1 Canonical Field Quantization
In this chapter we will see in full detail how to build a quantum field. We will focus on the simplest case of a real, scalar, non-relativistic field. As we will see, this will turn out to be a quantum theory of sound propagating through a crystal. This is the phonon field, whose particle excitations are called phonons.
References
N. W. Ashcroft and N. D. Mermin, Solid State Physics (Harcourt 1976), Appendix L
R. P. Feynman, Statistical Mechanics (Advanced Book Classics, 1972) Chapter 6
A. Altland and B. Simons, Condensed Matter Field Theory (Cambridge University Press, 2010), Chapter 1
T. Lancaster and S. J. Blundell, Quantum Field Theory for the Gifted Amateur (Oxford University Press, 2014), Section I
S. M. Girvin and K. Yang, Modern Condensed Matter Physics (Cambridge University Press, 2019), Section 6.2
1.1 Recap: Quantum Harmonic Oscillators
Recall the Time Independent Schroedinger Equation (TISE) for the quantum harmonic oscillator:
\[\hat{H}|n\rangle=\left(\frac{\hat{p}^{2}}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^{2}\right)|n\rangle=E_{n}|n\rangle.\]
The potential is shown in the position basis in Fig. 1.2.
An elegant way to find all eigenstates and eigenvalues is to use creation and annihilation operators (aka ladder operators, aka raising and lowering operators). The intuition developed in this simple QM problem forms the entire basis of QFT.
1.1.1 Creation and Annihilation Operators
The raising operator is
\[\boxed{\hat{a}^{\dagger}\triangleq\sqrt{\frac{m\omega}{2}}\left(\hat{x}-\frac{i}{m\omega}\hat{p}\right).} \tag{1.1}\]
It is non-Hermitian. Its hermitian conjugate is the lowering operator
\[\boxed{\hat{a}=\sqrt{\frac{m\omega}{2}}\left(\hat{x}+\frac{i}{m\omega}\hat{p}\right).} \tag{1.2}\]
This gives
\[\begin{aligned} \hat{a}^{\dagger}\hat{a} & =\frac{m\omega}{2}\left(\hat{x}^{2}+\frac{\hat{p}^{2}}{m^{2}\omega^{2}}+\frac{i}{m\omega}\left(\hat{x}\hat{p}-\hat{p}\hat{x}\right)\right). \end{aligned}\]
The term in nested parentheses is just the commutator
\[\left[\hat{x},\hat{p}\right]=i\hat{\mathbb{I}}\]
where \(\hat{\mathbb{I}}\) is the identity operator. Hence,
\[\hat{a}^{\dagger}\hat{a}=\frac{m\omega}{2}\hat{x}^{2}+\frac{\hat{p}^{2}}{2m\omega}-\frac{1}{2}\hat{\mathbb{I}}\]
and so
\[\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hat{\mathbb{I}}\right)=\hat{H}.\]
Therefore the TISE can be written
\[\boxed{\hat{H}|n\rangle=\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hat{\mathbb{I}}\right)|n\rangle=E_{n}|n\rangle.} \tag{1.3}\]
1.1.2 Commutation relations
The commutator of the creation and annihilation operators is
\[\begin{aligned} \left[\hat{a},\hat{a}^{\dagger}\right] & =\frac{m\omega}{2}\left[\hat{x}+\frac{i}{m\omega}\hat{p},\hat{x}-\frac{i}{m\omega}\hat{p}\right]\\ & =-i\left[\hat{x},\hat{p}\right] \end{aligned}\]
and so \[\boxed{\left[\hat{a},\hat{a}^{\dagger}\right]=\hat{\mathbb{I}}}\]
The commutator of the operators with the Hamiltonian is therefore:
\[\begin{aligned} \left[\hat{H},\hat{a}^{\dagger}\right] & =\left[\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hat{\mathbb{I}}\right),\hat{a}^{\dagger}\right]\\ & =\omega\left(\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger}-\hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a}\right)\\ & =\omega\hat{a}^{\dagger}\left[\hat{a},\hat{a}^{\dagger}\right] \end{aligned}\]
giving
\[\boxed{\left[\hat{H},\hat{a}^{\dagger}\right]=\omega\hat{a}^{\dagger}.} \tag{1.4}\]
Similarly,
\[\left[\hat{H},\hat{a}\right]=-\omega\hat{a}.\]
1.1.3 Energy eigenstates and eigenvalues
Using Eq. 1.4 we can see the effect of \(\hat{a}^{\dagger}\) on the TISE of Eq. 1.3:
\[\begin{aligned} \hat{H}|n\rangle & =E_{n}|n\rangle\\ & \downarrow\nonumber \\ \hat{a}^{\dagger}\hat{H}|n\rangle & =E_{n}\hat{a}^{\dagger}|n\rangle\\ & \downarrow\nonumber \\ \left(\hat{H}\hat{a}^{\dagger}-\left[\hat{H},\hat{a}^{\dagger}\right]\right)|n\rangle & =E_{n}\hat{a}^{\dagger}|n\rangle\\ & \downarrow\nonumber \\ \left(\hat{H}\hat{a}^{\dagger}-\omega\hat{a}^{\dagger}\right)|n\rangle & =E_{n}\hat{a}^{\dagger}|n\rangle \end{aligned}\]
and the final result
\[\boxed{\hat{H}\left(\hat{a}^{\dagger}|n\rangle\right)=\left(E_{n}+\omega\right)\left(\hat{a}^{\dagger}|n\rangle\right).}\]
That is, if \(|n\rangle\) is an eigenstate of the harmonic oscillator with eigenvalue \(E_{n}\), then \(\hat{a}^{\dagger}|n\rangle\) is an eigenstate with eigenvalue \(E_{n}+\omega\). Repeating the process \(l\) times we find
\[\hat{H}\left(\left(\hat{a}^{\dagger}\right)^{l}|n\rangle\right)=\left(E_{n}+l\omega\right)\left(\left(\hat{a}^{\dagger}\right)^{l}|n\rangle\right).\]
This tells us that the energy levels are evenly spaced, and that
\[\left(\hat{a}^{\dagger}\right)^{l}|n\rangle\propto|n+l\rangle. \tag{1.5}\]
Similarly, we find
\[\hat{H}\left(\hat{a}|n\rangle\right)=\left(E_{n}-\omega\right)\left(\hat{a}|n\rangle\right).\]
The creation and annihilation operators move the state up or down the rungs of an energy ladder with evenly spaced rungs. While there exist an infinite number of rungs, the energies do not stretch down to negative energies. To see this, first note that
\[\left|\hat{a}|n\rangle\right|^{2}\ge0\]
because the thing on the left, whatever it is, is the square modulus of something, and that is always \(\ge0\). Expanding it out we have
\[\begin{aligned} \langle n|\hat{a}^{\dagger}\hat{a}|n\rangle & \ge0\\ \langle n|\frac{1}{\omega}\hat{H}-\frac{1}{2}|n\rangle & \ge0\\ \langle n|\frac{1}{\omega}E_{n}-\frac{1}{2}|n\rangle & \ge0\\ E_{n} & \ge\frac{\omega}{2}. \end{aligned}\]
Therefore there is a lowest-energy state, a lowest rung to the ladder. This is the ground state which we denote \(|0\rangle\). This is just a convenient label for a ket. It is not the number zero! You can think of it as shorthand for \(|\psi_{n=0}\rangle\). The ground state has the property that
\[\boxed{\hat{a}|0\rangle=0}\]
where the right hand side really is the number \(0\), so that any further action of lowering operators continues to return \(0\). We can find the energy of this state straightforwardly:
\[\begin{aligned} \hat{H}|0\rangle & =\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hat{\mathbb{I}}\right)|0\rangle\\ & =\frac{\omega}{2}|0\rangle \end{aligned}\]
so \(E_{0}=\omega/2\). This is the ground state energy, also called the zero-point energy. Combining the results, we see that
\[\hat{H}|n\rangle=\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\right)|n\rangle=\omega\left(n+\frac{1}{2}\right)|n\rangle\]
for integer \(n\). Defining the ‘number operator’
\[\hat{n}\triangleq\hat{a}^{\dagger}\hat{a}\]
we see that
\[\hat{n}|n\rangle=n|n\rangle.\]
1.1.4 Eigenstates
To find the energy eigenstates in the position basis we can use the definition of the lowering operator from Eq. 1.2:
\[\begin{aligned} \sqrt{\frac{m\omega}{2}}\left(\hat{x}+\frac{i}{m\omega}\hat{p}\right)|0\rangle & =0\\ & \downarrow\nonumber \\ \left(x+\frac{1}{m\omega}\frac{\text{d}}{\text{d}x}\right)\phi_{0}\left(x\right) & =0. \end{aligned}\]
This is a first order linear ODE which has the solution
\[\phi_{0}\left(x\right)\propto\exp\left(-\frac{m\omega}{2}x^{2}\right).\]
All excited states can then be found by acting the creation operator on this state repeatedly. For example,
\[\begin{aligned} \phi_{1}\left(x\right) & \propto\hat{a}^{\dagger}\phi_{0}\left(x\right)\\ & =\left(x-\frac{1}{m\omega}\frac{\text{d}}{\text{d}x}\right)\exp\left(-\frac{m\omega}{2}x^{2}\right)\\ & =2x\exp\left(-\frac{m\omega}{2}x^{2}\right). \end{aligned}\]
1.2 Second quantization
Note Added: within single-particle quantum mechanics, the following description of the nth excited state of the harmonic oscillator as n bosons is really only an analogy. It striclty becomes precise only when we move to QFT and allow multi-particle states.
