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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 5 months |
| seen | Feb 14 at 16:49 | |
| stats | profile views | 9 |
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Jan 23 |
awarded | Enthusiast |
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Dec 31 |
answered | How exactly do superstrings reduce the number of dimensions in bosonic string theory from 26 to 10 and remove the tachyons? |
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Dec 30 |
awarded | Supporter |
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Dec 27 |
comment |
The Double Slit Experiment - delayed measurement Ah, thank you. Done. |
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Dec 27 |
awarded | Editor |
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Dec 27 |
revised |
The Double Slit Experiment - delayed measurement added 3 characters in body |
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Dec 27 |
comment |
The Double Slit Experiment - delayed measurement I probably should have been clearer in my wording. I intended to define the term "well constructed" in the context of this particular question as an experiment such that the existence of path information completely destroys the interference pattern. Indeed, this was to avoid the exact situation that your link describes. Obviously, the interference pattern need not constitute a perfect momentum measurement nor need the path information constitute a perfect position measurement so it is feasible to get both. Perhaps I should have said "well constructed quantum eraser experiment." |
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Dec 27 |
answered | The Double Slit Experiment - delayed measurement |
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Dec 27 |
comment |
Why do many people say vector fields describe spin-1 particle but omit the spin-0 part? For our purposes you should completely disregard what the (1/2,1/2) irrep decomposes into under SO(3). Yes, it is true that this decomposition must always contain the correct spin representation, but that is not necessary to understand the spin. The point is that if you take a (1/2,1/2) field like the photon and calculate the creation and annihilation operators you will find that these operators always create spin 1 single particle states and never spin 0 states. That's why we say that the (1/2,1/2) irrep is spin 1. |
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Dec 26 |
comment |
Why do many people say vector fields describe spin-1 particle but omit the spin-0 part? So, you can see that a massive vector field (Lorentz group) is spin 1 because it creates single particle states that transform in the spin 1 representation of the little group SO(3), not because it decomposes into the spin 1 representation under the SO(3) subgroup. |
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Dec 26 |
comment |
Why do many people say vector fields describe spin-1 particle but omit the spin-0 part? Ok, I see what is going on here. You have to differentiate between representations of the Lorentz group and representations of the Poincare group. Of course you are right that the spin is defined by the representation of the little group. However, to determine the spin of a field you do not decompose the representation of the Lorentz group into representations of the little group. The little group is used to construct representations of the Poincare group in which single particle states transform. |
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Dec 25 |
awarded | Teacher |
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Dec 25 |
answered | Why do many people say vector fields describe spin-1 particle but omit the spin-0 part? |
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Dec 25 |
answered | Are branes in 4D-spacetime moving, or are they static? |
