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Radial Schrödinger and Dirac Equations

Variational Formulation of the Schrödinger equation

Lagrangian is:

\L(\psi) = \half (\nabla \psi)^2 + V(x) \psi^2(x)

Subject to the normalization constrain:

N[\psi] = \int|\psi(x)|^2 \d^3 x - 1 = 0

The action is:

S[\psi] = \int \L \d^3 x

Variating it (subject to the normalization condition) we get:

0 = \delta (S - \epsilon N) = \delta\int\half (\nabla \psi)^2 + V(x) \psi^2(x) \d^3x - \epsilon \left(\int|\psi(x)|^2 \d^3 x - 1\right) = = \int (\nabla \psi)\cdot(\nabla\delta\psi) + 2 V \psi \delta \psi -2\epsilon\psi\delta\psi\d^3 x = 2\int \left(-\half\nabla^2 \psi + V \psi - \epsilon\psi\right) \delta \psi \d^3 x + \int ({\bf n}\cdot\nabla\psi) \delta \psi\, \d^2 x

Which gives the Schrödinger equation assuming the surface integral vanishes.

Note: to apply the variation \delta correctly, one uses the definition:

\delta F[\psi] \equiv \left.{\d\over\d\epsilon}F[\psi + \epsilon \delta\psi] \right|_{\epsilon=0}

Weak Formulation

The weak formulation is obtained from the above by substituting \delta\psi \to v (the test function) so we get:

\int \half(\nabla \psi)\cdot(\nabla v) + V \psi v - \epsilon\psi v\,\d^3 x

Radial Schrödinger equation

There are two ways to obtain the radial Schrödinger equation. Either from the Lagrangian, or from the equation itself.

From the Equation

-{1\over2}\nabla^2\psi({\bf x})+V(r)\psi({\bf x})=E\psi({\bf x})

The way to solve it is to separate the equation into radial and angular parts by writing the Laplace operator in spherical coordinates as:

\nabla^2f = {\partial^2 f\over\partial\rho^2} +{2\over \rho}{\partial f\over\partial\rho} -{L^2 f\over \rho^2}

L^2= -{\partial^2\over\partial\theta^2} -{1\over\sin^2\theta}{\partial^2\over\partial\phi^2} -{1\over\tan\theta}{\partial\over\partial\theta}

Substituting \psi({\bf x})=R(\rho)Y(\theta,\phi) into the Schrödinger equation yields:

-{1\over2}\nabla^2(RY)+VRY=ERY

-{1\over2}R''Y-{1\over\rho}R'Y+{L^2RY\over2\rho^2}+VRY=ERY

Using the fact that L^2Y=l(l+1)Y we can cancel Y and we get the radial Schrödinger equation:

-{1\over2}R''-{1\over\rho}R'+{l(l+1)R\over2\rho^2}+VR=ER

Normalization:

1 = \int |\psi|^2 \d^3 x = \int R^2 |Y|^2 \d^3 x = \int R^2 |Y|^2 \rho^2\d\Omega\d \rho = \int R^2 \rho^2 \d\rho \int |Y|^2 \d\Omega = \int R^2 \rho^2 \d \rho

From the Lagrangian

We need to convert the Lagrangian to spherical coordinates. In order to easily make sure we do things covariantly, we start from the action (which is a scalar):

S[\psi] = \int \half (\nabla \psi)^2 + V(x) \psi^2(x) \, \d^3 x = = \int (\half (\nabla (RY))^2 + V (RY)^2 )\rho^2\d \rho \d\Omega = = \int (\half (R'^2Y^2 + R^2(\nabla Y)^2 + 2RR'({\bf\hat\rho}Y)\cdot\nabla Y) + V (RY)^2 )\rho^2\d \rho \d\Omega = = \int \left(\half \left(R'^2 + R^2{l(l+1)\over\rho^2}\right) + V R^2\right)\rho^2\d \rho = = \int \half \rho^2 R'^2 + (\rho^2 V + \half l(l+1)) R^2\,\d \rho =

where we used the following properties of spherical harmonics:

\int Y^2\d\Omega = 1 \int \rho^2 (\nabla Y)^2\d\Omega = l(l+1) (Y{\bf \hat \rho})\cdot(\rho \nabla Y) = 0