Equation 1.5 tells us we can create the \(n^{\text{th}}\) excited state of the harmonic oscillator by acting \(n\) raising operators on the ground state.
Since all rungs of the ladder are evenly spaced with spacing \(\omega\), we can also interpret the \(n^{\text{th}}\) excited state as the presence of \(n\) identical particles each of energy \(\omega\).
This is the origin of the name ‘creation and annihilation operators’: they create and annihilate particles in a second-quantised description.
‘First quantization’ is the realisation that classical particles exhibit wave-like properties in QM. We call the process of rewriting a problem in terms of creation and annihilation operators ‘second quantization’. It is the realisation that classical waves gain particle-like properties in QM. The descriptions are equivalent.
The fact that we can fit multiple particles into the same state identifies \(\hat{a}^{\dagger}\) as the creation operator of a boson.
1.3 \(N\) Isolated Quantum Harmonic Oscillators
Now consider \(N\) independent QHOs evenly spaced with spacing \(a\) along a chain. Assume periodic boundary conditions, so that site \(N+1\) is equivalent to site \(1\) (i.e. the chain is really a ring). The Hamiltonian is
\[\hat{H}=\sum^{N}_{n=1}\frac{\hat{p}^{2}_{n}}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^{2}_{n}. \tag{1.6}\]
We can rewrite this using creation and annihilation operators, each with a site label:
\[\hat{H}=\sum^{N}_{n=1}\omega\left(\hat{a}^{\dagger}_{n}\hat{a}_{n}+\frac{1}{2}\hat{\mathbb{I}}\right).\]
Here, \(\hat{a}^{\dagger}_{n}\) creates an excitation of energy \(\omega\) at site \(n\). We can also take advantage of periodicity (translational symmetry) to Fourier transform, using
\[\begin{aligned} \hat{a}^{\dagger}_{n} & =\frac{1}{\sqrt{N}}\sum^{N}_{k=1}\exp\left(2\pi i\frac{k}{N}n\right)\hat{a}^{\dagger}_{k}\\ \hat{a}^{\dagger}_{k} & =\frac{1}{\sqrt{N}}\sum^{N}_{n=1}\exp\left(-2\pi i\frac{k}{N}n\right)\hat{a}^{\dagger}_{n} \end{aligned}\]
to give
\[\begin{aligned} \hat{H} & =\sum_{n}\omega\hat{a}^{\dagger}_{n}\hat{a}_{n}+\omega\frac{1}{2}\hat{\mathbb{I}}N\\ & =\frac{1}{N}\sum_{n,k,q}\omega\exp\left(2\pi i\frac{k-q}{N}n\right)\hat{a}^{\dagger}_{k}\hat{a}_{q}+\omega\frac{1}{2}\hat{\mathbb{I}}N\nonumber \\ & =\sum_{k,q}\omega\delta_{kq}\hat{a}^{\dagger}_{k}\hat{a}_{q}+\omega\frac{1}{2}\hat{\mathbb{I}}N\\ & =\sum_{k}\omega\hat{a}^{\dagger}_{k}\hat{a}_{k}+\omega\frac{1}{2}\hat{\mathbb{I}}N \end{aligned}\]
where we used the useful expression
\[\frac{1}{N}\sum^{N}_{n=1}\exp\left(2\pi i\frac{n}{N}\left(k-q\right)\right)=\delta_{kq} \tag{1.7}\]
which follows since the sum is over equally distributed points around the unit circle. In fact we can also use this expression to rewrite as
\[\begin{aligned} \hat{H} & =\sum^{N}_{k}\omega\left(\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}\hat{\mathbb{I}}\right). \end{aligned}\]
In this case, \(\hat{a}^{\dagger}_{k}\) creates an excitation of energy \(\omega\) with wavevector \(k_{n}=\pi k/N\), where \(1\le k\le N\).
The Hamiltonian in Eq 1.6 is overly simplified: the QHOs are uncoupled, and so it is hard to imagine how one might physically create an excitation across multiple oscillators. It is a remarkable result, which we will derive shortly, that when interactions are turned on between the oscillators, provided translational invariance is preserved, the effect is merely to add a wavevector dependence to the angular frequency \(\omega\rightarrow\omega_{k}\). That is:
\[\hat{H}_{\text{interacting}}=\sum^{N}_{k}\omega_{k}\left(\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}\hat{\mathbb{I}}\right).\]
Before seeing how this works, let us first build intuition by looking at the classical problem.
1.4 \(N\) Coupled Classical Harmonic Oscillators
Consider a classical chain of balls connected by springs of stiffness \(K\) (Fig. 1.3) along an \(N\)-site chain. This is a chain of coupled classical harmonic oscillators. The balls oscillate about equilibrium positions \(an\) (where \(a\) is a lattice constant and \(n\in\left[1,N\right]\)). Let the position of ball \(n\) at time \(t\) be \(u_{n,t}\), and define the displacement from the equilibrium position to be \(x_{n,t}=an+u_{n,t}\). From now on we will only care about the small displacements \(u_{n,t}\).
1.4.1 Recap: forces on balls and springs
Wobbling one of the balls will lead to a vibration propagating through the chain. To find the dispersion relation of this propagating mode, consider the force on the \(n^{\text{th}}\) ball (using Newton’s 2nd law):
\[m\ddot{x}_{n,t}=K\left(x_{n+1,t}-x_{n,t}\right)-K\left(x_{n,t}-x_{n-1,t}\right). \tag{1.8}\]
We need the normal modes of oscillation. To find them, first take advantage of the translational symmetry by applying a Fourier transform:
\[x_{n,t}=\frac{1}{N}\sum^{N}_{k=1}\int\text{d}\omega\exp\left(2\pi i\left(kn/N-\omega t\right)\right)Q_{k,\omega}.\]
Let’s change variables so that \(k\) has dimensions of inverse length, so that it can be called a wavevector. To do this, change variables to \(k'a=2\pi k/N\) and \(\omega'=2\pi\omega\), and redefine \(Q_{k,\omega}\) to \(Q_{k',\omega'}\):
\[x_{n,t}=\frac{1}{2\pi N}\sum^{2\pi/a}_{k'=2\pi/Na}\int\text{d}\omega'\exp\left(i\left(k'an-\omega't\right)\right)Q_{k',\omega'}\]
and then relabel back to \(k'\rightarrow k\), \(\omega'\rightarrow\omega\):
\[x_{n,t}=\frac{1}{2\pi N}\sum^{2\pi/a}_{k=2\pi/Na}\int\text{d}\omega\exp\left(i\left(kan-\omega t\right)\right)Q_{k,\omega}.\]
Inserting the Fourier transform into the equations of motion gives:
\[\begin{aligned} m\sum_{k}\int\text{d}\omega\exp\left(i\left(kan-\omega t\right)\right)\left(-\omega^{2}\right)Q_{k,\omega}\nonumber \\ & =K\sum_{k}\int\text{d}\omega\exp\left(i\left(kan-\omega t\right)\right)Q_{k,\omega}\left(\left(\exp\left(ika\right)-1\right)-\left(1-\exp\left(-ika\right)\right)\right). \end{aligned}\]
We can inverse Fourier transform to find a separate solution for each pair of \(k\) and \(\omega\): \[-m\omega^{2}_{k}Q_{k,\omega}=2K\left(\cos\left(ka\right)-1\right)Q_{k,\omega}\]
where the notation \(\omega_{k}\) emphasises that this is a specific solution \(\omega\left(k\right)\). Hence
\[\begin{aligned} \omega_{k} & =\sqrt{\frac{2K}{m}\left(1-\cos\left(ka\right)\right)}\\ & =2\sqrt{\frac{K}{m}}\left|\sin\left(\frac{ka}{2}\right)\right| \end{aligned}\]
(noting that only the positive root is physical). Around \(ka\approx0\) (long wavelengths) the mode is linearly dispersing:
\[\omega_{k}\approx\sqrt{\frac{K}{m}}\left|ka\right|\]
which defines the speed of sound along the chain via
\[\omega_{k}=ck\]
giving
\[c=a\sqrt{\frac{K}{m}}.\]
1.4.2 The classical Lagrangian
Thanks to work carried out in Bristol by Aharonov and Bohm, we know that in quantum mechanics we must deal with potentials rather than forces. To set up the quantum problem we must rephrase the classical problem construct the Lagrangian, and from there the Hamiltonian, of the classical system. We will then quantize this in Section @subsec-Coupled-Quantum.
