I've been reading various documents about quantum field theory over the last several weeks, specifically about the canonical quantization of quantum fields. In doing so, I've come to realize that quantum mechanics has a lot of crazy math and even crazier physical interpretations, and I just took that for granted, but now those things are coming back to haunt me in quantum field theory. It is very hard for me to wrap my head around, and I feel like I could use a lot more help in visualizing and intuiting what certain concepts in canonical quantization mean. This will be the first of a few posts which are outlets for me to gather my thoughts and put them out there for you all to see and correct; this one will be about classical fields.

I feel like the easiest quantum field system to study is the phonon. It is a spin-0 bosonic system, so it can be described by a scalar field. Furthermore, said field can be restricted to one dimension, which simplifies the math even further. This means that taking the continuum limit becomes a bit easier than in three dimensions. Follow the jump to see how it goes.

Let us consider $N$ harmonic oscillators with nearest-neighbor couplings of the same strength. If the chain has length $l$ and nearest-neighbor equilibrium separation $l_0$ such that $l = Nl_0$ and obeys periodic boundary conditions, then the Lagrangian for this system is \[ L = \sum_{n = 1}^{N} \left(\frac{1}{2} m\dot{y}_{n}^{2} - \frac{1}{2} \kappa (y_{n + 1} - y_{n})^2 \right) \] and the corresponding Hamiltonian, found from \[ H = \sum_n p_n \dot{y}_n - L \] is \[ H = \sum_{n = 1}^{N} \left(\frac{p_{n}^{2}}{2m} + \frac{1}{2} \kappa (y_{n + 1} - y_{n})^2 \right) \] where the displacement of oscillator $n$ is $y_{n}$, $m$ is the mass of each oscillator, $\kappa$ is the force per unit length of each oscillator, and the momentum of oscillator $n$ in terms of its velocity is $p_n = m\dot{y}_n$ as found from the Lagrangian. How would we get to the continuum limit? We can take the limit of $N$ becoming very large and $l_0$ becoming very small, such that $l$ remains constant and becomes the only meaningful length scale. Through elementary wave analysis, it can be shown that the mass $m$ would be replaced by the density $\rho$, and the force per unit length $\kappa$ would be replaced by the force $f$ itself. This yields the continuum Lagrangian \[ L = \int \left(\frac{1}{2} \rho \dot{y}^2 (x) - \frac{1}{2} f \left(\frac{\partial y}{\partial x} \right)^2 \right) \, dx \] and the corresponding continuum Hamiltonian is \[ H = \int \left(\frac{p_y^2 (x)}{2\rho} + \frac{f}{2} \left(\frac{\partial y}{\partial x} \right)^2 \right) \, dx \] for this system.

How does this all come about? For the continuum system, the mass gets replaced by the density as $m = \int \rho \, dx$, and the force per unit length conversely gets replaced by the force as $\kappa = \frac{\partial f}{\partial x}$. The mass of the overall chain is finite, so the mass of a point is zero (and the mass of an infinitesimal chunk of the chain is infinitesimally small). Meanwhile, the force of the overall chain is finite, which actually means that the force over an infinitesimal length would be extremely large. That's why the mass had to be replaced by the density and the force per unit length had to be replaced by the overall force. That explains the prefactors in front of the kinetic and potential terms. The sums over the oscillator index $n$ get replaced by corresponding integrals over $x$. This means that $x$ is now a continuous index analogous to $n$ and is not a degree of freedom (i.e. cannot be quantized as an operator), because neither is $n$. In taking the limit of large $N$ and small $l_0$ to keep $l$ finite, the terms with $y_{n + 1} - y_{n}$ are multiplied and divided by $l_0$, such that $f = \kappa l_0$, and $y_{n + 1} - y_{n} = l_0 \frac{\partial y}{\partial x}$. Next, the momentum $p_n$ of each oscillator is replaced with the continuous momentum density $p_y (x) = \rho \dot{y}(x)$, where the index $n$ is replaced with $x$ (and the subscript $y$ clarifies what the conjugate variable is but does

But there's something more than just the Hamiltonian that can be found. It is true that in the discrete case each oscillator of displacement $y_n$ has a momentum $p_n$, and this momentum could be transferred to another particle that is not part of the oscillator system. However, because each oscillator is coupled to its nearest neighbors, part of the momentum transfer also arises by virtue of the fact that the displacements $y_n$ of nearest neighbors are generally different; this is how sinusoidal waves can travel on a lattice of oscillators, and waves have momentum by virtue of the de Broglie relation $p = \hbar k$. The momentum that can be carried overall by a discrete lattice (different from the momentum $p_n$ of each lattice point) is then something like \[ p = \sum_n p_n \frac{y_{n + 1} - y_{n - 1}}{2l_0} \] (though this might not be the exact relation — I am simply writing this in analogy to the more well-established continuum case coming up) and this can be transferred to other particles. Likewise, for a continuous system, the momentum becomes \[ p = \int p_y (x) \frac{\partial y}{\partial x} \, dx \] where the sum over $n$ has been replaced by an integral over $x$, and the gradient term which is a finite difference has been replaced by an actual partial derivative with respect to $x$ (and also note that the continuous $p_y (x)$ is a conjugate momentum

Finally, let us discuss the dispersion relation. The equation of motion from the continuum Lagrangian is \[ \frac{\partial^2 y}{\partial t^2} = \frac{f}{\rho} \frac{\partial^2 y}{\partial x^2} \] which is the wave equation where the propagation speed $v$ is given by $f = \rho v^2$. Plugging in the plane wave ansatz $y = y_0 e^{i(kx - \omega t)}$ yields \[ \omega(k) = v|k| \] (

I feel like the easiest quantum field system to study is the phonon. It is a spin-0 bosonic system, so it can be described by a scalar field. Furthermore, said field can be restricted to one dimension, which simplifies the math even further. This means that taking the continuum limit becomes a bit easier than in three dimensions. Follow the jump to see how it goes.

Let us consider $N$ harmonic oscillators with nearest-neighbor couplings of the same strength. If the chain has length $l$ and nearest-neighbor equilibrium separation $l_0$ such that $l = Nl_0$ and obeys periodic boundary conditions, then the Lagrangian for this system is \[ L = \sum_{n = 1}^{N} \left(\frac{1}{2} m\dot{y}_{n}^{2} - \frac{1}{2} \kappa (y_{n + 1} - y_{n})^2 \right) \] and the corresponding Hamiltonian, found from \[ H = \sum_n p_n \dot{y}_n - L \] is \[ H = \sum_{n = 1}^{N} \left(\frac{p_{n}^{2}}{2m} + \frac{1}{2} \kappa (y_{n + 1} - y_{n})^2 \right) \] where the displacement of oscillator $n$ is $y_{n}$, $m$ is the mass of each oscillator, $\kappa$ is the force per unit length of each oscillator, and the momentum of oscillator $n$ in terms of its velocity is $p_n = m\dot{y}_n$ as found from the Lagrangian. How would we get to the continuum limit? We can take the limit of $N$ becoming very large and $l_0$ becoming very small, such that $l$ remains constant and becomes the only meaningful length scale. Through elementary wave analysis, it can be shown that the mass $m$ would be replaced by the density $\rho$, and the force per unit length $\kappa$ would be replaced by the force $f$ itself. This yields the continuum Lagrangian \[ L = \int \left(\frac{1}{2} \rho \dot{y}^2 (x) - \frac{1}{2} f \left(\frac{\partial y}{\partial x} \right)^2 \right) \, dx \] and the corresponding continuum Hamiltonian is \[ H = \int \left(\frac{p_y^2 (x)}{2\rho} + \frac{f}{2} \left(\frac{\partial y}{\partial x} \right)^2 \right) \, dx \] for this system.

How does this all come about? For the continuum system, the mass gets replaced by the density as $m = \int \rho \, dx$, and the force per unit length conversely gets replaced by the force as $\kappa = \frac{\partial f}{\partial x}$. The mass of the overall chain is finite, so the mass of a point is zero (and the mass of an infinitesimal chunk of the chain is infinitesimally small). Meanwhile, the force of the overall chain is finite, which actually means that the force over an infinitesimal length would be extremely large. That's why the mass had to be replaced by the density and the force per unit length had to be replaced by the overall force. That explains the prefactors in front of the kinetic and potential terms. The sums over the oscillator index $n$ get replaced by corresponding integrals over $x$. This means that $x$ is now a continuous index analogous to $n$ and is not a degree of freedom (i.e. cannot be quantized as an operator), because neither is $n$. In taking the limit of large $N$ and small $l_0$ to keep $l$ finite, the terms with $y_{n + 1} - y_{n}$ are multiplied and divided by $l_0$, such that $f = \kappa l_0$, and $y_{n + 1} - y_{n} = l_0 \frac{\partial y}{\partial x}$. Next, the momentum $p_n$ of each oscillator is replaced with the continuous momentum density $p_y (x) = \rho \dot{y}(x)$, where the index $n$ is replaced with $x$ (and the subscript $y$ clarifies what the conjugate variable is but does

*not represent an index*). Finally, the Hamiltonian is found with the formula \[ H = \int p_y (x) \dot{y}(x) \, dx - L \] which is exactly analogous to the discrete oscillator case, where again the sum over the discrete index $n$ is replaced with an integral over the continuous index $x$.But there's something more than just the Hamiltonian that can be found. It is true that in the discrete case each oscillator of displacement $y_n$ has a momentum $p_n$, and this momentum could be transferred to another particle that is not part of the oscillator system. However, because each oscillator is coupled to its nearest neighbors, part of the momentum transfer also arises by virtue of the fact that the displacements $y_n$ of nearest neighbors are generally different; this is how sinusoidal waves can travel on a lattice of oscillators, and waves have momentum by virtue of the de Broglie relation $p = \hbar k$. The momentum that can be carried overall by a discrete lattice (different from the momentum $p_n$ of each lattice point) is then something like \[ p = \sum_n p_n \frac{y_{n + 1} - y_{n - 1}}{2l_0} \] (though this might not be the exact relation — I am simply writing this in analogy to the more well-established continuum case coming up) and this can be transferred to other particles. Likewise, for a continuous system, the momentum becomes \[ p = \int p_y (x) \frac{\partial y}{\partial x} \, dx \] where the sum over $n$ has been replaced by an integral over $x$, and the gradient term which is a finite difference has been replaced by an actual partial derivative with respect to $x$ (and also note that the continuous $p_y (x)$ is a conjugate momentum

*density*).Finally, let us discuss the dispersion relation. The equation of motion from the continuum Lagrangian is \[ \frac{\partial^2 y}{\partial t^2} = \frac{f}{\rho} \frac{\partial^2 y}{\partial x^2} \] which is the wave equation where the propagation speed $v$ is given by $f = \rho v^2$. Plugging in the plane wave ansatz $y = y_0 e^{i(kx - \omega t)}$ yields \[ \omega(k) = v|k| \] (

**UPDATE**: I added absolute value signs that were missing before) as the dispersion relation; that makes sense for this continuum limit (because in the discrete case, phonons typically have much more complicated dispersion relations, but this linear relation is also a limiting case for acoustic phonon modes). Overall, this review of the classical acoustic field wasn't too big of a deal, but it did put in place the definition for the field (not conjugate) momentum, made clear the relations between the sums and integrals, and made clear what the Hamiltonian is. In the next post, I will be explaining how to get from the fields to their Fourier components and then perform canonical quantization. Once again, I will frequently draw parallels between the discrete and continuum cases, because one may be easier to visualize than the other depending on the situation.
## No comments:

## Post a Comment