In my last post about intuiting and visualizing quantum field theory, I discussed the diagonalization of the Hamiltonian and overall momentum and how they become operators. In this post I'm going to discuss more the meanings of the operators and associated quantum states of this field. Follow the jump to see more.

In single-particle quantum mechanics, states are represented by complex vectors $|\psi\rangle$ in a Hilbert space, and observables like $x$, $p$, and $H$ become [Hermitian] operators acting on those vectors to produce other vectors in the same space: in general, $A|\psi\rangle = |\phi\rangle$ for some states $\psi$ & $\phi$ and some operator $A$ (which may or may not be Hermitian). In the special case that $A$ is an observable (i.e. a Hermitian operator) and $|A\rangle$ is its eigenstate, then said eigenstate is a quantum state for which measurement of that observable returns the value $A$ with certainty. This means that states $|x\rangle$, $|p\rangle$, and $|E\rangle$ correspond to eigenstates of $x$, $p$, and $H$ respectively and represent states of definite position, momentum, and energy respectively. For a general state $|\psi\rangle$ though, for an observable $A$, the average value $\langle A \rangle = \langle \psi |A|\psi \rangle$ will be meaningful (even if it is not a measurable eigenvalue of $A$), and there may be some nontrivial uncertainty $\Delta A \equiv \sqrt{\langle \psi |A^2 |\psi \rangle - \langle \psi |A|\psi \rangle^2}$ for the average associated with that state. This is true of $x$ and $p$ for the ground state of the single-particle harmonic oscillator satisfying $H|0\rangle = \frac{1}{2} \hbar \omega \cdot |0\rangle$.

What are the operators and states for our acoustic field then? Let us start with a general statement about states. In single-particle quantum mechanics, there is only one state $|\psi\rangle$ to consider. To make things a little easier to see for the continuous case, let us go to the case of discrete coupled oscillators. The oscillator at lattice index $n$ can be a certain single-particle state, but that state may differ from one lattice index to another. This means that the state of the system would need to be defined for each lattice point $|\psi_n \rangle$. The logical extension of this is that in the continuum system, the discrete index $n$ becomes the continuous index $x$, so the system is generally defined by the state $|\psi (x)\rangle$.

The first operator to consider is the first one that really matters for the continuous field itself, and that is the displacement $y(x)$. Its eigenstates are $|y(x)\rangle$ and they define states where the displacement is known with certainty at every spatial index $x$. Due to the commutation relation with the momentum density, an eigenstate of the displacement will have fully indefinite momentum density at each spatial index $x$.

The second operator to consider should therefore be $p_y (x)$. Its eigenstates $|p_y (x)\rangle$ define states where the momentum density is known with certainty at every spatial index $x$. Again, due to the commutation relation with the displacement, an eigenstate of the momentum density will have fully indefinite displacement at each spatial index $x$.

I mentioned earlier that a general quantum state $|\psi (x)\rangle$ does

This looks a little weirder than before. What is going on? Well, I mentioned before that the displacement is classically defined at every spatial index $x$, so its eigenstates need to be defined like that too. So what the wavefunctional represents is the probability amplitude that the quantum field state will be measured to be an eigenstate of displacement at every spatial index $x$, such that the measured displacement is the observable $y(x)$; this can be pictured as a waveform on a chain having a definite frozen shape. Why is it a wavefunctional and not just a wavefunction though? That's just because the displacement $y$ is defined as a "vector" of sorts (in its own Hilbert space, when boundary conditions matter) indexed continuously by $x$, and a functional takes a vector as its input and returns a scalar as its output. Furthermore, as shall be seen soon, calculus looks a little different for functionals of vectors (i.e. $\psi [y(x)]$) as compared to functions of a single variable (i.e. $\psi (x)$).

