# Why does the creation operator take a continuum value for the momentum?

Imagine that you have a lattice and a set of masses. Each mass at a lattice point. Now each two neighbouring masses are connected with spring.

Now in Classical Mechanics (CM) the ground state is the lattice when all masses are static and has zero energy. If you cause one mass to vibrate, all the other masses in the lattice would vibrate with the same frequency and the effect should propagate with a finite velocity.

Now replacing the springs with potentials and applying QM. The ground state is a vibrating lattice with non-zero energy. However the lattice should be allowed to vibrate at discrete frequencies.

I have the following questions:

1. Applying the continuum limit (QFT). Why does the creation operator take a continuum value for the momentum?

2. Shouldn't it take only discrete values?

3. In the discrete lattice are these quantized vibrations what are called phonones?

4. What are the factors that determines the speed at which these phonons propagate?

5. In the continuum limit, assuming that the field occupy a finite volume, what is the effect of the boundary conditions on the dynamics of the field?

6. In CM what should the entropy of this system?

7. and similarly the same question in QM?

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Hi ILovePhysics, and welcome to Physics Stack Exchange! You're asking a whole bunch of questions in one post, which we prefer not to have. Could you narrow your post down to just one question, or maybe a couple very closely related ones? – David Z May 16 '12 at 22:37
questions 1,2,3,4,5, are related, but 6 is not. You need to know the temperature or the total energy (and that it is randomly distributed) to answer questions about entropy. 7 is a nonsequitor, your question is in QM. – Ron Maimon May 17 '12 at 0:37

because the spatial volume is infinite--- the p values in a finite volume are discrete, but as you take the infinite volume limit, you get a continuum of p's. The p's are Fourier variables, they are inside-out with respect to the space variable.

To see this consider a periodic function in one dimension with period L:

$$f(x+L)=f(x)$$

This function can be written as a Fourier series:

$$f(x) = \sum_p e^{ipx} f_p$$

The condition of periodicity implies that the p's are discrete, since

$$e^{ip(x+L)} =e^{ipL} e^{ipx}$$

and this is periodic when $pL$ is an integer multiple of $2\pi$, so that the p's are integer multiples of $2\pi\over L$.

If you are looking at a function on a discrete domain, the p's are bounded. This is simply the fact that doing a Fourier transform twice (after complex conjugation) does nothing. So if a finite volume in x space means a discrete p-space, a discrete x-space means a finite volume p-space.

A finite volume discrete x-space has a finite volume discrete p-space. For a square lattice, the Fourier transform values are a square lattice of the same number of points, where the lattice spacing in p space is the reciprocal of the length of the size of the lattice in x-space (times $2\pi$), and the p-space lattice size is the reciprocal of the lattice spacing in x (times $2\pi$).

You should internalize the inverting properties of Fourier transform to understand the quantization of fields.

1. Yes, these are phonons. Phonons are the discrete energy states of lattice vibrations when they are quantized harmonic oscillators each with discrete equally spaced energy levels. They can be intepreted as particles, but their position doesn't make sense outside the lattice, they are effective particles. Also their localization is problematic using the usual spring law, in the same way that localizing photons is hard.

2. The reason that momentum is continuous is the infinite volume limit, not the continuum limit. If you make a continuum limit in a finite size box (a finite dimension of the spring-mass system), you still have discrete momenta. The continuum limit makes the maximum momenta run off to infinity.

3. the p's do take discrete values in finite volume.

4. there are two speeds, the phase-velocity and the group velocity. The phase velocity is the ratio of the frequency to the momentum, and the group velocity is the derivative of the frequency with respect to the momentum. The two speeds are equal for masses on linear springs.

### Logical development

When you have mechanical particles in square lattice which have position vector X given by:

$$X_{(i,j,k)} = (i+x,j+y,k+z)$$

Where x,y,z are each independent functions of the particle label, (i,j,k), the spring potential energy is

$${k\over 2} \sum_{<i,i'>} (x_i-x_{i'})^2 + (y_i - y_{i'})^2 + (z_i-z_{i'})^2$$

Which is the sum of the potential energy in the spring between nearest neighbors, the sum is over all nearest neighbor pairs i and i', where i and i' are two lattice labels (triplets of integers with two of the integers equal, and the third differing by one unit), and you make sure that each pair is counted only once, not twice.

The kinetic energy is the sum of the particle kinetic energies:

$${m\over 2}(\sum_i \dot{x_i}^2 + \dot{y_i}^2 + \dot{z_i}^2)$$

The total energy decomposes into a separate x energy,y energy, and z-energy. Each one is an independent oscillator system. Let me call one of these $\phi$, and choose units where m=1 and k=1.

$$H = {1\over 2}\sum_i P_i ^2 + {1\over 2} \sum_{<i,i'>}(\phi_i - \phi_{i'})^2$$

Where $P_i$ is the momentum of $\phi_i$, which is just the time derivative $\dot{\phi_i}$, To solve this quadratic system, you write $\phi$ as a Fourier transform. If the lattice goes from 0 to L-1 in each of the x,y,z directions:

$$\phi_{(i,j,k)} = \sum_p a_p e^{i(p_x i + p_y j + p_z k)}$$

Where the first "i" in the exponent is the square root of -1, and the next i is the x-coordinate integer label. The values of p are such that when going from 0 to L, the lattice ends, so the p's are also a discrete lattice of the same size: adding $2\pi$ to any component of p doesn't change the value of $\phi$ on the integer positions, and the boundary conditions at L and 0 determine the allowed values of p. The simplest boundary conditions you can impose are periodic boundary conditions $\phi(0,i,j)=\phi(L,i,j)$ (and similarly for the y and z direction). In this case, the whole thing is a translationally invariant torus.

To understand this, you should practice Fourier transforms and Fourier series on various domains, with various boundary conditions. The three energy conserving boundary conditions are

• Fixed (Dirichlet) boundary conditions: $\phi(0,i,j)=\phi(L,i,j)=0$ (and similarly for the y,z directions)
• Neumann boundary conditions: you extend the lattice to -1, and say $\phi(0,i,j)=\phi(-1,i,j)$ and $\phi(L,i,j)=\phi(L-1,i,j)$, in other words, the field has zero spatial derivative perpendicular to the boundaries.
• Periodic boundary conditions.

In periodic boundary conditions (the other two have the same qualitative structure, the same frequencies, although they are more annoying because they impose relations between the creation and annihilation operators), the potential energy in terms of the Fourier variables is

$${1\over 2} \sum_p |a_p|^2 (2(1-cos(p_x)) + 2(1-cos(p_y)) + 2(1-cos(p_z)))$$

This can be calculated explicity by substituting, while the kinetic energy is

$${1\over 2} \sum_p \dot{a_p}^2$$

So you have a collection of harmonic oscillators, one at each p. The discrete energy levels are given by the frequency of oscillations

$$\omega_p^2 = 2\cos(p_x) + 2\cos(p_y) + 2\cos(p_z)$$

Which for small $p_x,p_y,p_z$ (remember, their discreteness scale is the inverse of the lattice length), you get

$$\omega_p^2 \approx p_x^2 + p_y^2 + p_z^2$$

or the frequency is the length of the momentum vector (just like for photons).

The quantized version just has phonons at each p created/annihilated by the quantum operators $a^*_p$ and $a_p$, and when there are n phonons with quasi-momentum p (quasi-momentum means lattice analog of momentum), they have energy $n\omega_p$.

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