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A knowledgeable reader might be confused by our assignment of statistics such that an electron tied to an even (odd) number of flux quanta is a fermion (boson). In reality, an electron tied to an integer number of flux quanta is a fermion, whereas one tied to a halfodd-integer flux quanta is a boson. To see this, consider an electron tied to α flux quanta. Take it around another identical composite particle in a closed loop. The total phase for this process has three contributions: 2π due to the electron going around the electron 2nd due to the electron going around the flux tube; and 2nd due to the flux tube going around the electron. The total phase is thus (2α + l)2π. Since a complete loop is equivalent to two exchanges, the phase corresponding to an exchange is half of this, which leads to a phase factor of This gives a fermion when α is an integer, and a boson when it is half-odd-integer. How does one reconcile this with the discussion in the text? It is easy to see that the assignment of statistics of the composite particles in the text amounts to a convention in which the phase due to the flux going around the charge is not counted. This leads to correct results provided one is consistent. To show this explicitly, we reformulate below our argument in terms of composite fermions that are electrons tied to an integer number of flux quanta. In this case one starts with an IQHE state Φn and adds an integer number (say m) of fhix quanta to each electron. One needs to be careful in determining the effective magnetic field of this new state; it is determined by the Aharonov-Bohm phase of an electron taken around in a closed loop of area A. Adding all three contributions, it is easy to see that it corresponds to a filling factor p given by equation (21). For writing the trial states, the natural way of adding m flux quanta to each electron is to multiply the state by which is the same as This shows that the convention used in the text produces the correct results. We have preferred to use this convention because it is widely used in the mean-field approach, and because it gives a more intuitive way of obtaining the effective magnetic field and filling factor of the new state
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A knowledgeable reader might be confused by our assignment of statistics such that an electron tied to an even (odd) number of flux quanta is a fermion (boson). In reality, an electron tied to an integer number of flux quanta is a fermion, whereas one tied to a halfodd-integer flux quanta is a boson. To see this, consider an electron tied to α flux quanta. Take it around another identical composite particle in a closed loop. The total phase for this process has three contributions: 2π due to the electron going around the electron; 2nd due to the electron going around the flux tube; and 2nd due to the flux tube going around the electron. The total phase is thus (2α + l)2π. Since a complete loop is equivalent to two exchanges, the phase corresponding to an exchange is half of this, which leads to a phase factor of This gives a fermion when α is an integer, and a boson when it is half-odd-integer. How does one reconcile this with the discussion in the text? It is easy to see that the assignment of statistics of the composite particles in the text amounts to a convention in which the phase due to the flux going around the charge is not counted. This leads to correct results provided one is consistent. To show this explicitly, we reformulate below our argument in terms of composite fermions that are electrons tied to an integer number of flux quanta. In this case one starts with an IQHE state Φn and adds an integer number (say m) of fhix quanta to each electron. One needs to be careful in determining the effective magnetic field of this new state; it is determined by the Aharonov-Bohm phase of an electron taken around in a closed loop of area A. Adding all three contributions, it is easy to see that it corresponds to a filling factor p given by equation (21). For writing the trial states, the natural way of adding m flux quanta to each electron is to multiply the state by which is the same as This shows that the convention used in the text produces the correct results. We have preferred to use this convention because it is widely used in the mean-field approach, and because it gives a more intuitive way of obtaining the effective magnetic field and filling factor of the new state.
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