Operators can be brought into this discussion as well. In single-particle quantum mechanics, operators have more concrete forms when acting on wavefunctions. For example, $\langle x|x|\psi\rangle = x \cdot \psi (x)$ from the eigenstate relation, and $\langle x|p|\psi\rangle = -i\hbar \frac{\partial \psi (x)}{\partial x}$ from the commutation relation. Similarly, \[ \langle y(x)|y(x)|\psi\rangle = y(x) \cdot \psi [y(x)] \] describes the action of the displacement operator at each spatial index $x$ on the wavefunctional. What gets even weirder is that \[ \langle y(x)|p_y (x)|\psi\rangle = -i\hbar \frac{\delta \psi [y(x)]}{\delta y(x)} \] describes the action of the momentum density operator on the wavefunctional. This superficially looks like the action of the single-particle momentum operator on the single-particle wavefunction, but the replacement of $\partial$ by $\delta$ is indicative that multivariable calculus has been replaced by

The expression \[ H = l^2 \int \left(\frac{p_y (-k) p_y (k)}{2\rho} + \frac{\rho v^2 k^2}{2} y(-k) y(k) \right) \, dk \] is of the Hamiltonian for the system. There are two problems with this: we don't know how do deal with operators expressed in terms of the normal mode index $k$ rather than the spatial index $x$, and we don't know the eigenstates of the Hamiltonian. Thankfully, the first problem is very easy to resolve, and the second problem is still not too hard either. The expression \[ \psi [y(k)] \equiv \langle y(k)|\psi \rangle \] is of the wavefunctional when the displacement is expressed in terms of the normal mode index; this is why the wavefunctional is not a normal wavefunction, as the displacement can be expressed in terms of either the spatial index or the plane wave normal mode index. This means that \[ \langle y(k)|y(k)|\psi \rangle = y(k) \cdot \psi [y(k)] \] is the action of the displacement operator and \[ \langle y(k)|p_y (k)|\psi \rangle = -i\hbar \frac{\delta \psi [y(k)]}{\delta y(-k)} \] is the action of the momentum density operator when each is expressed in terms of the plane wave normal mode index. Note the negative sign in front on the normal mode index in the denominator of the functional derivative due to the flipped sign in the commutation relation. Furthermore, note that strictly speaking, $y(k)$ and $p_y (k)$ are not measurable as they are complex quantities (i.e. not Hermitian); that said, I wonder if Hermitian operators $\sqrt{y(k)y(-k)}$ or $\sqrt{p_y (k)p_y (-k)}$ can be constructed through power series expansions (but that is merely a digression).

The second problem can be resolved in the same way as the single-particle harmonic oscillator as we already know that the wave along a chain, being a continuous sum over coupled harmonic oscillators in space, can be recast as a continuous sum over uncoupled harmonic oscillators in normal modes. We can define creation and annihilation operators through the relations \[ y(k) = \sqrt{\frac{\hbar}{2l\rho v|k|}} \left(a^{\dagger} (-k) + a(k)\right) \] and \[ p_y (k) = i\sqrt{\frac{\hbar\rho v|k|}{2l}} \left(a^{\dagger} (-k) - a(k)\right) \] for each normal mode index $k$; these satisfy \[ [a(k), a^{\dagger} (k')] = \frac{1}{l} \delta(k - k') \] as their commutation relation, analogous to the single-particle system. Substituting these into the Hamiltonian yields \[ H = \hbar lv \int |k| \cdot \left(a^{\dagger} (k) a(k) + \frac{1}{2} \right) \, dk \] which makes clear how the continuous sum over harmonic oscillators becomes quantized. Doing similar math for the overall momentum operator yields \[ p = \hbar l \int k \cdot \left(a^{\dagger} (k) a(k) + \frac{1}{2} \right) \, dk \] as the final expression diagonalized in terms of the continuous plane wave normal mode index $k$. But at this point the Hamiltonian has simply been rewritten to make its relation to the quantum harmonic oscillator more obvious. So what are the states?

As in single-particle quantum mechanics, we can define a ground state that yields zero upon annihilation. In the single-particle case, there is only one pair of creation and annihilation operators to consider. Now, though, such pairs of operators exist for each and every value of the continuous plane wave normal mode index. This means that $a(k)|0\rangle = 0$ for every $k$. The quantum state $|0\rangle$ is generally a little harder to picture though than its probability amplitude/wavefunctional. What does that look like? Well, in the basis of the displacement eigenstates, $\langle y(k)|a(k)|\psi\rangle = \sqrt{\frac{\rho lv|k|}{2\hbar}} \left(y(k) + \frac{\hbar}{\rho lv|k|} \frac{\delta}{\delta y(-k)} \right)$ with the Hermitian adjoint defined accordingly. Defining the notation $\psi_0 [y(k)] \equiv \langle y(k)|0\rangle$, then the equation to solve is $\frac{\hbar}{\rho l^2 v|k|} \frac{\delta \psi_0 [y(k)]}{\delta y(-k)} = -y(k)\psi_0 [y(k)]$ for every $k$. This is a

