Relativistic Quantum Mechanics: Dirac Equation

Renan Cabrera
cabrer7@uwindsor.ca

Initialization

In[1]:=

Needs["TensorCalculus4`Tensorial`"]

In[2]:=

Needs["LinearAlgebra`MatrixManipulation`"]

In[3]:=

DeclareBaseIndices[{0, 1, 2, 3}]

In[4]:=

DefineTensorShortcuts[{{γ, x, ψ, α, P, c}, 1}, {{g, δ}, 2}]

labs = {x, δ, g, Γ} ;

In[6]:=

SetTensorValueRules[#, DiagonalMatrix[{1, -1, -1, -1}]] &/@{gdd[a, b], guu[a, b]} ;

SetTensorValues[δud[i, j], IdentityMatrix[NDim]] ;

Dirac's Gamma

First let's define the α matrices directly from the Pauli matrices

In[8]:=

In[9]:=

αu[μ]

%//ArrayExpansion[μ]

Out[9]=

α_μ^μ

Out[10]=

We also need β

In[11]:=

β = {{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, -1, 0}, {0, 0, 0, -1}}

Out[11]=

{{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, -1, 0}, {0, 0, 0, -1}}

They  anticommute

In[12]:=

αu[i] . αu[j] + αu[j] . αu[i]

%//ArrayExpansion[i, {1, 2, 3}]//ArrayExpansion[j, {1, 2, 3}]

Out[12]=

α_i^i . α_j^j + α_j^j . α_i^i

Out[13]=

In[14]:=

β . β

β . αu[j] + αu[j] . β == Array[0&, {NDim, NDim}]

%//ArrayExpansion[j, {1, 2, 3}]

Out[14]=

{{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}}

Out[15]=

Out[16]=

{True, True, True}

The following cell illustrates the commutation relation in detail. Just set the index equal to 1,2 or 3.

In[17]:=

index = 1 ;

HoldForm[β] . αu[j] + αu[j] . HoldForm[β] == 0

Print[SequenceForm["Case for j = ", index]]

%%/.{HoldForm[β] →MatrixForm[β], αu[j] →MatrixForm[αu[index]/.TensorValueRules[α]], 0→Array[0&, {NDim, NDim}]}

MapAt[HoldForm @@ {#/.MatrixForm→Identity} &, %, {{1, 1}, {1, 2}}]

%//ReleaseHold

Clear[index]

Out[18]=

β . α_j^j + α_j^j . β == 0

Case for j = 1

Out[20]=

Out[21]=

{{0, 0, 0, -1}, {0, 0, -1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}} + {{0, 0, 0, 1}, {0, 0, 1, 0}, {0, -1, 0, 0}, {-1, 0, 0, 0}} == {{0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}}

Out[22]=

True

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)

In[24]:=

SetTensorValueRules[γu[μ], {β, β . αu[1], β . αu[2], β . αu[3]}]

But it is also useful to define more components for γ and the identity matrix

In[25]:=

Out[25]=

In[26]:=

Format[Id] = "I" ;

Now let ' s test the anticommutation of  γ

In[27]:=

γu[i] . γu[j] + γu[j] . γu[i] == 2 * guu[i, j] Id

(%//ArrayExpansion[{i, j}])/.NewTensorValueRules[γ]/.TensorValueRules[g]

Out[27]=

γ_i^i . γ_j^j + γ_j^j . γ_i^i == 2 I g_ (ij)^(ij)

Out[28]=

{{True, True, True, True}, {True, True, True, True}, {True, True, True, True}, {True, True, True, True}}

Dirac Equation

Now let's define the Dirac Spinor ψ

In[29]:=

SetTensorValueRules[ Tensor[ψ], {ψd[0], ψd[1], ψd[2], ψd[3]}]

Then The Dirac equation is:

In[30]:=

I * γu[μ] . PartialD[Tensor[ψ], μ] - m^2 * Id . Tensor[ψ] == 0

Attributes[m] = {Constant} ;

Out[30]=

-m^2 I . ψ +  γ_μ^μ . ψ_ (, μ) == 0

Adding some additional rules.

In[32]:=

PartialD[_][Id, _] = 0 ;

PartialD[_][Tensor[γ, __], _] = 0 ;

Unprotect[Dot] ;

Dot[0, _] = 0 ;

Dot[_, 0] = 0 ;

Protect[Dot] ;

The Dirac equation with the coordinates explicitly:

In[38]:=

DiracEquation = (I * γu[μ] . PartialD[labs][Tensor[ψ], xu[μ]] - m^2 * Id . Tensor[ψ] == 0)

Out[38]=

-m^2 I . ψ +  γ_μ^μ . ∂ψ/∂x_μ^μ == 0

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

In[39]:=

DiracEquation //EinsteinSum[]

Out[39]=

Expanding and replacing values we get the following system of first order partial differential equations.