The classical balls and springs model is governed by the Lagrangian
\[L=\sum^{N}_{n=1}\frac{1}{2}m\dot{x}^{2}_{n}-\frac{1}{2}m\omega^{2}\left(x_{n+1}-x_{n}\right)^{2}-\frac{1}{2}m\omega^{2}\left(x_{n}-x_{n-1}\right)^{2}.\]
To check this is equivalent to Eq. 1.8, we can derive the classical equations of motion using the Euler Lagrange equations:
\[\left(\frac{\partial L}{\partial x_{n}}\right)_{\dot{x}_{n}}-\frac{\text{d}}{\text{d}t}\left(\frac{\partial L}{\partial\dot{x}_{n}}\right)_{x_{n}}=0\]
along with the identities
\[\left(\frac{\partial x_{m}}{\partial x_{n}}\right)_{\dot{x}_{m}}=\left(\frac{\partial\dot{x}_{m}}{\partial\dot{x}_{n}}\right)_{x_{m}}=\delta_{nm}\]
where we have introduced the
Kronecker delta:
\[\delta_{nm}\triangleq\begin{cases} \begin{array}{c} 0,\\ 1, \end{array} & \begin{array}{c} n\ne m\\ n=m. \end{array}\end{cases}\]
We find
\[\begin{aligned} \left(\frac{\partial L}{\partial x_{n}}\right)_{\dot{x}_{n}} & =\sum^{N}_{m=1}-m\omega^{2}\left(x_{m+1}-x_{m}\right)\left(\delta_{m+1,n}-\delta_{m,n}\right)-m\omega^{2}\left(x_{m}-x_{m-1}\right)\left(\delta_{m,n}-\delta_{m-1,n}\right)\\ & =-2m\omega^{2}\left(2x_{n}-x_{n-1}-x_{n+1}\right) \end{aligned}\]
and
\[\begin{aligned} \left(\frac{\partial L}{\partial\dot{x}_{n}}\right)_{x_{n}} & =m\dot{x}_{n} \end{aligned}\]
so that the Euler Lagrange equations read
\[\begin{aligned} -2m\omega^{2}\left(2x_{n}-x_{n-1}-x_{n+1}\right)-m\ddot{x}_{n} & =0\\ & \downarrow\\ m\ddot{x}_{n} & =2m\omega^{2}\left(x_{n+1}-x_{n}\right)-2m\omega^{2}\left(x_{n}-x_{n-1}\right) \end{aligned}\]
which, as expected, is Eq. 1.8 (with force constant \(K=2m\omega^{2}\)).
1.4.3 The Classical Hamiltonian
To find the Hamiltonian, we need the momentum canonically conjugate to the position. This is defined to be:
\[\begin{aligned} p_{n} & \triangleq\left(\frac{\partial L}{\partial\dot{x}_{n}}\right)_{x_{n}} \end{aligned}\]
which in this case gives
\[p_{n}=m\dot{x}_{n}.\]
The Hamiltonian is then the Legendre transform of the Lagrangian:
\[\begin{aligned} H & \triangleq\sum_{n}p_{n}\cdot\dot{x}_{n}-L. \end{aligned}\]
In this case we find
\[H=\sum_{n}\frac{1}{2m}p^{2}_{n}+\frac{1}{2}m\omega^{2}\left(x_{n+1}-x_{n}\right)^{2}+\frac{1}{2}m\omega^{2}\left(x_{n}-x_{n-1}\right)^{2}. \tag{1.9}\]
This is the classical Hamiltonian for the balls and springs model. It is a constant of motion corresponding to the total energy of the system.
1.5 \(N\) Coupled Quantum Harmonic Oscillators
Now let’s combine these ideas, with a chain of coupled quantum harmonic oscillators. Our aim will be to confirm the intuition that we can decompose the problem into one of \(N\) independent QHOs, one for each wavevector \(k\). To create the quantum problem we can ‘canonically quantize’ the classical Hamiltonian in Eq. 1.9. This simply means promoting observables to operators:
\[\hat{H}=\sum^{N}_{n=1}\frac{\hat{p_{n}}^{2}}{2m}+\frac{1}{2}m\omega^{2}\sum_{n}\left(\hat{x}_{n+1}-\hat{x}_{n}\right)^{2}+\left(\hat{x}_{n}-\hat{x}_{n-1}\right)^{2}\]
which obey the ‘canonical commutation relations’:
\[\begin{aligned} \left[\hat{x}_{n},\hat{p}_{m}\right] & =i\hat{\mathbb{I}}\delta_{nm}\\ \left[\hat{x}_{n},\hat{x}_{m}\right] & =0\\ \left[\hat{p}_{n},\hat{p}_{m}\right] & =0. \end{aligned}\]
Note that operators on different sites always commute. We can again take advantage of translational invariance to define
\[\begin{aligned} \hat{x}_{n} & =\frac{1}{\sqrt{N}}\sum^{2\pi/a}_{k=2\pi/Na}\exp\left(ikan\right)\hat{Q}_{k}\\ \hat{p}_{n} & =\frac{1}{\sqrt{N}}\sum_{k}\exp\left(-ikan\right)\hat{\Pi}_{k} \end{aligned}\]
and
\[\begin{aligned} \hat{Q}_{k} & =\frac{1}{\sqrt{N}}\sum^{N}_{n=1}\exp\left(-ikan\right)\hat{x}_{n}\\ \hat{\Pi}_{k} & =\frac{1}{\sqrt{N}}\sum^{N}_{n=1}\exp\left(ikan\right)\hat{p}_{n}. \end{aligned}\]
Note that these tranformed co-ordinates also obey canonical commutation relations:
\[\begin{aligned} \left[\hat{Q}_{k},\hat{\Pi}_{q}\right] & =\frac{1}{N}\sum_{n,m}\exp\left(ia\left(qm-kn\right)\right)\left[\hat{x}_{n},\hat{p}_{m}\right]\\ & =\frac{1}{N}\sum_{n,m}\exp\left(ia\left(qm-kn\right)\right)i\hat{\mathbb{I}}\delta_{nm}\\ & =\frac{1}{N}\sum_{n}\exp\left(ia\left(q-k\right)n\right)i\hat{\mathbb{I}}\\ & =i\hat{\mathbb{I}}\delta_{kq} \end{aligned}\]
using Eq 1.7 in the last line. Similarly,
\[\left[\hat{Q}_{k},\hat{Q}_{q}\right]=\left[\hat{\Pi}_{k},\hat{\Pi}_{q}\right]=0.\]
However, \(\hat{Q}_{k}\) and \(\hat{\Pi}_{k}\) are no longer Hermitian. Rather,
\[\hat{Q}^{\dagger}_{k}=\hat{Q}_{-k}.\]
In these new variables the Hamiltonian reads
\[\hat{H}=\frac{1}{2m}\sum_{k}\hat{\Pi}_{-k}\hat{\Pi}_{k}+4m\omega^{2}\sum_{k}\sin^{2}\left(\frac{ka}{2}\right)\hat{Q}_{-k}\hat{Q}_{k}\]
or
\[\begin{aligned} \hat{H} & =\sum_{k}\frac{1}{2m}\hat{\Pi}_{-k}\hat{\Pi}_{k}+\frac{1}{2}m\omega^{2}_{k}\hat{Q}_{-k}\hat{Q}_{k}\\ & =\sum_{k}\frac{1}{2m}\hat{\Pi}^{\dagger}_{k}\hat{\Pi}_{k}+\frac{1}{2}m\omega^{2}_{k}\hat{Q}^{\dagger}_{k}\hat{Q}_{k} \end{aligned} \tag{1.10}\]
where
\[\omega_{k}\triangleq\sqrt{8}\omega\left|\sin\left(\frac{ka}{2}\right)\right|.\]
Unfortunately Eq. 1.10 makes it clear that states at \(k\) and \(-k\) are coupled, which is not quite what we want. If we can decouple the modes, we can find the eigenstates and eigenvalues using creation and annihilation operators as for the single harmonic oscillator. A clever definition of creation operator allows us to do this while simultaneously decoupling \(k\) from \(-k\) (the two steps can also be done separately, with more algebra). We define a different creation operator for each mode \(k\):
\[\boxed{\hat{a}^{\dagger}_{k}\triangleq\sqrt{\frac{m\omega_{k}}{2}}\left(\hat{Q}_{-k}-\frac{i}{m\omega_{k}}\hat{\Pi}_{k}\right)}\]
giving
\[\begin{aligned} \hat{a}_{k} & =\sqrt{\frac{m\omega_{k}}{2}}\left(\hat{Q}^{\dagger}_{-k}+\frac{i}{m\omega_{k}}\hat{\Pi}^{\dagger}_{k}\right)\\ & =\sqrt{\frac{m\omega_{k}}{2}}\left(\hat{Q}_{k}+\frac{i}{m\omega_{k}}\hat{\Pi}_{-k}\right). \end{aligned}\]
Note that \(\hat{a}^{\dagger}_{k}\ne\hat{a}_{-k}\). These obey the commutation relations:
\[\begin{aligned} \left[\hat{a}_{k},\hat{a}^{\dagger}_{q}\right] & =\left[\sqrt{\frac{m\omega_{k}}{2}}\left(\hat{Q}_{k}+\frac{i}{m\omega_{k}}\hat{\Pi}^{\dagger}_{k}\right),\sqrt{\frac{m\omega_{q}}{2}}\left(\hat{Q}^{\dagger}_{q}-\frac{i}{m\omega_{q}}\hat{\Pi}_{q}\right)\right]\\ & =\frac{1}{2}m\sqrt{\omega_{k}\omega_{q}}\left[\hat{Q}_{k}+\frac{i}{m\omega_{k}}\hat{\Pi}^{\dagger}_{k},\hat{Q}^{\dagger}_{q}-\frac{i}{m\omega_{q}}\hat{\Pi}_{q}\right]\\ & =\frac{i}{2}\sqrt{\omega_{k}\omega_{q}}\left(\frac{1}{\omega_{k}}\left[\hat{\Pi}^{\dagger}_{k},\hat{Q}^{\dagger}_{q}\right]-\frac{1}{\omega_{q}}\left[\hat{Q}_{k},\hat{\Pi}_{q}\right]\right)\\ & =\frac{i}{2}\sqrt{\omega_{k}\omega_{q}}\left(-\frac{1}{\omega_{k}}i\delta_{k,q}\hat{\mathbb{I}}-\frac{1}{\omega_{q}}i\delta_{k,q}\hat{\mathbb{I}}\right)\\ & =\delta_{k,q}\hat{\mathbb{I}} \end{aligned}\]
and
\[\left[\hat{a}_{k},\hat{a}_{q}\right]=\left[\hat{a}^{\dagger}_{k},\hat{a}^{\dagger}_{q}\right]=0.\]
We can now re-arrange to get
