8.4. Quantum Mechanics¶

We use a metric with signature -2 in this section.

8.4.1. From QED to Quantum Mechanics¶

The QED Lagrangian density is

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

Plugging this Lagrangian into the Euler-Lagrange equation of motion for a field, we get:

$(i\hbar c\gamma^\mu D_\mu-mc^2)\psi=0$

$\partial_\nu F^{\nu\mu}=-ec\bar\psi\gamma^\mu\psi$

The first equation is the Dirac equation in the electromagnetic field and the second equation is a set of Maxwell equations ( $\partial_\nu F^{\nu\mu}=-ej^\mu$ ) with a source $j^\mu=c\bar\psi\gamma^\mu\psi$ , which is a 4-current comming from the Dirac equation.

The fields $\psi$ and $A^\mu$ are quantized. The first approximation is that we take $\psi$ as a wavefunction, that is, it is a classical 4-component field. It can be shown that this corresponds to taking the tree diagrams in the perturbation theory.

We multiply the Dirac equation by $\gamma^0$ from left to get:

$0=\gamma^0(i\hbar c\gamma^\mu D_\mu-mc^2)\psi= \gamma^0(i\hbar c\gamma^0(\partial_0+{i\over\hbar}eA_0)+ic\gamma^i (\partial_i+{i\over\hbar}eA_i)-mc^2)\psi=$

$= (i\hbar c\partial_0+i\hbar c\gamma^0\gamma^i\partial_i-\gamma^0mc^2-ceA_0 -ce\gamma^0\gamma^iA_i)\psi$

and we make the following substitutions (it’s just a formalism, nothing more): $\beta=\gamma^0$ , $\alpha^i=\gamma^0\gamma^i$ , $p_j=i\hbar\partial_j$ , $\partial_0={1\over c}{\partial\over\partial t}$ to get

$(i\hbar{\partial\over\partial t}+c\alpha^i p_i-\beta mc^2-ceA_0-ce\alpha^iA_i)\psi=0\,.$

or:

$i\hbar{\partial\psi\over\partial t}=(c\alpha^i(-p_i+eA_i) +\beta mc^2+ceA_0)\psi\,.$

This can be written as:

$i{\partial\psi\over\partial t}=H\psi\,,$

where the Hamiltonian is given by:

$H=c\alpha^i(-p_i+eA_i)+\beta mc^2+ceA_0\,,$

or introducing the electrostatic potential $\phi=cA_0$ and writing the momentum as a vector (see the appendix for all the details regarding signs):

$H=c{\boldsymbol\alpha}\cdot({\bf p}-e{\bf A})+\beta mc^2+e\phi\,.$

The right hand side of the Maxwell equations is the 4-current, so it’s given by:

$j^\mu=c\bar\psi\gamma^\mu\psi$

Now we make the substitution $\psi=e^{-imc^2t}\varphi$ , which states, that we separate the largest oscillations of the wavefunction and we get

$j^0=c\bar\psi\gamma^0\psi=c\psi^\dagger\psi=c\varphi^\dagger\varphi$

$j^i=c\bar\psi\gamma^i\psi=c\psi^\dagger\alpha^i\psi=c\varphi^\dagger\alpha^i\varphi$

Derivation of the Pauli Equation¶

We start from the Dirac equation:

$H\psi = W\psi$

where:

$H=c{\boldsymbol\alpha}\cdot({\bf p}-e{\bf A})+\beta mc^2+e\phi\,. W = E + mc^2$

$W$ is the relativistic energy, $E$ is the nonrelativistic energy, $e\phi=V$ is the potential. The matrices ${\boldsymbol\alpha}$ and $\beta$ are given by:

$\alpha^i = \gamma^0\gamma^i \beta = \gamma^0$

so written explicitly:

$\alpha = \begin{pmatrix} 0 & {\boldsymbol\sigma} \\ {\boldsymbol\sigma} & 0 \\ \end{pmatrix} \beta = \begin{pmatrix} \one & 0 \\ 0 & -\one \\ \end{pmatrix}$

And the Dirac equation is:

$\begin{pmatrix} V+mc^2 & c{\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) \\ c{\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) & V-mc^2 \\ \end{pmatrix} \begin{pmatrix} \psi^L \\ \psi^S \\ \end{pmatrix} = W \begin{pmatrix} \psi^L \\ \psi^S \\ \end{pmatrix}$

After introducing $E$ we get:

$\begin{pmatrix} V & c{\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) \\ c{\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) & V-2mc^2 \\ \end{pmatrix} \begin{pmatrix} \psi^L \\ \psi^S \\ \end{pmatrix} = E \begin{pmatrix} \psi^L \\ \psi^S \\ \end{pmatrix}$

We put everything on the left hand side:

$\begin{pmatrix} V -E & c{\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) \\ c{\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) & V-E-2mc^2 \\ \end{pmatrix} \begin{pmatrix} \psi^L \\ \psi^S \\ \end{pmatrix} = 0$

We put $c$ next to $\psi^S$ :

$\begin{pmatrix} V -E & {\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) \\ c{\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) & {V-E\over c}-2mc \\ \end{pmatrix} \begin{pmatrix} \psi^L \\ c\psi^S \\ \end{pmatrix} = 0$

And we divide the second equation by $c$ :

$\begin{pmatrix} V -E & {\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) \\ {\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) & {V-E\over c^2}-2m \\ \end{pmatrix} \begin{pmatrix} \psi^L \\ c\psi^S \\ \end{pmatrix} = 0$

Now we express $c\psi^S$ from the second equation:

$c\psi^S ={ {\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) \psi^L \over 2m - {V-E\over c^2}}$

And substitute into the first equation:

$\left( V - E + {\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) {1 \over 2m - {V-E\over c^2}} {\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) \right) \psi^L = 0$

So we get the following equation (so far this is an exact equation for the first two components of the Dirac equation, no approximation has been made):

$\left( {\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) {1 \over 2m - {V-E\over c^2}} {\boldsymbol\sigma}\cdot({\bf p}-e{\bf A}) + V \right) \psi^L = E \psi^L$

Note that the first operator ${\bf p}$ (on the left hand side) acts among other things on the $V$ in the denominator. By doing the nonrelativistic approximation ${V-E\over c^2} \ll 2m$ we obtain the Pauli equation:

(8.4.1.1)¶ $\left( {\left({\boldsymbol\sigma}\cdot({\bf p}-e{\bf A})\right)^2 \over 2m} + V \right) \psi^L = E \psi^L$

We can see, that the quantity

$M = m - {V-E\over 2c^2}$

can be interpreted as relativistic mass.

Using the relations between the Pauli matrices, we can further simplify:

$\left({\boldsymbol\sigma}\cdot({\bf p}-e{\bf A})\right)^2 = \left({\bf p}-e{\bf A}\right)^2+i{\boldsymbol\sigma} \cdot{({\bf p}-e{\bf A})\times({\bf p}-e{\bf A})} = = \left({\bf p}-e{\bf A}\right)^2+i{\boldsymbol\sigma} \cdot\left({\bf p}\times{\bf p}-e{\bf A}\times{\bf p} -e{\bf p}\times{\bf A}+e^2{\bf A}\times{\bf A}\right) = = \left({\bf p}-e{\bf A}\right)^2-ie{\boldsymbol\sigma} \cdot\left({\bf A}\times{\bf p} +{\bf p}\times{\bf A}\right) = = \left({\bf p}-e{\bf A}\right)^2-ie{\boldsymbol\sigma} \cdot\left({\bf A}\times{\bf p}-{\bf A}\times{\bf p} -i\hbar(\nabla\times{\bf A})\right) = = \left({\bf p}-e{\bf A}\right)^2-{e\hbar}{\boldsymbol\sigma} \cdot(\nabla\times{\bf A}) = = \left({\bf p}-e{\bf A}\right)^2-{e\hbar}{\boldsymbol\sigma} \cdot{\bf B}$

