Electric Susceptibility
Since we are going to apply time-dependent perturbations in the linear response regime, we need to remind ourselves, how the values of an observable vary when the perturbation comes into play. That is to say we need to remind ourselves of the Kubo formula.
{\displaystyle \langle {\hat {A}}(t)\rangle =\left\langle {\hat {A}}\right\rangle _{0}-{\frac {i}{\hbar }}\int _{t_{0}}^{t}dt'\left\langle \left[{\hat {A}}(t),{\hat {V}}{\mathord {\left(t'\right)}}\right]\right\rangle _{0}}
And thus in linear response theory, the change in expectation value of an operator \hat A is \delta\langle \hat A(t)\rangle = \frac{i}{\hbar}\int_{-\infty}^t dt' \, \langle [\hat H_I(t'), \hat A(t)] \rangle_0 , \tag{1}
with \hat H_I(t') the interaction Hamiltonian.
In our case we have a perturbation due to the interaction between electric polarization and the electric field.
\hat H(t) = H_0 - \int d\mathbf r'\,\hat{\mathbf P}(\mathbf r') \cdot \mathbf E(\mathbf r',t), \tag{2}
where \hat{\mathbf P}(\mathbf r) is the polarization operator.
For \hat A(t) = \hat P_i(\mathbf r,t),\ Equation 1 becomes: \delta \langle \hat P_i(\mathbf r,t)\rangle = \frac{i}{\hbar}\int_{-\infty}^t dt' \int d\mathbf r' \, \Big\langle \big[ -\hat P_j(\mathbf r',t') E_j(\mathbf r',t'), \, \hat P_i(\mathbf r,t) \big] \Big\rangle_0 .
Since E_j(\mathbf r',t') is a classical field (a c-number, not an operator), it commutes with operators and can be pulled out: = -\frac{i}{\hbar}\int_{-\infty}^t dt' \int d\mathbf r' \, E_j(\mathbf r',t') \, \langle [\hat P_j(\mathbf r',t'), \hat P_i(\mathbf r,t)] \rangle_0 .
Using [X,Y] = -[Y,X], this becomes \delta \langle \hat P_i(\mathbf r,t)\rangle = \frac{i}{\hbar}\int_{-\infty}^t dt' \int d\mathbf r' \, \langle [\hat P_i(\mathbf r,t), \hat P_j(\mathbf r',t')] \rangle_0 \, E_j(\mathbf r',t') .
We can now introduce the retarded susceptibility (response function): \chi_{ij}^R(\mathbf r,t;\mathbf r',t') = \frac{i}{\hbar} \, \Theta(t-t') \, \langle [\hat P_i(\mathbf r,t), \hat P_j(\mathbf r',t')] \rangle_0 .
Then the response takes the compact Green’s-function form: \delta \langle \hat P_i(\mathbf r,t)\rangle = \int d\mathbf r' \int_{-\infty}^\infty dt' \, \chi_{ij}^R(\mathbf r,t;\mathbf r',t') \, E_j(\mathbf r',t').
Susceptibility links Polarization and Electric Field Greenly:
So susceptibility is really a Green’s function and when a weak external electric field \mathbf E(\mathbf r,t) is applied, the system responds linearly with an induced polarization \mathbf P(\mathbf r,t). In the linear response regime, the most general linear relation between cause (field) and effect (polarization) is
P_i(\mathbf r,t) = \int d\mathbf r' \int_{-\infty}^\infty dt' \, \chi_{ij}(\mathbf r,t;\mathbf r',t') \, E_j(\mathbf r',t'). \tag{3}
Here \chi is the susceptibility tensor, has certain properties by default or by imposition:
Because the system is weakly perturbed, the polarization is a linear functional of the applied field.
A perturbation at (\mathbf r',t') can influence the polarization at (\mathbf r,t). This makes \chi depend on two spacetime points.
A result cannot precede its cause. Therefore, \chi_{ij}(\mathbf r,t;\mathbf r',t') = 0 \quad \text{for } t < t' .
But how to express susceptibility?
To find an expression for suceptibility, we need the following steps:
- Use the interaction hamiltonian to find the perturbed wavefunction.
- Find the elements of the polarization operator.
- Exploit the relationship between the current density and polarization operator.
- Take the statistical average on initial states to find the susceptibility.
For brownie points we then try to find frequency domain alternatives too.