# Reif about number of states.

1. Nov 30, 2006

### quasar987

There is another passage in Reif that got me really confused. It is in section 9.9 (Quantum states of a single particles).

We are considering an ideal gaz in a rectangular box of volume $L_xL_yL_z=V$ and we are making the approximation that the wave function of a particle of the gaz is that of a free particle that must have the lenghts of the box as a period. In other words, we say that

$$\psi(\vec{r})=e^{i\vec{k}\cdot \vec{r}}$$

and we demand that

$$k_i=\frac{2\pi}{L_i}n_i, \ \ \ \ \ i=x,y,z, \ \ \ \ \ n_i \in \mathbb{Z}$$

Reif then argues that for any macroscopic box, the L_i are large and thus, in a small interval [k_i,ki+dk_i] of the wave number, there are very many states of the particle (many n_i)...

But I look at $k_i=\frac{2\pi}{L_i}n_i$ and wheter the ratio $2\pi/L_i$ is small depends heavily on the choice of units we use to measure distances.

Last edited: Nov 30, 2006
2. Dec 1, 2006

### OlderDan

But the units of k and L are reciprocal. If L becomes numerically small by choice of units, k-space expands by the same factor and dk expands with it. Think of Ldk as a dimensionless invariant. If the argument is valid in one system of units, it is valid in any system.

3. Dec 1, 2006

### quasar987

$$dn_i=\frac{L_i}{2\pi}dk_i$$

But we're interested in showing that for a small interval dk_i, dn_i is quite larger than unity. And this is not units invariant. If you choose some unit system in which L_i/2pi is large and then choose a small value for d_k_i such that dn_i is still large, then as you change your units, d_n stays the same, true, but only because as L_i/2pi becomes smaller, dk_i becomes larger.

But we want dk_i to be small.

There is also the possibility that it is not true that for a small dk, there are many states in that interval. This does not change anything to the discussion to follow; we can still define a density of states by

$$\rho(k_x,k_y,k_z)dk_xdk_ydk_z=\left(\frac{L_x}{2\pi}\right)\left(\frac{L_y}{2\pi}\right)\left(\frac{L_z}{2\pi}\right)dk_xdk_ydk_z$$

who cares if the number of states is larger than one for some smallish set of dk_i?

What do you think?

Last edited: Dec 1, 2006
4. Dec 1, 2006

### OlderDan

I think we are agreed that dn is not unit dependent as long as we allow dk to "expand" numerically in reciprocal space inversely to the numerical contraction of L by choice of units. I think that we need to look at a bigger picture. The density of states or number of states per unit energy becomes important in determining the behavior of systems. For electrons in a solid, the macroscopic properties are determined by the electrons with the highest energies. Since electrons are Fermions, the low energy states get filled, and limited occupancy drives electrons to higher energy levels. The "Fermi surface" is a surface in k space occupied by the highest energy electrons, and their energy is called the "Fermi energy". The symmetry of the occupancy of the low level electrons keeps them from contributing to macroscopic properties. It is a redistribution of the eneries of the electrons very close to the Fermi surface in k-space that determines the interesting properties of solids. The number of such electrons in a bulk material is huge.

What makes dk big or small is how large it is in relation to the volume in k space of a single state. That is independent of the choice of units. For free electron energies like the particles in a box, the energy is kinetic and proportional to k². The volume of k space associated with a single state is a little cube surrounding one particular vector k value. Let's agreee that dk will be the length of one side of such a cube. For small k values, very few states have the same scalar k value. However, compared to the volume of a spherical shell of radius k and thickness dk, which is proportional to k²dk, the volume for a single state is tiny for the important higher k values. These are the states for which dn/dk ~ k² is high, independent of the units chosen to express L.

Last edited: Dec 1, 2006
5. Dec 1, 2006

### marcusl

Quasar, there should be a normalization somewhere that makes the units irrelevant. Let's see, I gave my copy of Reif to my son but I have a scarcely-browsed copy of tolman on my shelf...

ok, the energy of the n1,n2,n3 state is
E = h^2 / (8*m) * sum_over_i {(ni/Li)^2}.
Now we can see that L appears in comparison to h/sqrt(m), which is large for a macroscopic box regardless of which length units are chosen (since units for h will also change accordingly).

Tolman finishes the calc as follows:
The total number of states having energy <= E is
V = triple integral of dn1*dn2*dn3
= 4*pi/3 * (2*m*E/h^2)^(3/2) * v
where v=L1*L2*L3.
Then number of states between E and E+deltaE is
deltaV = dV/dE * deltaE = 4*pi*m*v/h^3 * sqrt(2*m*E) * deltaE

Again v*m^(3/2)/h^3 is big so the density of states is big. Hope this helps!

Edit: I think I'm saying the same thing as OlderDan...

6. Dec 2, 2006

### quasar987

Thanks a lot to you both for your replies.

I now have another question on the same topic. It is when Reif (9.13) talks about the volumic density of state per unit wavenumber in each directions (equation 9.13.7). He says that it is given by

$$2\frac{1}{\exp\left(\frac{\hbar ck}{k_BT}-1\right)}\frac{1}{(2\pi)^3}dk_xdk_ydk_z$$

So it's (number of independant polarization directions) x (mean number of photons in the state with wave number k) x (volumic density of states per unit wavenumber).

My question is, why does he include the different polarizations? As far as I know, polarization has nothing to do with the particle interpretation of light.

But since he does count the different diredtions of polarization as different states of the photons, it must be that there is a way to (experimentally) determine the direction of polarization of a photon. So my real question is, what is this way? When does it become apparent that two photons apparently in the same state really are not because they have different polarization?

Last edited: Dec 2, 2006
7. Dec 2, 2006

### OlderDan

Photons are polarized. Their polarization corresponds to an angular momentum state. They are spin 1 particles. See for example

http://www.mathpages.com/rr/s9-04/9-04.htm

Numerous quantum interactions have verified that photons have this property. The experiments on quantum entanglement are based on detection of phton polarizations.

8. Dec 2, 2006

### quasar987

I tought the photons had spin 0. I guess that is wrong. Isn't there a particle that has spin 0?

Do you know why is it they call the photon's spin polarization and not just spin?

Last edited: Dec 2, 2006
9. Dec 3, 2006

### OlderDan

As far as I know, photons exhibit all the polarization properties of classical EM waves. It is easier to describe macroscopic effects in terms of polarization rather than resorting to angular momentum considerations. If you pass a beam of light through a polarizing filter, it comes through polarized whether you think of the beam as a wave, or as photons. In particle interactions, angular momentum needs to be considered, so spin is the appropriate representation of polarization.

There is at least one spin zero particle thought to exist, the Higgs boson, though as far as I know it has not yet been observed. Some others have been hypothesized.

http://en.wikipedia.org/wiki/List_of_particles

10. Dec 3, 2006

### quasar987

How does a polarizing filter works?

11. Dec 3, 2006

### nrqed

They have spin 1.

There is no observed *fundamental* particle with spin 0. The Standard Model predicts one spin 0 particle: the Higgs particle, but it has not been observed yet. Of course, there are many ways to have bound states with spin 0. For example, pions are bound states of a quark and antiquark with a total spin of 0. You can get an electron and a positron in a bound state of total spin 0. And so on.
Usually, people use "polarization" for the classical wave description of the EM field, and simply "spin" when talking in terms of photons.

Maybe "spin polarization" is used to talk about the z component of the spin ($+ \hbar$ or $- \hbar$) but I don't recall having seen that expression.

Hope this helps