At the end, we have introduced the magnetic field ${\bf B} = {\nabla\times{\bf A}}$ . In the above, one has to be careful, because ${\bf p}$ and ${\bf A}$ don’t commute and also the operator ${\bf p}$ acts on everything on the right. We used the formula ${\bf p}\times{\bf A}=-{\bf A}\times{\bf p}-i\hbar(\nabla\times{\bf A})$ , that can be proven by:

$({\bf p}\times{\bf A} \psi)_i = \epsilon_{ijk}p_j A_k \psi = = -i\hbar\epsilon_{ijk}\partial_j (A_k \psi) = = -i\hbar\epsilon_{ijk}((\partial_j A_k)\psi + A_k\partial_j \psi) = = -i\hbar\epsilon_{ijk}((\partial_j A_k)\psi - A_j\partial_k \psi) = = \epsilon_{ijk}(-i\hbar(\partial_j A_k)\psi - A_j p_k \psi) = = -i\hbar((\nabla\times{\bf A})\psi)_i - ({\bf A}\times{\bf p} \psi)_i$

Putting this into the Pauli equation (8.4.1.1), we get:

(8.4.1.2)¶ $\left( {\left({\bf p}-e{\bf A}\right)^2 \over 2m} + V -{e\hbar\over 2m}{\boldsymbol\sigma}\cdot{\bf B} \right) \psi^L = E \psi^L$

We can expand $\left({\bf p}-e{\bf A}\right)^2$ as follows:

$\left({\bf p}-e{\bf A}\right)^2 = p^2 - e({\bf p}\cdot{\bf A}+{\bf A}\cdot{\bf p}) + e^2 A^2 = = p^2 - 2e {\bf A}\cdot{\bf p} + e^2 A^2 + ie\hbar(\nabla\cdot{\bf A})$

where we used:

${\bf p}\cdot{\bf A}\psi = -i\hbar\partial_i A^i\psi = -i\hbar A^i\partial_i \psi -i\hbar(\partial_i A^i)\psi = {\bf A}\cdot{\bf p} \psi -i\hbar(\nabla\cdot{\bf A})\psi$

and (8.4.1.2) becomes:

$\left( {p^2 \over 2m} - {e\over m} {\bf A}\cdot{\bf p} + {e^2\over 2m} A^2 + {ie\hbar\over 2m}(\nabla\cdot{\bf A}) + V -{e\hbar\over 2m}{\boldsymbol\sigma}\cdot{\bf B} \right) \psi^L = E \psi^L$

Using ${\bf p} = -i\hbar\nabla$ we get:

(8.4.1.3)¶ $\left( -{\hbar^2 \over 2m}\nabla^2 + {ie\hbar\over m} {\bf A}\cdot\nabla + {e^2\over 2m} A^2 + {ie\hbar\over 2m}(\nabla\cdot{\bf A}) + V -{e\hbar\over 2m}{\boldsymbol\sigma}\cdot{\bf B} \right) \psi^L = E \psi^L$

Note that $V=e\phi$ and $e=-|e|$ is the (negative) electron charge. In the Coulomb gauge the term $\nabla\cdot{\bf A}=0$ . Sometimes one can neglect the quadratic term $e^2A^2$ .

Example: velocity and length gauges¶

Let us assume a given spatially homogeneous time dependent electric field ${\bf E}(t)$ and no magnetic field ${\bf B}=0$ . Then one choice (gauge) of the electromagnetic potentials is:

$\phi = 0 {\bf A}(t) = -\int_0^t {\bf E}(t') \d t'$

The vector potential ${\bf A}$ is time dependent, but spatially homogeneous (constant). Let us first check that we are getting the correct ${\bf E}$ and ${\bf B}$ :

${\bf E} = -\nabla\phi-{\partial{\bf A}\over\partial t} = {\bf E}(t) {\bf B} = \nabla\times{\bf A} = 0$

We also have $\nabla\cdot{\bf A}=0$ , since ${\bf A}$ does not depend on coordinates. Substituting into (8.4.1.3) we get:

(8.4.1.4)¶ $\left( -{\hbar^2 \over 2m}\nabla^2 + {ie\hbar\over m} {\bf A}\cdot\nabla + {e^2\over 2m} A^2 \right) \psi = E \psi$

The $\nabla$ in the ${\bf A}\cdot\nabla$ term is a velocity, so this gauge is called a velocity gauge. We can apply a gauge transformation $\Lambda = -{\bf A}\cdot{\bf x}$ and we get:

$\phi \to \phi - {\partial\Lambda\over\partial t} = \phi + {\partial{\bf A}\over\partial t}\cdot{\bf x} = \phi - {\bf E}\cdot{\bf x} {\bf A} \to {\bf A} + \nabla\Lambda = {\bf A} - {\bf A} = 0$

Substituting this into (8.4.1.4) (and using $V=e\phi$ ), we obtain:

(8.4.1.5)¶ $\left( -{\hbar^2 \over 2m}\nabla^2 -e{\bf E}\cdot{\bf x} \right) \psi = E \psi$

Due to the ${\bf x}$ term in ${\bf E}\cdot{\bf x}$ , this gauge is called the length gauge. Note that $e=-|e|$ is negative for electrons.

Nonrelativistic Limit in the Lagrangian¶

We use the identity ${\partial\over\partial t}\left(e^{-imc^2t}f(t)\right)= e^{-imc^2t}(-imc^2+{\partial\over\partial t})f(t)$ to get:

$L=c^2\partial^\mu\psi^*\partial_\mu\psi-m^2c^4\psi^*\psi= {\partial\over\partial t}\psi^*{\partial\over\partial t}\psi -c^2\partial^i\psi^*\partial_i\psi-m^2c^4\psi^*\psi=$

$=(imc^2+{\partial\over\partial t})\varphi^* (-imc^2+{\partial\over\partial t})\varphi -c^2\partial^i\varphi^*\partial_i\varphi-m^2c^4\varphi^*\varphi=$

$=2mc^2\left[{1\over2}i(\varphi^*{\partial\varphi\over\partial t}- \varphi{\partial\varphi^*\over\partial t})- {1\over2m}\partial^i\varphi^*\partial_i\varphi +{1\over2mc^2}{\partial\varphi^*\over\partial t} {\partial\varphi\over\partial t}\right]$

The constant factor $2mc^2$ in front of the Lagrangian is of course irrelevant, so we drop it and then we take the limit $c\to\infty$ (neglecting the last term) and we get

$L={1\over2}i(\varphi^*{\partial\varphi\over\partial t}- \varphi{\partial\varphi^*\over\partial t})- {1\over2m}\partial^i\varphi^*\partial_i\varphi$

After integration by parts we arrive at the Lagrangian for the Schrödinger equation:

$L=i\varphi^*{\partial\varphi\over\partial t} -{1\over 2m}\partial^i\varphi^*\partial_i \varphi$

Klein-Gordon Equation¶

The Dirac equation implies the Klein-Gordon equation:

$0=(-i\hbar c\gamma^\mu D_\mu-mc^2)(i\hbar c\gamma^\nu D_\nu-mc^2)\psi= (\hbar^2c^2\gamma^\mu\gamma^\nu D_\mu D_\nu+m^2c^4)\psi=$

$=(\hbar^2c^2g^{\mu\nu}D_\mu D_\nu+m^2c^4)\psi =(\hbar^2c^2D^\mu D_\mu+m^2c^4)\psi$

Note however, the $\psi$ in the true Klein-Gordon equation is just a scalar, but here we get a 4-component spinor. Now:

$D_\mu D_\nu = (\partial_\mu+ieA_\mu)(\partial_\nu+ieA_\nu)= \partial_\mu\partial_\nu+ie(A_\mu\partial_\nu+A_\nu\partial_\mu+ (\partial_\mu A_\nu))-e^2A_\mu A_\nu$

$[D_\mu, D_\nu] = D_\mu D_\nu-D_\nu D_\mu=ie(\partial_\mu A_\nu)- ie(\partial_\nu A_\mu)$

We rewrite $D^\mu D_\mu$ :

$D^\mu D_\mu=g^{\mu\nu}D_\mu D_\nu= \partial^\mu\partial_\mu+ie((\partial^\mu A_\mu)+2A^\mu\partial_\mu) -e^2A^\mu A_\mu=$

$=\partial^\mu\partial_\mu+ ie((\partial^0 A_0)+2A^0\partial_0+(\partial^i A_i)+2A^i\partial_i) -e^2(A^0A_0+A^i A_i)=$

$=\partial^\mu\partial_\mu +i{1\over c^2}{\partial V\over\partial t}+ 2i{V\over c^2}{\partial\over\partial t} +ie(\partial^i A_i)+2ieA^i\partial_i -{V^2\over c^2}-e^2A^iA_i$

The nonrelativistic limit can also be applied directly to the Klein-Gordon equation:

$0=(\hbar^2c^2D^\mu D_\mu+m^2c^4)\psi=$

$=\left( \hbar^2c^2\partial^\mu\partial_\mu +i{\partial V\over\partial t} +2iV{\partial\over\partial t} +i\hbar ec^2(\partial^i A_i) +2i\hbar ec^2A^i\partial_i -V^2 -e^2c^2A^iA_i +m^2c^4 \right)e^{-{i\over\hbar}mc^2t}\varphi=$

$=\left( \hbar^2{\partial^2\over\partial t^2} -c^2\hbar^2\nabla^2 +2iV{\partial\over\partial t} +i{\partial V\over\partial t} +i\hbar ec^2(\partial^i A_i) +2i\hbar ec^2A^i\partial_i -V^2 -e^2c^2A^iA_i +m^2c^4 \right)e^{-{i\over\hbar}mc^2t}\varphi=$

$=e^{-{i\over\hbar}mc^2t}\left( \hbar^2(-{i\over\hbar}mc^2+{\partial\over\partial t})^2 -\hbar^2c^2\nabla^2 +2iV(-{i\over\hbar}mc^2+{\partial\over\partial t}) +i{\partial V\over\partial t} +i\hbar ec^2(\partial^i A_i) +2i\hbar ec^2A^i\partial_i -V^2+ \right.$

$\left. -e^2c^2A^iA_i +m^2c^4 \right)\varphi=$

$=e^{-{i\over\hbar}mc^2t}\left( -2i\hbar mc^2{\partial\over\partial t}+\hbar^2{\partial^2\over\partial t^2} -c^2\hbar^2\nabla^2 +2Vm{c^2\over\hbar} +2iV{\partial\over\partial t} +i{\partial V\over\partial t} +i\hbar ec^2(\partial^i A_i) +2i\hbar ec^2A^i\partial_i -V^2+ \right.$

$\left. -e^2c^2A^iA_i \right)\varphi=$

$= -2mc^2 e^{-{i\over\hbar}mc^2 t} \left(i\hbar{\partial\over\partial t}+\hbar^2{\nabla^2\over2m}-V -{1\over2mc^2}{\partial^2\over\partial t^2}-{i\over2mc^2}{\partial V\over\partial t}+{V^2\over2mc^2}-{iV\over mc^2}{\partial\over\partial t}+\right.$

$\left.-{i\hbar e\over2m}\partial^i A_i-{i\hbar e\over m}A^i\partial_i+{e^2\over2m}A^iA_i\right)\varphi$

Taking the limit $c\to\infty$ we again recover the Schrödinger equation:

$i\hbar{\partial\over\partial t}\varphi=\left(-\hbar^2{\nabla^2\over2 m}+V +{i\hbar e\over2m}\partial^i A_i +{i\hbar e\over m}A^i\partial_i -{e^2\over2m}A^iA_i \right)\varphi\,,$

we rewrite the right hand side a little bit:

$i\hbar{\partial\over\partial t}\varphi=\left({\hbar^2\over2 m} (\partial^i\partial_i +{i\over\hbar}e\partial^i A_i +2{i\over\hbar}eA^i\partial_i -{e^2\over\hbar^2}A^iA_i ) +V \right)\varphi\,,$

$i\hbar{\partial\over\partial t}\varphi=\left({\hbar^2\over2 m} (\partial^i+{i\over\hbar}eA^i)(\partial_i+{i\over\hbar}eA_i) +V \right)\varphi\,,$

$i\hbar{\partial\over\partial t}\varphi=\left({1\over2 m} \hbar^2D^iD_i +V \right)\varphi\,,$

Using (see the appendix for details):

$\hbar^2D^iD_i=-\hbar^2\delta_{ij}D^iD^j =-\hbar^2\left({i\over\hbar}({\bf p}-e{\bf A})\right)^2 =({\bf p}-e{\bf A})^2$

we get the usual form of the Schrödinger equation for the vector potential:

$i\hbar{\partial\over\partial t}\varphi=\left({({\bf p}-e{\bf A})^2\over2 m} +V \right)\varphi\,.$

A little easier derivation:

$0=(\hbar^2c^2 D^\mu D_\nu+m^2c^4)\psi=$

$=(\hbar^2c^2 D^0 D_0+\hbar^2c^2D^i D_i+m^2c^4)\psi=$

$=2mc^2\left({\hbar^2\over2m} D^0 D_0+{\hbar^2\over2m}D^i D_i+\half mc^2\right)\psi=$

$=2mc^2\left({\hbar^2\over2m} \left(\partial^0+{i\over\hbar}eA^0\right) \left(\partial_0+{i\over\hbar}eA_0\right)+\half mc^2+{\hbar^2\over2m}D^i D_i \right) e^{-{i\over\hbar}mc^2 t} \varphi=$

$=2mc^2\left({\hbar^2\over2m} \left(\partial^0+{i\over\hbar}eA^0\right) e^{-{i\over\hbar}mc^2 t} \left(\partial_0-{i\over\hbar}mc+{i\over\hbar}eA_0\right)+\half mc^2+{\hbar^2\over2m}D^i D_i \right) \varphi=$

$=2mc^2 e^{-{i\over\hbar}mc^2 t} \left({\hbar^2\over2m} \left(\partial^0-{i\over\hbar}mc+{i\over\hbar}eA^0\right) \left(\partial_0-{i\over\hbar}mc+{i\over\hbar}eA_0\right)+\half mc^2+{\hbar^2\over2m}D^i D_i \right) \varphi=$

$=2mc^2 e^{-{i\over\hbar}mc^2 t} \left( {\hbar^2\over2m}\partial^0\partial_0 -\half mc^2 -{e^2A^0A_0\over 2m} +ceA^0 +{\hbar^2\over m}{i\over\hbar}e(\partial^0 A^0+A^0\partial^0) -i\hbar c\partial_0 +\half mc^2+{\hbar^2\over2m}D^i D_i \right) \varphi=$

$=2mc^2 e^{-{i\over\hbar}mc^2 t} \left( -i\hbar {\partial\over\partial t} +{\hbar^2\over2m}D^i D_i +ceA^0 +{\hbar^2\over2mc^2}{\partial^2\over\partial t^2} -{e^2\phi^2\over 2mc^2} +{ie\hbar\over mc^2}({\partial\over\partial t} \phi + \phi{\partial\over\partial t}) \right) \varphi=$

$=2mc^2 e^{-{i\over\hbar}mc^2 t} \left( -i\hbar {\partial\over\partial t} +{({\bf p}-e{\bf A})^2\over2m} +e\phi +{\hbar^2\over2mc^2}{\partial^2\over\partial t^2} -{e^2\phi^2\over 2mc^2} +{ie\hbar\over mc^2}({\partial\over\partial t} \phi + \phi{\partial\over\partial t}) \right) \varphi$

and letting $c\to\infty$ we get the Schrödinger equation:

$i\hbar {\partial\over\partial t}\varphi= \left( {({\bf p}-e{\bf A})^2\over2m} +e\phi \right)\varphi$

8.4.2. Perturbation Theory¶

We want to solve the equation:

(8.4.2.1)¶ $i\hbar{\d \over\d t}\ket{\psi(t)}=H(t)\ket{\psi(t)}$

with $H(t) = H^0 + H^1(t)$ , where $H^0$ is time-independent part whose eigenvalue problem has been solved:

$H^0\ket{n^0}=E^0_n\ket{n^0}$

and $H^1(t)$ is a small time-dependent perturbation. $\ket{n^0}$ form a complete basis, so we can express $\ket{\psi(t)}$ in this basis:

(8.4.2.2)¶ $\ket{\psi(t)} = \sum_n d_n(t)e^{-{i\over\hbar}E^0_n t}\ket{n^0}$

Substituting this into (8.4.2.1), we get:

$\sum_n\left( i\hbar{\d\over\d t} d_n(t)+E^0_n d_n(t) \right)e^{-{i\over\hbar}E^0_n t}\ket{n^0} =\sum_n\left( E^0_n d_n(t) +H^1 d_n(t) \right)e^{-{i\over\hbar}E^0_n t}\ket{n^0}$

so:

$\sum_n i\hbar{\d\over\d t}\left( d_n(t)\right) e^{-{i\over\hbar}E^0_n t}\ket{n^0} =\sum_n d_n(t) e^{-{i\over\hbar}E^0_n t}H^1\ket{n^0}$

Choosing some particular state $\ket{f^0}$ of the $H^0$ Hamiltonian, we multiply the equation from the left by $\bra{f^0}e^{{i\over\hbar}E^0_f t}$ :

$\sum_n i\hbar{\d\over\d t}\left( d_n(t)\right)e^{i w_{fn} t} \braket{f^0|n^0} =\sum_n d_n(t) e^{i w_{fn} t}\braket{f^0|H^1|n^0}$

where $w_{fn}={E^0_f - E^0_n\over \hbar}$ . Using $\braket{f^0|n^0}=\delta_{fn}$ :

$i\hbar{\d\over\d t}d_f(t) =\sum_n d_n(t) e^{i w_{fn} t}\braket{f^0|H^1|n^0}$

we integrate from $t_1$ to $t$ :

$i\hbar\left((d_f(t)-d_f(t_1)\right) =\sum_n\int_{t_1}^t d_n(t') e^{i w_{fn} t'}\braket{f^0|H^1(t')|n^0} \d t'$

Let the initial wavefunction at time $t_1$ be some particular state $\ket{\psi(t_1)}=\ket{i^0}$ of the unperturbed Hamiltonian, then $d_n(t_1)=\delta_{ni}$ and we get:

(8.4.2.3)¶ $d_f(t) =\delta_{fi}-{i\over\hbar}\sum_n\int_{t_1}^t d_n(t') e^{i w_{fn} t'}\braket{f^0|H^1(t')|n^0} \d t'$

This is the equation that we will use for the perturbation theory.

In the zeroth order of the perturbation theory, we set $H^1(t)=0$ and we get:

$d_f(t)=\delta_{fi}$

In the first order of the perturbation theory, we take the solution $d_n(t)=\delta_{ni}$ obtained in the zeroth order and substitute into the right hand side of (8.4.2.3):

$d_f(t) = \delta_{fi} -{i\over\hbar}\int_{t_1}^{t} e^{i w_{fi} t'}\braket{f^0|H^1(t')|i^0}\d t'$

In the second order, we take the last solution, substitute into the right hand side of (8.4.2.3) again:

$d_f(t) = \delta_{fi}+ \left(-{i\over\hbar}\right)\int_{t_1}^{t} e^{i w_{fi} t'}\braket{f^0|H^1(t')|i^0}\d t' +$

$+ \left(-{i\over\hbar}\right)^2\sum_n \int_{t_1}^t\d t''\int_{t_1}^{t''}\d t' e^{iw_{fn}t''}\braket{f^0|H^1(t'')|n^0} e^{i w_{ni} t'}\braket{n^0|H^1(t')|i^0}$

And so on for higher orders of the perturbation theory — more terms will arise on the right hand side of the last formula, so this is our main formula for calculating the $d_n(t)$ coefficients.

Time Independent Perturbation Theory¶

As a special case, if $H^1$ doesn’t depend on time, the coefficients $d_n(t)$ simplify, so we calculate them in this section explicitly. Let’s take

$H(t) = H^0 + e^{t/\tau} H^1$

so at the time $t_1=-\infty$ the Hamiltonian $H(t)=H^0$ is unperturbed and we are interested in the time $t=0$ , when the Hamiltonian becomes $H(t) = H^0 + H^1$ (the coefficients $d_n(t)$ will still depend on the $\tau$ variable) and we do the limit $\tau\to\infty$ (this corresponds to smoothly applying the perturbation $H^1$ at the time negative infinity).

Let’s calculate $d_f(0)$ :

$d_f(0) = \delta_{fi}+ \left(-{i\over\hbar}\right)\int_{-\infty}^0 e^{i w_{fi} t'}e^{t\over\tau}\d t'\braket{f^0|H^1|i^0} +$

$+ \left(-{i\over\hbar}\right)^2\sum_n \int_{-\infty}^0\d t''\int_{-\infty}^{t''}\d t' e^{iw_{fn}t''} e^{i w_{ni} t'} e^{t''\over\tau} e^{t'\over\tau} \braket{f^0|H^1|n^0} \braket{n^0|H^1|i^0} =$

$= \delta_{fi}+ \left(-{i\over\hbar}\right) {1\over{1\over\tau}+i\omega_{fi}} \braket{f^0|H^1|i^0} +$

$+ \left(-{i\over\hbar}\right)^2\sum_n {1\over{1\over\tau}+i\omega_{ni}} {1\over{2\over\tau}+i\omega_{fn}+i\omega_{ni}} \braket{f^0|H^1|n^0} \braket{n^0|H^1|i^0}$

Taking the limit $\tau\to\infty$ :

$d_f(0) = \delta_{fi}+ \left(-{1\over\hbar}\right) {1\over\omega_{fi}} \braket{f^0|H^1|i^0} +$

$+ \left(-{1\over\hbar}\right)^2\sum_n {1\over\omega_{ni}} {1\over\omega_{fn}+\omega_{ni}} \braket{f^0|H^1|n^0} \braket{n^0|H^1|i^0} =$

$= \delta_{fi}- {\braket{f^0|H^1|i^0}\over E_f^0-E_i^0} +$

$+ \sum_n { \braket{f^0|H^1|n^0} \braket{n^0|H^1|i^0} \over (E_n^0-E_i^0)(E_f^0-E_i^0) }$

Substituting this into (8.4.2.2) evaluated for $t=0$ :

$\ket{\psi(0)}=\sum_n d_n(0) \ket{n^0}=$

$= \ket{i^0}- \sum_n {\ket{n^0}\braket{n^0|H^1|i^0}\over E_n^0-E_i^0} +$

$+ \sum_{n,m} {\ket{n^0} \braket{n^0|H^1|m^0} \braket{m^0|H^1|i^0} \over (E_m^0-E_i^0)(E_n^0-E_i^0) }$

The sum $\sum_n$ is over all $n\neq i$ , similarly for the other sum. Let’s also calculate the energy:

$E =\braket{\psi(0)|H|\psi(0)} =\braket{\psi(0)|H^0+H^1|\psi(0)} =$

$\left(\cdots- \sum_{n'\neq i} {\braket{i^0|H^1|n'^0}\bra{n'^0}\over E_{n'}^0-E_i^0} +\bra{i^0}\right) (H^0+H^1) \left(\ket{i^0}- \sum_{n\neq i} {\ket{n^0}\braket{n^0|H^1|i^0}\over E_n^0-E_i^0} +\cdots\right)$

To evaluate this, we use the fact that $\braket{i^0|H^0|i^0}=E_i^0$ and $\braket{i^0|H^0|n^0}=E_i^0\delta_{ni}$ :

$E = E_i^0 + \braket{i^0|H^1|i^0} - \sum_{n\neq i} {\braket{i^0|H^1|n^0}\braket{n^0|H^1|i^0}\over E_n^0-E_i^0}+\cdots =$

$= E_i^0 + \braket{i^0|H^1|i^0} - \sum_{n\neq i} {|\braket{n^0|H^1|i^0}|^2\over E_n^0-E_i^0}+\cdots$

Where we have neglected the higher order terms, so we can identify the corrections to the energy $E$ coming from the particular orders of the perturbation theory:

$E_i^0 = \braket{i^0|H^0|i^0}$

$E_i^1 = \braket{i^0|H^1|i^0}$

$E_i^2 = - \sum_{n\neq i} {|\braket{n^0|H^1|i^0}|^2\over E_n^0-E_i^0}$

8.4.3. Scattering Theory¶

The incoming plane wave state is a solution of

$H_0\ket{{\bf k}}=E_k\ket{{\bf k}}$

with $H_0={p^2\over 2m}$ . E.g.

$\braket{{\bf r}|{\bf k}}=e^{i{\bf r}\cdot{\bf k}}$

$E_k = {\hbar^2 k^2\over 2 m}$

We want to solve:

$(H_0+V)\ket{\psi}=E_k\ket{\psi}$

The solution of this is:

$\ket{\psi}=\ket{{\bf k}}+{1\over E_k-H_0}V\ket{\psi} =\ket{\bf{k}}+GV\ket{\psi}$

where

$G={1\over E_k-H_0}$

is the Green function for the Schrödinger equation. $G$ is not unique, it contains both outgoing and ingoing waves. As shown below, one can distinguish between these two by adding a small $i\epsilon$ into the denominator, that moves the poles of the Green functions above and below the $x$ -axis:

$G_+={1\over E_k-H_0+i\epsilon}$

$G_-={1\over E_k-H_0-i\epsilon}$

Both $G_+$ and $G_-$ are well-defined and unique. One can calculate both Green functions explicitly:

$G_+({\bf r}, {\bf r'}) = \braket{{\bf r}|G_+|{\bf r'}}=\bra{{\bf r}}{1\over E_k-H_0+i\epsilon}\ket{{\bf r'}}= =\int{\d^3k'\over(2\pi)^3} {\braket{{\bf r}|{\bf k'}}\braket{\bf{k'}|\bf{r'}}\over E_k-E_{k'}+i\epsilon} =\int{\d^3k'\over(2\pi)^3} {e^{i{\bf k'}\cdot({\bf r}-{\bf r'})}\over E_k-E_{k'}+i\epsilon} ={2m\over\hbar^2}\int{\d^3k'\over(2\pi)^3} {e^{i{\bf k'}\cdot({\bf r}-{\bf r'})}\over k^2-{k'}^2+i\epsilon}= ={4\pi m\over(2\pi)^3\hbar^2i|{\bf r}-{\bf r'}|} \int_{-\infty}^\infty\d^3k' k'{e^{i k'|{\bf r}-{\bf r'}|}\over k^2-{k'}^2+i\epsilon} ={4\pi m\over(2\pi)^3\hbar^2i|{\bf r}-{\bf r'}|} (2\pi i)k{e^{i k|{\bf r}-{\bf r'}|}\over 2k}= ={me^{i k|{\bf r}-{\bf r'}|}\over2\pi\hbar^2|{\bf r}-{\bf r'}|}$

Similarly:

$G_-({\bf r}, {\bf r'}) = \braket{{\bf r}|G_-|{\bf r'}} =\bra{{\bf r}}{1\over E_k-H_0-i\epsilon}\ket{{\bf r'}} =\cdots ={me^{-i k|{\bf r}-{\bf r'}|}\over2\pi\hbar^2|{\bf r}-{\bf r'}|}$

Assuming $|{\bf r'}|\ll|{\bf r}|$ , we can taylor expand $|{\bf r}-{\bf r'}|$ :

$|{\bf r}-{\bf r'}| =e^{-{\bf r'}\cdot\nabla}|{\bf r}| =\left(1-{\bf r'}\cdot\nabla+\left(-{\bf r'}\cdot\nabla\right)^2 +O\left(r'^3\right) \right)|{\bf r}| =|{\bf r}|-{\bf r'}\cdot\nabla|{\bf r}|+O\left(r'^2\right) = =r-{\bf r'}\cdot{\bf \hat r}+O\left(r'^2\right)$

so:

$e^{i k|{\bf r}-{\bf r'}|} \approx e^{ikr} e^{-i k{\bf r'}\cdot{\bf\hat r}} |{\bf r}-{\bf r'}| \approx r$

and simplify the result even further:

$G_+({\bf r}, {\bf r'}) ={m\over2\pi\hbar^2}{e^{ikr}\over r} e^{-i k{\bf r'}\cdot{\bf\hat r}} G_-({\bf r}, {\bf r'}) ={m\over2\pi\hbar^2}{e^{-ikr}\over r} e^{i k{\bf r'}\cdot{\bf\hat r}}$

Let’s get back to the solution of the Schrödinger equation:

$\ket{\psi}=\ket{\bf{k}}+G_+V\ket{\psi}$

It contains the solution $\ket{\psi}$ on both sides of the equation, so we express it explicitly:

$\ket{\psi}-G_+V\ket{\psi}=\ket{\bf{k}}$

$\ket{\psi}={1\over 1-G_+V}\ket{\bf{k}}$

and multiply by $V$ :

$V\ket{\psi}={V\over 1-G_+V}\ket{\bf{k}}=T\ket{\bf{k}}$

where $T$ is the transition matrix:

$T={V\over 1-G_+V}=V(1+G_+V + (G_+V)^2 + \cdots)=$

$=V+VG_+V + VG_+VG_+V + \cdots=$

$=V+V{1\over E_k-H_0+i\epsilon}V + V{1\over E_k-H_0+i\epsilon}V{1\over E_k-H_0+i\epsilon}V + \cdots$

Then the final solution is:

$\ket{\psi}=\ket{\bf{k}}+G_+V\ket{\psi}=\ket{\bf{k}}+G_+T\ket{{\bf k}}$

and in a coordinate representation:

$\psi({\bf r})=\braket{{\bf r}|\psi} =\braket{{\bf r}|\bf{k}}+\braket{{\bf r}|G_+T|{\bf k}} =\braket{{\bf r}|\bf{k}}+\int\d^3 r'\braket{{\bf r}|G_+|{\bf r'}} \braket{{\bf r'}|T|{\bf k}}=$

$=\braket{{\bf r}|\bf{k}}+\int\d^3 r'\d^3k'\braket{{\bf r}|G_+|{\bf r'}} \braket{{\bf r'}|{\bf k'}}\braket{{\bf k'}|T|{\bf k}}=$

$=e^{i{\bf k}\cdot{\bf r}} +\int\d^3 r'\d^3k' G_+({\bf r}, {\bf r'}) e^{i{\bf k'}\cdot{\bf r'}} \braket{{\bf k'}|T|{\bf k}}$

Plugging the representation of the Green function for $|{\bf r'}|\ll|{\bf r}|$ in:

$\psi({\bf r}) =e^{i{\bf k}\cdot{\bf r}} + {m\over2\pi\hbar^2}{e^{ikr}\over r} \int\d^3 r'\d^3k' e^{-i k{\bf r'}\cdot{\bf\hat r}} e^{i{\bf k'}\cdot{\bf r'}} \braket{{\bf k'}|T|{\bf k}}= =e^{i{\bf k}\cdot{\bf r}} + {m\over2\pi\hbar^2}{e^{ikr}\over r} \int\d^3 r'\d^3k' e^{i {\bf r'}\cdot({\bf k'}-k{\bf\hat r})} \braket{{\bf k'}|T|{\bf k}}= =e^{i{\bf k}\cdot{\bf r}} + {m\over2\pi\hbar^2}{e^{ikr}\over r} \int\d^3k' \delta({\bf k'}-k{\bf\hat r}) \braket{{\bf k'}|T|{\bf k}}= =e^{i{\bf k}\cdot{\bf r}} + {m\over2\pi\hbar^2}{e^{ikr}\over r} \braket{k{\bf\hat r}|T|{\bf k}}= =e^{i{\bf k}\cdot{\bf r}} + f(\theta,\phi)\, {e^{ikr}\over r}$

where the scattering amplitude $f(\theta,\phi)$ is:

$f(\theta,\phi)= {m\over2\pi\hbar^2} \braket{k{\bf\hat r}|T|{\bf k}} = {m\over2\pi\hbar^2} \braket{{\bf k'}|T|{\bf k}}$

Where ${\bf k'}=k{\bf\hat r}$ is the final momentum.

The differential cross section ${\d\sigma\over\d\Omega}$ is defined as the probability to observe the scattered particle in a given state per solid angle, e.g. the scattered flux per unit of solid angle per incident flux:

${\d\sigma\over\d\Omega} = {1\over|{\bf j}_i|}{\d n\over\d\Omega} = {r^2\over|{\bf j}_i|}{\d n\over r^2\d\Omega} = {r^2\over|{\bf j}_i|}{\d n\over \d S} = {r^2\over|{\bf j}_i|}\,{\bf j}_o\cdot {\bf n} = {r^2\over|{\bf j}_i|}\,{\bf j}_o\cdot {\bf \hat r} = = {r^2\over{\hbar k\over m}}\,{\hbar k\over m}\left({1\over r^2} +{i\over k r^3}\right)|f(\theta, \phi)|^2 = \left(1 +{i\over k r}\right)|f(\theta, \phi)|^2 \to |f(\theta, \phi)|^2$

where we used $|{\bf j}_i|={\hbar k\over m}$ and

${\bf j}_o\cdot {\bf \hat r} ={\hbar\over2 m i}\left( \psi^*\nabla\psi- \psi\nabla\psi^* \right)\cdot{\bf \hat r} ={\hbar\over2 m i}\left( \psi^*{\partial\over\partial r}\psi- \psi{\partial\over\partial r}\psi^* \right) =$

$={\hbar\over2 m i}\left( f^*(\theta, \phi){e^{-ikr}\over r}{\partial\over\partial r} \left(f(\theta, \phi){e^{ikr}\over r}\right)- f(\theta, \phi){e^{ikr}\over r}{\partial\over\partial r}\left(f^*(\theta, \phi){e^{-ikr}\over r}\right) \right)=$

$={\hbar k\over m}\left({1\over r^2}+{i\over k r^3} \right)|f(\theta, \phi)|^2$

Let’s write the explicit formula for the transition matrix:

$\braket{{\bf k'}|T|{\bf k}} =\int\d^3r\braket{{\bf k'}|{\bf r}}\braket{{\bf r}|V|{\bf k}} +\int\d^3r\d^3r'\braket{{\bf k'}|{\bf r}}\braket{{\bf r}|VG_+|{\bf r'}} \braket{{\bf r'}|V|{\bf k}}+\cdots=$

$=\int\d^3r e^{i({\bf k}-{\bf k'})\cdot{\bf r}}V({\bf r}) +\int\d^3r\d^3r'e^{-i{\bf k'}\cdot{\bf r}} V({\bf r}) {e^{i k|{\bf r}-{\bf r'}|}\over|{\bf r}-{\bf r'}|} V({\bf r'})e^{i{\bf k}\cdot{\bf r'}}+\cdots$

Born Approximation¶

The Born approximation is just the first term:

$\braket{{\bf k'}|T|{\bf k}} \approx\int\d^3r e^{i({\bf k}-{\bf k'})\cdot{\bf r}}V({\bf r}) = \int \d r\, \d\theta\,\d\phi\, e^{iqr\cos\theta}V(r) r^2\sin\theta = = 4\pi\int_0^\infty rV(r)\sin(qr)\,\d r$

We can also write it as:

$\braket{{\bf k'}|T|{\bf k}} \approx\int\d^3r e^{-i{\bf q}\cdot{\bf r}}V({\bf r}) = \tilde V({\bf q})$

where $\bf q=k'-k$ . Note that for $\bf |k'|\approx |k|$ we can write $|{\bf q}|$ using the angle $\theta$ between the vectors $\bf k'$ and $\bf k$ :

$|{\bf q}| = |{\bf k}' - {\bf k}| = \sqrt{k'^2 + k^2 - 2k'k\cos\theta} \approx \sqrt{k^2 + k^2 - 2k^2\cos\theta} = = \sqrt{2k^2 (1 -\cos\theta)} = \sqrt{4k^2 \sin^2 {\theta\over 2}} = 2k\sin\left(\theta\over2\right)$

Given the $\tilde V({\bf q})$ we can then calculate the scattering potential $V({\bf r})$ by the Fourier transform:

$V({\bf r}) = \int {\d^3 q\over (2\pi)^3} \tilde V({\bf q}) e^{i{\bf q}\cdot {\bf r}}$

Example 1:

$\tilde V({\bf q}) = - {g^2\over |{\bf q}|^2 + m_{\phi}^2} V({\bf r}) = \int {\d^3 q\over (2\pi)^3} {-g^2\over |{\bf q}|^2 + m_{\phi}^2} e^{i{\bf q}\cdot {\bf r}} = \cdots = - {g^2\over 4\pi} {1\over r} e^{-m_\phi r}$

Example 2:

$\tilde V({\bf q}) = {e^2\over |{\bf q}|^2} V({\bf r}) = \int {\d^3 q\over (2\pi)^3} {e^2\over |{\bf q}|^2} e^{i{\bf q}\cdot {\bf r}} = \cdots = {e^2\over 4\pi r}$

Example 3 — Yukawa potential in Born approximation:

$V(r) = -V_0 {e^{-\alpha r}\over r} \tilde V({\bf q}) = -{4\pi V_0\over |{\bf q}|^2 + \alpha^2} f(\theta,\phi) = {m\over2\pi\hbar^2} \braket{{\bf k'}|T|{\bf k}} = {m\over2\pi\hbar^2} \tilde V({\bf q}) = -{m\over2\pi\hbar^2} {4\pi V_0\over |{\bf q}|^2 + \alpha^2} = -{2m\over\hbar^2} {V_0\over |{\bf q}|^2 + \alpha^2} {\d\sigma\over\d\Omega} = |f(\theta, \phi)|^2 = \left(2mV_0\over \hbar^2\right)^2 {1\over\left(|{\bf q}|^2 + \alpha^2\right)^2} = \left(2mV_0\over \hbar^2\right)^2 {1\over\left(4k^2\sin^2\left(\theta\over2\right) + \alpha^2\right)^2} \sigma = \int {\d\sigma\over\d\Omega} \d\Omega = \int {\d\sigma\over\d\Omega} \sin\theta \d \theta\d\phi = = \left(2mV_0\over \hbar^2\right)^2\int {1\over\left(4k^2\sin^2\left(\theta\over2\right) + \alpha^2\right)^2} \sin\theta \d \theta\d\phi = = \left(2mV_0\over \hbar^2\right)^2 2\pi\int_0^\pi {\sin\theta\d\theta\over \left(4k^2\sin^2\left(\theta\over2\right) + \alpha^2\right)^2} = = \left(2mV_0\over \hbar^2\right)^2 2\pi\int_0^\pi {\sin\theta\d\theta\over \left(2k^2(1-\cos\theta) + \alpha^2\right)^2} = = \left(2mV_0\over \hbar^2\right)^2 2\pi\int_{-1}^1 {\d y\over \left(2k^2(1+y) + \alpha^2\right)^2} = = \left(2mV_0\over \hbar^2\right)^2 2\pi \int_{\alpha^2}^{4k^2+\alpha^2} {2k^2\d z\over z^2} = = \left(2mV_0\over \hbar^2\right)^2 2\pi 2k^2 \left({1\over\alpha^2} - {1\over 4k^2 + \alpha^2}\right)$

Example 4 — Coulomb potential in Born approximation:

$\alpha \to 0 {\d\sigma\over\d\Omega} = \left(2mV_0\over \hbar^2\right)^2 {1\over\left(4k^2\sin^2\left(\theta\over2\right)\right)^2} = \left(2mV_0\over 4\hbar^2k^2\right)^2 {1\over\sin^4{\theta\over2}} E = {p^2\over 2m} = {\hbar^2k^2\over 2m} {\d\sigma\over\d\Omega} = \left(V_0\over 4 E\right)^2 {1\over\sin^4{\theta\over2}} V_0 \to {Z Z' e^2 \over 4\pi \epsilon_0} = Z Z' \alpha \hbar c {\d\sigma\over\d\Omega} = \left(ZZ'\alpha\hbar c\over 4 E\right)^2 {1\over\sin^4{\theta\over2}}$

By setting $E=\half m v_0^2$ we obtain the classical Rutherford cross-section formula.

8.5. Systematic Perturbation Theory in QM¶

We have

$H = H_0 + e^{-\epsilon |t|} H_1$

where the ground state of the noninteracting Hamiltonian $H_0$ is:

$H_0\ket{0} = E_0\ket{0}$

and the ground state of the interacting Hamiltonian $H$ is:

$H\ket{\Omega} = E\ket{\Omega}$

Then:

$H\ket{\Omega} = (H_0 + H_1)\ket{\Omega} = E\ket{\Omega} \braket{0|H_0 + H_1|\Omega} = E\braket{0 | \Omega} E_0\braket{0|\Omega} + \braket{0|H_1|\Omega} = E\braket{0 | \Omega} E = E_0 + {\braket{0|H_1|\Omega}\over\braket{0 | \Omega}}$

We can also write

$\ket{\Omega} = \lim_{\epsilon\to0+} U_\epsilon(0, -\infty)\ket{0}$

where

$U_\epsilon(t, t_0) = T \exp\left(-{i\over\hbar}\int_{t_0}^t \d t' e^{-\epsilon|t'|} H_1(t')\right)$

Let’s write several common expressions for the ground state energy:

$\Delta E = E - E_0 = {\braket{0|H_1|\Omega}\over\braket{0 | \Omega}} = {\braket{0|H_1 U(0, -\infty)|0}\over\braket{0 |U(0, -\infty)|0}} = = \lim_{t\to0} {\braket{0|H_1 U(t, -\infty)|0}\over \braket{0 |U(t, -\infty)|0}} = \lim_{t\to0} {\braket{0|i\partial_t U(t, -\infty)|0}\over \braket{0 |U(t, -\infty)|0}} = \lim_{t\to0} {i\partial_t\braket{0| U(t, -\infty)|0}\over \braket{0 |U(t, -\infty)|0}} = = \lim_{t\to0} i\partial_t\log\braket{0| U(t, -\infty)|0} \equiv \lim_{t\to\infty(1-i\epsilon)} i{\d\over\d t}\log \braket{0| U(t, -\infty)|0}$

The last expression incorporates the $\epsilon$ dependence of $U_\epsilon$ explicitly. The vacuum amplitude is sometimes denoted by $R(t)$ :

$R(t) = \braket{0| U(t, -\infty)|0}$

The two point (interacting) Green (or correlation) function is:

$G(x, y) = \braket{\Omega|T\phi(x)\phi(y)|\Omega} = {\braket{0|T\phi(x)\phi(y)U(\infty, -\infty)|0}\over \braket{0|U(\infty, -\infty)|0}}$

The $\epsilon\to0$ limit of $U_\epsilon$ is tacitly assumed to make this formula well defined (sometimes the other way $t\to\infty(1-i\epsilon)$ of writing the same limit is used). Another way of writing the formula above for the Green function in QM is:

$G({\bf k}_1, {\bf k}_2, t_2-t_1) = i \braket{\Omega|T c_{{\bf k}_2}(t_2)c_{{\bf k}_1}^\dag(t_1)|\Omega} = i {\braket{0|T c_{{\bf k}_2}(t_2)c_{{\bf k}_1}^\dag(t_1) U(\infty, -\infty)|0}\over \braket{0|U(\infty, -\infty)|0}}$

Last type of similar expressions to consider is the scattering amplitude:

$\braket{f|U(\infty, -\infty)|i}$

where the initial state is let’s say a boson+fermion and the final state a boson+antifermion:

$\ket{i} = a_{\bf k}^\dag b_{\bf l}^{s\dag} \ket{0} \ket{f} = a_{\bf p}^\dag a_{\bf q}^{r\dag} \ket{0}$

This is just an example, the $\ket{i}$ and $\ket{f}$ states can contain any number of (arbitrary) particles.

8.6. Appendix¶

8.6.1. Units and Dimensional Analysis¶

The evolution operator is dimensionless:

$U(-\infty,\infty) = T\exp\left({i\over\hbar}\int_{-\infty}^{\infty}\d^4 x \L(x) \right)$

So:

$\left[\int_{-\infty}^{\infty}\d^4 x \L(x) \right] = [\hbar] = M^0$

where $M$ is an arbitrary mass scale. Length unit is $M^{-1}$ , so then

$[\L(x)] = M^4$

For the particular forms of the Lagrangians above we get:

$[m\bar ee] = [m^2 Z_\mu Z^\mu] = [m^2 H^2] = [i\bar e\gamma^\mu\partial_\mu e] = [\L] = M^4$

so $[\bar ee] = M^3$ , $[Z_\mu Z^\mu]=[H^2] = M^2$ and we get

$[e] = [\bar e] = M^{3\over2}$

$[Z_\mu] = [Z^\mu] = [H] = [\partial_\mu] = [\partial^\mu] = M^1$

Example: what is the dimension of $G_\mu$ in $\L = -{G_\mu\over\sqrt2} [\bar \psi_{\nu_\mu}\gamma^\mu (1-\gamma_5) \psi_\mu] [\bar \psi_e\gamma^\mu (1-\gamma_5) \psi_{\nu_e}]$ ? Answer:

$[\L] = [G_\mu \bar\psi\psi\bar\psi\psi]$

$M^4 = [G_\mu] M^{3\over2}M^{3\over2}M^{3\over2}M^{3\over2}$

$[G_\mu] = M^{-2}$

In order to get the above units from the SI units, one has to do the following identification:

$kg\to M^1$

$m\to M^{-1}$

$s\to M^{-1}$

$A\to M^1$

The SI units of the above quantities are:

$[\phi] = \rm V={kg\,m^2\over A\,s^3}=M [A_\mu]={[\phi]\over [c]}=\rm{V\,s\over m} = {kg\, m\over A\,s^2}=M [c]=\rm {m\over s} = 1 [e]=\rm C = A\, s=1 [\hbar]=\rm J\,s = {m^2\,kg\over s}=1 [\partial_\mu]=\rm {1\over m}=M [F_{\mu\nu}]=[\partial_\mu A_\nu]=\rm {kg\over A\,s^2}=M^2 [\L]=[F_{\mu\nu}]^2=\rm {kg^2\over A^2\,s^4}=M^4 [\psi]=\rm {kg^{1\over2}\over A\,m\,s}=M^{3\over2}$

The SI units are useful for checking that the $c$ , $e$ and $\hbar$ constants are at correct places in the expression.