We now minimize the action (subject to the normalization \int \rho^2 R^2\d\rho = 1) to obtain the radial equation:

0 = \delta (S - \epsilon N) = \delta \int \half \rho^2 R'^2 + (\rho^2 V + \half l(l+1)) R^2 - \epsilon \rho^2R^2 \,\d \rho = = 2\int \half \rho^2 R'(\delta R)' + (\rho^2 V + \half l(l+1)) R\delta R - \epsilon \rho^2 R\delta R \,\d \rho = = 2\int \left( (-\half \rho^2 R')' + (\rho^2 V + \half l(l+1)) R - \epsilon \rho^2 R\right)\delta R \,\d \rho + [\rho^2 R' \delta R]^a_b

So the radial equation is:

(1)(-\half \rho^2 R')' + (\rho^2 V + \half l(l+1)) R = \epsilon \rho^2 R

In agreement with the previous result.

Solving for u=rR

We can also make the substitution u = rR and solve for u:

R = {u\over r} R' = {u'\over r} - {u\over r^2}

and we substitute this to (1):

-\half \left(r^2\left({u'\over r} - {u\over r^2}\right)\right)'+ \left(V + {l(l+1)\over 2 r^2}\right) r u = \epsilon r u -\half ru''+ \left(V + {l(l+1)\over 2 r^2}\right) r u = \epsilon r u -\half u''+ \left(V + {l(l+1)\over 2 r^2}\right) u = \epsilon u

Perturbative Correction to Energy

We introduce P and Q by P(r) = u(r) and Q(r) = P'(r) = u'(r). The radial Schrödinger equation is then:

P'(r) = Q(r) Q'(r) = -2\left(E - V(r) - {l(l+1)\over 2 r^2}\right)P(r)

Let P_1 and Q_1 represent the radial wave function and its derivative at E_1 and P_2, Q_2 at E_2, so the following holds:

Q_1'(r) = -2\left(E_1 - V(r) - {l(l+1)\over 2 r^2}\right)P_1(r) Q_2'(r) = -2\left(E_2 - V(r) - {l(l+1)\over 2 r^2}\right)P_2(r)

Now we evaluate (Q_2 P_1 - P_2 Q_1)' using the relations above:

(Q_2 P_1 - P_2 Q_1)' = Q_2' P_1 + Q_2 P_1' - P_2'Q_1 - P_2 Q_1' = Q_2' P_1 + Q_2 Q_1' - Q_2'Q_1 - P_2 Q_1' = Q_2' P_1 - P_2 Q_1' = 2 (E_1 - E_2) P_1 P_2

We integrate the last formula on the intervals (0, a_c) and (a_c, \infty):

[Q_2 P_1 - P_2 Q_1]^{a_c}_0 = 2 (E_1 - E_2) \int^{a_c}_0 P_1(r) P_2(r) \,\d r [Q_2 P_1 - P_2 Q_1]^\infty_{a_c} = 2 (E_1 - E_2) \int^\infty_{a_c} P_1(r) P_2(r) \,\d r

On the interval (0, a_c) we know the exact solution corresponding to the energies E_1 and E_2 by integrating outwards (the solution will eventually diverge for large r except for the eigenvalues, but we only need it up to a_c) and we know that P_1(0) = P_2(0) = 0, so we get:

Q_2(a_c^-) P_1(a_c^-) - P_2(a_c^-) Q_1(a_c^-) = 2 (E_1 - E_2) \int^{a_c}_0 P_1(r) P_2(r) \,\d r

where a_c^- means that we need the values at a_c when integrating the equation from the left (the value will generally be different when integrating the equation from the right, unless the energy is an eigenvalue). Similarly on the other interval where P_1(\infty) = P_2(\infty)=0:

-(Q_2(a_c^+) P_1(a_c^+) - P_2(a_c^+) Q_1(a_c^+)) = 2 (E_1 - E_2) \int^\infty_{a_c} P_1(r) P_2(r) \,\d r

Taking the sum of the last two expressions:

2 (E_1 - E_2) \int^\infty_0 P_1(r) P_2(r) \,\d r = Q_2(a_c^-) P_1(a_c^-) - P_2(a_c^-) Q_1(a_c^-) -(Q_2(a_c^+) P_1(a_c^+) - P_2(a_c^+) Q_1(a_c^+))

