Find .

Solution:

## Archive for June, 2009

### Monday Math 78

June 29, 2009### Physics Friday 78

June 26, 2009One of the common geometric symmetries is rotational symmetry. One often speaks of n-fold rotational symmetry, meaning that it is unchanged by a rotation about a certain point (for 2D) or axis (3D) through an angle that is an integer multiple of 2π/*n*. For example, the face of the two of spades, below left, has 2-fold rotation symmetry, as a 180° rotation leaves it unchanged; while the triskele symbol below right has 3-fold rotation symmetry, being unchanged by a 120° (2π/3) rotation.

So, we have *n* so that the smallest rotation under which the shape is unchanged is 2π/*n*. Note, then, that *n*=1 corresponds to a rotationally asymmetric object; it requires a full rotation to be unchanged. Lastly, let us define *n*=0 for circular (or cylindrical or spherical) symmetry, where rotation through any angle is a symmetry. What would happen, then, if we have n=½? This would be an object which would not be symmetric under a full 360° rotation, but instead requires two full rotations to be returned to its initial state. While there is no simple figure like the above to serve as a geometric example of this, one can find a physical example in the Wikipedia article on orientation entanglement. One could also analogise to a machine drive belt in the shape of a Möbius strip, as is found in many cars; the belt returns to its original configuration only after two full revolutions. With this in mind, we can also gain some idea of *n*=3/2, and other half-integer *n* values.

What, though, does this have to do with our previous parts on indistinguishable particles and particle spin? The answer is that the above provides an alternate picture of spin^{1}. We identify the rotation symmetry number *n* with the spin number *s*. Again, we must note that a point particle has no “parts” to move under a rotation, so it does not exactly fit, but the behavior of the wavefunction under rotation is similar. In particular, a particle of integer spin has a wavefunction unchanged by a full rotation, while those of half-integer spin require two full rotations to be unchanged, and have a wavefunction that changes sign under a single full rotation.

Proving that this symmetry picture of spin matches to our intrinsic angular momentum picture requires both quantum field theory and special relativity. However, doing so allows one to prove the spin-statistics theorem. Recall in our discussion of indistinguisable particles the exchange operator, under which the joint wavefunction must be an eigenstate, with eigenvalue 1 or -1, for two indistinguishable particles. The key is that this exchange operator can be constructed from (180°) rotations; the result of the swapping of two particles is determined by the way one behaves under a full rotation (see here). This gives us the spin-statistics theorem: bosons, the particles which are symmetric under exchange, are those with integer spin; while fermions, which are antisymmetric under exchange, are those with half-integer spin.

1. The above description of spin-as-rotation-symmetry is taken from my vague memories of a similar description in a general-audience quantum mechanics book I read as a child; I do not recall which work.

### Monday Math 77

June 22, 2009Find .

Solution:

### Physics Friday 77

June 19, 2009**Part 2 of 4: Spin**

Fundamental particles have a number of intrinsic properties, which are fixed, and the same for all members of a species. These include examples such as the rest mass of a particle, and its electric charge. Another intrinsic property is one called ‘spin.’ The spin is a vector (of sorts), and its direction for particles of non-zero spin is an intrinsic degree of freedom.

Recall our discussion of the angular momentum in quantum mechanics, particularly the operators and their commutation relations, and the eigenvalue ladder. Spin is often described as being like the particles rotating around their axis; this fits only in that the previously mentioned mathematical description of quantized angular momentum also applies to spin, and that spin must be counted in the total angular momentum, with a few powerful differences; in analogy to the angular momentum numbers *l* and *m* (, ), we have *s* and *m _{s}* (, ). In one difference, the relationship between a charged particle’s spin and its magnetic moment involves a g-factor, and thus a gyromagnetic ratio, incompatible with classic physics (see here).

Another important difference is in the allowed values of

*s*. As we found here,

*m*ranges in integer steps from –

*l*to

*l*, so that for integer

*n*; similarly,

*m*ranges in integer steps from –

_{s}*s*to

*s*. When dealing with the eigenstates of angular momentum in position space, continuity in the azimuthal angular coordinate eliminated the half-integer values for

*l*. Spin, however, is under no such restriction;

*s*may be an integer or a half-integer.

