4.3: Translational Symmetry and Conservation of Momentum

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Consider a transformation operator in 1 dimension \(\hat{D}\) which acts on the coordinates of a system as a displacement \(\hat{D}[f(x)] = f(x+l)\). The eigenfunctions of \(\hat{D}\) satisfy \(\hat{D}|\phi (x) \rangle = d |\phi (x) \rangle = |\phi (x+l) \rangle\). The general solutions to this equation are \(\phi (x) = e^{ikx}u(x)\) where \(u(x)\) satisfies \(u(x) = u(x + l)\) and \(k\) is complex.

This kind of translational symmetry exists when we have a crystal structure. Now consider a 1D closed loop of N atoms: Uniqueness of the wavefunction requires that \(\phi (x) = \phi (x + Nl) \Rightarrow e^{ikx} = e^{ik(x+N l)}\). Thus possible wavefunctions must have real \(k\) and the form

\[\phi (x) = e^{2\pi nix/N l}u(x); \quad k = 2\pi n/Nl \nonumber\]

The momentum of this state is given by:

\[−i\hbar \int \phi^* (x) \frac{d\phi (x)}{dx} dx = \int \phi^* (x) \left[ \frac{2\pi \hbar n}{Nl} + \frac{u' (x)}{u(x)} \right] \phi (x)dx \nonumber\]

The RHS first term gives is the familiar \(\hbar k\), which we associate with the momentum. If \(u(x)\) has some definite parity, then \(u' (x)\) will have opposite parity and the second term will be the integral of an odd function (i.e. zero by symmetry). Thus \(k\) is a quantum number associated with translational symmetry, which in turn has an operator \(\hat{D}\) which commutes with the Hamiltonian and is thus a constant of the motion. Translational symmetry is associated with conservation of momentum.

For which the TISE, with the atom described by a potential \(V (x)\), and a particular value of \(k\), can be written

\[\hat{H}_k u_k (x) = \frac{\hbar^2}{2m} \left[ (k − i \frac{d}{dx})^2 + V (x) \right] u_k(x) = E_k u_k (x) \nonumber\]

since the phase has been eliminated, we simply have a particle in a fixed volume \(u_k(x) = u_k(x+l))\), which means a series of discrete energy levels (bands). Thus all states can be labelled by \(k\) and a band index \(n\).

We can write the semiclassical group velocity of the wavefunction as

\[v_g = \frac{d\omega}{dk} = \frac{1}{\hbar} \frac{dE}{dk} \nonumber\]

using \(E = \hbar\omega\). A formal proof using the velocity operator gives the same result for the velocity. Assuming that \(E\) does vary with \(k\), this means we have a time-independent state which nevertheless has a permanent, non-zero velocity through the lattice.