\[\begin{aligned} \hat{Q}_{k} & =\sqrt{\frac{1}{2m\omega_{k}}}\left(\hat{a}_{k}+\hat{a}^{\dagger}_{-k}\right)\\ \hat{\Pi}_{k} & =i\sqrt{\frac{m\omega_{k}}{2}}\left(\hat{a}^{\dagger}_{k}-\hat{a}_{-k}\right) \end{aligned}\]
and so
\[\begin{aligned} \hat{H} & =\sum_{k}\frac{1}{2m}\hat{\Pi}^{\dagger}_{k}\hat{\Pi}_{k}+\frac{1}{2}m\omega^{2}_{k}\hat{Q}^{\dagger}_{k}\hat{Q}_{k}\\ & =\sum_{k}\frac{\omega_{k}}{4}\left\{ \left(\hat{a}_{k}-\hat{a}^{\dagger}_{-k}\right)\left(\hat{a}^{\dagger}_{k}-\hat{a}_{-k}\right)+\left(\hat{a}^{\dagger}_{k}+\hat{a}_{-k}\right)\left(\hat{a}_{k}+\hat{a}^{\dagger}_{-k}\right)\right\} \\ & =\sum_{k}\frac{\omega_{k}}{4}\left(\hat{a}_{k}\hat{a}^{\dagger}_{k}+\hat{a}^{\dagger}_{-k}\hat{a}_{-k}+\hat{a}^{\dagger}_{k}\hat{a}_{k}+\hat{a}_{-k}\hat{a}^{\dagger}_{-k}\right)\\ & =\sum^{2\pi/a}_{k=2\pi/Na}\omega_{k}\left(\frac{1}{2}\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}\hat{a}^{\dagger}_{-k}\hat{a}_{-k}+\frac{1}{2}\hat{\mathbb{I}}\right) \end{aligned}\]
where I have used the commutators to put creation operators to the left of annihilation operators. The modes are now nicely decoupled (\(k\) and \(-k\) terms appear separately). Separating the \(-k\) sum:
\[\hat{H}=\sum^{2\pi/a}_{k=2\pi/Na}\omega_{k}\left(\frac{1}{2}\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}\hat{\mathbb{I}}\right)+\sum^{2\pi/a}_{k=2\pi/Na}\omega_{k}\left(\frac{1}{2}\hat{a}^{\dagger}_{-k}\hat{a}_{-k}\right)\]
we can change variables in the second term to \(k'=-k\) to give:
\[\hat{H}=\sum^{2\pi/a}_{k=2\pi/Na}\omega_{k}\left(\frac{1}{2}\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}\hat{\mathbb{I}}\right)+\sum^{-2\pi/a}_{k'=-2\pi/Na}\omega_{-k'}\left(\frac{1}{2}\hat{a}^{\dagger}_{k'}\hat{a}_{k'}\right).\]
As before we can relabel \(k'\rightarrow k\):
\[\hat{H}=\sum^{2\pi/a}_{k=2\pi/Na}\omega_{k}\left(\frac{1}{2}\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}\hat{\mathbb{I}}\right)+\sum^{-2\pi/a}_{k=-2\pi/Na}\omega_{-k}\left(\frac{1}{2}\hat{a}^{\dagger}_{k}\hat{a}_{k}\right).\]
Since \(\omega_{k}=\omega_{-k}\), this is
\[\hat{H}=\sum^{2\pi/a}_{k=2\pi/Na}\omega_{k}\left(\frac{1}{2}\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}\hat{\mathbb{I}}\right)+\sum^{-2\pi/a}_{k=-2\pi/Na}\omega_{k}\left(\frac{1}{2}\hat{a}^{\dagger}_{k}\hat{a}_{k}\right).\]
Finally, we note that since \(n\in\left[1,N\right]\) is periodic (i.e. \(N+1\equiv1\)), \(k\) must also be periodic. Hence the sum must be over exactly the same set of \(k\) points. Hence, both \(k\) sums are the same, and we have
\[\hat{H}=\sum^{2\pi/a}_{k=2\pi/Na}\omega_{k}\left(\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}\hat{\mathbb{I}}\right).\]
Hence, using the Fourier transform we have written the coupled QHOs in terms of independent creation and annihilation operators. The number operator
\[\hat{n}_{k}\triangleq\hat{a}^{\dagger}_{k}\hat{a}_{k}\]
counts the number of particles in mode \(k\), i.e. with wavevector \(k\).
1.5.1 Excitations of a Quantum Chain: Fock Space
Each operator \(\hat{a}^{\dagger}_{k}\) works exactly like a creation operator acts on a single QHO. Acting it on the ground state creates an excitation with a well defined wavevector. There are \(N\) independent \(k\) modes. We previously labelled these \(-\pi/a\le k\le\pi/a\), but for notational simplicity let’s rescale to \(1\le k\le N\). The ground state, or vacuum state, \(|\Omega\rangle\) is a product of \(N\) independent modes:
\[|\Omega\rangle=|0_{k=1}\rangle\otimes|0_{k=2}\rangle\otimes|0_{k=3}\rangle\otimes|0_{k=4}\rangle\otimes\ldots\]
which can be written compactly as
\[|\Omega\rangle=|0000\ldots\rangle.\]
Acting a creation operator with wavevector \(k\) raises mode \(k\), at a cost of \(\omega_{k}\). Then
\[\hat{a}^{\dagger}_{k=3}|\Omega\rangle=|00100\ldots\rangle\]
or
\[\left(\hat{a}^{\dagger}_{k=3}\right)^{3}\left(\hat{a}^{\dagger}_{k=2}\right)^{4}\left(\hat{a}^{\dagger}_{k=1}\right)^{2}|\Omega\rangle=|24300\ldots\rangle\]
and so on. There’s no problem with exciting any given mode multiple times, just as there’s no problem with exciting a single QHO to its \(n^{\text{th}}\) energy level. Note that \(|24300\ldots\rangle\) is just shorthand for a tensor product of \(N\) separate kets:
\[|24300\ldots\rangle\triangleq|k=1,n=2\rangle\otimes|k=2,n=4\rangle\otimes|k=3,n=3\rangle\otimes|k=4,n=0\rangle\otimes\ldots\]
In single-particle QM we consider kets living in a Hilbert space (with some caveats!). We say that this new object lives in a Fock Space. This is the space of all Hilbert spaces, each of which has a fixed particle number. You can equally well create an excitation at a specific position \(an\) by acting the creation operator
\[\hat{a}^{\dagger}_{n}=\frac{1}{\sqrt{N}}\sum^{N}_{k=1}\exp\left(2\pi ikn/N\right)\hat{a}^{\dagger}_{k}\]
on the vacuum state:
\[\begin{aligned} \hat{a}^{\dagger}_{n}|\Omega\rangle & =\frac{1}{\sqrt{N}}\sum^{N}_{k=1}\exp\left(2\pi ikn/N\right)\hat{a}^{\dagger}_{k}|\Omega\rangle\\ & =\frac{1}{\sqrt{N}}\exp\left(2\pi in/N\right)\hat{a}^{\dagger}_{k=1}|000\ldots\rangle+\exp\left(4\pi in/N\right)\hat{a}^{\dagger}_{k=2}|000\ldots\rangle+\ldots\\ & =\frac{1}{\sqrt{N}}\exp\left(2\pi in/N\right)|100\ldots\rangle+\frac{1}{\sqrt{N}}\exp\left(4\pi in/N\right)|010\ldots\rangle+\ldots \end{aligned}\]
You can think by analogy of plucking a guitar string at a given position along its length. The effect is to excite all the harmonics of the string with different amplitudes.
1.6 Classical Field Theory
So far we have worked with discrete positions \(n\). A field is a continuum, so to proceed we must smooth out our description. Our motivation is that we are typically only interested in lengthscales / wavelengths significantly larger than a lattice spacing, and so we can ‘coarse grain’ or smooth over the lattice details.
1.6.1 Classical Lagrangian Field Theory
In the classical balls and springs model we can make a conceptual leap by approximating our discrete positions by a continuum field. This amounts to making the change
\[x_{n,t}\rightarrow\varphi_{\boldsymbol{x},t}. \tag{1.11}\]
I have introduced bold \(\boldsymbol{x}\) to emphasise that this works in any spatial dimension, and because we will shortly need to differentiate between 3-vectors \(\boldsymbol{x}\) and 4-vectors \(x^{\mu}=\left(t,\boldsymbol{x}\right)^{\mu}\). Formula 1.11 is the defining expression of field theory. While conceptually simple, the consequences are profound. Rather than individual balls oscillating about discrete sites \(n\) with displacements \(x_{n,t}\), we instead have a continuous field defined at all positions \(\boldsymbol{x}\) and times \(t\). You can think of it as a flexible membrane, like the surface of a drum. In crystals the result leads to simplified mathematics. In the standard model, however, there are no individual sites: only the smooth field exists, and it does so across all of spacetime. It is worth taking the time to think about Eq. 1.11 conceptually before proceeding.