Now we use the fact that P_1(a_c^-) = P_1(a_c^+) and P_2(a_c^-) = P_2(a_c^+), because we match the two solutions from the left and right, so that the function is continuous (it’s derivative will have a jump though):

2 (E_1 - E_2) \int^\infty_0 P_1(r) P_2(r) \,\d r = P_1(a_c) (Q_2(a_c^-)-Q_2(a_c^+)) - P_2(a_c)(Q_1(a_c^-)-Q_1(a_c^+))

By requiring, that the energy E_2 is an eigenvalue, it follows that there is no jump in the derivative, so we set Q_2(a_c^-)=Q_2(a_c^+) and we get:

2 (E_1 - E_2) \int^\infty_0 P_1(r) P_2(r) \,\d r = -P_2(a_c)(Q_1(a_c^-)-Q_1(a_c^+))

that gives us an exact formula for the eigenvalue E_2:

E_2 = E_1 + {P_2(a_c)(Q_1(a_c^-)-Q_1(a_c^+))\over 2\int^\infty_0 P_1(r) P_2(r) \,\d r}

We approximate the value of P_2(a_c) by P_1(a_c) as well as the integral \int^\infty_0 P_1(r) P_2(r) \,\d r by \int^\infty_0 P_1^2(r) \,\d r and we get an approximation for the eigenenergy:

E_2 \approx E_1 + {P_1(a_c)(Q_1(a_c^-)-Q_1(a_c^+))\over 2\int^\infty_0 P_1^2(r) \,\d r}

We use this approximation iteratively until the convergence is achieved (the discontinuity in Q(r) at r=a_c is small enough, or equivalently, the correction to the energy is small enough).

For Dirac equation, one obtains a similar formula:

E_2 \approx E_1 + c {P_1(a_c)(Q_1(a_c^-)-Q_1(a_c^+))\over \int^\infty_0 P_1^2(r) +Q_1^2(r) \,\d r}

So it is just the previous formula multiplied by 2c and the normalization is calculated using both P and Q (as usual for the Dirac equation).

Weak Formulation

The weak formulation is obtained from the action above by substituting \delta R \to v (the test function) so we get:

\int \half \rho^2 R'v' + (\rho^2 V + \half l(l+1)) Rv\,\d\rho = \epsilon \int \rho^2 Rv \,\d \rho

We can also start from the equation itself, multiply by a test function v:

(-\half \rho^2 R')'v + (\rho^2 V + \half l(l+1)) Rv = \epsilon \rho^2 Rv

We integrate it. Normally we need to be using \rho^2\d\rho in order to integrate covariantly, but the above equation was already multiplied by \rho^2 (i.e. strictly speaking, it is not coordinate independent anymore), so we only integrate by \d\rho:

\int (-\half \rho^2 R')'v + (\rho^2 V + \half l(l+1)) Rv \d\rho = \epsilon \int \rho^2 Rv \d\rho

After integration by parts:

\int \half \rho^2 R'v' + (\rho^2 V + \half l(l+1)) Rv \d\rho -\half[\rho^2R'v]_0^a = \epsilon \int \rho^2 Rv \d\rho

Where a is the end of the domain (the origin is at 0). The boundary term is zero at the origin, so we get:

\int \half \rho^2 R'v' + (\rho^2 V + \half l(l+1)) Rv \d\rho +\half\rho^2R'(a)v(a) = \epsilon \int \rho^2 Rv \d\rho

We usually want to have the boundary term \half\rho^2R'(a)v(a) equal to zero. This is equivalent to either letting R'(a) = 0 (we prescribe the zero derivative of the radial wave function at a) or we set v(a)=0 (which corresponds to zero Dirichlet condition for R, i.e. setting R(a)=0).

