phys212ln9

# phys212ln9 - Physics 212 Statistical mechanics II Fall 2006...

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Unformatted text preview: Physics 212: Statistical mechanics II, Fall 2006 Lecture IX Let us work out a simple example of the linear-response formula from the last lecture. Suppose that the starting Hamiltonian describes a single spin-half in a magnetic field along the ˆ z axis: H = Δ ¯ h S z , (1) which has two eigenstates: spin down with energy- Δ / 2, and spin up with energy Δ / 2. We know from the derivation last time that a perturbation Hamiltonian proportional to S z will not induce any transitions and hence no linear response. Instead let us try H 1 ( t ) = λ ( t ) S x , (2) where λ ( t ) has units of frequency (inverse time) Then the response in a measurement of S x at time t to the perturbation is δ h S x i = Z t-∞ dt h S x ( t- t ) ,S x λ ( t ) ¯ h i . (3) We can use the unperturbed Hamiltonian to obtain S x ( t- t ) = e iH ( t- t ) / ¯ h S x e- iH ( t- t ) / ¯ h = e i Δ( t- t ) / (2¯ h ) e- i Δ( t- t ) / (2¯ h ) S x e- i Δ( t- t ) / (2¯ h ) e i Δ( t- t ) / (2¯ h ) . (4) Then the matrix multiplication gives S x ( t- t ) = ¯ h 2 e i Δ( t- t ) / ¯ h e- i Δ( t- t ) / ¯ h = cos(Δ( t- t ) / ¯ h ) S x- sin(Δ( t- t ) / ¯ h ) S y , (5) where we have used S x = ¯ h 2 1 1 , S y = ¯ h 2- i i (6) which satisfy [ S x ,S y ] = i ¯ hS z . Now only the S y term in S x ( t- t ) gives a nonzero result in the commutator with S x . Since ( S y ,S x ) =- i [ S y ,S x ] =- ¯ hS z , we obtain the simple form δ h S x i = Z t-∞ dt h S z sin(Δ( t- t ) / ¯ h ) λ ( t ) i = Z t-∞ dt sin(Δ( t- t ) / ¯ h ) λ ( t ) h S z i . (7) The statistical average of S z is time-independent, and is just h S z i = ¯ h 2 e β Δ / 2- e- β Δ / 2 e β Δ / 2- e- β Δ / 2 = ¯ h 2 tanh( β Δ / 2) . (8) Combining everything, the response is δ h S x i = tanh( β Δ / 2) Z t-∞ dt sin(Δ( t- t ) / ¯ h ) λ ( t ) . (9) 1 The fact that in this example temperature just gives a prefactor that is independent of the time dependence of λ ( t ) is special: it results from the simple commutators of the spin operators. We see that the response goes to zero at high temperatures kT Δ. The periodic form of the integrand makes sense in a picture of classical precession of the spin, and is consistent with what we could have derived with ordinary time-dependent perturbation theory. Clearly a weak constant-in-time perturbation will have less effect than an equally weak perturbation at the resonant frequency Δ / ¯ h . The frequency-space version of the linear-response formula can be used to write a compact rela- tionship known as the fluctuation-dissipation theorem between response functions and correlation functions (cf. problem set). We now introduce a quick rewriting of the Kubo formula from the last lecture, then move on to scaling theories of equilibrium systems. The Kubo formula derived last time was δ h ˆ O 2 i = λ Z t-∞ dt h ˆ O 2 ( t- t ) , ˆ O 1 ( t ) i , (10) where ( A,B ) =- i [ A,B ], and remember that here ˆ O 1 ( t ) carries a physical time dependence, while...
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## This note was uploaded on 08/01/2008 for the course PHYSICS 212 taught by Professor Moore during the Fall '06 term at Berkeley.

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phys212ln9 - Physics 212 Statistical mechanics II Fall 2006...

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