The following might help (or it might do the opposite!). In classical mechanics it is common to denote generalised co-ordinates as \(q_{n,t}\) rather than \(x_{n,t}\). In this alternative notation, when we coarse grain and smooth out, we simply switch \(q_{n,t}\rightarrow q_{\boldsymbol{x},t}\), interpolating the discrete variable \(n\) with continuous positions \(\boldsymbol{x}\). Then we just relabel \(q_{\boldsymbol{x},t}\rightarrow\varphi_{\boldsymbol{x},t}\) as a matter of convention.
The Lagrangian density coarse-grains as follows:
\[\begin{aligned} \mathcal{L} & =\frac{1}{2}m\dot{x}^{2}_{n}-\frac{1}{2}m\omega^{2}\left(x_{n+1}-x_{n}\right)^{2}-\frac{1}{2}m\omega^{2}\left(x_{n}-x_{n-1}\right)^{2}\\ & \downarrow\text{coarse-grain}\\ \mathcal{L} & =\frac{1}{2}m\dot{\varphi}{}^{2}_{\boldsymbol{x}}-\frac{1}{2}m\omega^{2}\left(\varphi_{\boldsymbol{x}+\boldsymbol{a}}-\varphi_{\boldsymbol{x}}\right)^{2}-\frac{1}{2}m\omega^{2}\left(\varphi_{\boldsymbol{x}}-\varphi_{\boldsymbol{x}-\boldsymbol{a}}\right)^{2}\\ & \approx\frac{1}{2}m\dot{\varphi}{}^{2}_{\boldsymbol{x}}-\frac{1}{2}m\omega^{2}\left(a\nabla\varphi_{\boldsymbol{x}}\right)^{2}-\frac{1}{2}m\omega^{2}\left(a\nabla\varphi_{\boldsymbol{x}}\right)^{2}\\ & =\frac{1}{2}m\dot{\varphi}{}^{2}_{\boldsymbol{x}}-m\omega^{2}a^{2}\left(\nabla\varphi_{\boldsymbol{x}}\right)^{2}. \end{aligned}\]
In the intermediate step we made a Taylor expansion
\[\varphi_{\boldsymbol{x}+\boldsymbol{a}}=\varphi_{\boldsymbol{x}}+\boldsymbol{a}\cdot\nabla\varphi_{\boldsymbol{x}}+\mathcal{O}\left(a^{2}\right)\]
and kept only the lowest term in \(a\), since we are considering coarse-grained (smooth) fields. Let us redefine the dimension of our field (or the units in which it is measured) to absorb the \(m\), and rescale lengths to absorb the lattice constant so that
\[\mathcal{L}=\frac{1}{2}\dot{\varphi}{}^{2}_{\boldsymbol{x}}-\omega^{2}\left(\nabla\varphi_{\boldsymbol{x}}\right)^{2}.\]
Both are common conventions. The removal of \(m\) is helpful, since the mass of the field appears differently in field theory: specifically a field with mass \(m\) has the Lagrange density
\[\mathcal{L}=\frac{1}{2}\dot{\varphi}{}^{2}_{\boldsymbol{x}}-\omega^{2}a^{2}\left(\nabla\varphi_{\boldsymbol{x}}\right)^{2}-m^{2}\varphi^{2}_{x}.\]
We will see more of this later. Let’s also write things in a general spatial dimension \(D\), as it is no more difficult that using \(D=1\) and it will allow us to use bold \(\boldsymbol{x}\) (which will be helpful when we go relativistic). As before, neglect the \(t\) label since everything is evaluated at the same time.
The Lagrangian is
\[L=\int\text{d}^{D}\boldsymbol{x}\mathcal{L}\left(\varphi_{\boldsymbol{x}},\dot{\varphi}_{\boldsymbol{x}}\right)\]
and action
\[S=\int\text{d}tL=\int\text{d}^{D+1}\boldsymbol{x}\mathcal{L}.\]
This notation already emphasises that the action is a Lorentz invariant scalar. We can find the Euler Lagrange equations using the principle of least action:
\[\begin{aligned} S\left[\varphi+\lambda\epsilon\right] & =\int\text{d}t\int\text{d}^{D}\boldsymbol{x}\left\{ \frac{1}{2}\left(\dot{\varphi_{\boldsymbol{x}}}+\lambda\dot{\epsilon}_{\boldsymbol{x}}\right){}^{2}-\omega^{2}\left(\nabla\varphi_{\boldsymbol{x}}+\lambda\nabla\epsilon_{\boldsymbol{x}}\right)^{2}\right\} \\ & \Downarrow\\ \left.\left(\frac{\partial S\left[\varphi+\lambda\epsilon\right]}{\partial\lambda}\right)_{\varphi,\epsilon}\right|_{\lambda=0} & =\int\text{d}t\int\text{d}^{D}\boldsymbol{x}\left\{ \dot{\varphi_{\boldsymbol{x}}}\dot{\epsilon}_{\boldsymbol{x}}-2\omega^{2}\nabla\varphi_{\boldsymbol{x}}\nabla\epsilon_{\boldsymbol{x}}\right\} \\ & =\int\text{d}t\int\text{d}^{D}\boldsymbol{x}\epsilon_{\boldsymbol{x}}\left\{ -\ddot{\varphi}_{\boldsymbol{x}}+2\omega^{2}\nabla^{2}\varphi_{\boldsymbol{x}}\right\} \end{aligned}\]
using integration by parts with respect to time in the first term, and space in the second. Setting this equal to zero for all \(\epsilon\) requires that
\[\ddot{\varphi}_{\boldsymbol{x}}=c^{2}\nabla^{2}\varphi_{\boldsymbol{x}}\]
where \(c=\sqrt{2}\omega\). This is the classical wave equation. As hoped, we still have our classical sound mode propagating in our coarse-grained theory.
NB you will need to understand the principle of coarse graining, but will not need to memorise the particular case detailed above.
1.6.2 Classical Hamiltonian Field Theory
As before, we need the momentum conjugate to the field. This is given by
\[\pi_{\boldsymbol{x},t}\triangleq\left(\frac{\partial\mathcal{L}}{\partial\dot{\varphi}_{\boldsymbol{x},t}}\right)_{\varphi_{\boldsymbol{x},t}}\]
where we will continue to neglect the \(t\) label. In our case we find
\[\pi_{\boldsymbol{x}}=\dot{\varphi}_{\boldsymbol{x}}.\]
The Hamiltonian is again the Legendre transform of the Lagrangian:
\[H\triangleq\left(\int\text{d}^{D}\boldsymbol{x}\pi_{\boldsymbol{x}}\dot{\varphi}_{\boldsymbol{x}}\right)-L\]
and in our case this gives
\[H=\int\text{d}^{D}\boldsymbol{x}\frac{1}{2}\pi{}^{2}_{\boldsymbol{x}}+\omega^{2}\int\text{d}^{D}\boldsymbol{x}\left(\nabla\varphi_{\boldsymbol{x}}\right)^{2}.\]
1.7 Quantum Field Theory of the Harmonic Chain
1.7.1 Canonical Field Quantization
Finally we are ready to create our first quantum field. To do so, we simply promote our classical fields to non-commuting quantum operators:
\[\begin{aligned} \varphi_{\boldsymbol{x}} & \rightarrow\hat{\varphi}_{\boldsymbol{x}}\\ \pi_{\boldsymbol{x}} & \rightarrow\hat{\pi}_{\boldsymbol{x}}. \end{aligned}\]
We are now explicitly justified in neglecting the \(t\) label since, in the Schroedinger picture, operators are time-independent. The quantum fields obey the canonical commutation relations
\[\left[\hat{\varphi}_{\boldsymbol{x}},\hat{\varphi}_{\boldsymbol{y}}\right]=\left[\hat{\pi}_{\boldsymbol{x}},\hat{\pi}_{\boldsymbol{y}}\right]=0\]
and
\[\left[\hat{\varphi}_{\boldsymbol{x}},\hat{\pi}_{\boldsymbol{y}}\right]=i\delta^{D}\left(\boldsymbol{x}-\boldsymbol{y}\right)\hat{\mathbb{I}}.\]
Note that in this description, unlike in single-particle quantum mechanics, positions are not themselves operators. Rather, positions should be thought of as like the fixed discrete lattice sites \(n\), while the field is like \(x_{n}\), the displacement of the ball from that site.