Weak Formulation for u

\int \half u'v' + \left(V + {l(l+1)\over 2\rho^2}\right) uv \,\d\rho -\half\left[u'v\right]_0^R = \epsilon \int uv \,\d\rho

We prescribe u(0) = u(R) = 0, so we get:

\int \half u'v' + \left(V + {l(l+1)\over 2\rho^2}\right) uv \,\d\rho = \epsilon \int uv \,\d\rho

Dirac Notation

We can also write all the formulas using the Dirac notation:

\one = \int \d\rho \rho^2 \ket{\rho}\bra{\rho} \braket{\rho|\rho'} = {\delta(\rho-\rho')\over \rho^2} \braket{\rho|R} = R(\rho) \braket{\rho|\hat H|R} = {1\over\rho^2}(-\half \rho^2 R')' + (V + \half {l(l+1)\over\rho^2}) R \hat H \ket{R} = E\ket{R}

Then normalization is:

\braket{R|R} = \int \d\rho \rho^2 \braket{R|\rho}\braket{\rho|R} = \int \d\rho \rho^2 R^2(\rho)

The operator \hat H can be written as:

\braket{\rho|\hat H|\rho'} = \braket{\rho|\rho'}\left( -\half{1\over\rho^2}{\d\over\d\rho}\left(\rho^2{\d\over\d\rho}\right) + (V + \half {l(l+1)\over\rho^2}) \right)

so to recover the above formula, we do:

\braket{\rho|\hat H| R} = \int\d\rho'\rho'^2\braket{\rho|\hat H|\rho'} \braket{\rho'|R}= =\int\d\rho'\rho'^2 {\delta(\rho-\rho')\over \rho^2} \left( -\half{1\over\rho^2}{\d\over\d\rho}\left(\rho^2{\d\over\d\rho}\right) + (V + \half {l(l+1)\over\rho^2}) \right) R(\rho') = {1\over\rho^2}(-\half \rho^2 R')' + (V + \half {l(l+1)\over\rho^2}) R

Operator \hat H is symmetric, because:

\int f{1\over\rho^2}(\rho^2 g')' \rho^2\d\rho = \int {1\over\rho^2}(\rho^2 f')'g \rho^2\d\rho

The weak formulation is:

\braket{v|H|R} = E\braket{v|R} \int\d\rho\rho^2 \braket{v|\rho}\braket{\rho|H|R} = E \int\d\rho\rho^2\braket{v|\rho}\braket{\rho|R} \int\d\rho\rho^2 v(\rho)\left( {1\over\rho^2}(-\half \rho^2 R')' + (V + \half {l(l+1)\over\rho^2}) R \right) = E \int\d\rho\rho^2v(\rho)R(\rho)

and we obtain the FE formulation by expanding \ket{R} =\sum_j R_j \ket{j} (note that the basis \ket{j} is not orthogonal, so in particular \sum_j \ket{j}\bra{j} \neq 1):

\sum_j\braket{i|H|j}R_j = E\sum_j\braket{i|j}R_j

This is a generalized eigenvalue problem. In the special case of an orthonormal basis, \braket{i|j} = \delta_{ij} (which FE is not), we get:

\sum_j\braket{i|H|j}R_j = R_i R_i = \braket{i|R}

Which is an eigenvalue problem.

Variational Formulation of the Dirac equation

The QED Lagrangian density is

\L=\bar\psi(i\hbar c\gamma^\mu D_\mu-mc^2)\psi-{1\over4}F_{\mu\nu}F^{\mu\nu}

where:

D_\mu=\partial_\mu+{i\over \hbar}eA_\mu F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu

We will treat the fields as classical fields, so we get the classical wave Dirac equation, after plugging this Lagrangian into the Euler-Lagrange equation of motion:

(i\hbar c\gamma^\mu D_\mu-mc^2)\psi=0 \partial_\nu F^{\nu\mu}=-ec\bar\psi\gamma^\mu\psi

Notice that the Lagrangian happens to be zero for the solution of Dirac equation (e.g. the extremum of the action). This has nothing to do with the variational principle itself, it’s just a coincindence.

In this section we are only interested in the Dirac equation, so we write the Lagrangian as:

\L=\bar\psi(i\hbar c\gamma^\mu D_\mu-mc^2)\psi = =\psi^\dag\gamma^0(i\hbar c\gamma^\mu D_\mu-mc^2)\psi= =\psi^\dag\gamma^0(i\hbar c\gamma^0(\partial_0+{i\over\hbar}eA_0)+ic\gamma^i (\partial_i+{i\over\hbar}eA_i)-mc^2)\psi= =\psi^\dag(i\hbar c\partial_0+i\hbar c\gamma^0\gamma^i\partial_i-\gamma^0mc^2-ceA_0 -ce\gamma^0\gamma^iA_i)\psi= =\psi^\dag(i\hbar{\partial\over\partial t}+c\alpha^i p_i-\beta mc^2-ceA_0-ce\alpha^iA_i)\psi= =-\psi^\dag(-i\hbar{\partial\over\partial t}+c\alpha^i (-p_i+eA_i)+\beta mc^2+ceA_0)\psi= =-\psi^\dag(-i\hbar{\partial\over\partial t}+c{\boldsymbol\alpha}\cdot({\bf p}-e{\bf A})+\beta mc^2+V)\psi

where we introduced the potential by V = c e A_0. We also could have done the same manipulation to the dirac equation itself and we would get the same expression:

