Separating Hamiltonians in spontaneous symmetry breaking investigation

Question

I read a book about spontaneous symmetry breaking. In the book, the author says:

Using a Fourier transformation, it's always possible to divide the Hamiltonian into two parts, one is collective part with zero wave number, or center-of-mass, the other is part with finite wave number which gives description of internal freedom. To describe the breaking of a global spontaneous symmetry, we only need to consider the collective part.

And the author gives an example:

Considering a harmonious crystal, whose Hamiltonian is:

$$H=\sum_{x,\sigma} \frac{P^{2}(x)}{2m}+\frac{1}{2}m\omega^{2}_{0}(X(x)-X(x+\sigma))^{2}$$

where $X$ and $P$ are respectively position and momentum operators.

Then the collective part of the Hamiltonian is:

$$H_{CoM}=\frac{P^{2}_{tot}}{2mN}$$

$$\text{and} $$

$$P_{tot}=\sum P(x)$$

Thus, the Hamiltonian can be divided as:

$$H=H_{CoM}+\sum_{k\neq 0}H_{int}(k)$$

So I can understand $P_{tot}$ is the center-of-mass, since this operator describes the momentum as a whole. Yet I cannot see how it's related to wave number zero, nor figure out why $k\neq 0$ is describing internal freedom.

To me, $P_{tot}$ can have eigenstates corresponding to nonzero eigenvalues, so $k$ here seems not to be the wave number proportional to whole momentum. Besides, $H$ is in general a multi-particle Hamiltonian, so $k$ seems not to be proportional to individual momentum either.

Could any body help me?

Answer 1

I think that this is basically an issue of Fourier transform. When you do a FT of any function (let's stay in $1$d, for convenience) you have

$$ F(k) = \int\! dx e^{-ikx}f(x)$$

As you can see, the $k=0$ is a special point where you just sum over all contributions equally. It means that $F(k=0)$ doesn't contain any information about the spatial structure of $f(x)$, and any internal modulations or fluctuations in it will not be evident in it. The information about them will only be present via the $k\neq 0$ components. So $P(k=0)$ is the center of mass momentum because it is a straight-forward summation over all momentum at all points.

Another way to see it is that the spatial derivatives of $f(x)$ will transform to $(-ik)^nF(k)$, so $k=0$ is a value that has no information about the derivatives, i.e. about the internal structure.