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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?

EDIT Here is a link to what I mean

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"?

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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
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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.

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Is there a difference in quantisation if the nanotube is armchair or zig-zag? –  Josh Apr 10 '12 at 19:04
    
In a jellium model, where you have a positively charged homogenous background no. A detailed calculation including the carbon lattice gives you for each nanotube a different band structure but the essential features are still the same. –  DaniH Apr 10 '12 at 21:11
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it really makes sense if you take a slice of the bunch of nanofibers in any axial plane, and you'll notice that the fibers form roughly a regular lattice. A regular lattice implies that the longest wavelength not severely absorbed by the carbon walls are of the order of the lattice separation. Also the integer multiples of that wavelength will satisfy the same boundary conditions at the lattice walls.

Obviously, such rule cannot be anything else than a raw approximation. In real fibers, the nanotube separation will nearly never been completely regular

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