In[40]:=

(DiracEquation[[1]] //EinsteinSum[])/.TensorValueRules[ ψ]

ExpandedDiracEquation = Thread @ Equal[%/.NewTensorValueRules[ γ], 0]

Out[40]=

Out[41]=

Particle in the Rest Frame

In[42]:=

ExpandedDiracEquation

Out[42]=

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.

In[43]:=

TrialSolutionRF = ψd[i_] →cd[i] Exp[-I * Pd[0] * xu[0]//SumExpansion[ρ]]

Format[Pd[0]] := "E" ;

Out[43]=

ψ_i_^i_→^(- P_0^0 x_0^0) c_i^i

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.

In[45]:=

PartialD[_][cd[_], _] = 0 ;

PartialD[_][Pd[_], _] = 0 ;

PartialD[_][Pu[_], _] = 0 ;

PartialD[_][gdd[__], _] = 0 ;

Plugging in this trial solution into the actual Dirac equation we get.

In[49]:=

ExpandedDiracEquation/.TrialSolutionRF//Simplify

TrialExpandedDiracEquationRF = %/.Exp[_] →1

Out[49]=

Out[50]=

{c_0^0 (-m^2 + E) == 0, c_1^1 (-m^2 + E) == 0, c_2^2 (m^2 + E) == 0, c_3^3 (m^2 + E) == 0}

And after some simplification:

In[51]:=

MRF = First @ LinearEquationsToMatrices[TrialExpandedDiracEquationRF, cd[i]//ToArrayValues[]]

Out[51]=

{{-m^2 + E, 0, 0, 0}, {0, -m^2 + E, 0, 0}, {0, 0, m^2 + E, 0}, {0, 0, 0, m^2 + E}}

The system has a non trivial solution only if the determinat is zero. (Not all the equations are independent)

In[52]:=

Det[MRF]//Simplify

Out[52]=

(m^4 - (E)^2)^2

If we let the Energy or P_0^0 to be a free parameter we get two solutions that will make the determinant zero.

In[53]:=

Solve[ Evaluate[Simplify @ Det[MRF]] == 0, Pd[0]]

Out[53]=

{{E→ -m^2}, {E→ -m^2}, {E→m^2}, {E→m^2}}

One corresponds to a positive energy (particle?)and the other to a negative energy (antiparticle?)

In[54]:=

Eigensystem[MRF]

Out[54]=

{{-m^2 + E, -m^2 + E, m^2 + E, m^2 + E}, {{0, 1, 0, 0}, {1, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 0}}}

Recalling the trial solution, this means that the solutions are four; the first two corresponding to the positive energy.

In[55]:=

{1, 0, 0, 0} * Exp[-I * m * xu[0]]

Out[55]=

{^(- m x_0^0), 0, 0, 0}

In[56]:=

{0, 1, 0, 0} * Exp[-I * m * xu[0]]

Out[56]=

{0, ^(- m x_0^0), 0, 0}

And the last two corresponding to the negative energy

In[57]:=

{0, 0, 1, 0} * Exp[I * m * xu[0]]

Out[57]=

{0, 0, ^( m x_0^0), 0}

In[58]:=

{0, 0, 0, 1} * Exp[I * m * xu[0]]

Out[58]=

{0, 0, 0, ^( m x_0^0)}

Free Particle Solution

Let P to be the 4momentum vector. Let's try a solution for the free particle with the following plane wave.

In[59]:=

TrialSolution = ψd[i_] →cd[i] Exp[-I * gdd[ρ, s] Pu[s] * xu[ρ]//ToArrayValues[] ]

Out[59]=

ψ_i_^i_→^(- P_0^0 x_0^0 +  P_1^1 x_1^1 +  P_2^2 x_2^2 +  P_3^3 x_3^3) c_i^i

Where we  define the tensor obejct c as a constant. The same for the components of the momentum.

In[60]:=

PartialD[_][cd[_], _] = 0 ;

PartialD[_][Pd[_], _] = 0 ;

PartialD[_][Pu[_], _] = 0 ;

PartialD[_][gdd[__], _] = 0 ;

Format[Pu[0]] := "E" ;

Plugging in this trial solution into the actual Dirac equation we get.

In[65]:=

ψd[i]/.TrialSolution

Out[65]=

^(- E x_0^0 +  P_1^1 x_1^1 +  P_2^2 x_2^2 +  P_3^3 x_3^3) c_i^i

In[66]:=

(ExpandedDiracEquation/.TrialSolution//Simplify)

%/.Exp[_] →1

TrialExpandedDiracEquation = Map[Collect[#, cd[_]] &, %, {2}]

Out[66]=

Out[67]=

Out[68]=

And after some simplification:

In[69]:=

TrialExpandedDiracEquation//MatrixForm

Out[69]//MatrixForm=

Extracting a matrix from the system of equations

In[70]:=

M = First @ LinearEquationsToMatrices[TrialExpandedDiracEquation, cd[i]//ToArrayValues[]]

Out[70]=

{{-m^2 + E, 0, -P_3^3, -P_1^1 +  P_2^2}, {0, -m^2 + E, -P_1^1 -  P_2^2, P_3^3}, {P_3^3, P_1^1 -  P_2^2, -m^2 - E, 0}, {-P_1^1 -  P_2^2, P_3^3, 0, m^2 + E}}

The system has a non trivial solution only if the determinant is zero. (Not all the equations are independent)

In[71]:=

Det[M]//Simplify

Out[71]=

-(m^4 - (E)^2 + (P_1^1)^2 + (P_2^2)^2 + (P_3^3)^2)^2

If we let the Energy or P_0^0 to be a free parameter we get two solutions that will make the determinat zero.

In[72]:=

Energies = Union @ Flatten[Solve[ Evaluate[Simplify @ Det[M]] == 0, Pu[0]]]

Out[72]=

{E→ -(m^4 + (P_1^1)^2 + (P_2^2)^2 + (P_3^3)^2)^(1/2), E→ (m^4 + (P_1^1)^2 + (P_2^2)^2 + (P_3^3)^2)^(1/2)}

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:

In[73]:=

Thread[M . {u1, u2, u3, u4} == {0, 0, 0, 0}]

SolPositive = Solve[ %[[{2, 3}]], {u3, u4}][[1]]//Cancel

Out[73]=

Out[74]=

{u4→ -(-m^4 u2 + u2 (E)^2 - u2 (P_1^1)^2 - u2 (P_2^2)^2 - u1 P_1^1 P_3^3 -  u1 P_2^2 P_3^3)/((m^2 + E) P_3^3), u3→ -(-u2 P_1^1 +  u2 P_2^2 - u1 P_3^3)/(m^2 + E)}

The first eigenvector is:

In[75]:=

EigenVectorPositiveE1 = {u1, u2, u3, u4}/.SolPositive/.{u2→0, u1→1}//Cancel

Out[75]=

{1, 0, P_3^3/(m^2 + E), (P_1^1 +  P_2^2)/(m^2 + E)}

The second eigenvector is:

In[76]:=

EigenVectorPositiveE1 = {u1, u2, u3, u4}/.SolPositive/.{u1→0, u2→1}/.Pu[0]^2→Pu[1]^2 + Pu[2]^2 + Pu[3]^2 + m^2

Out[76]=

{0, 1, -(-P_1^1 +  P_2^2)/(m^2 + E), -(m^2 - m^4 + (P_3^3)^2)/((m^2 + E) P_3^3)}

Now for the eigenvectors of negative energy

In[77]:=

Thread[M . {u1, u2, u3, u4} == {0, 0, 0, 0}]

SolNegativeE = Solve[ %[[{1, 2}]], {u1, u2}][[1]]//Cancel

Out[77]=

Out[78]=

{u1→ (-u4 P_1^1 +  u4 P_2^2 - u3 P_3^3)/(m^2 - E), u2→ (-u3 P_1^1 -  u3 P_2^2 + u4 P_3^3)/(m^2 - E)}

In[79]:=

EigenVectorNegativeE1 = EigenVectorNegativeE1 = {u1, u2, u3, u4}/.SolNegativeE/.{u3→1, u4→0}

Out[79]=

{-P_3^3/(m^2 - E), (-P_1^1 -  P_2^2)/(m^2 - E), 1, 0}

In[80]:=

EigenVectorNegativeE2 = EigenVectorNegativeE1 = {u1, u2, u3, u4}/.SolNegativeE/.{u3→0, u4→1}

Out[80]=

{(-P_1^1 +  P_2^2)/(m^2 - E), P_3^3/(m^2 - E), 0, 1}

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


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