(-i\hbar{\partial\over\partial t}+c{\boldsymbol\alpha}\cdot({\bf p}-e{\bf A})+\beta mc^2+V)\psi = 0

The corresponding eigenvalue problem is:

(c{\boldsymbol\alpha}\cdot({\bf p}-e{\bf A})+\beta mc^2+V)\psi = W\psi

Radial Dirac equation

As for the Schrödinger equation, there are two ways to obtain the radial Dirac equation. Either from the Lagrangian, or from the equation itself.

From the Equation

The manipulations are well known, one starts by writing the Dirac spinors using the spin angular functions and radial components P and Q:

\psi = \left(\begin{array}{c}{P\over\rho}\chi^{j_3}_\kappa\\ i{Q\over\rho}\chi^{j_3}_{-\kappa}\end{array}\right) \psi^\dag = \left(\begin{array}{cc}{P\over\rho}\chi^{j_3}_\kappa & -i{Q\over\rho}\chi^{j_3}_{-\kappa}\end{array}\right)

and putting this into the Dirac equation one obtains:

\left(\begin{array}{cc} \left(-\hbar c \left({\d\over\d\rho} - {\kappa\over\rho}\right)Q + (V+mc^2-W)P\right) & 0\\ 0 & \left(\hbar c \left({\d\over\d\rho} + {\kappa\over\rho}\right)P + (V-mc^2-W)Q\right) \end{array}\right) \left( \begin{array}{c} {1\over \rho}\chi^{j_3}_\kappa \\ i{1\over\rho}\chi^{j_3}_{-\kappa} \end{array} \right) =0

So one obtains the following radial equations:

-\hbar c \left({\d\over\d\rho} - {\kappa\over\rho}\right)Q + (V+mc^2-W)P=0 \hbar c \left({\d\over\d\rho} + {\kappa\over\rho}\right)P + (V-mc^2-W)Q=0

From the Lagrangian

We can reuse the calculations from the previous sections, because the Lagrangian happens to be zero for the solution of the Dirac equation:

\L=\bar\psi(i\hbar c\gamma^\mu D_\mu-mc^2)\psi = =-\psi^\dag(-i\hbar{\partial\over\partial t}+c{\boldsymbol\alpha}\cdot({\bf p}-e{\bf A})+\beta mc^2+V)\psi= = \left(\begin{array}{cc}{P\over\rho}\chi^{j_3}_\kappa & -i{Q\over\rho}\chi^{j_3}_{-\kappa}\end{array}\right) \left(\begin{array}{cc} \left(-\hbar c \left({\d\over\d\rho} - {\kappa\over\rho}\right)Q + (V+mc^2)P\right) & 0\\ 0 & \left(\hbar c \left({\d\over\d\rho} + {\kappa\over\rho}\right)P + (V-mc^2)Q\right) \end{array}\right) \left( \begin{array}{c} {1\over \rho}\chi^{j_3}_\kappa \\ i{1\over\rho}\chi^{j_3}_{-\kappa} \end{array} \right) = = {1\over\rho^2} P \left(-\hbar c \left({\d\over\d\rho} - {\kappa\over\rho}\right)Q + (V+mc^2)P\right) \chi^{j_3}_\kappa\chi^{j_3}_\kappa + {1\over\rho^2} Q \left(\hbar c \left({\d\over\d\rho} + {\kappa\over\rho}\right)P + (V-mc^2)Q\right) \chi^{j_3}_{-\kappa}\chi^{j_3}_{-\kappa}

We can now write the action:

S = \int \L \,\rho^2 \,\d\rho\d\Omega

the spin angular functions integrate to 1:

\int \chi^{j_3}_\kappa\chi^{j_3}_\kappa \d\Omega = 1 \int \chi^{j_3}_{-\kappa}\chi^{j_3}_{-\kappa} \d\Omega = 1

the \rho^2 cancels out and we get:

S[P, Q] = \int P \left(-\hbar c \left({\d\over\d\rho} - {\kappa\over\rho}\right)Q + (V+mc^2)P\right) + Q \left(\hbar c \left({\d\over\d\rho} + {\kappa\over\rho}\right)P + (V-mc^2)Q\right) \,\d\rho= =\int -\hbar c(PQ' - QP') + \hbar c {2\kappa\over\rho} PQ + V(P^2+Q^2) + m c^2 (P^2 - Q^2) \d\rho

the normalization condition is:

N = \int P^2 + Q^2 \d\rho - 1 = 0

and we can variate the action, we also shift the energy W=\epsilon + mc^2:

0 = \delta (S - W N) = \delta (S - \epsilon N - mc^2N)

which effectively adds -mc^2(P^2+Q^2) into the Lagrangian, which changes the term mc^2(P^2 - Q^2) into -2mc^2 Q^2. We can now variate the (constrained) action:

0=\delta\int -\hbar c(PQ' - QP') + \hbar c {2\kappa\over\rho} PQ + V(P^2+Q^2) - 2m c^2 Q^2 \d\rho= = 2\int \left(-\hbar c((\delta P)Q' - P'\delta Q) + \hbar c{\kappa\over\rho} ((\delta P)Q + P\delta Q)) + (P\delta P + Q\delta Q)V -2mc^2Q\delta Q - \epsilon(P\delta P + Q\delta Q)\right)\d\rho +[P\delta Q - Q\delta P]^R_0 = = 2\int \delta P \left(-\hbar c Q' + \hbar c{\kappa\over\rho}Q + PV - \epsilon P \right)+ \delta Q \left(\hbar c P' + \hbar c{\kappa\over\rho}P + QV - 2mc^2Q - \epsilon Q \right)\d\rho +[P\delta Q - Q\delta P]^R_0 =

which gives the two radial equations:

-\hbar c Q' + \hbar c{\kappa\over\rho}Q + PV = \epsilon P \hbar c P' + \hbar c{\kappa\over\rho}P + QV - 2mc^2Q = \epsilon Q

Weak Formulation

The weak formulation can be obtained by substituting \delta P \to v_1 and \delta Q\to v_2 into the action above (and separating the integrals) and omitting the the boundary term:

\int -\hbar c Q'v_1 + \hbar c{\kappa\over\rho}Qv_1 + PVv_1\d\rho = \epsilon \int Pv_1 \d\rho \int \hbar c P'v_2 + \hbar c{\kappa\over\rho}Pv_2 + QVv_2 -2mc^2Qv_2\d\rho = \epsilon \int Q v_2 \d\rho

We can also start from the radial equations themselves to get the same result. If we start from the equations themselves (which is the most elementary approach), there are no boundary terms (because we didn’t integrate by parts). We can separate the integrals according to the matrix elements that they contribute to:

\int PVv_1 \d\rho + \int -\hbar c Q'v_1 + \hbar c{\kappa\over\rho}Qv_1 \d\rho = \epsilon \int Pv_1 \d\rho \int \hbar c P'v_2 + \hbar c{\kappa\over\rho}Pv_2 + \int (V -2mc^2)Qv_2 \d\rho = \epsilon \int Q v_2 \d\rho

To show that this problem generates a symmetric matrix, it is helpful to write the radial equations in the following form:

\hat H \ket{P, Q} = \epsilon \ket{P, Q}

where:

\ket{P, Q} = \left(\begin{array}{c} P(\rho) \\ Q(\rho)\end{array}\right) \hat H = \left(\begin{array}{cc} V(\rho) & \hbar c \left(-{\d\over\d\rho}+{\kappa\over\rho}\right) \\ \hbar c \left({\d\over\d\rho}+{\kappa\over\rho}\right) & V(\rho) - 2mc^2 \\ \end{array}\right)

the operator \hat H is Hermitean (\hat H^\dag = \hat H), because \left(-{\d\over\d\rho}\right)^\dag = {\d\over\d\rho}:

\int f{\d\over\d\rho}g \d\rho = \int \left(-{\d\over\d\rho}\right)f g \d\rho

and all the other quantities are just scalars.

