While trying to solve an exercise, I ran into what looks like a contradiction. I'm sure I'm making some kind of mistake, but I couldn't spot it. I'm not asking for help in solving the exercise, which I already solved, but I need to clarify my ideas.

The exercise goes like this: use linear response theory to calculate the magnetization $\vec M$ (magnetic moment per unit volume) induced in a zero-temperature system of noninteracting spin-$\frac{1}{2}$ fermions, which is perturbed by a uniform magnetic field $\vec H$. Hence, derive Pauli paramagnetism: $\vec M = \chi_P \vec H$ with $\chi_P = \frac{3 \mu_B^2 n}{2 \epsilon_F}$, where $n$ is the density of particles (the rest of the notation should be self-explanatory).

Very brief summary of the standard procedure

I have no problem in carrying out the standard linear response calculation, which involves the calculation of the retarded correlator: $$i D_{ij}^R(\vec x t, \vec x' t') = \theta(t-t') \langle \left[ \delta \hat M_i (\vec x t), \delta \hat M_j (\vec x' t') \right] \rangle$$ where the mean value is carried out on the ground state of the unperturbed system (i.e. the state in which fermions fill the Fermi sphere) and the operators are time-evolved in the Heisenberg picture using the unperturbed hamiltonian.

It turns out that we can express this correlator in terms of the zero-order retarded polarization function $\Pi^{(0)R}$ and at the end of the day one finds that the magnetization is given by (in momentum space): $$\vec M = \left( -\mu_b^2 \lim_{k \rightarrow 0} \Pi^{(0)R}\left(\vec k, \omega = 0\right) \right) \vec H$$ By carrying out the limit, one recovers Pauli paramagnetism. The reason why we use the polarization at $\omega = 0$ and in the limit $\vec k \rightarrow 0$ is that $\vec H$ is static and uniform.

My problem

My problem is that from the following reasoning I conclude that there should be no magnetization at all!

The perturbed hamiltonian is given by $$ \mathcal{H} = \mathcal{H}_0 + \delta \mathcal{H}$$ where:

  • $\mathcal{H}_0$ is just the kinetic energy $\hat T$ of the system;
  • $\delta \mathcal{H} = - \vec H \cdot \int \hat{\vec M} (\vec x) \space d_3x = - \vec H \cdot \hat{\vec M}$;
  • the density of magnetic moment $\hat{\vec M} (\vec x)$ in $\delta \mathcal{H}$ is defined by ($\vec \sigma$ is the vector of Pauli matrices, and summation on repeated indexes is intended): $\hat{\vec M} (\vec x) = -\mu_B \hat{\psi}_\alpha^\dagger (\vec x) \vec \sigma_{\alpha \beta} \hat{\psi}_\beta (\vec x)$.

From the expression of $\delta \mathcal{H}$ it easily follows that the component of the total magnetization that is parallel to the external magnetic field $\vec H$ is a constant of motion.

When carrying out the linear response calculation, we're imagining that in the distant past the system was in the unperturbed ground state and then the external magnetic field was switched on. Indeed, when deriving the linear response theory, we make use of the interaction picture to deal with the perturbation.

The component of $\hat{\vec M}$ parallel to $\vec H$ is a constant of motion and so its mean value should stay constant along the process of switching on the magnetic field. But in the unperturbed ground state it was zero! So, how is it possible that we find a finite paramagnetic effect in the end?



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