In my previous post I discussed the intuition behind the classical acoustic field in one dimension. Now I'm going to talk about diagonalizing the Hamiltonian and making the step into quantum field theory. Follow the jump to see what it's like.

The Hamiltonian for the discrete system of nearest-neighbor coupled oscillators is \[ H = \sum_{n = 1}^{N} \left(\frac{p_n^2}{2m} + \frac{\kappa}{2} (y_{n + 1} - y_n)^2 \right) \] and this becomes \[ H = \int \left(\frac{p_y^2 (x)}{2\rho} + \frac{f}{2} \left(\frac{\partial y}{\partial x}\right)^2 \right) \, dx \] for the continuous system. It is clear in the discrete case that the Hamiltonian is not diagonal: for each $n$, both $y_n$ and $y_{n + 1}$ come into play due to the coupling of nearest neighbors. This difference becomes the derivative $\frac{\partial y}{\partial x}$ in the continuous case. But why does the Hamiltonian need to be diagonalized anyway? Here's why: even in classical mechanics, only by diagonalizing the Hamiltonian for a system of coupled oscillators can the general motion of the system in phase space be found as a superposition of normal modes. The simplest nontrivial such case is of 2 coupled oscillators: the normal modes are independent oscillators themselves, corresponding to either both oscillators moving with the same amplitude in the same direction or both oscillators moving with the same amplitude in opposite directions. As more oscillators are added, the normal modes start to resemble traveling plane waves more, and indeed traveling plane waves are the normal modes of a long continuous chain. Therefore, we need to diagonalize the Hamiltonian by considering plane waves.

Let us define the Fourier expansion as \[ g(x) = \frac{l}{\sqrt{2\pi}} \int g(k) e^{ikx} \, dk \] and the Fourier transform as \[ g(k) = \frac{1}{\sqrt{2\pi} l} \int g(x) e^{-ikx} \, dx \] with the factors of the length scale $l$ added so that $g(x)$ and $g(k)$ have the same dimensions regardless of the function $g$ and with the factors of $(2\pi)^{-\frac{1}{2}}$ added to make the expansion and transform symmetric. The other assumption implicit in this is that $l$ is long enough that $k$ can take on values along the real continuum rather than being restricted to a discrete set of values as it might be if boundary conditions mattered. Another assumption is that any function $g$ that we use is square-integrable along $x$ or $k$. Why do we care about the Fourier transform? Well, it is essentially the plane wave expansion coefficient of any function for wavevector $k$, which is good because we know that plane waves of any wavevector $k$ are the normal modes of this continuous chain. Hence, we now care more about the components of $y$ and $p_y$ in the plane wave expansion rather than their values at any position index $x$. Because the Fourier transform is complex, even if $g(x)$ is real, $g(k)$ will be complex. However, if $g^{\star} (x) = g(x)$, then $g^{\star} (k) = g(-k)$, and this will be true for both $y$ and $p_y$. This means that the derivatives $\frac{\partial}{\partial x}$ turn into multiplication by $ik$ when performing the Fourier expansion as plane waves are eigenfunctions of spatial differentiation. After doing the gritty math, \[ H = l^2 \int \left(\frac{p_y^{\star} (k) p_y (k)}{2\rho} + \frac{\rho}{2} v^2 k^2 y^{\star} (k) y(k) \right) \, dk \] is the Hamiltonian as expressed in terms of the plane wave normal modes rather than in terms of the continuous lattice position; this relies on the substitutions $f = \rho v^2$ and $\omega (k) = v|k|$ (

What just happened here? The Hamiltonian used to describe a continuum of coupled oscillators indexed by their position $x$ along the chain. The chain has plane wave normal modes. By considering any wave along the chain to be a superposition of such normal modes, the Hamiltonian now describes a continuum of

Now it is time to quantize this system. Considering the discrete system described by $(y_n, p_n)$ in terms of the discrete equilibrium lattice positions $n$, it is possible to exactly know either the displacement or momentum of a given lattice site oscillator but not both simultaneously. However, it is OK to measure eigenstates of $y_n$ and $p_{n' \neq n}$ simultaneously. This means that the displacement and momentum at each lattice site become operators at each lattice site satisfying the commutation relation $[y_m, p_n] = i\hbar \delta_{mn}$. For the continuous system, the momentum of each oscillator is now a momentum density as the discrete index $n$ becomes the continuous index $x$, so \[ [y(x), p_y (x')] = i\hbar \delta(x - x') \] is the new commutation relation for the displacement and momentum density at different points along the chain; $y(x)$ and $p_y (x)$ are now

The Hamiltonian for the discrete system of nearest-neighbor coupled oscillators is \[ H = \sum_{n = 1}^{N} \left(\frac{p_n^2}{2m} + \frac{\kappa}{2} (y_{n + 1} - y_n)^2 \right) \] and this becomes \[ H = \int \left(\frac{p_y^2 (x)}{2\rho} + \frac{f}{2} \left(\frac{\partial y}{\partial x}\right)^2 \right) \, dx \] for the continuous system. It is clear in the discrete case that the Hamiltonian is not diagonal: for each $n$, both $y_n$ and $y_{n + 1}$ come into play due to the coupling of nearest neighbors. This difference becomes the derivative $\frac{\partial y}{\partial x}$ in the continuous case. But why does the Hamiltonian need to be diagonalized anyway? Here's why: even in classical mechanics, only by diagonalizing the Hamiltonian for a system of coupled oscillators can the general motion of the system in phase space be found as a superposition of normal modes. The simplest nontrivial such case is of 2 coupled oscillators: the normal modes are independent oscillators themselves, corresponding to either both oscillators moving with the same amplitude in the same direction or both oscillators moving with the same amplitude in opposite directions. As more oscillators are added, the normal modes start to resemble traveling plane waves more, and indeed traveling plane waves are the normal modes of a long continuous chain. Therefore, we need to diagonalize the Hamiltonian by considering plane waves.

