# Quantization of momentum in nanotubes

I'm reading about carbon nanotubes and how the momentum (lets call it $k_x$) is quantized along the circumferential direction and not along the cylindrical (call this $k_y$). I can follow the maths okay, but what I don't understand is the physical reason why $k_x$ CANNOT take any value and that it must be quantized?

It's much more simple than what you thought, is it to do with the modes the electron can take? So if it is not one of these values it would interfere with itself around the circumferance like the "particle in the box"?

• Can you give a reference, as I have no idea what you're talking about. A quick Google gave hits referring to dark excitons. If this is what you mean I assume the length in the y direction is too long for the quantisation to be seen i.e. the level spacing is less than kT. – John Rennie Apr 10 '12 at 17:43
• Sure, I've edited in a link above. – Josh Apr 10 '12 at 17:55
• Ah OK, I think it's as i guessed above i.e. the circumference is much shorter than the length so there is only noticable quatisation around the circumference. – John Rennie Apr 10 '12 at 17:58
• So is it like what I said with the particle in a box and the electron being confined leading to the modes? – Josh Apr 10 '12 at 19:02
• Yes, sort of. The electron isn't confined in the sense it hits a potential barrier as in a box, but the fact it loops round the cylinder imposes similar boundary conditions, i.e. $\psi(x) = \psi(x + 2\pi r)$ so you get similar quantisation. – John Rennie Apr 10 '12 at 19:24

The reason is that you have periodic boundary conditions in the azimuthal direction while there are no special constraints along the cylinder axis (note that, as in the radial direction we have the $\pi$ bonds of the carbon lattice the electron's wavefunction must be strongly confined). Other way to see this, in the azimuthal direction you must have an integer number of electron wavelengths along the perimeter by continuity of the wavefunction $\psi$. As a consequence, at room temperature your only degree of freedom comes along the cylinder axis and your nanotube is a quasi 1D system.