McNair Scholars Research Journal
Volume 1 Article 3
2014
Solving the Dirac Equation: Using Fourier Transform
Vincent P. Bell Embry-Riddle Aeronautical University, [email protected]
Follow this and additional works at: https://commons.erau.edu/mcnair
Recommended Citation Bell, Vincent P. (2014) "Solving the Dirac Equation: Using Fourier Transform," McNair Scholars Research Journal: Vol. 1 , Article 3. Available at: https://commons.erau.edu/mcnair/vol1/iss1/3
This Article is brought to you for free and open access by the Journals at Scholarly Commons. It has been accepted for inclusion in McNair Scholars Research Journal by an authorized administrator of Scholarly Commons. For more information, please contact [email protected]. Bell: Solving the Dirac Equation: Using Fourier Transform
Vincent Bell
Solving the Dirac Equation: Using Fourier Transform
Vincent Bell, Embry-Riddle Aeronautical University Abstract When looking at the spin of a particle, the Dirac Equation is used to explain the significance of the spin. Here we solve Dirac Equation using the Fourier Transform, as well as explain and define the terms of the Dirac Equation. In addition, the Dirac Hamiltonian changes in Quantum Mechanics are examined. There will be a discussion of what the alpha matrices and beta matrix are made up of. The end result is defining the Dirac Equation for a particle in an electromagnetic field and solving this equation. Introduction
The Dirac Equation is a relativistic quantum wave equation which was formulated by Paul Adrien Maurice Dirac in 1928. The Dirac Equation is used for the description of elementary spin – ½ particles, for example electrons. The equation demands the existence of antiparticles and actually predated their experimental discovery. This made the discovery of the positron, the antiparticle of the electron, one of the greatest triumphs of modern theoretical physics. The Dirac Equation looks as follows:
D Ψ = EH Ψ (1)
Where the H D is Dirac’s Hamiltonian. This Hamiltonian is given by: r 2 (2) D ˆ K +•= βα mcPcH K = 1, 2, 3 Where the alpha matrix,αˆ , and the beta matrix, β , are 4 x 4 matrices. The αˆ K K r consists of the Pauli Matrices. Pauli Matrices will be discussed later. The momentum, p , is a 3-D vector in the X, Y, Z direction, normally seen in the Cartesian coordinate system. The momentum is given by: r (3) = [ ,, pppP zyx ] The momentum in quantum mechanics is represented as operator and then becomes:
Published by Scholarly Commons, 2014 1 McNair Scholars Research Journal, Vol. 1 [2014], Art. 3
r r iP h∇−→ (4) r ∂ ∂ ∂ where =∇ ,, ∂ ∂ ∂zyx And where h is Dirac’s constant, which is Planck’s constant divided by 2π. For relativistic effects to occur the particle has to be travelling close to or a fraction of the speed of light, which is designated by c. And, m is the mass of the particle. The
momentum is a vector and the dot product is being done between the momentum andαˆ K .
Also αˆ K is in 3 dimensions just like the momentum. Now Dirac’s Hamiltonian takes the form of: r h 2 (5) D ciH ˆ K +∇•−= βα mc ∂ ∂ ∂ (6) h ˆˆˆ 2 D −= ciH 1 + 2 + 3 +βααα mc ∂ ∂ ∂zyx Now the steady state Dirac Equation is denoted by:
∂ ∂ ∂ h ˆˆˆ 2 − ci 1 + 2 + 3 + βααα mc Ψ (7) ∂ ∂ ∂zyx = EΨ The above equation is also known as time independent Dirac Equation. But the time dependent Dirac Equation, the total energy, E, becomes an operator in quantum mechanics. The operator looks as follows:
∂ →iE h (8) ∂t From using equations 7 and 8, the time dependent Dirac Equation becomes:
∂ ∂ ∂ h ˆˆˆ 2 (9) − ci + + 321 +βααα mc Ψ ∂ ∂ ∂zyx ∂Ψ = ih ∂t
Pauli Matrices and Beta Matrix
Since αˆ K consist of Pauli Matrices, the Pauli Matrices are:
10 0 − i (10) σ1 = σ 2 = 01 i 0 01 σ 3 = −10
https://commons.erau.edu/mcnair/vol1/iss1/3 2 Bell: Solving the Dirac Equation: Using Fourier Transform
Theαˆ K has 3 parts, as shown in equation 2, and is constructed from the Pauli Matrices. So, the 3 α-matrices are as follows:
O σ O σ ˆ 1 ˆ 2 (11) α1 = α 2 = σ1 O σ 2 O O σ ˆ 3 α3 = σ 3 O Where O represents the null matrix that consists of a 2 x 2 matrix with all zeros. This makes the alpha matrices a 4 x 4 matrix, which is governed by:
1000 000 −i (12) 0100 i 000 αˆ1 = αˆ = 0010 2 − i 000 0001 i 000 0100 −1000 αˆ = 3 0001 − 0010 Similarly to the 3 α-matrices, beta matrix, β , consists of the identity matrix.
OI (13) β = − IO 0001 (14) 0010 β = − 0100 −1000 Since the alpha matrices and beta matrix are both 4 x 4 matrices, therefore the wave function, Ψ , has to be a 4-component spinor. This is because the wave function consists of the particle spin – ½ up and down, made up of two components. But the other two components are made up of the anti-particle spin, – ½ up and down. The 4- component spinor is:
ψ 1 (15) ψ 2 =Ψ ψ 3 ψ 4 Electromagnetic Field Dirac Equation
When a particle is in an electromagnetic field, there are terms that must be accounted for. Due to the interaction of the electromagnetic fields, the energy and
Published by Scholarly Commons, 2014 3 McNair Scholars Research Journal, Vol. 1 [2014], Art. 3
momentum are changed slightly. The generalized Dirac Equation is obtained by making the following substitution:
∂ →iE h − qφ (16) r r ∂t h −∇−→ AqiP (17) r Where q is the charge of the particle, φ is the electric scalar potential, and A is the magneticr vector potential. This vector has X, Y, and Z components. The dot product of the A and the alpha matrices is just like the dot product of momentum and the alpha matrices as explained before. This is so that the first alpha matrix goes with the X components of the momentum andr the X component of the magnetic vector potential. The magnetic vector potential, A , is known as: r (18) = [ ,, AAAA zyx ] With this change the Dirac Equation for a particle in an electromagnetic field becomes: r r h 2 [ ci α K ˆˆ K βα mcAcq ]Ψ+•−∇•− (19) ∂Ψ = ih qφΨ− ∂t Equation 19 can be changed to where the Dirac Equation becomes:
r ∂Ψ h 2 h [ ci ˆ K βα ] =Ψ+Ψ+∇•− iVmc (20) r ∂t ( αφ ˆ K •−= AcqV ) (21)
Solution Using Fourier Transform
From looking at the steady state (time independent) Dirac Equation for a particle in an electromagnetic field, you can use Fourier transform to get the solution. Starting with: r h 2 [ ciEV ˆ K βα mc ]Ψ−∇•+=Ψ (22) The 3-D Fourier transform then becomes:
1 ∞ r r (23) χ r = ψ ⋅− xsi 3xde sj )( 2/3 ∫∫ ∫ j ()2π ∞− where j = 1,2,3,4
https://commons.erau.edu/mcnair/vol1/iss1/3 4 Bell: Solving the Dirac Equation: Using Fourier Transform
Next the gradient of the 3-D Fourier transform is developed into: r 3 r r r r r χ sj )( ( ) sj =−=⋅∇ sisi χχ sj )()( (24) From plugging equation 24 into equation 22, the Dirac Equation then can be converted into: r h ˆ 2 r [ ( K )−•− mcscE ]χβα sj )( = sF )( (25) 1 ∞ r r sF )( = ⋅− xsi 3 xdev (26) 2/3 ∫∫ ∫ j ()2π ∞− r Since the wave function is the 4 component spinor, then χ sj )( and sF )( also have to be 4 components. There then is a lot of algebra that ultimately leads to:
−mcE 2 χ χ ( ) 1 1 (27) 2 3 ()−mcE χ O σK r χ2 2 −hc •s 2 ∑ K ()+mcE χ3 K=1σK O χ3 2 ()+mcE χ4 χ4 F 1 s)( F = 2 s)( F3 s)( F 4 s)( r r This system can be simplified down to a D matrix times χ sj )( , which looks as follows:
χ F 1 1 s)( (28) rχ F D 2 = 2 s)( F χ3 3 s)( χ F 4 4 s)( 2 ( −mcE ) (29) 2 3 r ()−mcE O σK r −hc • = Ds 2 ∑ K ()+mcE K=1σK O ()+mcE 2 r From taking the inverse D matrix, equation 28 becomes:
Published by Scholarly Commons, 2014 5 McNair Scholars Research Journal, Vol. 1 [2014], Art. 3
χ F 1 1 s)( (30) χ r F 2 = D−1 2 s)( F χ3 3 s)( χ F 4 4 s)(
r r For D to have an inverse, the determinant of D must exist. If not then this solution does not work. Using the inverse 3-D Fourier transform, the solution can be turned into:
F ψ1 1 s )( (31) ∞ r r r F ψ2 1 2 s )( = xsi De −⋅ 1 3 sd 2/3 ∫∫ ∫ F ψ3 ()2π 3 s )( ∞− ψ F 4 4 s )(
Conclusion
The Dirac Equation for a particle in an electromagnetic field shows us how the particle is affected by the electric field and the magnetic field. Also the Dirac equation tells how the field affects the spin and the polarity of the particle itself. From this solution, one can model the particle and place different potential energies into the equation. There are models that are currently being built to model how the particle would react in Gaussian potential energy. r A reminder ris that the D matrix has to have a determinant for the Fourier solution to work at all. If D does notr have a determinant, then the solution would have to be found another way. But D should always be able to have a determinant; therefore, a solution will always be able to be obtained. A Couple of things that are next are to look at some of the numerical solutions of, first, the 1-D Fourier solution, then 2-D and 3-D numerical solutions, as well as looking at the Dirac Equation in spherical coordinates instead of Cartesian coordinates, as looked at here. Acknowledgements
First want to thank Dr. Bereket Berhane and Dr. Timothy Smith for the guidance and training on the topic. Also for putting in the time to help me when I need and being there for me during this research experience. I would also like to thank Embry-Riddle Aeronautical University McNair Program and the McNair Staff, Dr. Miranda McBride and Ms. Paula Reed, for allowing me to take part in this special program. This program has and continues to allow me to go farther than what I could while in undergraduate.
https://commons.erau.edu/mcnair/vol1/iss1/3 6 Bell: Solving the Dirac Equation: Using Fourier Transform
Also thanks to the other physics teacher at Embry-Riddle that put in time and/or allowed me to use their books for this research.
References 1.) Greiner, W., Relativistic Quantum Mechanics. Wave Equations, Springer; 3rd Edition, Berlin, Germany, 1994.
2.) Shu, F. H., The Physics of Astrophysics Volume I: Radiation, University Science Books, Mill Valley, CA, April 1991.
3.) Smith, T. A., “Some Notes and Theorems on Modern Relativistic Quantum Theory,” Master Thesis, Department of Mathematics and Statistics, West Florida University, Pensacola, FL, 2002.
Published by Scholarly Commons, 2014 7