Our potential term has translational symmetry (as it only cares about the gradient of the field). We can again take advantage of this by working in the Fourier basis:
\[\begin{aligned} \hat{\varphi}_{\boldsymbol{x}} & =\sum_{\boldsymbol{k}}\exp\left(2\pi i\boldsymbol{k}\cdot\boldsymbol{x}\right)\hat{\Phi}_{\boldsymbol{k}}\\ \hat{\pi}_{\boldsymbol{x}} & =\sum_{\boldsymbol{k}}\exp\left(-2\pi i\boldsymbol{k}\cdot\boldsymbol{x}\right)\hat{\Pi}_{\boldsymbol{k}} \end{aligned} \tag{1.12}\]
and the inverse transforms
\[\begin{aligned} \hat{\Phi}_{\boldsymbol{k}} & =\int\text{d}^{D}\boldsymbol{x}\exp\left(-2\pi i\boldsymbol{k}\cdot\boldsymbol{x}\right)\hat{\varphi}_{\boldsymbol{x}}\\ \hat{\Pi}_{\boldsymbol{k}} & =\int\text{d}^{D}\boldsymbol{x}\exp\left(2\pi i\boldsymbol{k}\cdot\boldsymbol{x}\right)\hat{\pi}_{\boldsymbol{x}}. \end{aligned} \tag{1.13}\]
Important point: in going to the field description, we have moved to describing real space with a continuum, meaning we use integrals over \(\boldsymbol{x}\) instead of a discrete sum over lattice sites. However, in the case of a chain with periodic boundary conditions, \(\boldsymbol{k}\) remains discrete. This is because the boundary conditions require you to fit an integer number of half-wavelengths around the system. Hence it remains a sum over \(\boldsymbol{k}\). In general, whether \(\boldsymbol{k}\) is discrete or continuous depends on the boundary conditions.
The new fields obey the canonical commutation relations
\[\left[\hat{\Phi}_{\boldsymbol{k}},\hat{\Phi}_{\boldsymbol{q}}\right]=\left[\hat{\Pi}_{\boldsymbol{k}},\hat{\Pi}_{\boldsymbol{q}}\right]=0\]
and
\[\begin{aligned} \left[\hat{\Phi}_{\boldsymbol{k}},\hat{\Pi}_{\boldsymbol{q}}\right] & =\int\text{d}^{D}\boldsymbol{x}\int\text{d}^{D}\boldsymbol{y}\exp\left(2\pi i\left(\boldsymbol{q}\cdot\boldsymbol{y}-\boldsymbol{k}\cdot\boldsymbol{x}\right)\right)\left[\hat{\varphi}_{\boldsymbol{x}},\hat{\pi}_{\boldsymbol{y}}\right]\\ & =\int\text{d}^{D}\boldsymbol{x}\int\text{d}^{D}\boldsymbol{y}\exp\left(2\pi i\left(\boldsymbol{q}\cdot\boldsymbol{y}-\boldsymbol{k}\cdot\boldsymbol{x}\right)\right)i\delta^{D}\left(\boldsymbol{x}-\boldsymbol{y}\right)\hat{\mathbb{I}}\\ & =\int\text{d}^{D}\boldsymbol{x}\exp\left(2\pi i\left(\boldsymbol{q}-\boldsymbol{k}\right)\cdot\boldsymbol{x}\right)i\hat{\mathbb{I}}\\ & =i\delta^{D}\left(\boldsymbol{k}-\boldsymbol{q}\right)\hat{\mathbb{I}} \end{aligned}\]
using the useful relation
\[\int\text{d}^{D}\boldsymbol{x}\exp\left(2\pi i\left(\boldsymbol{q}-\boldsymbol{k}\right)\cdot\boldsymbol{x}\right)=\delta_{\boldsymbol{k},\boldsymbol{q}}.\]
As before, the quantum fields are not Hermitian:
\[\begin{aligned} \hat{\Phi}{}^{\dagger}_{\boldsymbol{k}} & =\hat{\Phi}_{\boldsymbol{-k}}\\ \hat{\Pi}{}^{\dagger}_{\boldsymbol{k}} & =\hat{\Pi}_{\boldsymbol{-k}}. \end{aligned}\]
1.7.2 The Hamiltonian of the Quantum Field
We canonically quantize our field Hamiltonian as
\[\begin{aligned} H & =\int\text{d}^{D}\boldsymbol{x}\frac{1}{2}\pi{}^{2}_{\boldsymbol{x}}+\omega^{2}\int\text{d}^{D}\boldsymbol{x}\left(\nabla\varphi_{\boldsymbol{x}}\right)^{2}\\ & \downarrow\text{canonically quantize}\nonumber \\ \hat{H} & =\int\text{d}^{D}\boldsymbol{x}\frac{1}{2}\hat{\pi}{}^{2}_{\boldsymbol{x}}+\omega^{2}\int\text{d}^{D}\boldsymbol{x}\left(\nabla\hat{\varphi}_{\boldsymbol{x}}\right)^{2} \end{aligned}\]
and Fourier transforming using Eqs. 1.12 gives
\[\begin{aligned} \hat{H} & =\int\text{d}^{D}\boldsymbol{x}\sum_{\boldsymbol{k}}\sum_{\boldsymbol{q}}\left\{ \exp\left(-2\pi i\left(\boldsymbol{k}+\boldsymbol{q}\right)\cdot\boldsymbol{x}\right)\frac{1}{2}\hat{\Pi}_{\boldsymbol{k}}\hat{\Pi}_{\boldsymbol{q}}+\omega^{2}\nabla\left(\exp\left(2\pi i\boldsymbol{k}\cdot\boldsymbol{x}\right)\hat{\Phi}_{\boldsymbol{k}}\right)\cdot\nabla\left(\exp\left(2\pi i\boldsymbol{q}\cdot\boldsymbol{x}\right)\hat{\Phi}_{\boldsymbol{q}}\right)\right\} \\ & =\int\text{d}^{D}\boldsymbol{x}\sum_{\boldsymbol{k}}\sum_{\boldsymbol{q}}\left\{ \exp\left(-2\pi i\left(\boldsymbol{k}+\boldsymbol{q}\right)\cdot\boldsymbol{x}\right)\frac{1}{2}\hat{\Pi}_{\boldsymbol{k}}\hat{\Pi}_{\boldsymbol{q}}-\exp\left(2\pi i\left(\boldsymbol{k}+\boldsymbol{q}\right)\cdot\boldsymbol{x}\right)\left(2\pi\omega\right)^{2}\boldsymbol{k}\cdot\boldsymbol{q}\hat{\Phi}_{\boldsymbol{k}}\hat{\Phi}_{\boldsymbol{q}}\right\} \\ & =\sum_{\boldsymbol{k}}\sum_{\boldsymbol{q}}\delta_{\boldsymbol{k},-\boldsymbol{q}}\left\{ \frac{1}{2}\hat{\Pi}_{\boldsymbol{k}}\hat{\Pi}_{\boldsymbol{q}}-\left(2\pi\omega\right)^{2}\boldsymbol{k}\cdot\boldsymbol{q}\hat{\Phi}_{\boldsymbol{k}}\hat{\Phi}_{\boldsymbol{q}}\right\}\\ & =\sum_{\boldsymbol{k}}\left\{ \frac{1}{2}\hat{\Pi}_{-\boldsymbol{k}}\hat{\Pi}_{\boldsymbol{k}}+\left(2\pi\omega\right)^{2}\boldsymbol{k}^{2}\hat{\Phi}_{-\boldsymbol{k}}\hat{\Phi}_{\boldsymbol{k}}\right\} \\ & =\sum_{\boldsymbol{k}}\left\{ \frac{1}{2}\hat{\Pi}^{\dagger}_{\boldsymbol{k}}\hat{\Pi}_{\boldsymbol{k}}+\frac{1}{2}\omega^{2}_{\boldsymbol{k}}\hat{\Phi}^{\dagger}_{\boldsymbol{k}}\hat{\Phi}_{\boldsymbol{k}}\right\} \end{aligned} \tag{1.14}\]
where we have defined
\[\omega_{\boldsymbol{k}}\triangleq\sqrt{8}\pi\omega\left|\boldsymbol{k}\right|\]
in the last line.