Stricly speaking, the exact Dirac notation (that is coordinate/representation independent) would be the following (notice the missing \rho^2 in the completeness relation, which is different to the radial Schrödinger equation):

\hat H \ket{P, Q} = \epsilon \ket{P, Q} \one = \int \d\rho \ket{\rho}\bra{\rho} \braket{\rho|\rho'} = \delta(\rho-\rho') \int \braket{\rho|\hat H|\rho'}\braket{\rho'|P, Q}\d\rho' = \epsilon \braket{\rho|P, Q} \braket{\rho|P, Q} = \left(\begin{array}{c} P(\rho) \\ Q(\rho)\end{array}\right) \braket{\rho|\hat H|\rho'} = \delta(\rho-\rho') \left(\begin{array}{cc} V(\rho) & \hbar c \left(-{\d\over\d\rho}+{\kappa\over\rho}\right) \\ \hbar c \left({\d\over\d\rho}+{\kappa\over\rho}\right) & V(\rho) - 2mc^2 \\ \end{array}\right)

The normalization is:

\braket{P, Q| P, Q} = \int \d\rho \braket{P, Q|\rho}\braket{\rho|P, Q} = \int \d\rho (P^2+Q^2) = 1

The weak formulation is:

\braket{v|\hat H|P, Q} = \epsilon \braket{v|P, Q}

where the test function \ket{v} is one of:

\ket{v} = \begin{cases} \ket{v_1}\left(\begin{array}{c}1\\0\\\end{array}\right) \cr \ket{v_2}\left(\begin{array}{c}0\\1\\\end{array}\right) \cr \end{cases}

The FE formulation is then obtained by expanding \ket{P, Q} = \sum_k q_k \ket{k}:

\sum_l \braket{k|\hat H|l}q_l = \epsilon \sum_l\braket{k|l}q_l

The basis \ket{k} can be for example the FE basis, some spline basis set, or gaussians. The basis has actually 2n base functions and it enumerates each equation like this:

\ket{k} = \begin{cases} \ket{i}\left(\begin{array}{c}1\\0\\\end{array}\right) & \mbox{for } i=k < n\cr \ket{i}\left(\begin{array}{c}0\\1\\\end{array}\right) & \mbox{for } i=k >= n\cr \end{cases}

So at the end of the day, the \braket{k|\hat H|l} matrix looks like this:

\braket{k|\hat H|l} = \left(\begin{array}{cc} \braket{i|V(r)|j} & \hbar c \braket{i|-{\d\over\d\rho}+{\kappa\over\rho}|j} \\ \hbar c \braket{i|{\d\over\d\rho}+{\kappa\over\rho}|j} & \braket{i|V(r) - 2mc^2|j} \\ \end{array}\right)

The matrix is 2n \times 2n, composed of those 4 matrices n \times n. The \braket{k|l} matrix looks like this:

\braket{k|l} = \left(\begin{array}{cc} \braket{i|j} & 0 \\ 0 & \braket{i|j} \\ \end{array}\right)

We can also write the matrix elements explicitly. Let \ket{i} = B_i(\rho), then:

\braket{i|j} = \int B_i B_j \,\d\rho \braket{i|V|j} = \int B_i V B_j \,\d\rho \braket{i|V-2mc^2|j} = \int B_i (V-2mc^2) B_j \,\d\rho \hbar c \braket{i|{\d\over\d\rho}+{\kappa\over\rho}|j} = \hbar c\int B_i B_j' + B_i {\kappa\over\rho} B_j \,\d\rho \hbar c \braket{i|-{\d\over\d\rho}+{\kappa\over\rho}|j} = \hbar c\int -B_i B_j' + B_i {\kappa\over\rho} B_j \,\d\rho

Other Forms of Dirac Equations

The radial Dirac equations are:

\hat H \ket{P, Q} = \epsilon \ket{P, Q} \ket{P, Q} = \left(\begin{array}{c} P(\rho) \\ Q(\rho)\end{array}\right) \hat H = \left(\begin{array}{cc} V(\rho) & \hbar c \left(-{\d\over\d\rho}+{\kappa\over\rho}\right) \\ \hbar c \left({\d\over\d\rho}+{\kappa\over\rho}\right) & V(\rho) - 2mc^2 \\ \end{array}\right)