We can turn back to the case of discrete uncoupled oscillators. The ground state wavefunction for each oscillator looks like $\psi_0 (x_j) \propto \mathrm{exp}\left(-\frac{\sqrt{m}}{\hbar} \omega_j x_j^{\star} x_j \right)$. The overall ground state wavefunction is then the product of all these oscillator ground state wavefunctions: $\psi_0 (x_1, x_2, \ldots) \propto \prod_j \psi_0 (x_j)$. The problem is that a discrete product like that can't easily be converted for a continuous index $x$. What do we do? Thankfully we can use the property that the product of exponentials is the exponential of the sum of exponents: $\psi_0 (x_1, x_2, \ldots) \propto \mathrm{exp}\left(-\frac{\sqrt{m}}{\hbar} \sum_j \omega_j x_j^{\star} x_j \right)$. Of course a continuous sum is just an integral, so the continuous functional differential equation can be solved by analogy as $\psi_0 [y(k)] \propto \mathrm{exp}\left(-\frac{\rho^{\frac{1}{2}} l^{\frac{3}{2}} v}{\hbar} \int k y(-k) y(k) \, dk \right)$. This is the wavefunctional in terms of the displacement in the basis of the normal mode index rather than the spatial index; to switch to the spatial index, the Fourier transform needs to be carried out in the integral, but that is too much tedious work for me right now. Classically, this corresponds to having no displacement anywhere. Quantum mechanically, this corresponds to every infinitesimal oscillator along the continuum being in its ground state.

What is the energy and overall momentum of the ground state? Because $a(k)|0\rangle = 0$ for all $k$, the overall momentum is easy to find: $\int k \, dk = 0$ as $k$ is an odd function. Said more plainly, the ground state of the continuous chain has equal contributions of positive and negative momentum $\hbar k$ to cancel out. However, the energy gives a weird result, because $\int |k| \, dk$ diverges; said more plainly, the ground state energy of every infinitesimal uncoupled normal mode oscillator is nonnegative, so adding all of those energies gives an infinitely large value. This is known as an ultraviolet catastrophe because normal modes of arbitrarily high wavevector (and therefore energy $E = \hbar v|k|$) are contributing to the total energy. The solution to this should be an upper cutoff function (either a hard cutoff or a smoother function), but for now we can sweep that under the rug and discuss only the differences in energy and overall momentum present.

What are the excited states? Excitation of the ground state once in the single-particle case is easy: $|1\rangle \equiv a^{\dagger} |0\rangle$. In the continuous case, there is a continuum of $k$ to choose from for excitation, and the first excited state is then $|k\rangle \equiv a^{\dagger} (k) |0\rangle$ for a particular value of $k$; the corresponding wavefunctional is found by acting on $\psi_0 [y(k)]$ with the displacement field-space representation of $a^{\dagger} (k)$. Classically, this corresponds to and can be pictured as the continuous chain taking on a normal mode (traveling plane wave) with the given wavevector $k$. Quantum mechanically, the plane wave normal mode oscillator of continuous index $k$ has been excited once above the ground state.

This is quite a lot for one post, so I'll continue the discussion in the next post. In that, I might discuss a little more the intuition behind the creation and annihilation operators along with introducing...particles!

