雷曼表示 Lehmann Representation

多体理论

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参考 Reference

• Alexander L. Fetter & John Dirk Walecka, Quantum Theory of Many-Particle Systems: Chapter 3.
• Gerald D. Mahan, Many-Particle Physics, 3rd Edition: Chapter 2.
• Piers Coleman, Introduction to Many Body Physics: Chapter 6.

传播子 Propagator

in Heisenberg picture, a particle propagates from (x', t') to (x, t), or say creates at (x', t') and annihilates at (x, t).

$$\left\langle x t \mid x' t' \right\rangle = \left\langle 0 \right| \psi \psi^\dagger \left| 0 \right\rangle$$

The Green Function for a single particle is defined as:

$$iG = \left\langle 0 \right| T (\psi \psi^\dagger) \left| 0 \right\rangle$$

where $T$ is the time ordering operator, for fermion operators:

$$T (\psi \psi^\dagger) = \theta(t - t') \psi \psi^\dagger - \theta(t' - t)\psi^\dagger \psi$$

Suppose the system has translational symmetry,

$$\psi(xt) = e^{-ip \cdot x} e^{i Ht} \psi(0) e^{-i Ht} e^{i p \cdot x}$$

0 in $\psi(0)$ means time $t=0$, and position $x=0$, $e^{i p \cdot x}$ is a translational operator along x direction, $x$ is the parameter for translation.

Matrix Element

Now we are ready to calculate the matrix element $\left\langle \psi \psi^\dagger \right\rangle$ and $\left\langle \psi^\dagger \psi \right\rangle$.

$$\left\langle \psi \psi^\dagger \right\rangle = \sum\limits_n \left\langle 0 \right| \psi \left| n \right\rangle \left\langle n \right| \psi^\dagger \left| 0 \right\rangle$$

index $n$ denotes a complete set, $\left| n \right\rangle$ is the state ket for $H$ and $P$:

$$H \left| n \right\rangle = E_n \left| n \right\rangle$$

$$P \left| n \right\rangle = P_n \left| n \right\rangle$$

In general, we calculate the matrix element,

$$\left\langle a \right| \psi(x,t) \left| b \right\rangle = \left\langle a \right| e^{-ip \cdot x} e^{i Ht} \psi e^{-i Ht} e^{i p \cdot x} \left| b \right\rangle$$

returns,

$$\left\langle a \right| \psi \left| b \right\rangle e^{i (p_b - p_a )x } e^{ -i (E_b - E_a) t }$$

For ground state $\left| 0 \right\rangle$, suppose there are N fermions form a FS (Fermi sphere).

$$H \left| 0 \right\rangle = E(N) \left| 0 \right\rangle$$

we use $E(N)$ denotes the ground state energy for N fermions, and

$$P \left| 0 \right\rangle = 0$$

means the momentum for ground state is 0.

Now,

$$\left\langle 0 \right| \psi(xt) \left| n \right\rangle = \left\langle 0 \right| \psi \left| n \right\rangle e^{i p_n x - i(E_n(N+1) - E(N))t}$$

since $\psi$ is annihilation operator and ground state has N particles, the excited state $\left| n \right\rangle$ should has N+1 particles.

Likewise,

$$\left\langle n \right| \psi^\dagger (x' t') \left| 0 \right\rangle = \left\langle n \right| \psi^\dagger \left| 0 \right\rangle e^{ - i p_n x' + i (E_n(N+1) - E(N)) t'}$$

Now, for $t > t'$,

$$\left\langle \psi \psi^\dagger \right\rangle = \sum\limits_n \left\langle 0 \right| \psi \left| n \right\rangle \left\langle n \right| \psi^\dagger \left| 0 \right\rangle e^{i p_n(x-x') - i (E_n(N+1) - E(N))(t - t')}$$

for $t < t'$,

$$\left\langle \psi^\dagger \psi \right\rangle = \sum\limits_n \left\langle 0 \right| \psi^\dagger \left| n \right\rangle \left\langle n \right| \psi \left| 0 \right\rangle e^{- i p_n(x-x') + i (E_n(N - 1) - E(N))(t - t')}$$

