Sunday, 23 May 2010

The Moon

This is the best image from fifteen photographs of the Moon that I took this evening. My flat has a skylight thing in the roof, which opens and tilts in various ways (it would be great for using a telescope, actually...) so I decided to set up my camera and tripod and take some photos of the moon, since it was a clear night.

Thursday, 4 March 2010

The Road To Quantum Field Theory: Updated Post List

Once again, I'm posting an index, essentially, of the posts in the `Road to QFT' series, so far. The previous list was here, but I'll reproduce all of the links in this post too.

Pre-Series Posts
These were posts on related topics (special relativity, quantum mechanics and particle physics)
The `Road To Quantum Field Theory' Series
Posts in the `Road to QFT' series, in chronological order.
Out-of-Series Posts
Posts on related topics which were not intended as part of the `Road to QFT' series, but may be of interest anyway.
Roadmap
As before, I'll post a rough list of things still to be covered. By comparing the previous roadmap to the most recent two posts, you'll notice I've done things slightly out of the order I mentioned there, so this is really a very approximate guide.
  • Relativistic QM: The Klein-Gordon Equation & Spin-Zero Particles
  • Relativistic QM 2: The Dirac Equation & Spin-Half Particles
  • Quantum Electrodynamics (QED)
  • Quantum Chromodynamics (QCD)
  • SU(2) and Electroweak Unification
  • Quark Flavour Mixing
  • Spontaneous Symmetry Breaking and the Higgs Mechanism
  • Beyond the Standard Model: Neutrino Mass Terms & Neutrino Mixing
This is still a very ambitious set of things to cover. In particular, the posts so far have only brushed the surface, providing the background needed to understand relativistic quantum mechanics and gauge theories. The real work is yet to come! Note also that I am not a theorist, so my coverage of some of the more advanced topics may be less than completely thorough! I've added sections on quark flavour mixing and neutrino masses & mixing to the roadmap since they are of particular interest to me.

    Tuesday, 2 March 2010

    Gauge Invariance in Quantum Mechanics

    This post continues the 'Road to Quantum Field Theory' series.

    The Lorentz force law for a non-relativistic particle of charge q, moving with velocity v in both electric (E) and magnetic (B) fields, is given by:
    \mathbf{F} = q\mathbf{E} + q\mathbf{v}\times\mathbf{B}
    This can be derived from Hamilton's equations using the classical Hamiltonian given below.
    H = \frac{1}{2m}\left(\mathbf{p}-q\mathbf{A}\right)^2 + qV
    The Schrödinger equation for a charged particle in an EM field is,
    \left[\frac{1}{2m}\left(-i\nabla - q\mathbf{A}\right)^2 + qV \right] \psi(\mathbf{x},t) = i\frac{\partial \psi(\mathbf{x},t)}{\partial t}
    obtained from the Hamiltonian through the substitution
    \mathbf{p} \rightarrow -i\nabla
    as is usual for the quantum mechanical momentum operator. We can identify operator combinations:
    \mathbf{D} = \nabla - iq\mathbf{A}
    D^0 = \frac{\partial}{\partial t} + iqV
    which replace the operators
    \nabla, \frac{\partial}{\partial t}
    when we move from the free-particle Schrödinger equation to the electromagnetic field case.

    The solutions for the wavefunctions of the Schrödinger equation describe completely the behaviour of a particle under the influence of the potentials A and V, but these potentials are not unique, as I showed in the previous post in this series. Instead, they can be changed by a gauge transformation:
    \mathbf{A}\rightarrow\mathbf{A}^\prime=\mathbf{A}+\nabla\chi
    V\rightarrow V^\prime = V - \frac{\partial\chi}{\partial t}
    Maxwell's equations for E and B will remain invariant under these transformations (this was the topic of the previous post), but will the physics described by the Schrödinger equation be the same if we make these changes to the potentials there?

    The answer is no! The Schrödinger equation is not gauge invariant, since the same wavefunction cannot satisfy both the original and transformed versions! All is not lost, however. The wavefunction itself is not directly observable, whereas E and B are. If we do not require the wavefunction to remain invariant when the potentials undergo a transformation, then we are free to change the wavefunction such that we retain invariance of physical observables. Write the Schrödinger equation in terms of some transformed wavefunction,
    \psi\rightarrow\psi^\prime
    i.e.
    \left[\frac{1}{2m}\left(-i\nabla - q\mathbf{A}^\prime\right)^2 + qV^\prime\right] \psi^\prime(\mathbf{x},t) = i\frac{\partial \psi^\prime (\mathbf{x},t)}{\partial t}

    Now, the form in which we have written the equation above is identical to that of the original Schrödinger equation, except that we are writing it in terms of primed quantities (psi', A', V') instead of unprimed (psi, A, V). Both equations describe the same physics, so if we can find such a psi', then the Schrödinger equation is gauge covariant; that is, it maintains the same form under a gauge transformation.

