!) we arrive at:

    H = \frac{\hbar^2}{2m} \left( -\Delta + K^4 y^2 + 2 K^2 y \cdot i \partial _x \right) 

where we have called K^2 = \frac{e H}{\hbar c}. We then, as usual, try to find a base of eigenstates[2] which make Schrödinger's equation separable:

    \Psi(x,y) = X(x) \cdot Y(y)
    H[\Psi(x,y)] = E \cdot \Psi(x,y)

We have:

    \partial _x[\Psi] = \frac{X'}{X} \cdot \Psi
    \partial _x^2[\Psi] = \frac{X''}{X} \cdot \Psi

So:

    H[\Psi] = \frac{\hbar^2}{2m} \cdot \left(- \frac{X''}{X} - \frac{Y''}{Y} + K^4 y^2 + 2 K^2 y \cdot i \frac{X'}{X} \right) \Psi

The last term inside the parenthesis imposes that the equation will only be separable if X´/X is constant. The only such base that doesn't diverge at infinite is Fourier's base:

     X(x) = e^{i k \cdot x}
     X'(x) = i k \cdot X(x) 
     X''(x) = -k^2 \cdot X(x) 

We then substitute our solution for X in Schrödinger's equation:

    \frac{\hbar^2}{2m} \cdot \left(k^2 - \frac{Y''}{Y} + K^4 y^2 - 2 K^2 k y \right) \Psi = E \Psi

Now the parenthesis is not an operator, but a scalar[3], so we can drop the \Psi. Each base function Y must satisfy:

    \frac{\hbar^2}{2m} \cdot \left(k^2 - \frac{Y''}{Y} + K^4 y^2 - 2 K^2 k y \right) = E
    k^2 - \frac{Y''}{Y} + K^4 y^2 - 2 K^2 k y = \frac{2mE}{\hbar^2}
    \frac{Y''}{Y} = K^4 y^2 - 2 K^2 k y + \left(k^2 - \frac{2mE}{\hbar^2}\right)

Which has a smell of Quantum Harmonic Oscillator (id est, Hermite's equation). We can easily transform it algebraically by converting the right hand side to a perfect square (through a change of variable):

See also[edit]

What are you talking about?[edit]

  1. The k vector is the unit vector along the Z axis. The Z axis is chosen here (as is standard practice) just for the sake of giving a name to the direction of the magnetic field. X (the i vector) and Y (the j vector) are thus whatever two axes are perpendicular to the magnetic field here and follow the w:right-hand rule.
  2. I.e., states of the system, wavefunctions if you like, for which the time independent Schroedinger equation holds.
  3. I.e., a number. Operators can be thought of as matrices that multiply the wavefunctions, which would be vectors. That means that in Schroedinger's equation (H*psi) would be a vector, and (E*psi) too, but you cannot drop the psi because H is a matrix while E is a number. In our case up here, we have found a vector for which the Hamiltonian operator is effectivly just multiplying it by a number. Thus that number is equal to E, and we can from that calculate the value of E. This method of solving matricial differential equations is analogous to Diagonalization. The value obtained for E will be the energy of that particular state and, as is common e.g. in Fourier problems, is restricted to a discrete set of values (the energy is quantizised).