1.7.3 Creation and Annihilation Operators
The procedure now follows Section @subsec-Coupled-Quantum exactly. Defining the field creation operator
\[\hat{a}^{\dagger}_{\boldsymbol{k}}\triangleq\sqrt{\frac{\omega_{\boldsymbol{k}}}{2}}\left(\hat{\Phi}_{-\boldsymbol{k}}-\frac{i}{\omega_{\boldsymbol{k}}}\hat{\Pi}_{\boldsymbol{k}}\right)\]
we find
\[\begin{aligned} \left[\hat{a}_{\boldsymbol{k}},\hat{a}^{\dagger}_{\boldsymbol{q}}\right] & =\delta_{\boldsymbol{k},\boldsymbol{q}}\hat{\mathbb{I}} \end{aligned}\]
and
\[\left[\hat{a}_{\boldsymbol{k}},\hat{a}_{\boldsymbol{q}}\right]=\left[\hat{a}^{\dagger}_{\boldsymbol{k}},\hat{a}^{\dagger}_{\boldsymbol{q}}\right]=0.\]
We can now re-arrange to get
\[\begin{aligned} \hat{\Phi}_{\boldsymbol{k}} & =\sqrt{\frac{1}{2\omega_{\boldsymbol{k}}}}\left(\hat{a}_{\boldsymbol{k}}+\hat{a}^{\dagger}_{-\boldsymbol{k}}\right)\\ \hat{\Pi}_{\boldsymbol{k}} & =i\sqrt{\frac{\omega_{\boldsymbol{k}}}{2}}\left(\hat{a}^{\dagger}_{\boldsymbol{k}}-\hat{a}_{-\boldsymbol{k}}\right) \end{aligned}\]
and so the Hamiltonian is
\[\begin{aligned} \hat{H} & =\sum_{\boldsymbol{k}}\left\{ \frac{1}{2}\hat{\Pi}^{\dagger}_{\boldsymbol{k}}\hat{\Pi}_{\boldsymbol{k}}+\frac{1}{2}\omega^{2}_{\boldsymbol{k}}\hat{\Phi}^{\dagger}_{\boldsymbol{k}}\hat{\Phi}_{\boldsymbol{k}}\right\} \\ & =\sum_{\boldsymbol{k}}\left\{ \frac{\omega_{\boldsymbol{k}}}{4}\left(\hat{a}_{\boldsymbol{k}}-\hat{a}^{\dagger}_{-\boldsymbol{k}}\right)\left(\hat{a}^{\dagger}_{\boldsymbol{k}}-\hat{a}_{-\boldsymbol{k}}\right)+\frac{\omega_{\boldsymbol{k}}}{4}\left(\hat{a}^{\dagger}_{\boldsymbol{k}}+\hat{a}_{-\boldsymbol{k}}\right)\left(\hat{a}_{\boldsymbol{k}}+\hat{a}^{\dagger}_{-\boldsymbol{k}}\right)\right\} \\ & =\sum_{\boldsymbol{k}}\frac{\omega_{\boldsymbol{k}}}{4}\left(\hat{a}^{\dagger}_{\boldsymbol{k}}\hat{a}_{\boldsymbol{k}}+\hat{a}_{\boldsymbol{k}}\hat{a}^{\dagger}_{\boldsymbol{k}}+\hat{a}^{\dagger}_{-\boldsymbol{k}}\hat{a}_{-\boldsymbol{k}}+\hat{a}_{-\boldsymbol{k}}\hat{a}^{\dagger}_{-\boldsymbol{k}}\right)\\ & =\sum_{\boldsymbol{k}}\left\{ \frac{\omega_{\boldsymbol{k}}}{4}\left(\hat{a}^{\dagger}_{\boldsymbol{k}}\hat{a}_{\boldsymbol{k}}+\left(\hat{a}^{\dagger}_{\boldsymbol{k}}\hat{a}_{\boldsymbol{k}}+\left[\hat{a}_{\boldsymbol{k}},\hat{a}^{\dagger}_{\boldsymbol{k}}\right]\right)+\hat{a}^{\dagger}_{-\boldsymbol{k}}\hat{a}_{-\boldsymbol{k}}+\left(\hat{a}^{\dagger}_{-\boldsymbol{k}}\hat{a}_{-\boldsymbol{k}}+\left[\hat{a}_{-\boldsymbol{k}},\hat{a}^{\dagger}_{-\boldsymbol{k}}\right]\right)\right)\right\} \\ & =\sum_{\boldsymbol{k}}\frac{\omega_{\boldsymbol{k}}}{2}\left(\hat{a}^{\dagger}_{\boldsymbol{k}}\hat{a}_{\boldsymbol{k}}+\hat{a}^{\dagger}_{-\boldsymbol{k}}\hat{a}_{-\boldsymbol{k}}+\delta\left(0\right)\hat{\mathbb{I}}\right). \end{aligned} \tag{1.15}\]
Unfortunately, by going to the continuum, the zero point energy has now become \(\sum_{\boldsymbol{k}}\omega_{\boldsymbol{k}}\delta\left(0\right)/2\). This is a countably infinite sum over an uncountably infinite quantity! However, there is a standard method for dealing with this infinity, which we turn to now.
1.7.4 Normal Ordering
The infinity in Eq 1.15 is only the first of many infinities one will encounter in QFT. This one is perfectly harmless, since the absolute energy is meaningless: it is only differences in energies that have meaning. This infinity is also quite natural: we have placed a separate QHO at every position in space. Each QHO has a finite ground state energy. There are an uncountably infinite number of positions, and so the ground state energy must also be uncountably infinite. We can remove this infinity by redefining our zero point energy (by an infinite amount).
There is a more systematic way to do this, however. Note that the infinity came about because we had to use the commutator \(\left[\hat{a}_{\boldsymbol{k}},\hat{a}^{\dagger}_{\boldsymbol{k}}\right]\). If we had written all creation operators to the left of all annihilation operators, we would not have needed to do this. In fact, this infinity comes from an ambiguity introduced in canonical quantization: classical commuting observables become quantum non-commuting operators. Hence, the order can switch classically without causing a change, but quantum mechanically this changes the total energy. For example, if you look back at Eq 1.14, you will see that we did the \(\boldsymbol{k}\) integral first, setting \(\boldsymbol{k}\) to \(-\boldsymbol{q}\), then we relabelled \(\boldsymbol{k}\) to \(\boldsymbol{q}\). You can instead do the \(\boldsymbol{q}\) integral first, but this switches the order of the raising and lowering operators! Either choice should give the same physical results, and it is this ambiguity which is causing an apparent issue.
We define the normal ordering of a chain of operators to be such that all creation operators are to the left of all annihilation operators. We denote the normal ordering of an operator \(\hat{\mathcal{O}}\) to be \(:\hat{\mathcal{O}}:\). In our case
\[\hat{H}=\sum_{\boldsymbol{k}}\frac{\omega_{\boldsymbol{k}}}{4}\left(\hat{a}^{\dagger}_{\boldsymbol{k}}\hat{a}_{\boldsymbol{k}}+\hat{a}_{\boldsymbol{k}}\hat{a}^{\dagger}_{\boldsymbol{k}}+\hat{a}^{\dagger}_{-\boldsymbol{k}}\hat{a}_{-\boldsymbol{k}}+\hat{a}_{-\boldsymbol{k}}\hat{a}^{\dagger}_{-\boldsymbol{k}}\right)\]
and so its normal ordering, placing all creation operators to the left (without worrying about the resulting commutators) is
\[:\hat{H}:=\sum_{\boldsymbol{k}}\frac{\omega_{\boldsymbol{k}}}{2}\left(\hat{a}^{\dagger}_{\boldsymbol{k}}\hat{a}_{\boldsymbol{k}}+\hat{a}^{\dagger}_{-\boldsymbol{k}}\hat{a}_{-\boldsymbol{k}}\right). \tag{1.16}\]
This means that using normal ordered operators instead of the operators themselves, we subtract exactly the infinite term we had trouble with:
\[:\hat{H}:=\hat{H}-\sum_{\boldsymbol{k}}\frac{\omega_{\boldsymbol{k}}}{2}\delta\left(0\right)\hat{\mathbb{I}}.\]
In general, normal ordering will remove this particular type of infinity, introduced from the ordering ambiguity of quantum operators under canonical quantization. It is an infinite shift of the zero point energy, placing the ground state at zero energy. Eq 1.16 is now a totally reasonable (normal-ordered) Hamiltonian, in which the \(\boldsymbol{k}\) and \(-\boldsymbol{k}\) modes are decoupled. As before, by noting the periodicity in \(\boldsymbol{k}\) both terms now contribute an equal amount. Hence,
\[:\hat{H}:=\sum_{\boldsymbol{k}}\omega_{\boldsymbol{k}}\hat{a}^{\dagger}_{\boldsymbol{k}}\hat{a}_{\boldsymbol{k}}.\]
1.7.5 Excitations of a Quantum Field: Particles
Now let’s take a step back and inspect our handywork, as we did for the quantum chain in Section @subsec-Creating-excitations-chain (you might like to re-read that section quickly). Each operator \(\hat{a}^{\dagger}_{\boldsymbol{k}}\) works exactly like a creation operator acts on a single QHO. Acting it on the ground state creates an excitation with a well defined wavevector. The ground state, or vacuum state, \(|\Omega\rangle\) is a product of \(N\rightarrow\infty\) independent modes. In 1D:
\[|\Omega\rangle=|0_{k=1}\rangle\otimes|0_{k=2}\rangle\otimes|0_{k=3}\rangle\otimes|0_{k=4}\rangle\otimes\ldots \tag{1.17}\]
Note that, while positions have become continuous variables (forming an uncountably infinite set), the possible wavevectors remain countable. Our boundary conditions quantize the possible wavevectors. This is quite natural: a guitar string exists at an uncountably infinite number of positions between its two ends, yet its harmonics are standing waves which fit along the string. Hence the notation in Eq. 1.17 is acceptable. Let’s look at what happens when we act the field creation operator on the vacuum: \[\begin{aligned} \hat{\Phi}^{\dagger}_{\boldsymbol{k}}|\Omega\rangle & =\sqrt{\frac{1}{2\omega_{\boldsymbol{k}}}}\left(\hat{a}^{\dagger}_{\boldsymbol{k}}+\hat{a}_{-\boldsymbol{k}}\right)|\Omega\rangle\\ & =\frac{1}{\sqrt{2E_{\boldsymbol{k}}}}\hat{a}^{\dagger}_{\boldsymbol{k}}|\Omega\rangle\\ & =\frac{1}{2E_{\boldsymbol{k}}}\sqrt{2E_{\boldsymbol{k}}}|\boldsymbol{k}\rangle\\ & =\frac{1}{2E_{\boldsymbol{k}}}|k\rangle. \end{aligned} \tag{1.18}\]
In the last line we have adopted the standard notation
\[|k\rangle=\sqrt{2E_{\boldsymbol{k}}}|\boldsymbol{k}\rangle \tag{1.19}\]
which guarantees the Lorentz invariance of objects such as \(\langle k|q\rangle=2E_{\boldsymbol{k}}\delta\left(\boldsymbol{k}-\boldsymbol{q}\right)\) (a fact we established in Section @subsec-Relativistic-conventions). We are currently looking at a non-relativistic theory, but since QFT is so frequently applied in relativistic settings, notation such as Eq 1.19 has been established with this in mind.
