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I
{\displaystyle I}
E
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N
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{\displaystyle E_{\uparrow }(k)=\epsilon (k)+I{\frac {N_{\uparrow }-N_{\downarrow }}{N}},\qquad E_{\downarrow }(k)=\epsilon (k)-I{\frac {N_{\uparrow }-N_{\downarrow }}{N}}}
be the other way around? If we for example assume a fully polarized state with
N
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N
,
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0
{\displaystyle N_{\uparrow }=N,N_{\downarrow }=0}
E
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I
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,
{\displaystyle E_{\uparrow }(k)=\epsilon (k)+I,\qquad E_{\downarrow }(k)=\epsilon (k)-I,}
with the total energy
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I
{\displaystyle E_{T}=\sum _{k,\sigma }E_{\sigma }(k)=\sum _{k}E_{\uparrow }(k)=E_{kin}+NI}
In general we have
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2
{\displaystyle E_{T}=\sum _{k,\sigma }E_{\sigma }(k)=\sum _{k,\sigma }\epsilon (k)+\sum _{\sigma }\sigma N_{\sigma }IP=E_{kin}+NIP^{2}}
But that means that the polarization will always increase the total energy. Only a negative
I
{\displaystyle I}
Jan Krieg (talk ) 14:55, 9 April 2014 (UTC) Reply
67.198.37.16 (talk ) 17:09, 15 May 2024 (UTC) Reply 
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