Gás Ideal de Fermi - Artigo sobre calculo de energia de férmios. Mecânica quântica. PDF

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Artigo sobre calculo de energia de férmios.
Mecânica quântica. ...

Description

ln Z(T, V, µ) =

X

hnj i =

j

ln {1 + exp [−β (ǫj − µ)]}

1 exp [β (ǫj − µ) + 1]

Φ 1 Φ(T, V, µ) = − ln Z(T, V, µ) β

Φ = −pV

1 ln Z(T, V, µ) V →∞ V

⇒ p(T, µ) = kB T lim

U = hEi =

X

N = hN i =

j

ǫj hnj i =

X j

hnj i =

X j

X j

ǫj exp [β (ǫj − µ)] + 1

1 exp [β (ǫj − µ)] + 1

µ = µ(T, V, N )

Hmetal = TL + Te + HLL + Hee + HeL TL HLL Te

Hee HeL

n

Hmetal = {TL + HLL } + Te + Tee +

o

(0) HeL

n

+ HeL −

(0) HeL

(0)

Hel = Te + Hee + HeL M N N X X X −~2 2 1 X e2 = + + ∇ 2m i 2 i6=j |~ri − ~rj | i=1 α=1 i=1

Z´ e2     ~ r − R ~ i α

o

1023

N

Hel Ψ(~r1 , ..., ~rN ) =

−~2 X 2 ∇ (~r1 , ..., ~rN )Ψ(~r1 , ..., ~rN ) = Et Ψ(~r1 , ..., ~rN ) 2m i=1 i

He ψ(~r ) =

−~ 2 ∇ ψ(~r ) = ǫψ(~r) 2m

1 ~ ψ(~r ; ~σ ) = √ eik·~ru~σ V u~σ

~σ

ǫ~k,~σ =

~2 k 2 2m

~k = 2π (lx , ly , lz ) L lx , ly , lz = 0, ±1, ±2, ... N N Y 1 X p (−1) ψpi (~ri , ~σi ) Ψ= N! p i=1

i ~k

~σ

2V ln Z(T, V, µ) = (2π )3

ˆ

3

  2 2  ~ k k ln 1 + exp −β −µ 2m 

F

2V hF i = (2π )3

ˆ

3

2V kF (k) = (2π )3

ˆ

3/2 ˆ  2V 1 2m k4πk F (k) = 4π (2π )3 2 ~2 2

ǫǫ

1/2

F (ǫ) =

D(ǫ)

ˆ

ǫD(ǫ)F (ǫ)

V 2π D(ǫ) = 2 (2π )3



2m ~2

3/2

ǫ

1/2

1 = 2 4π



2m ~2

3/2

ǫ1/2 ǫ

V m 1 g(ǫ) = D(ǫ) = 2 2 V π ~

r

2mǫ ~2

g g(E) E N

g(ǫ) N =1

ǫ(~k)

Ln D(ǫ) = γ (2π )n =

γ

γ

ˆ

Ln (2π )n

ǫ=

ˆ

~ǫ S    ∇  ~ ~k ǫ(~k)   n ~ k ) kδ ǫ − ǫ(

γ = 2 ~ǫ S

n n −1

n

L

~k

ǫ

ln Z(T, V, µ) =



ˆ

∞

ǫD(ǫ) ln {1 + exp [−β(ǫ − µ)]}

0

 n~k =

1   ~2 k2  exp β 2m − µ + 1   n~k → f (ǫ)

f (ǫ) =

1 exp [β(ǫ − µ)] + 1

   2 2  −1 ˆ ~ k V 3 k exp β −µ +1 N = 2 (2π )3 2m ˆ ∞ = ǫD(ǫ)f (ǫ) 0

 −1    2 2 ˆ 2 2 V ~ k 3 ~ k −µ +1 U = 2 k exp β (2π )3 2m 2m ˆ ∞ ǫD (ǫ)ǫf (ǫ) = 0

~k = 0

ln Z = = = = =

ˆ

∞

ǫD(ǫ) ln {1 + exp [−β(ǫ − µ)]} 3/2 ˆ ∞  1 2m ǫǫ1/2 ln {1 + exp [−β(ǫ − µ)]} 2 2 ~ 4π 0 3/2 3/2   ˆ ∞ 1 1 2 3/2 2 2m 2m ∞ ǫǫ3/2 f (ǫ) βV − 2 ǫ βV f (ǫ)| 0 + 2 3 4π 4π ~2 ~2 3 0 3/2  ˆ ∞ 1 2 2m ǫǫ3/2 f (ǫ) βV 3 4π 2 ~2 0 ˆ ∞ 2 V 2 U ǫD (ǫ)ǫf (ǫ) = 3 kB T 0 3 kB T 0

