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I
{\displaystyle I}
E
↑
(
k
)
=
ϵ
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k
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+
I
N
↑
−
N
↓
N
,
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↓
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k
)
=
ϵ
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−
I
N
↑
−
N
↓
N
{\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
↑
=
N
,
N
↓
=
0
{\displaystyle N_{\uparrow }=N,N_{\downarrow }=0}
E
↑
(
k
)
=
ϵ
(
k
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+
I
,
E
↓
(
k
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=
ϵ
(
k
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−
I
,
{\displaystyle E_{\uparrow }(k)=\epsilon (k)+I,\qquad E_{\downarrow }(k)=\epsilon (k)-I,}
with the total energy
E
T
=
∑
k
,
σ
E
σ
(
k
)
=
∑
k
E
↑
(
k
)
=
E
k
i
n
+
N
I
{\displaystyle E_{T}=\sum _{k,\sigma }E_{\sigma }(k)=\sum _{k}E_{\uparrow }(k)=E_{kin}+NI}
In general we have
E
T
=
∑
k
,
σ
E
σ
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k
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=
∑
k
,
σ
ϵ
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k
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+
∑
σ
σ
N
σ
I
P
=
E
k
i
n
+
N
I
P
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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