Let us define the Fourier expansion as \[ g(x) = \frac{l}{\sqrt{2\pi}} \int g(k) e^{ikx} \, dk \] and the Fourier transform as \[ g(k) = \frac{1}{\sqrt{2\pi} l} \int g(x) e^{-ikx} \, dx \] with the factors of the length scale $l$ added so that $g(x)$ and $g(k)$ have the same dimensions regardless of the function $g$ and with the factors of $(2\pi)^{-\frac{1}{2}}$ added to make the expansion and transform symmetric. The other assumption implicit in this is that $l$ is long enough that $k$ can take on values along the real continuum rather than being restricted to a discrete set of values as it might be if boundary conditions mattered. Another assumption is that any function $g$ that we use is square-integrable along $x$ or $k$. Why do we care about the Fourier transform? Well, it is essentially the plane wave expansion coefficient of any function for wavevector $k$, which is good because we know that plane waves of any wavevector $k$ are the normal modes of this continuous chain. Hence, we now care more about the components of $y$ and $p_y$ in the plane wave expansion rather than their values at any position index $x$. Because the Fourier transform is complex, even if $g(x)$ is real, $g(k)$ will be complex. However, if $g^{\star} (x) = g(x)$, then $g^{\star} (k) = g(-k)$, and this will be true for both $y$ and $p_y$. This means that the derivatives $\frac{\partial}{\partial x}$ turn into multiplication by $ik$ when performing the Fourier expansion as plane waves are eigenfunctions of spatial differentiation. After doing the gritty math, \[ H = l^2 \int \left(\frac{p_y^{\star} (k) p_y (k)}{2\rho} + \frac{\rho}{2} v^2 k^2 y^{\star} (k) y(k) \right) \, dk \] is the Hamiltonian as expressed in terms of the plane wave normal modes rather than in terms of the continuous lattice position; this relies on the substitutions $f = \rho v^2$ and $\omega (k) = v|k|$ (

**UPDATE**: I added absolute value signs that should have been there). For the record, the overall field momentum \[ p = \int p_y (x) \left(\frac{\partial y}{\partial x}\right) \, dx \] has the same issue in terms of the continuous lattice position due to the presence of $\frac{\partial y}{\partial x}$, but after the plane wave expansion, \[ p = l^2 \int k p_y (k) y^{\star} (k) \, dk \] is now diagonal.What just happened here? The Hamiltonian used to describe a continuum of coupled oscillators indexed by their position $x$ along the chain. The chain has plane wave normal modes. By considering any wave along the chain to be a superposition of such normal modes, the Hamiltonian now describes a continuum of

*uncoupled*oscillators, but now the index is of the plane wave normal mode wavevector $k$ rather than of the position along the chain. For a while I was getting confused by what people meant when they said things like "quantum fields are infinite sums of harmonic oscillators" or "each field normal mode is an oscillator". Now I understand, thanks to the toy model of a continuum acoustic field in one dimension: the continuum of coupled oscillators in position space is equivalent to a continuum of uncoupled oscillators in wavevector space, because each plane wave normal mode of a given wavevector evolves in time as an uncoupled oscillator whose frequency is given by the dispersion relation.Now it is time to quantize this system. Considering the discrete system described by $(y_n, p_n)$ in terms of the discrete equilibrium lattice positions $n$, it is possible to exactly know either the displacement or momentum of a given lattice site oscillator but not both simultaneously. However, it is OK to measure eigenstates of $y_n$ and $p_{n' \neq n}$ simultaneously. This means that the displacement and momentum at each lattice site become operators at each lattice site satisfying the commutation relation $[y_m, p_n] = i\hbar \delta_{mn}$. For the continuous system, the momentum of each oscillator is now a momentum density as the discrete index $n$ becomes the continuous index $x$, so \[ [y(x), p_y (x')] = i\hbar \delta(x - x') \] is the new commutation relation for the displacement and momentum density at different points along the chain; $y(x)$ and $p_y (x)$ are now

*quantum field operators*because they are operator-valued functions of a spatial index $x$, but they can act on states just like the operators $x$, $p$, $L_z$, and others from single-particle quantum mechanics. I will get more into the details of these operators in the next post, but I will end this post with a statement about the quantum Hamiltonian. The Hamiltonian as expressed in terms of the position index is unchanged in form. That said, enforcing the commutation relation between the displacement and momentum density at different positions along the chain yields the commutation relation \[ [y(k), p_y (k')] = \frac{i\hbar}{l^2} \delta(k + k') \] for the displacement and momentum density as functions of the normal mode wavevector index, which makes sense thanks to the relations $y^{\dagger} (k) = y(-k)$ and $p_y^{\dagger} (k) = p_y (-k)$ where the complex conjugates have been replaced by Hermitian adjoints as the classical observables have been replaced by quantum operators. Thus, \[ H = l^2 \int \left(\frac{p_y (-k) p_y (k)}{2\rho} + \frac{\rho}{2} v^2 k^2 y(-k) y(k) \right) \, dk \] is the Hamiltonian operator for this system. The classical overall momentum does not become a Hermitian quantum operator if the displacement and momentum density operators are substituted blindly (as those operators are not Hermitian and do not commute), but making the replacement $AB \rightarrow \frac{1}{2} \left(AB + BA\right)$ yields \[ p = \frac{l^2}{2} \int k \left(p_y (k) y(-k) + y(-k) p_y (k) \right) \, dk \] as the (Hermitian) overall momentum operator for the field. Both of these have now been diagonalized. In the next post, I will discuss the states of this quantum field and their meanings, along with further operator methods.
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