Simple Quantum Paramagnet, Canonical Ensemble Origin of magnetic - PowerPoint PPT Presentation

Simple Quantum Paramagnet, Canonical Ensemble Origin of magnetic moments: Electron spin and orbital angular momentum � + L � ≡ J � � S µ � = g µ J µ ≡ e ¯ h/ 2 m e c J B B Nuclear angular momentum � � I µ � = g µ I µ ≡ e ¯ h/ 2 m p c I N N 8.044 L 16 B1

∂ m = − gµ H m B 2J+1 m = J, J − 1 , · · · − J g µ B H sinh[( J + 1 J J 2 ) � ] � ) m − ∂ m /kT � � Z 1 ( T, H ) = e = ( e = 1 � ] sinh[ 2 m = − J m = − J gµ H level spacing B � ≡ = kT kT Z = Z 1 ( � ) N = Z ( � ) at fixed N Note Z 1 = Z 1 ( � ) 8.044 L 16 B2 a

∂ m = − gµ H m B 2J+1 m = J, J − 1 , · · · − J g µ B H sinh[( J + 1 J J 2 ) � ] � ) m − ∂ m /kT � � Z 1 ( T, H ) = e = ( e = 1 � ] sinh[ 2 m = − J m = − J gµ H level spacing B � ≡ = kT kT Z = Z 1 ( � ) N = Z ( � ) at fixed N Note Z 1 = Z 1 ( � ) 8.044 L 16 B2 b

− ∂ m /kT /Z 1 = e � m /Z 1 p ( m ) = e ( gµ m ) e � m    1 �Z 1 B < µ > = � = gµ  B Z 1 Z 1 �� m ≡ gµ JB ( � ) M = N < µ > = gµ NJB ( � ) B J B J   1  1 �Z 1 B ( � ) =  J J Z 1 �� 1 1 1 1 1 � � = ( J + ) coth[( J + ) � ] − coth[ � ] J 2 2 2 2 This is called the “Brillouin Function”. 8.044 L 16 B3

J + 1 1 x lim B ( η ) = η coth x � + x « 1 J 3 η � 0 3 x − η e − 2 x lim B ( η ) = 1 − coth x � 1 + 2 e x » 1 J η �⇐ J ( gµ ) 2 J ( J + 1) H B M � N High T (Curie Law) 3 kT � � 1 − η � N gµ J 1 − Low T (Energy Gap) e B J 8.044 L 16 B4

MAGNETIZATION OF A QUANTUM PARAMAGNET 1 0.8 0.6 0.4 0.2 1 2 3 4 1 0.8 0.6 0.4 y c 0.2 1 2 3 4 x 8.044 L16B5

( gµ ) 2 J � � � � ∂M ∂η = Ngµ J B → ( η ) B → ( η ) B χ = N ≡ T B J J ∂H ∂H kT T T gµ H B Note: T and H enter only through η ≡ kT � � � � ∂η η ∂η η = = − ∂H H ∂T T T H We now show that this � U = 0. 8.044 L 16 B 6

dU = TdS + HdM �� � � � � �� � � � � ∂S ∂S ∂M ∂M = dT + + H dT + T dH dH ∂T ∂H ∂T ∂H H T H T � � � � � � � � � � � � ∂S ∂M ∂S ∂M = T + H dT + T + H dH ∂T ∂T ∂H ∂H H H T T � �� � � �� � 0 0 = 0 for all paths ⇒ U = 0 8.044 L 16 B 7

� � � � � � � � ∂S ∂M ∂S ∂S T + H = T + H ∂T ∂T H ∂T ∂H H H T � �� � � �� � S → ( η )( − η/T ) S → ( η )( η/H ) = − ηS → ( η ) + ηS → ( η ) = 0 A similar expansion shows that the other term is also zero. 8.044 L 16 B 8

⇒ Internal Energy dU = d /Q + d /W = TdS + HdM dU ≡ adiabatic ( d /Q = 0) work /Q = TdS, d /Q = 0 � dS = 0 � dM = 0 � dU = 0 d dU = 0 for any change: U = 0 for this model But E ≡ N < ∂ > = − HM = 0 !! 8.044 L 16 B 9

� Energy = energy to create H field 1 � + energy to assemble M 2 + energy to move M into H � 3 8.044 L 16 B 10

� 1 does not appear when using d /W = HdM . � 2 We did not create the � µ . They do not interact. � U = 0 in the current example. � � · M � , energy of macroscopic moment in H � 3 − H equals N < ∂ > in the current example. 8.044 L 16 B 11

