If we solve the time-independent Schrödinger equation in 1D

$$\frac{d^2\psi(x)}{dx^2} + \frac{2m}{\hbar^2} \left[E - V(x)\right] \psi(x) = 0$$

for a periodic potential $V(x)$ with wave function $\psi(x)$ subject to a periodic boundary condition: $\psi(x) = \psi(x + Ga)$, where $a$ is the period of $V(x)$ and $G$ is a positive integer so that the lattice lenght $L = G a$, then the general form of $\psi(x)$ is given by $$ \psi_k(x) = e^{ikx} u_k(x), $$ with $u_k(x) = u_k(x+a)$ and $k = \frac{2\pi g}{Ga}, g \in \mathbb{Z}$. This is the famous Bloch theorem.

I am reading a textbook The Wave Mechanics of Electrons in Metals by Stanley Raimes where the author then introduces a new quantum number $l$, relabel $\psi$ and $u$ as $\psi_{kl}(x)$ and $u_{kl}(x)$ and plug $\psi_{kl}(x) = e^{ikx} u_{kl}(x)$ in the Schrödinger equation. Then he takes complex comnugates of the Schrödinger equaiton and compare it with the same equation but $k$ replaed by $-k$. These operations yield the following familiar equations:

$$u_{-kl}(x) = u_{kl}^*(x),$$


$$E_l(-k) = E_l(k).$$


Why do we need to introduce a new quantum number $l$ in the first place? If we wanted to solve the Schrödinger equation for a specified potential $V(x)$, then we would have seen how a new parameter $l$ makes its way into the wave function and energy expression, if needed. But here we are not doing that, so what is the reason?

  • $\begingroup$ I do not have access to the book, but could not $l$ be the label of the individual energy bands? The Bloch wavenumber $k$ is not enough to label all the states. You need another discrete index to label the bands. $\endgroup$
    – mike stone
    Commented Dec 26, 2019 at 16:20
  • $\begingroup$ You are right, but how do we know this information in the first place? Is that an assumption? $\endgroup$
    – rainman
    Commented Dec 26, 2019 at 16:24
  • 1
    $\begingroup$ It's not an assumption, but something one can deduce from looking at how, for any energy $E$, the linearly independent pair of solutions $\psi_1(x)$ with boundary conditions $\psi_1(0)=1$, $\psi'(0)=0$ and $\psi_2(x)$ with $\psi_2(0)=0$, $\psi'_2(0)=1$ map under the shift $x\to x+a$. If we set $\Psi=(\psi_1,\psi_2)^{\rm transpose}$ we must have $\Psi(x+a) = T \Psi(x)$ for some transfer matrix $T$. Consideration of ${\rm det}(T)$ gives you the bands. This is described in many books $\endgroup$
    – mike stone
    Commented Dec 26, 2019 at 17:09
  • $\begingroup$ Sorry! I meant ${\rm tr}(T)$ not ${\rm det}(T)$. The latter is alwyas unity. $T$ depends on $E$ and ${\rm tr}(T(E))= 2\cos k$ gives the relation between $E$ and $k$. So the energy range where ${\rm tr}(T(E))\le 2$ are the energy bands and the range where ${\rm tr}(T(E))\ge 2$ are the band gaps. $\endgroup$
    – mike stone
    Commented Dec 27, 2019 at 15:11

1 Answer 1


We have the Bloch's theorem to get the ansatz for the wavefunction, $ψ_{k}(x)=e^{ikx}u_{k}(x)$ to solve the Schrodinger's equation. We know that $u_{k}(x)$ is a periodic function in x with a period equal to the lattice constant, but we are still left with the task of finding what exactly $u_{k}(x)$ is. Substituting our ansatz into the Schrodinger equation now gives us a differential equation in $u_{k}(x)$ : $$ \left[ \frac{1}{2m}\left(\frac{ℏ}{i} \nabla + ℏk \right)^{2} + V(x) \right] u_{k}(x) = E u_{k}(x)$$

Now for this equation, for the given potential $V(x)$ and the given boundary condition, we can have different functions $u_{k}(x)$ that satisfy it, giving a spectrum of energies. Hence, we need a new quantum number to label these different states, which correspond to eigenstates of different energies. In a solid crystal, these correspond to the different energy bands.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.