In single-particle quantum mechanics, states are represented by complex vectors $|\psi\rangle$ in a Hilbert space, and observables like $x$, $p$, and $H$ become [Hermitian] operators acting on those vectors to produce other vectors in the same space: in general, $A|\psi\rangle = |\phi\rangle$ for some states $\psi$ & $\phi$ and some operator $A$ (which may or may not be Hermitian). In the special case that $A$ is an observable (i.e. a Hermitian operator) and $|A\rangle$ is its eigenstate, then said eigenstate is a quantum state for which measurement of that observable returns the value $A$ with certainty. This means that states $|x\rangle$, $|p\rangle$, and $|E\rangle$ correspond to eigenstates of $x$, $p$, and $H$ respectively and represent states of definite position, momentum, and energy respectively. For a general state $|\psi\rangle$ though, for an observable $A$, the average value $\langle A \rangle = \langle \psi |A|\psi \rangle$ will be meaningful (even if it is not a measurable eigenvalue of $A$), and there may be some nontrivial uncertainty $\Delta A \equiv \sqrt{\langle \psi |A^2 |\psi \rangle - \langle \psi |A|\psi \rangle^2}$ for the average associated with that state. This is true of $x$ and $p$ for the ground state of the single-particle harmonic oscillator satisfying $H|0\rangle = \frac{1}{2} \hbar \omega \cdot |0\rangle$.

What are the operators and states for our acoustic field then? Let us start with a general statement about states. In single-particle quantum mechanics, there is only one state $|\psi\rangle$ to consider. To make things a little easier to see for the continuous case, let us go to the case of discrete coupled oscillators. The oscillator at lattice index $n$ can be a certain single-particle state, but that state may differ from one lattice index to another. This means that the state of the system would need to be defined for each lattice point $|\psi_n \rangle$. The logical extension of this is that in the continuum system, the discrete index $n$ becomes the continuous index $x$, so the system is generally defined by the state $|\psi (x)\rangle$.

**This does not mean that the state of the system is defined by a wavefunction**$\psi (x)$**!**What it means is that the Hilbert space state is conveniently defined by the continuous index $x$. That said, in the discrete case, the overall state of the system would be defined as a product of some sort (perhaps a tensor product, but I am not sure on this point) of the state of each oscillator indexed by $n$; furthermore, superpositions are possible, allowing for entangled states. In the continuous case, well, I'm not sure of any sort of continuous product operation, so I will stick to simply calling states of the quantum field as $|\psi\rangle$.The first operator to consider is the first one that really matters for the continuous field itself, and that is the displacement $y(x)$. Its eigenstates are $|y(x)\rangle$ and they define states where the displacement is known with certainty at every spatial index $x$. Due to the commutation relation with the momentum density, an eigenstate of the displacement will have fully indefinite momentum density at each spatial index $x$.

The second operator to consider should therefore be $p_y (x)$. Its eigenstates $|p_y (x)\rangle$ define states where the momentum density is known with certainty at every spatial index $x$. Again, due to the commutation relation with the displacement, an eigenstate of the momentum density will have fully indefinite displacement at each spatial index $x$.

I mentioned earlier that a general quantum state $|\psi (x)\rangle$ does

*not*correspond to a wavefunction $\psi (x)$ despite the deceptive similarity to single-particle quantum mechanics. What are the analogues of wavefunctions then? Here we can again turn to single-particle quantum mechanics for help. The wavefunction in position space is defined by $\psi (x) \equiv \langle x|\psi \rangle$, meaning that the wavefunction is the probability amplitude of measuring a state $|\psi \rangle$ to be a state of definite position with value $x$. For our field, the analogue of single-particle position is the position of each infinitesimal oscillator at spatial index $x$, which is the displacement $y(x)$. This means that \[ \psi [y(x)] \equiv \langle y(x)|\psi\rangle \] defines the wave*functional*of the system.This looks a little weirder than before. What is going on? Well, I mentioned before that the displacement is classically defined at every spatial index $x$, so its eigenstates need to be defined like that too. So what the wavefunctional represents is the probability amplitude that the quantum field state will be measured to be an eigenstate of displacement at every spatial index $x$, such that the measured displacement is the observable $y(x)$; this can be pictured as a waveform on a chain having a definite frozen shape. Why is it a wavefunctional and not just a wavefunction though? That's just because the displacement $y$ is defined as a "vector" of sorts (in its own Hilbert space, when boundary conditions matter) indexed continuously by $x$, and a functional takes a vector as its input and returns a scalar as its output. Furthermore, as shall be seen soon, calculus looks a little different for functionals of vectors (i.e. $\psi [y(x)]$) as compared to functions of a single variable (i.e. $\psi (x)$).