Φ = −kB T ln Z = −pV

3 U = pV 2

T →0

f (ǫ) → Θ(µ − ǫ) T = 0 ǫF

N = =

X

f (~k; ~σ ) =

~ ˆk~σǫF

ˆ

∞

ǫD(ǫ)f (ǫ)

0

ǫD(ǫ)

0

1 = 2V 2 4π 1 = 2V 2 4π





2m ~2 2m ~2

3/2 ˆ

ǫF

ǫǫ1/2

0

3/2

n=

2 1/2 2 3/2 2 1/2 ǫF = ǫF D(ǫF ) = V ǫF g(ǫF ) 3 3 3

N 2 g(ǫF ) = ǫ1/2 V 3 F

 2/3 ~2 2/3 ǫF = 3π 2 n 2m

ǫF =

~2 k 2F 2m

kF

 1/3 kF = 3π 2 n TF

TF = n ≈ 1023 T →0 kB T (300K) = 25meV

T ≈0

ǫF kB

TF ≈ 104 K T(

) ≪ TF

GaAs

n

ǫc (~k, ~σ ) =

~2 k 2 2m

mc

GaAs mc = 0, 067 m0

m0 1018 cm−3 meV

T ≪ TF

T ≪ TF

T =0 D(ǫ)f (ǫ)

kB T

∆N ≈ g (ǫF )V kB T

∆U ≈ kB T ∆N = V g (ǫF )(kB T )2

cV

1 = N



∂U ∂T T = 3kB TF



V,N

≈2

V g(ǫF )kB2 T N

cV = 3kB

TF = ǫF /kB

5 × 104 K

n(ǫ) = D(ǫ)f (ǫ)

2kB T

µ

K

cV = γT + δT 3 γ

δ

γ cV /T × T 2

T2 = 0 Na

Ag

γ

Na

Ag

T =0 TF

T ≪ TF

I= φ(ǫ) = Aǫn A

ˆ

∞

ǫf (ǫ)φ(ǫ) 0

n ≥ 1/2

T ≪ TF

f (ǫ) ǫ=µ

I=

f (ǫ)ψ(ǫ)|0∞ −

ˆ

ψ(ǫ) =

ˆ

∞

ǫψ(ǫ)f ′ (ǫ) 0

ǫ

ǫ′ φ(ǫ′ )

0

ǫ=0

ψ(ǫ) = 0

f (ǫ)

I=−

ˆ

ǫ→∞

∞

ǫψ(ǫ)f ′ (ǫ) 0

f ′ (ǫ)

ǫ=µ

∞

X 1 kψ ψ ψ(ǫ) = ψ(µ) + |ǫ=µ (ǫ − µ)k |ǫ=µ (ǫ − µ) + ...+ = k! ǫk ǫ k=0

Ik = −

ˆ

∞

ǫ(ǫ − µ)k f ′ (ǫ) =

0

1 βk

ˆ

∞

x −βµ

ex xk (ex + 1)2

k = 0, 1, 2, ...

−∞ exp(−βǫF ) Ik =

1 βk

ˆ

∞

−∞

k

x

ex xk (ex + 1)2

+ 0[exp(−βǫF )] k

I0 = 1 π2 3β 2

I2 =

I=

U = 2V

ˆ

ˆ

µ

ǫφ(ǫ) + 0

φ π2 (kB T )2 |ǫ=µ + ... 6 ǫ

∞

ǫCǫ

f (ǫ) = 2V C



f (ǫ) = 2V C



3/2

0

N = 2V

ˆ

∞

ǫCǫ

1/2

0

 2 5/2 π 2 2 1/2 µ + (kB T ) µ + ... 5 4

 2 3/2 π 2 2 −1/2 (kB T ) µ + ... µ + 12 3

g(ǫ) = Cǫ1/2

3/2 ǫF

3/2

=µ

(

π2 1+ 8



kB T µ

2

)

+ ...

N/V

µ = ǫF

(

π2 1− 12



T TF

2

)

+ ...

3 U = N ǫF 5

(

5π 2 1+ 12

cV =



T TF

π2 T kB 2 TF

2

)

+ ......