We shall introduce chemical potential $\mu$ and excitation energy $\omega_n$ to rewrite the expression in exponent,

$$E_n(N+1) - E(N) = E_n(N+1) - E(N+1) + E(N+1) - E(N)$$

$$E_n(N-1) - E(N) = E_n(N-1) - E(N-1) + E(N-1) - E(N)$$

returns:

$$E_n(N+1) - E(N) = \omega_n (N+1) + \mu$$

$$E_n(N-1) - E(N) = \omega_n (N-1) - \mu$$

Here we suppose N is large, $E(N+1) - E(N) = E(N) - E(N-1) = \mu$

Fourier transformation

Remind,

$$G(x) = \left\langle x \mid G \right\rangle = \sum\limits_k \left\langle x \mid k \right\rangle \left\langle k \mid G \right\rangle = \sum\limits_k \frac{1}{V} e^{i k x } G(k)$$

we define the FT, $G(xt)$ expressed $G(k \omega)$, as:

$$G(x-x',t-t') = \frac{1}{V}\sum\limits_k \int \frac{d \omega } { 2 \pi} G(k \omega) e^{i k (x-x') - i \omega(t - t')}$$

and the IFT, $G(k \omega)$ expressed $G(xt)$, as:

$$G(k \omega) = \int d^3 (x-x') \int d(t-t') G(x-x',t-t') e^{- i k(x-x') + i \omega (t - t')}$$

Remember,

$$\left\langle k \mid k' \right\rangle = \int d^3 x \left\langle k \mid x \right\rangle \left\langle x \mid k' \right\rangle = \frac{1}{V} \int d^3 x e^{ i (k' - k) x}$$

for $t>t'$,

$$\int_0^\infty dt e^{i (\omega + i \eta ) t} = - \frac{1}{i ( \omega + i \eta )}$$

likewise, for $t < t'$

$$\int_{-\infty}^0 dt e^{ i (\omega - i \eta ) t } = \frac{1}{ i (\omega - i \eta)}$$

Now,

$$G(k, \omega) = V \sum\limits_n \frac{\left\langle 0 \right| \psi \left| n,k \right\rangle \left\langle n,k \right| \psi^\dagger \left| 0 \right\rangle}{ \omega - \omega_{n,k} (N+1) - \mu + i \eta } + \frac{\left\langle 0 \right| \psi^\dagger \left| n, -k \right\rangle \left\langle n, -k \right| \psi \left| 0 \right\rangle}{ \omega + \omega_{n, -k} (N-1) - \mu - i \eta }$$

if the distribution of $\{ E_n \}$ is very dense, we can define:

$$\sum\limits_n V \left| \left\langle n,k \right| \psi^\dagger \left| 0 \right\rangle \right|^2 = \int_0^\infty d \omega' A(k, \omega' )$$

$$\sum\limits_n V \left| \left\langle n, -k \right| \psi \left| 0 \right\rangle \right|^2 = \int_0^\infty d \omega' B(k, \omega' )$$

and get the Lehmann representation:

$$G(k, \omega) = \int_0^\infty d \omega' \left[ \frac{A(k, \omega')}{ \omega - \omega' - \mu + i \eta } + \frac{B(k, \omega')}{ \omega + \omega' - \mu - i \eta } \right]$$

Poles

poles in upper plane:

$$\mu - \omega' + i \eta$$

poles in lower plane:

$$\mu + \omega' - i \eta$$

Retarded and Advanced GF

The GF is not analytic in either upper plane or lower plane. We can define the Retarded and Advanced GF so that they are analytic in upper or lower plane.

$$iG^R(xt,x't') = \theta(t - t') \left\langle \{ \psi(xt), \psi^\dagger(x't') \} \right\rangle$$

$$iG^A(xt,x't') = \theta(t' - t) \left\langle \{ \psi(xt), \psi^\dagger(x't') \} \right\rangle$$

Then,

$$G^{R,A} (k, \omega) = \int_0^\infty d \omega' \left[ \frac{A(k, \omega')}{ \omega - \mu - \omega' \pm i \eta } + \frac{B(k, \omega') }{ \omega - \mu + \omega' \pm i \eta } \right]$$

