Relativistic Quantum Mechanics: Dirac Equation
Renan Cabrera
cabrer7@uwindsor.ca
Initialization
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Dirac's Gamma
First let's define the α matrices directly from the Pauli matrices
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We also need β
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They anticommute
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The following cell illustrates the commutation relation in detail. Just set the index equal to 1,2 or 3.
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The convenient Dirac's Gamma are defined from the α and β matrices suitable for the metric we defined.(other shapes of the metric would require silight changes)
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But it is also useful to define more components for γ and the identity matrix
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Dirac Equation
Now let's define the Dirac Spinor ψ
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Then The Dirac equation is:
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Adding some additional rules.
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The Dirac equation with the coordinates explicitly:
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where m is the mass and its sign can be either positive or negative. Dirac interpretations says that they represent a particle and its antiparticle.
Expanding over the γ 's
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Expanding and replacing values we get the following system of first order partial differential equations.
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Particle in the Rest Frame
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Let P to be the 4momentum vector. Let's try a solution for a free particle in a rest frame system. This means that the only component of the momentum that survives is the zero one.
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Where we define the tensor object c as a constant. The same for the components of the momentum and we format the zero component as E refering as the enegy.
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Plugging in this trial solution into the actual Dirac equation we get.
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And after some simplification:
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The system has a non trivial solution only if the determinat is zero. (Not all the equations are independent)
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If we let the Energy or to be a free parameter we get two solutions that will make the determinant zero.
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One corresponds to a positive energy (particle?)and the other to a negative energy (antiparticle?)
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Recalling the trial solution, this means that the solutions are four; the first two corresponding to the positive energy.
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And the last two corresponding to the negative energy
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Free Particle Solution
Let P to be the 4momentum vector. Let's try a solution for the free particle with the following plane wave.
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Where we define the tensor obejct c as a constant. The same for the components of the momentum.
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Plugging in this trial solution into the actual Dirac equation we get.
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And after some simplification:
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Extracting a matrix from the system of equations
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The system has a non trivial solution only if the determinant is zero. (Not all the equations are independent)
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If we let the Energy or to be a free parameter we get two solutions that will make the determinat zero.
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With positive and negative possibilities for the energy.
There are only two different eigenvalues, this means that for each one there is a 2D plane space, included into the original 4D space. In other words for each energy we can choose two arbitrary parameters to span a 2D plane and we can choose any pair of equations to solve the parameters provided that a the particular pair of equations are not parallel (linearly dependent)
I played numerically with the equations and it seems that no pair of equations are ever parallel for any case of the autovalues, but conventionally there is a particular way to choose them as follows.
For the positive energy:
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The first eigenvector is:
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The second eigenvector is:
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Now for the eigenvectors of negative energy
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References
1) Kane, Gordon. Modern Elementary Particel Physics. (1993) Addison-Wesley. Section 5.1
2) Greiner, Walter & Reinhardt, Jochim. Field Quantization. (1996) Springer. Section 5
3) Doughty, Noel E. Lagrangian Interaction. (1990) Addison-Wesley Section 20.3
Created by Mathematica (November 22, 2007) |