Operators can be brought into this discussion as well. In single-particle quantum mechanics, operators have more concrete forms when acting on wavefunctions. For example, $\langle x|x|\psi\rangle = x \cdot \psi (x)$ from the eigenstate relation, and $\langle x|p|\psi\rangle = -i\hbar \frac{\partial \psi (x)}{\partial x}$ from the commutation relation. Similarly, \[ \langle y(x)|y(x)|\psi\rangle = y(x) \cdot \psi [y(x)] \] describes the action of the displacement operator at each spatial index $x$ on the wavefunctional. What gets even weirder is that \[ \langle y(x)|p_y (x)|\psi\rangle = -i\hbar \frac{\delta \psi [y(x)]}{\delta y(x)} \] describes the action of the momentum density operator on the wavefunctional. This superficially looks like the action of the single-particle momentum operator on the single-particle wavefunction, but the replacement of $\partial$ by $\delta$ is indicative that multivariable calculus has been replaced by

*functional*calculus; the rules aren't exactly the same, but they are similar.The expression \[ H = l^2 \int \left(\frac{p_y (-k) p_y (k)}{2\rho} + \frac{\rho v^2 k^2}{2} y(-k) y(k) \right) \, dk \] is of the Hamiltonian for the system. There are two problems with this: we don't know how do deal with operators expressed in terms of the normal mode index $k$ rather than the spatial index $x$, and we don't know the eigenstates of the Hamiltonian. Thankfully, the first problem is very easy to resolve, and the second problem is still not too hard either. The expression \[ \psi [y(k)] \equiv \langle y(k)|\psi \rangle \] is of the wavefunctional when the displacement is expressed in terms of the normal mode index; this is why the wavefunctional is not a normal wavefunction, as the displacement can be expressed in terms of either the spatial index or the plane wave normal mode index. This means that \[ \langle y(k)|y(k)|\psi \rangle = y(k) \cdot \psi [y(k)] \] is the action of the displacement operator and \[ \langle y(k)|p_y (k)|\psi \rangle = -i\hbar \frac{\delta \psi [y(k)]}{\delta y(-k)} \] is the action of the momentum density operator when each is expressed in terms of the plane wave normal mode index. Note the negative sign in front on the normal mode index in the denominator of the functional derivative due to the flipped sign in the commutation relation. Furthermore, note that strictly speaking, $y(k)$ and $p_y (k)$ are not measurable as they are complex quantities (i.e. not Hermitian); that said, I wonder if Hermitian operators $\sqrt{y(k)y(-k)}$ or $\sqrt{p_y (k)p_y (-k)}$ can be constructed through power series expansions (but that is merely a digression).

The second problem can be resolved in the same way as the single-particle harmonic oscillator as we already know that the wave along a chain, being a continuous sum over coupled harmonic oscillators in space, can be recast as a continuous sum over uncoupled harmonic oscillators in normal modes. We can define creation and annihilation operators through the relations \[ y(k) = \sqrt{\frac{\hbar}{2l\rho v|k|}} \left(a^{\dagger} (-k) + a(k)\right) \] and \[ p_y (k) = i\sqrt{\frac{\hbar\rho v|k|}{2l}} \left(a^{\dagger} (-k) - a(k)\right) \] for each normal mode index $k$; these satisfy \[ [a(k), a^{\dagger} (k')] = \frac{1}{l} \delta(k - k') \] as their commutation relation, analogous to the single-particle system. Substituting these into the Hamiltonian yields \[ H = \hbar lv \int |k| \cdot \left(a^{\dagger} (k) a(k) + \frac{1}{2} \right) \, dk \] which makes clear how the continuous sum over harmonic oscillators becomes quantized. Doing similar math for the overall momentum operator yields \[ p = \hbar l \int k \cdot \left(a^{\dagger} (k) a(k) + \frac{1}{2} \right) \, dk \] as the final expression diagonalized in terms of the continuous plane wave normal mode index $k$. But at this point the Hamiltonian has simply been rewritten to make its relation to the quantum harmonic oscillator more obvious. So what are the states?