    We know the relationship between A, V and A', V', so we can write down the transformation for psi':
    \psi^\prime (\mathbf{x},t) = e^{iq\chi(\mathbf{x},t)}\psi(\mathbf{x},t)
    where
    \chi
    is the same space- and time- dependent function appearing in the transformations of A and V. We can verify that this results in the gauge covariance of the Schrödinger equation:
    (-i\nabla - q\mathbf{A}^\prime)\psi^\prime=\left[-i\nabla-q\mathbf{A}-q(\nabla\chi)\right]e^{iq\chi}\psi
    =q(\nabla\chi)e^{iq\chi}\psi + e^{iq\chi}(-i\nabla\psi) + e^{iq\chi}(-q\mathbf{A}\psi) - q(\nabla\chi)e^{iq\chi}\psi
    The first and last terms cancel, leaving:
    (-i\nabla - q\mathbf{A}^\prime)\psi^\prime = e^{iq\chi}(-i\nabla - qA)\psi
    which can be written as:
    (-i\mathbf{D}^\prime \psi^\prime) = e^{iq\chi}(-i\mathbf{D}\psi)


    The space-time dependent phase factor feels the action of
    \nabla
    , but passes through the combined D' operator, converting it to D, so it becomes clear that D'psi' is related to Dpsi in the same way psi' is related to psi! Similarly,
    (iD^{0\prime}\psi^\prime = e^{iq\chi}(iD^0\psi)

    We can now erite:
    \frac{1}{2m}(-i\mathbf{D}^\prime)^2 \psi\prime = e^{iq\chi}\frac{1}{2m}(-i\mathbf{D})^2 \psi = e^{iq\chi}iD^0\psi = iD^{0\prime}\psi^\prime
    which demonstrates the correct relationship between psi and psi'. The question remains whether the same physics is described by both psi and psi'. We can see that it is in a number of ways:

    1. The probability density is given by
    |\psi|^2 = \psi^\dagger \psi
    , which is equivalent to
    |\psi\prime|^2 = \psi^{\prime\dagger} \psi^\prime = \psi^\dagger e^{-iq\chi}e^{+iq\chi}\psi = \psi^\dagger \psi

    2. The probability current,
    \psi^\dagger(\nabla\psi) - (\nabla\psi)\psi^\dagger
    is not invariant under a gauge transformation, but when we replace
    \nabla \rightarrow \mathbf{D} ~~,~~ \frac{\partial}{\partial t} \rightarrow D^0
    then,
    \psi^\dagger\prime(\mathbf{D}\psi^\prime = \psi^\dagger e^{-iq\chi} e^{iq\chi}(\mathbf{D}\psi) = \psi^\dagger\mathbf{D}\psi
    (and similarly for the other term).

    Hence, identical physics is described by both the wavefunction psi, and the gauge transformed wavefunction psi'; the gauge invariance of Maxwell's equations presents as a gauge covariance in quantum mechanics, provided that we transform not only the potentials, but the wavefunction also:
    \mathbf{A}\rightarrow \mathbf{A}^\prime = \mathbf{A} + \nabla\chi
    V\rightarrow V^\prime = V - \frac{\partial \chi}{\partial x}
    \psi\rightarrow \psi^\prime = i^{iq\chi}\psi

    Finally, we note that the new differential operators,
    \mathbf{D} = \nabla - iq\mathbf{A} ~~,~~D^0 = \frac{\partial}{\partial t} + iqV
    can be written in a Lorentz covariant form:
    D^\mu = \partial^\mu + iqA^\mu

    This allows us to write:
    -iD^{\prime\mu}\psi^\prime = e^{iq\chi}(-iD^\mu \psi)
    and it follows that an equation can involving the operator
    \partial^\mu
    can be made to be gauge invariant by under the combined transformations:
    A^\mu \rightarrow A^{\prime\mu} = A^\mu - \partial^\mu \chi
    \psi \rightarrow \psi^\prime = e^{iq\chi}\psi
    provided the minimal substitution is also made:
    \partial^\mu \rightarrow D^\mu = \partial^\mu + iqA^\mu

    This provides a simple mechanism to get the wave equation for a particle in an electromagnetic field from the equation for a free particle - by making the above substitution. This forms the basis of the gauge principle, i.e. the form of an interaction is determined by an insistence on local gauge invariance.

    That's enough for now; next time we'll move on to study the Klein-Gordon equation for relativistic spin-zero particles.

    Monday, 8 February 2010

    Farscape Intro Randomiser

    While watching Farscape Season 2, Episode 4 (Crackers Don't Matter) last night, I wrote a Python script to randomise elements of the Farscape intro. Here's an example:

    My name is John Crichton, a wormhole hit and I got shot through a military commander. Now I'm lost in some distant part of the universe on an insane radiation wave, full of escaped prisoners. I'm being hunted by a living ship. Doing everything I can. I'm just looking for a way home.



    import random
    
    adjectives = [
     ('insane', 'an'),
     ('living', 'a'),
     ('escaped', 'an'),
     ]
    
    nouns = [
     ('ship', 'a'),
     ('radiation wave', 'a'),
     ('military commander', 'a'),
     ('prisoner', 'a'),
     ('wormhole', 'a')
     ]
    
    
    parts = []
    parts.append(random.choice(nouns))
    nouns.remove(parts[0])
    parts.append(random.choice(nouns))
    nouns.remove(parts[1])
    parts.append(random.choice(adjectives))
    adjectives.remove(parts[2])
    parts.append(random.choice(nouns))
    nouns.remove(parts[3])
    parts.append(random.choice(adjectives))
    adjectives.remove(parts[4])
    parts.append(random.choice(nouns))
    nouns.remove(parts[5])
    parts.append(random.choice(adjectives))
    adjectives.remove(parts[6])
    parts.append(random.choice(nouns))
    nouns.remove(parts[7])
    
    result = 'My name is John Crichton, '
    result += parts[0][1] + ' ' + parts[0][0]
    result += " hit and I got shot through "
    result += parts[1][1] + ' ' + parts[1][0]
    result += ". Now I'm lost in some distant part of the universe on "
    result += parts[2][1] + ' ' + parts[2][0] + ' ' 
    result += parts[3][0]
    result += ", full of "
    result += parts[4][0] + ' ' + parts[5][0] + 's'
    result += ". I'm being hunted by "
    result += parts[6][1] + ' ' + parts[6][0] + ' ' + parts[7][0]
    result += ". Doing everything I can. I'm just looking for a way home."
    
    print result