2013-08-27

Particles in the Continuous Quantum Field

The last thing I discussed in the last post was about the energy eigenstates of the continuous field. The ground state $|0\rangle$ classically corresponds to there being no displacement in the chain at any spatial index $x$ and quantum mechanically corresponds to each oscillator for each normal mode index $k$ being in its ground state, while the first excited state $|k\rangle = a^{\dagger} (k)|0\rangle$ for a given $k$ classically corresponds to a traveling plane wave normal mode of wavevector $k$ and quantum mechanically corresponds to only the oscillator at the given normal mode index $k$ being in its first excited state (and all others being in their ground states). The excited state $|k\rangle$ has energy $E = \hbar v|k|$ above the ground state and overall momentum $p = \hbar k$ above the ground state. This post will discuss what the second and higher excited states are. Follow the jump to see more.

In the single-particle harmonic oscillator, there is only one second excited state $|2\rangle = 2^{-\frac{1}{2}} a^{\dagger 2} |0\rangle$, which is easy. But how can another excitation come about in the continuous chain when quantized?

This is where we have to leave classical intuition behind to simply follow the formalism of quantum mechanics. There are two ways a second excitation can be done. The first is that \[ |k_1 , k_2 \rangle = a^{\dagger} (k_1) a^{\dagger} (k_2) |0\rangle \] represents excitation of an oscillator with normal mode index $k_1$ once and of another oscillator with normal mode index $k_2$ once. Note that because \[ [a(k), a^{\dagger} (k')] = \frac{1}{l} \delta(k - k') \] then \[ |k_1 , k_2 \rangle = |k_2 , k_1 \rangle \] must be true. This allows for the second type of excitation \[ |k, k\rangle = 2^{-\frac{1}{2}} a^{\dagger 2} (k) |0\rangle \] which corresponds to exciting the single oscillator of normal mode index $k$ twice. Additionally, due to the symmetry of exchanging normal mode indices, these excitations are bosons.

Why do we need to leave classical intuition behind? Well, classically, what does the state $|k_1 , k_2 \rangle$ represent? It can't represent a normal mode of wavevector $k_1 + k_2$, because that would represent the state $|k_1 + k_2 \rangle$ which is a single excitation of total normal mode index $k_1 + k_2$. It can't be a superposition of normal modes $k_1$ and $k_2$ because that would not be a classical normal mode and therefore can't correspond to a quantum energy eigenstate either. No, this is something completely new to the quantum analysis of this system. Each quantum excitation of the continuous chain of normal mode index $k$ brings a discrete energy $\hbar v|k|$ and discrete overall momentum $\hbar k$. These excitations interact with other particles to transfer energy and momentum in those discrete chunks just like particles in the usual sense of the word might. (Because our continuous chain exists only in 1 dimension, angular momentum does not play a role, meaning that the excitations of this chain are spin-0. However, the electromagnetic field can impart or take up angular momentum in discrete chunks like a regular particle as it exists in 3 spatial dimensions by necessity.) Thus we call the discrete excitations of this continuous chain particles.

What else can we do about this new conception of particles? Well, you'll notice that in an earlier post, I mentioned that the state space of this quantum field is a Hilbert space of its own. Strictly speaking, that is not true. While it is true that the state space for a given number of particles is a Hilbert space, it is also possible for the overall state of the field to be a superposition of states of different particle numbers; formalizing this means that the overall state space is a direct sum of the $n$-particle Hilbert spaces for every $n$, and the result is called a Fock space. This allowance of superpositions of states of different particle numbers is the reason that quantum field theory is called a theory of varying particles (as particle number conservation is no longer paramount and is only present if the number operator commutes with the Hamiltonian. This is in general not true for many photon systems, which explains things like absorption and emission of photons by electrons.)

There is one more bit of formalism I'd like to introduce before getting into another cool set of states of this continuous chain. That bit of formalism is of the occupation number representation. Because the excitations of the continuous chain are bosons, multiple excitations can occur for the same normal mode index $k$. This means that it is more convenient to list the numbers of excitations for each normal mode index $k$ than to list every normal mode index $k$ excited that many times. For example, the state $|n(k_1) = 1, n(k_2) = 1\rangle = |k_1 , k_2 \rangle$. The state $|n(k') = 2\rangle = |k', k'\rangle$. Where it really shines is for a state like $|n(k_1) = 2, n(k_2) = 4, n(k_3) = 3 \rangle = |k_3, k_1, k_3, k_2, k_1, k_2, k_3, k_2, k_2 \rangle$ where listing each normal mode index $k$ that many times gets tedious. Plus, it more easily shows the relation to the harmonic oscillator: a state like $|n(k_1) = 2, n(k_2) = 4, n(k_3) = 3\rangle$ has the oscillator at normal mode index $k_1$ in the state $|2\rangle$, the oscillator at normal mode index $k_2$ in the state $|4\rangle$, and the oscillator and normal mode index $k_3$ in the state $|3\rangle$.

With this, we can move on to the coherent state. In single-particle quantum mechanics, a coherent state is defined by \[ |\alpha\rangle = e^{-\frac{|\alpha|^2}{2}} \sum_n \frac{\alpha^n}{\sqrt{n!}} |n\rangle \] satisfying \[ a|\alpha\rangle = \alpha \cdot |\alpha\rangle \] meaning that a coherent state is an eigenstate of the annihilation operator. (Because the annihilation operator is not Hermitian, there is no guarantee as there is for observables that coherent states form a complete and orthonormal basis. In fact, coherent states are not necessarily orthogonal, and the basis is overcomplete.) What does the coherent state mean? Well, the value $\alpha_0 = \frac{1}{\sqrt{2\hbar}} \left(\sqrt{m\omega} x_0 + \frac{i}{\sqrt{m\omega}} p_0 \right)$ defines a point in classical phase space $(x_0, p_0)$ as a complex number (as any two-component real vector can be written as a complex number). The time evolution of a coherent state is given by $|\alpha(t)\rangle = |e^{-i\omega t}\alpha_0 \rangle$, meaning that the initial point in phase space goes around in an ellipse, which exactly corresponds to classical behavior. Due to the uncertainty principle, though, there is some uncertainty in the mean position, momentum, and energy. That said, the coherent state is the closest approximation of classical behavior for this quantum oscillator, as the mean position and momentum vary sinusoidally over time while the mean energy is constant over time.

What does this have to do with our continuous chain? Well, coherent states can be defined for it too: \[ |\alpha (k)\rangle = e^{-\frac{|\alpha (k)|^2}{2}} \sum_{n(k)} \frac{\alpha^{n(k)}}{\sqrt{(n(k))!}} |n(k)\rangle \] defines a coherent state for the oscillator at normal mode index $k$. Because now the energy eigenvalue $n$ defines the number of excitations (particles) at normal mode index $k$, a coherent state does not have a definite number of particles. However, its behavior is also the closest to the classical situation. How does that work? I said that classically, the state $|k\rangle$ would correspond to a traveling wave of wavevector $k$, but quantum mechanically, there is nothing traveling at all; as an energy eigenstate, this simply evolves in a stationary manner with the phase $e^{-iv|k|t}$, such that the mean displacement $\langle k|y(x)|k\rangle$ and momentum density $\langle k|p_y (x)|k\rangle$ remain zero (with uncertainty in each). By contrast, the coherent state $|\alpha (k)\rangle$ evolves such that the mean displacement $\langle \alpha (k)|y(x)|\alpha (k)\rangle$ and momentum density $\langle \alpha (k)|p_y (x)|\alpha (k)\rangle$ form traveling plane waves as in the classical case and the mean energy is constant (with uncertainty in all of these quantities). Furthermore, the relation $|\langle n(k)|\alpha (k)\rangle|^2 = e^{-\langle \alpha (k)|n(k)|\alpha (k)\rangle} \frac{\langle \alpha (k)|n(k) \alpha (k) \rangle^{n(k)}}{(n(k))!}$ doesn't bring a whole lot to the table in the single-particle oscillator where $n$ is simply the energy eigenvalue, but in the continuous field where $n(k)$ is the number of excitations of the oscillator at normal mode index $k$ in the coherent state $|\alpha (k)\rangle$, this relation says that the probability of finding $n(k)$ excitations (particles) at that oscillator follows a Poisson distribution with mean $\langle \alpha (k)|n(k)|\alpha (k)\rangle$. This adds yet more credence to the notion that quantum field theory is a theory of varying particle number. Additionally, the value $\alpha (k)$ itself represents a point in classical phase space just as in the single-particle case, though here it represents a point for each given value of $k$: $\alpha_0 (k) = \frac{1}{\sqrt{2\hbar}} \left(\sqrt{\rho lv|k|} y_0 (k) + i\sqrt{\frac{l}{\rho v|k|}} p_{y_0} (k)\right)$ where $y_0$ and $p_{y_0}$ represent the configuration of the continuous chain as specified by the displacement and momentum density when represented as functions of the normal mode index $k$.

I would like to end this series of posts with an open question of mine. I have seen that some people define position eigenstates of the field by simply borrowing the definition from single-particle quantum mechanics and using the eigenstates $|k\rangle$ from quantum field theory in the same way: using the single-particle definition \[ \langle k|x\rangle = (2\pi)^{-\frac{1}{2}} e^{ikx} \] then the position eigenstate \[ |x\rangle = (2\pi)^{-\frac{1}{2}} \int e^{ikx} |k\rangle \, dk \] is said to hold even for a quantum field with excitations $|k\rangle$. But why should this be the case? Classically it doesn't make sense to say that the field exists at a certain position $x$, though it does make sense to say it is excited as a traveling wave of wavevector $k$. Of course, classically the field shouldn't have multiple discrete excitations, so that argument can be thrown out. Even so, though, we used that argument to say that our degrees of freedom are the displacement $y$ and momentum density $p_y$, while $x$ merely plays the role of an index and should not be considered an observable. So why can we now say all of a sudden that the quantum field can have excitations localized at a given position $x$? Yet, if we approach quantum field theory not from the angle of classical field theory but instead from the angle of multiparticle quantum mechanics, the latter framework definitely allows for countable discrete particles to each be in position eigenstates, and if all of those particles are in different position eigenstates, then the transformation of the overall state under the parity operator determines whether the particles are fermionic, bosonic, or distinguishable. At some point, though, when there becomes an infinite number of degrees of freedom, the position ceases to be a degree of freedom and merely labels an index, especially if the particles are indistinguishable. So at what point can $x$ as a degree of freedom for multiple particles be reconciled with $x$ as an index for a classical field to yield position eigenstates $|x\rangle$ in a quantum field? Is this just a quirk of mathematics, or can there really be excitations of a quantum field not with a particular wavevector but instead with a particular position? This is the one big open question I have about canonical quantization of a field.

Unless I think of other stuff, this is basically all I have to write for the moment about quantizing a classical field. Doing so has been immensely helpful for organizing my thoughts and answering some questions that were harder to answer when my thoughts weren't written down. I hope this is at least somewhat helpful to someone who may be looking for the same things that I was; note though that some of the formulas may be incorrect as I have barely scratched the surface of canonically quantizing a field. Now I'll get back to my usual posts (which for this week likely won't be too much more, but more will come next week).

No comments:

Post a Comment