What is \(|k\rangle\) physically? It is nothing other than a particle! We call this particle a phonon. It has crystal-momentum \(\boldsymbol{k}\), and energy \(E_{\boldsymbol{k}}=\omega_{\boldsymbol{k}}\).
Note that Eq 1.18 means we can create a particle with momentum \(\boldsymbol{k}\) using either \(\hat{a}^{\dagger}_{\boldsymbol{k}}\) or \(\hat{\Phi}^{\dagger}_{\boldsymbol{k}}\). The latter is more natural in a relativistic setting. In textbooks you are more likely to see particles created with the field operator than with the creation operator itself, although the reason is not always discussed.
At this point you might well ask why we would expect our theory to be relativistic, given that we started from balls and springs with absolute positions \(an\) and absolute times \(t\). Essentially, the coarse-grained theory is relativistic even though the microscopic theory is not. We will return to this point in the next chapter.
We can denote multi-particle states as
\[|k_{1},k_{2}\rangle=\hat{\Phi}^{\dagger}_{\boldsymbol{k}_{2}}\hat{\Phi}^{\dagger}_{\boldsymbol{k}_{1}}|\Omega\rangle\]
and so on. You can equally well create a particle at a specific position \(\boldsymbol{x}\) by acting the real-space field creation operator on the vacuum. For notational convenience, let’s switch to the case that \(\boldsymbol{k}\) is continuous (requiring different boundary conditions on the space). Then we can write
\[\begin{aligned} \hat{\varphi}^{\dagger}_{\boldsymbol{x}}|\Omega\rangle & =\int\text{d}^{D}\boldsymbol{k}\exp\left(-2\pi i\boldsymbol{k}\cdot\boldsymbol{x}\right)\hat{\Phi}^{\dagger}_{\boldsymbol{k}}|\Omega\rangle\\ & =\frac{1}{\sqrt{2}}\int\text{d}^{D}\boldsymbol{k}\frac{1}{2E_{\boldsymbol{k}}}\exp\left(-2\pi i\boldsymbol{k}\cdot\boldsymbol{x}\right)|k\rangle\\ & =\frac{1}{\sqrt{8}}\int\text{d}^{D+1}k\delta\left(E^{2}-m^{2}-\boldsymbol{k}^{2}\right)\exp\left(-2\pi i\boldsymbol{k}\cdot\boldsymbol{x}\right)|k\rangle \end{aligned}\]
which we conventionally denote
\[\hat{\varphi}^{\dagger}_{\boldsymbol{x}}|\Omega\rangle=|x\rangle.\]
Hence, acting the field creation operator at position \(\boldsymbol{x}\) creates a particle at spacetime position \(x\). Again, you can think by analogy of plucking a guitar string at a given position. The effect is to excite all the harmonics of the string, with different weights. In this case, ‘plucking’ means creating a particle from the vacuum, and the guitar string is the universe itself (here, a quantum crystal).
All of this intuition will carry across when you come to creating photons rather than phonons. However, photons are particles that do not require a medium through which to travel. This should again be natural, since photons are quantized particles of light, which requires no medium through which to propagate, while phonons are quantized particles of sound (lattice vibrations in a crystal) and sound does require a medium.
1.8 Time dependence
Single-particle non-relativistic QM amounts to solving the time dependent Schroedinger equation (TDSE):
\[i\frac{\text{d}|\psi\rangle}{\text{d}t}=\hat{H}|\psi\rangle.\]
Assuming the Hamiltonian is time-independent we have the general solution
\[|\psi\left(t\right)\rangle=\exp\left(-i\hat{H}t\right)|\psi\left(0\right)\rangle\]
which you can confirm by acting \(i\text{d}/\text{d}t\) on both sides. That is, if we have the solution at one time (which we obtain by solving the time independent Schroedinger equation, TISE), we automatically have the solution at all subsequent times. It is convenient to define
the time evolution operator:
\[\hat{U}\left(t\right)\triangleq\exp\left(-i\hat{H}t\right).\]
Since \(\hat{H}\) is Hermitian, \(\hat{U}\) is unitary:
\[\hat{U}\hat{U}^{\dagger}=\hat{U}^{\dagger}\hat{U}=\hat{\mathbb{I}}.\]
For our canonically quantized field operators
\[\hat{\Phi}_{\boldsymbol{k}}=\sqrt{\frac{1}{2\omega_{\boldsymbol{k}}}}\left(\hat{a}_{\boldsymbol{k}}+\hat{a}^{\dagger}_{-\boldsymbol{k}}\right)\]
we have so far not written any explicit time dependence. These operators are understood to be written in the Heisenberg picture, in which operators are time dependent and states are time independent. To obtain the operator \(\hat{\mathcal{O}}_{t}\) at time \(t\) from that at time \(0\) we use the time evolution operator:
\[\hat{\mathcal{O}}_{t}=\hat{U}^{\dagger}_{t}\hat{\mathcal{O}}_{0}\hat{U}_{t}\]
or
\[\hat{\mathcal{O}}_{t}=\exp\left(i\hat{H}t\right)\hat{\mathcal{O}}_{0}\exp\left(-i\hat{H}t\right). \tag{1.20}\]
To add back in the explicit time dependence of our field operators \(\hat{\Phi}_{\boldsymbol{x}}\) we can therefore use:
\[\hat{\varphi}_{x}\triangleq\hat{\varphi}_{\boldsymbol{x},t}=\hat{U}^{\dagger}_{t}\hat{\varphi}_{\boldsymbol{x}}\hat{U}_{t}=\exp\left(i\hat{H}t\right)\hat{\varphi}_{\boldsymbol{x}}\exp\left(-i\hat{H}t\right).\]
Explicitly, our field operators are given by:
\[\hat{\varphi}_{\boldsymbol{x}}=\sum_{\boldsymbol{k}}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\left\{ \hat{a}^{\phantom{\dagger}}_{\boldsymbol{k}}\exp\left(i\boldsymbol{k}\cdot\boldsymbol{x}\right)+\hat{a}^{\dagger}_{\boldsymbol{k}}\exp\left(-i\boldsymbol{k}\cdot\boldsymbol{x}\right)\right\} .\]
To add the field operators’ time dependence, we need to understand the effect of time evolution on the creation and annihilation operators. But this is simple: by definition, these operators create an evenly spaced ladder of energy eigenstates. From Eq 1.1:
\[\begin{aligned} \left[\hat{H}_{\boldsymbol{k}},\hat{a}^{\dagger}_{\boldsymbol{k}}\right] & =\omega_{\boldsymbol{k}}\hat{a}^{\dagger}_{\boldsymbol{k}}\\ \left[\hat{H}_{\boldsymbol{k}},\hat{a}_{\boldsymbol{k}}\right] & =-\omega_{\boldsymbol{k}}\hat{a}_{\boldsymbol{k}} \end{aligned}\]
where
\[\hat{H}=\sum_{\boldsymbol{k}}\hat{H}_{\boldsymbol{k}}.\]
In the Schroedinger picture states evolve unitarily in time according to the time dependent Schroedinger equation. The equivalent in the Heisenberg picture is the unitary evolution of operators according to the Heisenberg equation of motion. For an operator \(\hat{A}_{H}\left(t\right)\) in the Heisenberg picture,
\[i\frac{\text{d}\hat{A}_{H}}{\text{d}t}=\left[\hat{A}_{H},\hat{H}\right].\]
For the raising operator this gives
\[i\frac{\text{d}\hat{a}^{\dagger}_{\boldsymbol{k}}}{\text{d}t}=-\omega_{\boldsymbol{k}}\hat{a}^{\dagger}_{\boldsymbol{k}}\]
which is solved by
\[\hat{a}^{\dagger}_{\boldsymbol{k}}\left(t\right)=\exp\left(i\omega_{\boldsymbol{k}}t\right)\hat{a}^{\dagger}_{\boldsymbol{k}}\left(0\right).\]
From Eq 1.20 this gives
\[\hat{U}^{\dagger}_{t}\hat{a}^{\dagger}_{\boldsymbol{k}}\hat{U}_{t}=\exp\left(i\omega_{\boldsymbol{k}}t\right)\hat{a}^{\dagger}_{\boldsymbol{k}}.\]
The lowering operator gets the opposite sign in the exponent. Finally this gives \[\hat{\varphi}_{\boldsymbol{x},t}=\sum_{\boldsymbol{k}}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\left\{ \hat{a}^{\phantom{\dagger}}_{\boldsymbol{k}}\exp\left(-i\omega_{\boldsymbol{k}}t\right)\exp\left(i\boldsymbol{k}\cdot\boldsymbol{x}\right)+\hat{a}^{\dagger}_{\boldsymbol{k}}\exp\left(i\omega_{\boldsymbol{k}}t\right)\exp\left(-i\boldsymbol{k}\cdot\boldsymbol{x}\right)\right\}\]
or
\[\hat{\varphi}_{x}=\sum_{\boldsymbol{k}}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\left\{ \hat{a}^{\phantom{\dagger}}_{\boldsymbol{k}}\exp\left(-ik\cdot x\right)+\hat{a}^{\dagger}_{\boldsymbol{k}}\exp\left(ik\cdot x\right)\right\}\]
(the \(+---\) signature has flipped the sign in the exponents). So overall including the time dependence explicitly just amounts to switching \(\boldsymbol{k}\cdot\boldsymbol{x}\rightarrow-k\cdot x\) in the exponents!