I stated earlier that spin is an intrinsic property of a particle; in particular, the value of

*s*is fixed for a given particle species. For example, electrons, protons, neutrons, and all flavors of quarks are all spin-½, which means

*s*=½ for these particles. Photons are spin-1 (associated with polarization), while the delta baryons have a spin of 3/2. The hypothetical graviton, if it exists, must be spin-2 (as conservation of mass and momentum require that gravitational waves by quadrupole waves), and the hypothetical Higgs boson would be the only elementary particle with a spin of zero.

### Monday Math 76

June 15, 2009Find .

Solution:

### Physics Friday 76

June 12, 2009**Part 1 of ?: Distinguishable and Indistinguishable particles**

Suppose one has a pair of objects which interact with each other, such as two colliding billiard balls. For classical objects, we can distinguish between them. That is to say, if we name them ‘object 1’ and ‘object 2’, then following whatever set of interactions we have, we can look at one of the objects and, in theory at least, know whether it is object 1 or 2.

There are two ways we can distinguish the objects. The first is differences in the objects themselves; differences in mass, physical marks, etc. Two macroscopic objects are almost certain to be at least slightly different. The second way is to watch the objects as they interact and track their paths; you can follow 1 and 2 from their initial to final positions.

Now, let us compare to this the problem of a pair of microscopic particles of the same kind, such as two electrons, or two protons. Here, the physical properties of the two particles are identical; all electrons have the exact same mass, charge, etc. So the first method fails. For the second method, the principles of quantum physics tell us that between measurements, the particles do not have definite positions, instead being described probabilistically by the wavefunctions. As there are no definite trajectories, we cannot track the particles.

Together, this means that identical particles are *indistinguishable*; we cannot, even in principle, tell particles of the same species apart. This has important consequences for the joint wavefunctions of multiple particles, and for statistical mechanics.

In particular, suppose we have a pair of identical particles. We describe them with a joint wavefunction. Both particles’ state vectors occupy identical Hilbert spaces *H*, so that the joint wavefunction is in the Hilbert space formed by the tensor product *H*⊗*H*. If one of the particles is in a state *a*, and the other in state *b*, one might assume the resulting state is , as this is the standard for such a tensor product space. However, this implies that we can identify the particle in state *a* as ‘particle 1’, and the other as ‘particle 2’; would have them swapped. This violates what we said earlier about indistinguishability.

Let us consider a linear operator *P*, which swaps the values of the two state vectors; this is equivalent to swapping the two arbitrary labels we’ve attached to the particles (and their single-particle spaces):

.

This operator is its own inverse, and is Hermitian and unitary. Indistinguishability means that we should expect no physical observable result from the application of *P* to our state *Ψ*; at most, *P* can cause a change in the complex phase:

, with |*λ*|=1.

This means that valid states under indistinguishability are eigenvectors of *P*. But the fact that *P* is its own inverse means that its eigenvalues obey *λ*^{2}=1, so that *λ*=±1. Thus, the joint wavefunction must be either symmetric or antisymmetic under particle exchange; we have

for a symmetric state, or

for an antisymmetric state (the *C*s are normalization constants). For three or more particles, this still holds, with the wavefunction either symmetric or antisymmetric under the exchange of any two of the particles.

The question remaining is when are the wavefunctions symmetric, and when are they antisymmetric? The answer is that it depends on the particle species; there are two kinds of particles: one type, called bosons, which are always symmetric under exchange; and one type, called fermions, which are always antisymmetric. Which of these two groups a particle species belongs to is determined by an intrinsic particle property called “spin,” which we will discuss in our next part.

### Monday Math 75

June 8, 2009We previously showed that . What, then, is ?

Note first that . Thus:

.

Reversing order of integration,

.

Now, as , , and so

Now, I previously showed that ; thus

.

### Physics Friday 75

June 5, 2009Previously, I discussed the heat capacity ratio, also called the adiabatic index, *γ*, and derived the adiabatic condition, *PV ^{γ}*=const. Now, let us consider what happens to the temperature when an ideal gas changes volume via an adiabatic process. Solving the ideal gas law for pressure, we get . Substituting this into the adiabatic condition, we see that

As

*γ*>1,

*γ*-1>0, and so the temperature decreases when the gas expands adiabatically, and increases when the gas is compressed; these are known as adiabatic cooling and adiabatic heating, respectively.

### Monday Math 74

June 1, 2009Find the definite integral .

Solution: