Dirac equation: Difference between revisions

From PhyWiki
Jump to navigation Jump to search
 
(15 intermediate revisions by 3 users not shown)
Line 87: Line 87:
==Free particle solution==
==Free particle solution==


Substituting <math>\psi(x,y,z,t)=\psi (\vec r, t)=\psi _{0}(\vec r)e^{(-i/\hbar)Et}</math> into (), we get time-dependent Dirac equation:
Substituting <math>\psi(x,y,z,t)=\psi (\vec r, t)=\psi _{0}(\vec r)e^{(-i/\hbar)Et}</math> into (5), we get time-dependent Dirac equation:


<math>E \psi_{0}(\vec r)=(c \vec \alpha \vec p +mc^2 \beta)\psi_{0}(\vec r)</math>
<math>E \psi_{0}(\vec r)=(c \vec \alpha \vec p +mc^2 \beta)\psi_{0}(\vec r)</math>
Line 117: Line 117:
<math>\left|\begin{array}{cc}E-mc^2&-c \vec \sigma \vec p\\-c \vec \sigma \vec p&E+mc^2\end{array}\right|=0</math>
<math>\left|\begin{array}{cc}E-mc^2&-c \vec \sigma \vec p\\-c \vec \sigma \vec p&E+mc^2\end{array}\right|=0</math>


<math>\Rightarrow E^2=c^2 \vec p \; ^{2} +m^2c^4 \Rightarrow E_{\pm} = \pm c \sqrt {\vec p \; ^{2} +m^2c^2}</math>
<math>\Rightarrow E^2=c^2 \vec p \; ^{2} +m^2c^4 \Rightarrow E_{\pm} = \pm c \sqrt {\vec p \; ^{2} +m^2c^2} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (14)</math>


So, for a given value of momentum there are two values of energy one with positive sign the other with negative sign.
So, for a given value of momentum there are two values of energy one with positive sign the other with negative sign.
Line 135: Line 135:
General solution for free particles is as follows:
General solution for free particles is as follows:


<math>\psi (\vec r,t)= \int u_{E_{-}}(A,B)e^{i(-Et+\vec p \vec r) \hbar}d \vec p + \int u_{E_{+}}(D,F)e^{i(Et+\vec p \vec r) \hbar}d \vec p</math>
<math>\psi (\vec r,t)= \int u_{E_{-}}(A,B)e^{i(-Et+\vec p \vec r) \hbar}d \vec p + \int u_{E_{+}}(D,F)e^{i(Et+\vec p \vec r) \hbar}d \vec p \qquad \qquad \qquad \qquad \qquad (15)</math>


A,B,C,D can be determined from initial conditions.
A,B,C,D can be determined from initial conditions.
Line 155: Line 155:


==Spin operators==
==Spin operators==
Let the spin matrix <math>\vec{\sigma} \prime</math> be defined as
<math>\vec{\sigma} \prime = \left(\begin{array}{cc}  \vec{\sigma}&0\\ 0& \vec{\sigma}\end{array}\right)</math>
and the spin-1/2 operator <math>\vec S</math> be defined as
<math>\vec S = \frac{\hbar}{2} \vec{\sigma} \prime</math>
Using the spin-1/2 operator, we can determine many things from the Dirac equation. One of the more important things we can deduce is the g-factor that arises in the presence of a magnetic field. Starting with the time-independent Dirac equation in an electromagnetic field we can obtain
<math>[(E - e\phi)^2 - (c\vec p - e\vec A)^2 - m^2 c^4 + e\hbar c \vec{\sigma} \prime \cdot \vec H + ... ]\psi = 0</math>
where <math>\vec H = \vec{\nabla} \times \vec A</math>. Now let <math>E\prime = E - mc^2</math>. This means that <math>(E - e\phi )^2 - m^2 c^4 \approx 2mc^2 (E\prime - e\phi)</math> plus other terms much less than <math>\displaystyle{mc^2}</math>. Thus,
<math>\Rightarrow E\prime \psi \approx [\frac{1}{2m}(\vec p - \frac{e}{c} \vec A )^2 +e\phi -2 \frac{e}{2mc} \vec S \cdot \vec H ] \psi</math>
We can see here that since <math>\frac{e}{2mc} = \mu _B</math>, the value of 2 is given to us as the gyromagnetic ratio.
Another thing we can determine from the Dirac equation is the properties of spin-orbit coupling. First, let <math>e\phi (\vec r) \rightarrow V(\vec r)</math>. Simplifying our problem by allowing <math> \psi = \left(\begin{array}{c} \psi _1 \\ \psi _2 \end{array}\right)</math>, where both <math>\displaystyle{\psi _1}</math> and <math>\displaystyle{\psi _2}</math> have two entries each, and using <math>E = E\prime + mc^2</math>, the Dirac equation tells us that
<math>(E\prime - V)\psi _1 - c\vec{\sigma} \cdot \vec p \psi _2 = 0</math>
<math>(E\prime + 2mc^2 - V)\psi _2 - c\vec{\sigma} \cdot \vec p \psi _1 =0</math>
<math>\Rightarrow \psi _2 = \frac{c\vec{\sigma} \cdot \vec p}{E\prime +2mc^2 -V} \psi _1</math>
<math>\Rightarrow E\prime \psi _1 = \frac{1}{2m}(\vec{\sigma} \cdot \vec p)[1+\frac{E\prime -V}{2mc^2}]^{-1}(\vec{\sigma} \cdot \vec p) \psi _1 + V \psi _1</math>
Now lets simplify this expression a bit. We can expand the inverse factor here so that <math>[1+\frac{E\prime - V}{2mc^2}]^{-1} \approx 1-\frac{E\prime - V}{2mc^2}</math>. We also know that <math>\vec p V = V\vec p - i\hbar \vec{\nabla V}</math>, <math>(\vec{\sigma} \cdot \vec p)(\vec{\sigma} \cdot \vec p) = p^2</math>, and <math>(\vec{\sigma} \cdot \vec{\nabla V})(\vec{\sigma} \cdot \vec{p}) = \vec{\nabla V} \cdot \vec{p} + i\vec{\sigma} \cdot [\vec{\nabla V} \times \vec{p}]</math>. Thus
<math>\Rightarrow E\prime \psi _1 = [(1-\frac{E\prime - V}{2mc^2})\frac{p^2}{2m} + V]\psi _1 -\frac{\hbar ^2}{4m^2 c^2} \vec{\nabla V} \cdot \vec{\nabla \psi _1} + \frac{\hbar}{4m^2 c^2}\vec{\sigma} \cdot [\vec{\nabla V} \times \vec{p} \psi _1]</math>
Assuming <math>\displaystyle{V(\vec{r})}</math> is a central potential, we can further simplify this expression by noting that <math>\vec{\nabla V} \cdot \vec{\nabla} = \frac{dV}{dr}\frac{\partial}{\partial r}</math> and <math>\vec{\nabla V} = \frac{1}{r}\frac{dV}{dr}\vec{r}</math>. We also take note that <math>\displaystyle{E\prime - V \sim \frac{p^2}{2m}}</math>. Thus
<math>\Rightarrow E\prime \psi _1 = [\frac{p^2}{2m} - \frac{p^4}{8m^3 c^2} + V - \frac{\hbar ^2}{4m^2 c^2}\frac{dV}{dr}\frac{\partial}{\partial r} + \frac{\hbar}{4m^2 c^2}\frac{1}{r}\frac{dV}{dr}\vec{\sigma} \cdot (\vec{r} \times \vec{p})]\psi _1</math>
Finally, noting that <math>\vec{\sigma} \cdot (\vec{r} \times \vec{p}) = \frac{2}{\hbar}\vec{S_0} \cdot \vec{L}</math>, where <math>\displaystyle{\vec{S_0}}</math> is the spin-1/2 operator for <math>\displaystyle{\vec{\sigma}}</math>, we can finally see that
<math>\Rightarrow E\prime \psi _1 = [...+\frac{1}{2m^2 c^2}\frac{1}{r}\frac{dV}{dr}\vec{S_0} \cdot \vec{L}]\psi _1</math>
Thus, the Dirac equation tells us how spin and the angular momentum of the particle interact in the relativistic limit.


==Dirac hydrogen atom==
==Dirac hydrogen atom==
As with the Klein-Gordon equation, the Dirac equation can be solved exactly for the Coulomb potential. Starting with definitions, for the Coloumb potential, <math>\displaystyle{V(\vec{r}) = -\frac{Ze^2}{r}}</math>. Also, let <math>p_r = -i\hbar \frac{1}{r}\frac{\partial}{\partial r} r</math>, <math>\alpha _r = (\vec{\alpha} \cdot \frac{\vec{r}}{r})</math>, <math>\hbar k = \beta(\vec{\sigma} \prime \cdot \vec{L} + \hbar )</math>, and <math>\vec{\alpha} \cdot \vec{p} = \alpha _r p_r + \frac{i\hbar}{r} \alpha _r \beta k</math>. Using the condition that <math>\displaystyle{E^2 = c^2 p^2 + m^2 c^4}</math> in order to find conditions for these new components, we see that <math>\alpha _r ^2 = \beta ^2 = 1</math> and <math>\displaystyle{\{ \alpha _r , \beta \} = 0}</math>. Thus,
<math>H_D = c\alpha _r p_r + \frac{i\hbar c}{r} \alpha _r \beta k + \beta mc^2 + V</math>
It is easy to see that <math>\displaystyle{[k, H_D] = 0}</math>. Also, notice that <math>\hbar ^2 k^2 =(\vec{\sigma}\prime \cdot \vec{L})^2 + 2\hbar (\vec{\sigma}\prime \cdot \vec{L}) + \hbar ^2 = (\vec{L} \cdot \vec{S})^2 +\frac{\hbar ^2}{4} = \hbar ^2 j(j+1) +\frac{\hbar ^2}{4} = \hbar ^2 (j+\frac{1}{2})^2</math>, which implies that <math>\displaystyle{|k|=j+\frac{1}{2}}</math>. With this in mind, we use the simplification <math>\psi = \left(\begin{array}{c} \psi _1 \\ \psi _2 \end{array} \right)</math>, for which the radial components of <math>\displaystyle{\psi _1}</math> and <math>\displaystyle{\psi _2}</math> are <math>\frac{F(\rho )}{\rho}</math> and <math>\frac{G(\rho )}{\rho}</math>, where <math>\rho = \frac{1}{\hbar c}(m^2 c^4 - E^2 )^{\frac{1}{2}}r</math>. We solve these radial equations in the same manner as the Schrodinger or Klein-Gordon equation: finding their asymptotic properties(<math>e^{-\frac{\rho}{2}}</math> for <math>\displaystyle{\rho \rightarrow \infty}</math> and <math>\displaystyle{\rho ^s}</math> for <math>\displaystyle{\rho \rightarrow 0}</math>, where <math>\displaystyle{s=\sqrt{k^2 - \gamma ^2}}</math> and <math>\gamma = \frac{Ze^2}{\hbar c}</math>), applying these asymptotic properties to a power series expansion, using these adjusted power series expansions in the radial equation, and finding a recursion relation between the coefficients of the power series expansion. In the end, we find that the energies are given by
<math>E = mc^2 [1+\frac{\gamma ^2}{(s+N)^2}]^{-\frac{1}{2}}</math>, where N is a constant.
Expanding the root in this energy, we find
<math>E = mc^2 - \frac{Ry}{n^2} - \frac{mc^2 \gamma ^4}{2n^4}(\frac{n}{j+\frac{1}{2}} - \frac{3}{4}) + ...</math>, where <math>\displaystyle{n=N+j+\frac{1}{2}}</math>. Thus, we find that the Dirac equation yields the rest mass energy, non-relativistic energy, and the correct fine-structure correction, in contrast to the Klein-Gordon equation. (For more detail in the evaluation of this problem, see "[[Klein-Gordon_equation]] with Coulomb potential" as an example.)
==Dirac's Hole Theory==
From the solution of the Dirac Eqn for a free particle we find that the energy eigen value can be both positive and negative,
E = <math>\pm \sqrt{p^{2}c^{2} + m^{2}c^{4}}</math> = <math>\pm E_{p}</math>
So in the presence of an external electric field an electron can make a transition to the negative energy state. Sonce there is no ground state the electron can goto E=<math> - \infty </math> state. If this is so, we shall have a continuous emission of radtion which is not not observed physically. Therefore the theory so far developed is in difficulty to understand.
To remove these difficulties Dirac introduced some excellent ideas which are known as the Dirac's Hole Theory. According to the hole theory the vacuum or empty state is a sea of negative energy states which are completely filled with negative energy electrons. So due to the Paul's exclusion principle electrons cannot jump to the lower states, so no radiation will be seen.
Since the sea of negative energy states are completely by the negative energy electrons, the charge density would be infinite but is not observed physically. So in the 2nd Postulate Dirac states that electrons in the negative energy states will not exibhit its charge, momentum, field, intrinsic spin etc. But it can interact with external field.
Now according to the hole theory the positive and negative energy states are placed symmetrically about E = 0 and an electron in the negative state will be excited to the positive state if its energy is E= <math>\geq  2mc^{2}</math> when an electron from the negative energy goes to the positive energy state it will exhibit all of its properties such charge, momentum, field, intrinsic spin etc. At the same time it will create a hole in the negative energy state. After transition the charge and energy of the vacuum would be,
<math>Q= Q_{vac} - (-e)</math> = <math>Q_{vac} +|e|</math>
<math>E= E_{vac} - (-E_{p})</math> = <math>E_{vac} +E_{p}</math>
So it is now clear that the hole has a charge of <math>|e|</math> and energy <math>E_{p}</math>.
That is a transition of an electron from a negative energy state creates a particle pf charge <math>|e|</math> and energy of <math>E_{p}</math>. The particle is known as the hole or positron.
On the other hand if an energy level in the negative energy sea remains empty, an electron from the positive energy state can jump into it and both the hole and the electron will disappear. But a radiation would be emitted due to the transition from the higher energy to the lower energy state. Therefore pair production and annihilation can be explained with the help of this hole theory.
A perfect vacuum is region where all the states of positive energy are unoccupied and all those of negative energy are occupied. In a perfect vacuum, Maxwell's equation
<math>\bigtriangledown .E</math>= 0
must of course be valid. This means that the infinite distribution of negative energy electrons does not contribute to the electron field, Only the departure from the distribution in a vacuum will contribute to the electric density <math>\rho_{0}</math> in the Maxwell's equation,
<math>\bigtriangledown .E</math>= <math>\rho_{0}</math>
Thus there will be a contribution -e for each occupied state of positive energy and contribution of e for each unoccupied state of negative energy.
So the Hole theory results to a new fundamental symmetry in nature''To each particle there is an antiparticle and in particular the existence of electrons implies the existence of positrons.''

Latest revision as of 03:05, 22 April 2010

How to construct

Starting from the relativistic relation between energy and momentum:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^2=\vec p \; ^{2}c^2+m^2c^4}

or Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=c\sqrt{p^2+m^2c^2}}

From this equation we can not directly replace Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E, \vec p} by the corresponding operators since we don't have the definition for the square root of an operator. Therefore, first we need to linearize this equation as follows:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=c\sqrt{p^2+m^2c^2}=c\sqrt{(p_{x}^2+p_{y}^2+p_{z}^2)+m^2c^2}=c(\alpha _{x}p_{x}+\alpha _{y}p_{y}+\alpha _{z}p_{z})+\beta mc^2}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \alpha _{x},\alpha _{y},\alpha _{z}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \beta} are some operators independent of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec p} .

From this it follows that:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c^2(p_{x}^2+p_{y}^2+p_{z}^2+m^2c^2)=[c(\alpha _{x}p_{x}+\alpha _{y}p_{y}+\alpha _{z}p_{z})+\beta mc^2] . [c(\alpha _{x}p_{x}+\alpha _{y}p_{y}+\alpha _{z}p_{z})+\beta mc^2]}

Expanding the right hand side and comparing it with the left hand side, we obtain the following conditions for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \alpha _{x},\alpha _{y},\alpha _{z}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \beta}  :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha _{i}^2=\beta ^2=1 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ (1)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \alpha_ {i}\alpha_ {j}+\alpha_ {j}\alpha_ {i}=\{\alpha_ {i},\alpha_ {j}\}=2\delta_{ij} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \ \ \ (2)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \alpha_ {i} \beta+\beta \alpha_ {i}=\{\alpha_ {i},\beta\}=0 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \ \ \ \ (3)}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i=1,2,3} corresponds to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x, y, z}

In order to describe both particle (positive energy state) and antiparticle (negative energy state); spin-up state and spin-down state, the wave function must have 4 components and all operators acting on such states correspond to 4x4 matrices. Therefore, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \alpha _{x},\alpha _{y},\alpha _{z}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \beta} are 4x4 matrices. It is convention that these matrices are given as follows (in the form of block matrices for short):

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha_{x}=\left(\begin{array}{cc}0& \sigma_{x}\\ \sigma_{x}&0\end{array}\right)} ; Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \qquad \alpha_{y}=\left(\begin{array}{cc}0& \sigma_{y}\\ \sigma_{y}&0\end{array}\right)} ; Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \qquad \alpha_{z}=\left(\begin{array}{cc}0& \sigma_{z}\\ \sigma_{z}&0\end{array}\right)} ; Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \qquad \beta= \left(\begin{array}{cc}1&0\\0&-1\end{array}\right) \qquad (4)}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma_{x}, \;\sigma_{y}, \;\sigma_{z}} are 2 by 2 Pauli matrices.

Let's define:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec \alpha=\alpha _{x} \hat x+\alpha _{y} \hat y+\alpha _{z} \hat z}

Then we can write:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=c \vec \alpha \vec p +\beta mc^2}

Substituting all quantities by their corresponding operators, we obtain Dirac equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i \hbar \frac {\partial \psi}{\partial t}=H_{D} \psi \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (5)}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_{D}=c \vec \alpha \vec p + \beta mc^2}

Dirac equation can also be written explicitly as follows:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i \hbar \frac {\partial \psi_{1}}{\partial t}=c(p_{x}-ip_{y}) \psi _{4}+cp_{z} \psi _{3} + mc^2 \psi _{1} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (6)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i \hbar \frac {\partial \psi_{2}}{\partial t}=c(p_{x}+ip_{y}) \psi _{3}-cp_{z} \psi _{4} + mc^2 \psi _{2} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (7)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i \hbar \frac {\partial \psi_{3}}{\partial t}=c(p_{x}-ip_{y}) \psi _{2}+cp_{z} \psi _{1} - mc^2 \psi _{3} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (8)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i \hbar \frac {\partial \psi_{4}}{\partial t}=c(p_{x}+ip_{y}) \psi _{1}-cp_{z} \psi _{2} - mc^2 \psi _{4} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (9)}

In the present of electromagnetic field, Dirac equation becomes:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (i \hbar \frac {\partial }{\partial t} -e \phi) \psi = \left [ c \vec \alpha (\frac {\hbar}{i} \vec \nabla - \frac {e}{c} \bold A)+\beta mc^2 \right ]\psi \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \ \ \ \ (10)}

Continuity equation

Dirac equation and its adjoint equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i \hbar \frac {\partial \psi}{\partial t}=(-i \hbar c \vec \alpha \vec \nabla + mc^2 \beta) \psi}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -i \hbar \frac {\partial \psi ^{\dagger}}{\partial t}=(i \hbar c \vec \nabla \psi ^{\dagger} \vec \alpha + mc^2 \psi ^{\dagger} \beta )}

Multiplying Dirac equation by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi ^{\dagger}} from the left and the adjoint equation by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \psi} from the right, we get:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i \hbar \psi ^{\dagger} \frac {\partial \psi}{\partial t}=-i \hbar c \psi ^{\dagger} \vec \alpha \vec \nabla \psi+ mc^2 \psi ^{\dagger} \beta \psi}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -i \hbar \frac {\partial \psi ^{\dagger}}{\partial t} \psi=i \hbar c \vec \nabla \psi ^{\dagger} \vec \alpha \psi+ mc^2 \psi ^{\dagger} \beta \psi}

Subtracting one from the other, we get:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i \hbar \left ( \psi ^{\dagger} \frac {\partial \psi}{\partial t} + \frac {\partial \psi}{\partial t} \psi ^{\dagger} \right )=-i \hbar c \left [ \psi ^{\dagger} \vec \alpha \vec \nabla \psi + \vec \nabla \psi ^{\dagger} \vec \alpha \psi \right ] = -i \hbar c \vec \nabla \left ( \psi ^{\dagger} \vec \alpha \psi \right )}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \frac {\partial}{\partial t} \left ( \psi ^{\dagger} \psi \right )+ \vec \nabla \left ( c \psi ^{\dagger} \vec \alpha \psi \right ) = 0 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (11)}

Therefore, we can define:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho = \psi ^{\dagger} \psi} as probability density (12)

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec j = c \left ( \psi ^{\dagger} \vec \alpha \psi \right )} as probability current density (13)

Free particle solution

Substituting Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x,y,z,t)=\psi (\vec r, t)=\psi _{0}(\vec r)e^{(-i/\hbar)Et}} into (5), we get time-dependent Dirac equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E \psi_{0}(\vec r)=(c \vec \alpha \vec p +mc^2 \beta)\psi_{0}(\vec r)}

Let's seek for the plane wave solutions with momentum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec p} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{0}(\vec r)=u e^{(i/ \hbar) \vec p \vec r}}

u satisfies the following equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Eu=(c \vec \alpha \vec p +mc^2 \beta)u}

u can be written as follows:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u= \left[\begin{array}{cc}u_{1}\\u_{2}\\u_{3}\\u_{4}\end{array}\right]=\left(\begin{array}{cc}W\\W'\end{array}\right)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W= \left(\begin{array}{cc}u_{1}\\u_{2}\end{array}\right) \qquad W'= \left(\begin{array}{cc}u_{3}\\u_{4}\end{array}\right)}

The equation for u can be rewritten as:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\left(\begin{array}{cc}W\\W'\end{array}\right)= \left [ \left(\begin{array}{cc}0&c \vec \sigma \vec p \\c \vec \sigma \vec p & 0\end{array}\right)+\left(\begin{array}{cc}mc^2&0 \\0 & -mc^2\end{array}\right)\right ]\left(\begin{array}{cc}W\\W'\end{array}\right)}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{\begin{array}{cc}(E-mc^2)W-c \vec \sigma \vec p W'=0&\\-c \vec \sigma \vec p W+(E+mc^2)W'=0\end{array}\right.}

Condition for non-trivial solutions:

So, for a given value of momentum there are two values of energy one with positive sign the other with negative sign.

Substituting the value of energy into equation for W and W' yields :

Choosing the momentum along z direction, the wave functions can be written as follows:

General solution for free particles is as follows:

A,B,C,D can be determined from initial conditions.

Nonrelativistic limit

In this limit and:

Four independent solutions of Dirac equation can be chosen as follows:

Spin operators

Let the spin matrix be defined as

and the spin-1/2 operator be defined as

Using the spin-1/2 operator, we can determine many things from the Dirac equation. One of the more important things we can deduce is the g-factor that arises in the presence of a magnetic field. Starting with the time-independent Dirac equation in an electromagnetic field we can obtain

where . Now let . This means that plus other terms much less than . Thus,

We can see here that since , the value of 2 is given to us as the gyromagnetic ratio.

Another thing we can determine from the Dirac equation is the properties of spin-orbit coupling. First, let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e\phi (\vec r) \rightarrow V(\vec r)} . Simplifying our problem by allowing Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi = \left(\begin{array}{c} \psi _1 \\ \psi _2 \end{array}\right)} , where both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle{\psi _1}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle{\psi _2}} have two entries each, and using Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E = E\prime + mc^2} , the Dirac equation tells us that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (E\prime - V)\psi _1 - c\vec{\sigma} \cdot \vec p \psi _2 = 0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (E\prime + 2mc^2 - V)\psi _2 - c\vec{\sigma} \cdot \vec p \psi _1 =0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \psi _2 = \frac{c\vec{\sigma} \cdot \vec p}{E\prime +2mc^2 -V} \psi _1}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow E\prime \psi _1 = \frac{1}{2m}(\vec{\sigma} \cdot \vec p)[1+\frac{E\prime -V}{2mc^2}]^{-1}(\vec{\sigma} \cdot \vec p) \psi _1 + V \psi _1}

Now lets simplify this expression a bit. We can expand the inverse factor here so that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [1+\frac{E\prime - V}{2mc^2}]^{-1} \approx 1-\frac{E\prime - V}{2mc^2}} . We also know that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec p V = V\vec p - i\hbar \vec{\nabla V}} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\vec{\sigma} \cdot \vec p)(\vec{\sigma} \cdot \vec p) = p^2} , and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\vec{\sigma} \cdot \vec{\nabla V})(\vec{\sigma} \cdot \vec{p}) = \vec{\nabla V} \cdot \vec{p} + i\vec{\sigma} \cdot [\vec{\nabla V} \times \vec{p}]} . Thus

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow E\prime \psi _1 = [(1-\frac{E\prime - V}{2mc^2})\frac{p^2}{2m} + V]\psi _1 -\frac{\hbar ^2}{4m^2 c^2} \vec{\nabla V} \cdot \vec{\nabla \psi _1} + \frac{\hbar}{4m^2 c^2}\vec{\sigma} \cdot [\vec{\nabla V} \times \vec{p} \psi _1]}

Assuming Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle{V(\vec{r})}} is a central potential, we can further simplify this expression by noting that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{\nabla V} \cdot \vec{\nabla} = \frac{dV}{dr}\frac{\partial}{\partial r}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{\nabla V} = \frac{1}{r}\frac{dV}{dr}\vec{r}} . We also take note that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle{E\prime - V \sim \frac{p^2}{2m}}} . Thus

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow E\prime \psi _1 = [\frac{p^2}{2m} - \frac{p^4}{8m^3 c^2} + V - \frac{\hbar ^2}{4m^2 c^2}\frac{dV}{dr}\frac{\partial}{\partial r} + \frac{\hbar}{4m^2 c^2}\frac{1}{r}\frac{dV}{dr}\vec{\sigma} \cdot (\vec{r} \times \vec{p})]\psi _1}

Finally, noting that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{\sigma} \cdot (\vec{r} \times \vec{p}) = \frac{2}{\hbar}\vec{S_0} \cdot \vec{L}} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle{\vec{S_0}}} is the spin-1/2 operator for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle{\vec{\sigma}}} , we can finally see that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow E\prime \psi _1 = [...+\frac{1}{2m^2 c^2}\frac{1}{r}\frac{dV}{dr}\vec{S_0} \cdot \vec{L}]\psi _1}

Thus, the Dirac equation tells us how spin and the angular momentum of the particle interact in the relativistic limit.

Dirac hydrogen atom

As with the Klein-Gordon equation, the Dirac equation can be solved exactly for the Coulomb potential. Starting with definitions, for the Coloumb potential, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle{V(\vec{r}) = -\frac{Ze^2}{r}}} . Also, let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_r = -i\hbar \frac{1}{r}\frac{\partial}{\partial r} r} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha _r = (\vec{\alpha} \cdot \frac{\vec{r}}{r})} , , and . Using the condition that in order to find conditions for these new components, we see that and . Thus,

It is easy to see that . Also, notice that , which implies that . With this in mind, we use the simplification , for which the radial components of and are and , where . We solve these radial equations in the same manner as the Schrodinger or Klein-Gordon equation: finding their asymptotic properties( for and for , where and ), applying these asymptotic properties to a power series expansion, using these adjusted power series expansions in the radial equation, and finding a recursion relation between the coefficients of the power series expansion. In the end, we find that the energies are given by

, where N is a constant.

Expanding the root in this energy, we find

, where . Thus, we find that the Dirac equation yields the rest mass energy, non-relativistic energy, and the correct fine-structure correction, in contrast to the Klein-Gordon equation. (For more detail in the evaluation of this problem, see "Klein-Gordon_equation with Coulomb potential" as an example.)


Dirac's Hole Theory

From the solution of the Dirac Eqn for a free particle we find that the energy eigen value can be both positive and negative,

E = =

So in the presence of an external electric field an electron can make a transition to the negative energy state. Sonce there is no ground state the electron can goto E= state. If this is so, we shall have a continuous emission of radtion which is not not observed physically. Therefore the theory so far developed is in difficulty to understand.

To remove these difficulties Dirac introduced some excellent ideas which are known as the Dirac's Hole Theory. According to the hole theory the vacuum or empty state is a sea of negative energy states which are completely filled with negative energy electrons. So due to the Paul's exclusion principle electrons cannot jump to the lower states, so no radiation will be seen.

Since the sea of negative energy states are completely by the negative energy electrons, the charge density would be infinite but is not observed physically. So in the 2nd Postulate Dirac states that electrons in the negative energy states will not exibhit its charge, momentum, field, intrinsic spin etc. But it can interact with external field.

Now according to the hole theory the positive and negative energy states are placed symmetrically about E = 0 and an electron in the negative state will be excited to the positive state if its energy is E= when an electron from the negative energy goes to the positive energy state it will exhibit all of its properties such charge, momentum, field, intrinsic spin etc. At the same time it will create a hole in the negative energy state. After transition the charge and energy of the vacuum would be,

=

=

So it is now clear that the hole has a charge of and energy .

That is a transition of an electron from a negative energy state creates a particle pf charge and energy of . The particle is known as the hole or positron.

On the other hand if an energy level in the negative energy sea remains empty, an electron from the positive energy state can jump into it and both the hole and the electron will disappear. But a radiation would be emitted due to the transition from the higher energy to the lower energy state. Therefore pair production and annihilation can be explained with the help of this hole theory.

A perfect vacuum is region where all the states of positive energy are unoccupied and all those of negative energy are occupied. In a perfect vacuum, Maxwell's equation

= 0

must of course be valid. This means that the infinite distribution of negative energy electrons does not contribute to the electron field, Only the departure from the distribution in a vacuum will contribute to the electric density in the Maxwell's equation,

=

Thus there will be a contribution -e for each occupied state of positive energy and contribution of e for each unoccupied state of negative energy.

So the Hole theory results to a new fundamental symmetry in natureTo each particle there is an antiparticle and in particular the existence of electrons implies the existence of positrons.

Retrieved from "http://wiki.physics.fsu.edu/index.php?title=Dirac_equation&oldid=7759"