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Quantization of angular momentum

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Imagine a semi-classical birdcage of radius R with N regularly spaced bars individually separated by a spacing a. Now imagine there is a linear light source centered along the cylinder's axis z.

Use the dual wave/particle nature of light to show that angular momentum is quantized in the scattered light leaving the cage.


I'm having trouble starting this problem up. I think I'll have to solve some version of Schrodinger's equation, but I'm not sure. I thought of using Bragg's law, nλ = 2dsinθ, where I would set d = a in this case and try to figure out the case where n = 1. Can I assume a is quantized and have a = 2πR/N? I'm kind of lost so any help starting this problem up would be of great help.
 

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  • #2
TSny
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Think of the bars as forming a diffraction grating. The light travels from the z-axis radially outward toward the grating. What are the possible angles that the light can be diffracted relative to the radial direction?

Consider a single photon diffracted at one of those angles. How much linear momentum does the photon carry? How much angular momentum does this linear momentum produce about the z-axis?
 
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Think of the bars as forming a diffraction grating. The light travels from the z-axis radially outward toward the grating. What are the possible angles that the light can be diffracted relative to the radial direction?

Consider a single photon diffracted at one of those angles. How much linear momentum does the photon carry? How much angular momentum does this linear momentum produce about the z-axis?
Would the possible angles be where sinθ = nλ/a?

The linear momentum of the photon is p = E/c so if energy is quantized could I say that for
n = 1, p = 1.5ħω/c?

I don't know how to calculate the angular momentum of a photon.
 
  • #4
TSny
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Would the possible angles be where sinθ = nλ/a?
Yes, that look's good.
The linear momentum of the photon is p = E/c so if energy is quantized could I say that for
n = 1, p = 1.5ħω/c?

I don't know how to calculate the angular momentum of a photon.
You'll want to express the linear momentum of the photon in terms of the wavelength rather than the energy so that you can relate it to your formula for sinθ.

When the photon heads out in one of the allowed directions θ, it will carry linear momentum along that direction. From classical mechanics, we know how to calculate the angular momentum of a particle in terms of its linear momentum and the "lever arm" defined by the perpendicular distance from the origin to the line of motion of the particle. You'll probably want to draw a picture showing the line along which the photon travels after it exists the bird cage and use the picture to find an expression for the lever arm.
 
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Yes, that look's good.


You'll want to express the linear momentum of the photon in terms of the wavelength rather than the energy so that you can relate it to your formula for sinθ.

When the photon heads out in one of the allowed directions θ, it will carry linear momentum along that direction. From classical mechanics, we know how to calculate the angular momentum of a particle in terms of its linear momentum and the "lever arm" defined by the perpendicular distance from the origin to the line of motion of the particle. You'll probably want to draw a picture showing the line along which the photon travels after it exists the bird cage and use the picture to find an expression for the lever arm.
Okay so p = h/λ, therefore sinθ = nh/ap. For the angular momentum wouldn't the lever arm just be the radius of the birdcage? So L = Rp = Rnh/asinθ?
 
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TSny
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No, the line along which the photon travels will not be tangent to the bird cage. So, the lever arm will not equal R.
 
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No, the line along which the photon travels will not be tangent to the bird cage. So, the lever arm will not equal R.
The line it travels is radially outward so I don't know what to use as the perpendicular since it's trajectory is parallel to the radius of the cage.
 
  • #8
TSny
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The line it travels is radially outward so I don't know what to use as the perpendicular since it's trajectory is parallel to the radius of the cage.
The light travels radially from the center until it reaches the bars (i.e., the "grating"). The light will exit at one of the allowed angles θ relative to the radial direction. For θ = 0, the photon continues in the radial direction so that its line of travel has zero lever arm relative to the center of the cage (no angular momentum). But for θ ≠ 0, the line of travel after exiting the cage will not be radial.
 
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The light travels radially from the center until it reaches the bars (i.e., the "grating"). The light will exit at one of the allowed angles θ relative to the radial direction. For θ = 0, the photon continues in the radial direction so that its line of travel has zero lever arm relative to the center of the cage (no angular momentum). But for θ ≠ 0, the line of travel after exiting the cage will not be radial.
Okay, so I got the lever arm as being Rsinθ using the law of sines but in this case θ is in degrees rather than radians.
 
  • #10
TSny
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Yes, the lever arm is Rsinθ. (It doesn't matter whether you're using degrees or radians.) So, what do you get for the angular momentum carried away by the photon?
 
  • #11
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Yes, the lever arm is Rsinθ. (It doesn't matter whether you're using degrees or radians.) So, what do you get for the angular momentum carried away by the photon?
So L = Rsinθp; sinθ = nh/ap so p = nh/asinθ. That gives L = Rsinθnh/asinθ, so L = Rnh/a.
Which is the solution, right?
 
  • #12
TSny
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That's what I got. You can write the answer in a little neater form by using the fact that the circumference of the cage should be Na and expressing the result in terms of h-bar rather than h.
 
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That's what I got. You can write the answer in a little neater form by using the fact that the circumference of the cage should be Na and expressing the result in terms of h-bar rather than h.
Okay, so L = Nnh-bar. Thank you for all your help.
 

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