GF for free Fermions

Hamiltonian for Free fermions:

$$H = H_0 - \mu N = \sum\limits_{k} \hbar \omega_k c_k^\dagger c_k$$

suppose $\hbar = 1$, define the GF as:

$$iG(k, t-t') = \left\langle T c_k(t) c_k^\dagger (t') \right\rangle$$

$$\left\langle T c_k(t) c_k^\dagger (t') \right\rangle = \theta(t-t') \left\langle c_k(t) c_k^\dagger (t') \right\rangle - \theta(t' - t) \left\langle c_k^\dagger (t') c_k(t) \right\rangle$$

in Heisenberg Picture:

$$c_k(t)=e^{iHt} c_k e^{-iHt}$$

$$c_k^\dagger(t)=e^{iHt} c_k^\dagger e^{-iHt}$$

we can verify that:

$$e^{iHt} c_k e^{-iHt} \left| { } \right\rangle = e^{-i \omega_k t} c_k \left| { } \right\rangle$$

$$e^{iHt} c_k^\dagger e^{-iHt} \left| { } \right\rangle = e^{ i \omega_k t} c_k^\dagger \left| { } \right\rangle$$

yields,

$$\left\langle c_k(t) c_k^\dagger (t') \right\rangle = (1-n_k)e^{-i\omega_k(t-t')}$$

$$\left\langle c_k^\dagger (t') c_k(t) \right\rangle = n_k e^{-i\omega_k(t-t')}$$

Then,

$$G(k,t-t') = -i \left[ (1-n_k) \theta(t-t') - n_k \theta(t'-t) \right] e^{-i \omega_k (t-t')}$$

consider:

$$t-t' \to t$$

$$G(k,\omega) = -i \int_{-\infty}^\infty dt e^{i (\omega - \omega_k) t } e^{- \eta t } \left[ \theta(k-k_F) \theta(t) - \theta(k_F -k )\theta(-t) \right]$$

Finally,

$$G(k, \omega) = \left[ \frac{ \theta(k - k_F )}{ \omega - \omega_k + i \eta } + \frac{ \theta(k_F - k )}{ \omega - \omega_k - i \eta } \right]$$

Or, can be written as:

$$G^0(k, \omega) = \frac{1}{\omega - \omega_k + i \eta_k}$$

where,

$$\eta_k = \eta \text{sign}(k - k_F)$$

Meaning of GF

$t > t'$, $\left\langle \psi(xt) \psi^\dagger(x't') \right\rangle$, describes the propagation of particle, created at time $t'$;

$t < t'$, $\left\langle \psi^\dagger(x't') \psi(xt) \right\rangle$, describes the propagation of hole, created at time $t$ (annihilate a particle $\psi$ is equivalent to create a hole) .

Free Phonons

$$H_0 = \frac{1}{2} \sum\limits_{k \sigma } \{ P_{-k \sigma }P_{k \sigma } + \omega_\sigma^2 (k) Q_{-k \sigma}Q_{k \sigma} \} = \frac{1}{2} \sum\limits_{k \sigma} \hbar \omega_\sigma(k) \left( a_{k \sigma}^\dagger a_{k \sigma} + \frac{1}{2} \right)$$

The subscripts $\sigma$ refer to the polarization of the phonons.

Phonon fields:

$$Q_{k \sigma} = \sqrt{\frac{\hbar}{2 \omega_\sigma (k) }} \left( a_{k \sigma} + a_{-k, \sigma}^\dagger \right)$$

$$Q^\dagger_{k \sigma} = Q_{-k \sigma}$$

The GF for phonons is defined as:

$$i D_{k \sigma} (t-t') = \left\langle T \left(Q_{k \sigma}(t)Q_{-k \sigma }(t') \right) \right\rangle$$

The result is:

$$D^0(k, \omega) = \frac{\hbar}{2m \omega_q} \left[ \frac{1}{ \omega - \omega_q + i \eta} - \frac{1}{\omega + \omega_q - i \eta} \right] = \frac{\hbar}{2m \omega_q} \left[ \frac{2 \omega_q} {\omega^2 - \omega_q^2 + i \eta} \right]$$

Read

• Fetter & Walecka Book: page 72;
• Mahan Book: page 73.
• Coleman Book: page 102.

Problems

• Prove $[ G^R(k, \omega) ]^* = G^A (k, \omega)$

• Use the relation $\frac{1}{\omega \pm i \eta } = P \frac{1}{\omega} \mp i \pi \delta(\omega)$, verify that:

$$\Re G^{R,A}(k, \omega) = \mp P \int_{-\infty}^{\infty} \frac{d \omega'}{\pi} \frac{ G^{R,A}(k, \omega' ) }{ \omega - \omega'}$$

• For $| \omega | \to \infty$, verify that:

$$G(k, \omega) = G^R (k, \omega) = G^A (k, \omega)$$

$$\sim \frac{1}{\omega} \int_0^\infty d \omega' [A(k, \omega') + B(k, \omega')]$$

$$\sim \frac{1}{\omega}$$

JI Yanjiang @季燕江