After substitution S = f(r)P and T = f(r)Q, we get:

\hat H \ket{P, Q} = \epsilon \ket{P, Q} \hat H {1\over f} \ket{S, T} = \epsilon {1\over f}\ket{S, T} f\hat H {1\over f} \ket{S, T} = \epsilon \ket{S, T}

where:

f\hat H {1\over f}= f\left(\begin{array}{cc} V(\rho) & \hbar c \left(-{\d\over\d\rho}+{\kappa\over\rho}\right) \\ \hbar c \left({\d\over\d\rho}+{\kappa\over\rho}\right) & V(\rho) - 2mc^2 \\ \end{array}\right){1\over f} = = \left(\begin{array}{cc} V(\rho) & \hbar c \left(-f{\d\over\d\rho}{1\over f}+{\kappa\over\rho}\right) \\ \hbar c \left(f{\d\over\d\rho}{1\over f}+{\kappa\over\rho}\right) & V(\rho) - 2mc^2 \\ \end{array}\right)

and after using:

f{\d\over\d\rho}{1\over f} = {\d\over\d\rho} - {f'\over f}

we get:

f\hat H {1\over f} = \left(\begin{array}{cc} V(\rho) & \hbar c \left(-{\d\over\d\rho}+{\kappa\over\rho}+{f'\over f}\right) \\ \hbar c \left({\d\over\d\rho}+{\kappa\over\rho}-{f'\over f}\right) & V(\rho) - 2mc^2 \\ \end{array}\right)

Example I

In order to obtain equations for g and f, related to P and Q by:

P = \rho g Q = \rho f

so f(r) = {1\over \rho} and

{f'\over f} = -{1\over \rho}

and we get the radial Dirac equation for g and f:

f\hat H {1\over f} = \left(\begin{array}{cc} V(\rho) & \hbar c \left(-{\d\over\d\rho}+{\kappa-1\over\rho}\right) \\ \hbar c \left({\d\over\d\rho}+{\kappa+1\over\rho}\right) & V(\rho) - 2mc^2 \\ \end{array}\right)

Example II

For f(r) = \rho we get

{f'\over f} = {1\over \rho}

and

f\hat H {1\over f} = \left(\begin{array}{cc} V(\rho) & \hbar c \left(-{\d\over\d\rho}+{\kappa+1\over\rho}\right) \\ \hbar c \left({\d\over\d\rho}+{\kappa-1\over\rho}\right) & V(\rho) - 2mc^2 \\ \end{array}\right)

Example III

For f(r) = \rho^n we get

{f'\over f} = {n\over \rho}

and

f\hat H {1\over f} = \left(\begin{array}{cc} V(\rho) & \hbar c \left(-{\d\over\d\rho}+{\kappa+n\over\rho}\right) \\ \hbar c \left({\d\over\d\rho}+{\kappa-n\over\rho}\right) & V(\rho) - 2mc^2 \\ \end{array}\right)

Example I is just a special case for n=-1, Example II for n=1.

Example IV

For f(r) = \tanh \rho we get

{f'\over f} = {1\over \sinh\rho\cosh\rho}

and

f\hat H {1\over f} = \left(\begin{array}{cc} V(\rho) & \hbar c \left(-{\d\over\d\rho}+{\kappa\over\rho}+{1\over \sinh\rho\cosh\rho}\right) \\ \hbar c \left({\d\over\d\rho}+{\kappa\over\rho}-{1\over \sinh\rho\cosh\rho}\right) & V(\rho) - 2mc^2 \\ \end{array}\right)

Example V

For f(r) = \tanh^n k\rho we get

{f'\over f} = {nk\over \sinh k\rho\cosh k\rho}

and

f\hat H {1\over f} = \left(\begin{array}{cc} V(\rho) & \hbar c \left(-{\d\over\d\rho}+{\kappa\over\rho}+{nk\over \sinh k\rho\cosh k\rho}\right) \\ \hbar c \left({\d\over\d\rho}+{\kappa\over\rho}-{nk\over \sinh k\rho\cosh k\rho}\right) & V(\rho) - 2mc^2 \\ \end{array}\right)

Example IV is just a special case for n=1, k=1.

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