So what’s the result ? < H > S. M. ⇒ = U Thermo = U assembly + ( − � H · � M ) position (for d /W = HdM ) �     1 1 dZ ∂Z ∂η < H > = − = − = − HM     Z ∂β Z dη ∂β H H � �� � � �� � M/gµ B gµ B H 8.044 L 16 B 12

� � � � d /Q ∂S C M ≡ = T = 0 since S = S ( M ) dT ∂T M M � � � � � � d /Q 1 ∂M C = dU − HdM = − H ≡ H dT T ���� ∂T H H 0 H = NkJη 2 B � ( η ) J 8.044 L16B13

HEAT CAPACITY OF A QUANTUM PARAMAGNET 8.044 L16B14

Entropy of a Quantum Paramagnet • When is − kT ln Z ⇒ = F ? • How is a paramagnet like a sponge? 8.044 L 16 B 15

HIGH AND LOW TEMPERATURE BEHAVIOR OF A QUANTUM PARAMAGNET kT >> g µ B H kT << g µ B H g µ B H ENERGY LEVELS ALMOST MOMENT ALMOST MAXIMUM, EQUALLY POPULATED, ENERGY GAP BEHAVIOR CURIE LAW BEHAVIOR 8.044 L16B16

S ( kT « gµ H ) � Nk ln(1) = 0 B S ( kT » gµ H ) � Nk ln(2 J + 1) B J η ) m � Z 1 = ( e η ≡ gµ H/kT B m = − J Try − kT ln Z = F = U − TS � S = k ln Z = Nk ln Z 1 ���� 0 8.044 L 1 6B 17

S ( kT « gµ H ) � Nk ln(1) = 0 B S ( kT » gµ H ) � Nk ln(2 J + 1) Nk ln(2 J + 1) O.K. B J η ) m � Z 1 = ( e η ≡ gµ H/kT B m = − J Try − kT ln Z = F = U − TS � S = k ln Z = Nk ln Z 1 ���� 0 8.044 L 1 6B 18 •

S ( kT « gµ H ) � Nk ln(1) = 0 NkJ ( gµ H/kT ) wrong! B B S ( kT » gµ H ) � Nk ln(2 J + 1) Nk ln(2 J + 1) O.K. B J η ) m � Z 1 = ( e η ≡ gµ H/kT B m = − J Try − kT ln Z = F = U − TS � S = k ln Z = Nk ln Z 1 ���� ���� � �� � 0 wrong wrong � 8.044 L 1 6B 19

In the derivation of the canonical ensemble we found − kT ln Z = < E 1 > − TS 1 where < E 1 > = < H ( { p, q } ) > Then we set < E 1 > = U . But in the paramagnet < E 1 > = U − HM , thus − kT ln Z = U − HM − TS = G ( T, H ) for our model. ⇒ S = k ln Z − HM/T Nk ln Z 1 − H ( Ngµ B J ) /T → � ���� T → 0 8.044 L16B20

In general for magnetic systems, even when U = 0, dG = − SdT − MdH G ( T, H ) = − k B T ln Z � � ∂G M ( T, H ) = − ∂H T � � ∂G S ( T, H ) = − ∂T H 8.044 L16B21

ENTROPY OF A QUANTUM PARAMAGNET 8.044 L16B22

ADIABATIC DEMAGNETIZATION (MAGNETIC COOLING) INITIAL COOLING STAGE T initial 1 H = H initial THERMAL LINK 1 S M ~ 0 1 S total 1 S S high MAGNET ⊗ 1 T S = T initial kT << g µ B H SAMPLE 8.044 L16B23

ADIABATIC DEMAGNETIZATION (MAGNETIC COOLING) INITIAL COOLING STAGE T initial H ~ 0 S M ~ Nk ln(2J+1) ⊗ 2J+1 S total S S low kT >> g µ B H T S << T initial SAMPLE 8.044 L16B24

Electronic Example, CMN Cerium Mangesium Nitrate 2Ce(NO 3 ) 3 · 3Mg(NO 3 ) 2 · 24H 2 O Ce +++ J=5/2 T ordering ∼ 1.9 mK Cools 3 He and samples therein to ∼ 2 mK. 8.044 L 1 6B 25

Nuclear Example, Cu Metallic Copper Cu I=3/2 T ordering ∼ 1 µ K Cools Cu electrons and lattice to ∼ 10 µ K. 8.044 L 1 6B 26

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