As in single-particle quantum mechanics, we can define a ground state that yields zero upon annihilation. In the single-particle case, there is only one pair of creation and annihilation operators to consider. Now, though, such pairs of operators exist for each and every value of the continuous plane wave normal mode index. This means that $a(k)|0\rangle = 0$ for every $k$. The quantum state $|0\rangle$ is generally a little harder to picture though than its probability amplitude/wavefunctional. What does that look like? Well, in the basis of the displacement eigenstates, $\langle y(k)|a(k)|\psi\rangle = \sqrt{\frac{\rho lv|k|}{2\hbar}} \left(y(k) + \frac{\hbar}{\rho lv|k|} \frac{\delta}{\delta y(-k)} \right)$ with the Hermitian adjoint defined accordingly. Defining the notation $\psi_0 [y(k)] \equiv \langle y(k)|0\rangle$, then the equation to solve is $\frac{\hbar}{\rho l^2 v|k|} \frac{\delta \psi_0 [y(k)]}{\delta y(-k)} = -y(k)\psi_0 [y(k)]$ for every $k$. This is a

*functional differential equation*; how can we have any hope of solving this?We can turn back to the case of discrete uncoupled oscillators. The ground state wavefunction for each oscillator looks like $\psi_0 (x_j) \propto \mathrm{exp}\left(-\frac{\sqrt{m}}{\hbar} \omega_j x_j^{\star} x_j \right)$. The overall ground state wavefunction is then the product of all these oscillator ground state wavefunctions: $\psi_0 (x_1, x_2, \ldots) \propto \prod_j \psi_0 (x_j)$. The problem is that a discrete product like that can't easily be converted for a continuous index $x$. What do we do? Thankfully we can use the property that the product of exponentials is the exponential of the sum of exponents: $\psi_0 (x_1, x_2, \ldots) \propto \mathrm{exp}\left(-\frac{\sqrt{m}}{\hbar} \sum_j \omega_j x_j^{\star} x_j \right)$. Of course a continuous sum is just an integral, so the continuous functional differential equation can be solved by analogy as $\psi_0 [y(k)] \propto \mathrm{exp}\left(-\frac{\rho^{\frac{1}{2}} l^{\frac{3}{2}} v}{\hbar} \int k y(-k) y(k) \, dk \right)$. This is the wavefunctional in terms of the displacement in the basis of the normal mode index rather than the spatial index; to switch to the spatial index, the Fourier transform needs to be carried out in the integral, but that is too much tedious work for me right now. Classically, this corresponds to having no displacement anywhere. Quantum mechanically, this corresponds to every infinitesimal oscillator along the continuum being in its ground state.

What is the energy and overall momentum of the ground state? Because $a(k)|0\rangle = 0$ for all $k$, the overall momentum is easy to find: $\int k \, dk = 0$ as $k$ is an odd function. Said more plainly, the ground state of the continuous chain has equal contributions of positive and negative momentum $\hbar k$ to cancel out. However, the energy gives a weird result, because $\int |k| \, dk$ diverges; said more plainly, the ground state energy of every infinitesimal uncoupled normal mode oscillator is nonnegative, so adding all of those energies gives an infinitely large value. This is known as an ultraviolet catastrophe because normal modes of arbitrarily high wavevector (and therefore energy $E = \hbar v|k|$) are contributing to the total energy. The solution to this should be an upper cutoff function (either a hard cutoff or a smoother function), but for now we can sweep that under the rug and discuss only the differences in energy and overall momentum present.

What are the excited states? Excitation of the ground state once in the single-particle case is easy: $|1\rangle \equiv a^{\dagger} |0\rangle$. In the continuous case, there is a continuum of $k$ to choose from for excitation, and the first excited state is then $|k\rangle \equiv a^{\dagger} (k) |0\rangle$ for a particular value of $k$; the corresponding wavefunctional is found by acting on $\psi_0 [y(k)]$ with the displacement field-space representation of $a^{\dagger} (k)$. Classically, this corresponds to and can be pictured as the continuous chain taking on a normal mode (traveling plane wave) with the given wavevector $k$. Quantum mechanically, the plane wave normal mode oscillator of continuous index $k$ has been excited once above the ground state.

This is quite a lot for one post, so I'll continue the discussion in the next post. In that, I might discuss a little more the intuition behind the creation and annihilation operators along with introducing...particles!