8.6.2. Atomic Units¶

Hartree atomic units are defined using the relations:

$\hbar = m = e = 4\pi\epsilon_0 = 1$

so for example for the Bohr radius we get:

$a_0 = {4\pi\epsilon_0 \hbar^2 \over m e^2} = 1$

for fine structure constant ( $\alpha=1/137.036...$ ) we get:

$\alpha = {e^2\over 4\pi\epsilon_0 \hbar c} = {1\over c}$

from which we calculate the speed of light $c$ in atomic units as:

$c = {1\over\alpha}$

Energy is measured in Hartrees, one Hartree being

$1{\rm\,Ha} = {\hbar^2\over m a_0^2} = 1{\rm\,(a.u.)} = 27.211\rm\,eV$

Hamiltonian and the corresponding spectrum of the Hydrogen atom:

$H = -{\hbar^2\over 2m} \nabla^2 - {1\over 4\pi\epsilon_0} {e^2\over r} E_n = -{\hbar^2\over m a_0^2} {1\over 2n^2}$

become in atomic units:

$H = -{1\over 2} \nabla^2 - {1\over r} E_n = -{1\over 2n^2}$

Poisson equation (Gauss’s law)

$\nabla^2\phi = -{\rho\over\epsilon_0}$

becomes:

$\nabla^2\phi = -{4\pi\rho}$

8.6.3. Tensors in Special Relativity and QFT¶

In general, the covariant and contravariant vectors and tensors work just like in special (and general) relativity. We use the metric $g_{\mu\nu}={\rm diag}(1, -1, -1, -1)$ (e.g. signature -2, but it’s possible to also use the metric with signature +2). The four potential $A^\mu$ is given by:

$A^\mu=\left({\phi\over c}, {\bf A}\right) = (A^0, A^1, A^2, A^3)$

where $\phi$ is the electrostatic potential. Whenever we write $\bf A$ , the components of it are given by the upper indices, e.g. ${\bf A}=(A^1, A^2, A^3)$ . The components with lower indices can be calculated using the metric tensor, so it depends on the signature convention:

$A_\mu=g_{\mu\nu}A^\nu=(A^0, -{\bf A}) = (A^0, -A^1, -A^2, -A^3)$

In our case we got $A_0=A^0$ and $A_i = -A^i$ (if we used the other signature convention, then the sign of $A_0$ would differ and $A_i$ would stay the same). The length (squared) of the vector is:

$A^2 = A_\mu A^\mu = \left(A^0\right)^2 - \left| {\bf A} \right|^2 = \left(A^0\right)^2 - {\bf A}^2$

where ${\bf A}^2 \equiv |{\bf A}|^2 = (A^1)^2+(A^2)^2+(A^3)^2$ .

The position 4-vector is (in any metric):

$x^\mu = (ct, {\bf x})$

Gradient is defined as (in any metric):

$\partial_\mu = (\partial_0, \partial_1, \partial_2, \partial_3) = {\partial\over\partial x^\mu}= \left({1\over c}{\partial\over\partial t},{\partial\over\partial x},{\partial\over\partial y},{\partial\over\partial z}\right)$

the upper indices depend on the signature, e.g. for -2:

$\partial^\mu = (\partial^0, \partial^1, \partial^2, \partial^3)= \left({1\over c}{\partial\over\partial t},-{\partial\over\partial x},-{\partial\over\partial y},-{\partial\over\partial z}\right)$

and +2:

$\partial^\mu = (\partial^0, \partial^1, \partial^2, \partial^3)= \left(-{1\over c}{\partial\over\partial t},{\partial\over\partial x},{\partial\over\partial y},{\partial\over\partial z}\right)$

The d’Alembert operator is:

$\partial^2 \equiv \partial_\mu \partial^\mu$

the 4-velocity is (in any metric):

$v^\mu = {\d x^\mu\over\d\tau} = {\d t\over\d\tau}{\d x^\mu\over\d t} = \gamma(c, {\bf v})$

where $\tau$ is the proper time, $\gamma={\d t\over\d\tau}={1\over\sqrt{1 - {{\bf v}^2\over c^2}}}$ and ${\bf v}={\d {\bf x}\over\d t}$ is the velocity in the coordinate time $t$ . In the metric with signature +2:

$v^2 = v_\mu v^\mu = g_{\mu\nu}v^\mu v^\nu = -\gamma^2 c^2 + \gamma^2{\bf v}^2 = {-c^2 + {\bf v}^2 \over 1 - {{\bf v}^2\over c^2}} = -c^2$

With signature -2 we get $v^2 = c^2$ . The 4-momentum is (in any metric)

$p^\mu = m v^\mu = m\gamma(c, {\bf v})$

where $m$ is the rest mass. The fluid-density 4-current is (in any metric):

$j^\mu = \rho v^\mu = \rho\gamma(c, {\bf v})$

where $\rho$ is the fluid density at rest. For example the vanishing 4-divergence (the continuity equation) is written as (in any metric):

$0 = \partial_\mu j^\mu = {1\over c}{\partial\over\partial t} (\rho\gamma c) + \nabla \cdot (\rho\gamma {\bf v}) = {\partial\over\partial t} (\rho\gamma) + \nabla \cdot (\rho{\bf v}\gamma) = {\partial\over\partial t}\left(\rho\over \sqrt{1-{{\bf v}^2\over c^2}}\right) + \nabla \cdot\left(\rho{\bf v}\over \sqrt{1-{{\bf v}^2\over c^2}}\right)$

Momentum ( ${\bf p}=-i\hbar\nabla$ ) and energy ( $E=i\hbar{\partial\over\partial t}$ ) is combined into 4-momentum as

$p^\mu = \left({E\over c},{\bf p}\right) = i\hbar\left({1\over c}{\partial\over\partial t},-\nabla\right) = i\hbar\left(\partial_0,-\partial_j\right) = i\hbar\left(\partial^0,\partial^j\right) = i\hbar\partial^\mu p_\mu = g_{\mu\nu}p^\nu = i\hbar g_{\mu\nu}\partial^\nu = i\hbar\partial_\mu$

For the signature $+2$ we get $p^\mu = -i\hbar\partial^\mu$ and $p_\mu = -i\hbar\partial_\mu$ .

For $p^2$ we get (signature -2):

$p^2 = p_\mu p^\mu = (p^0)^2 - {\bf p}^2 = (p_0)^2 - {\bf p}^2 = {E^2\over c^2} - {\bf p}^2 p^2 = p_\mu p^\mu = m^2 v_\mu v^\mu = m^2 c^2$

comparing those two we get the following useful relations (valid in any metric):

${E^2\over c^2} - {\bf p}^2 = m^2 c^2 E^2 = m^2 c^4 + {\bf p}^2 c^2 E = \sqrt{m^2c^4 + {\bf p}^2c^2} = mc^2\sqrt{1 + {{\bf p}^2\over m^2c^2}} = mc^2\left(1 + {{\bf p}^2\over 2m^2c^2} + O\left({p^4\over m^4c^4}\right) \right) = = mc^2 + {{\bf p}^2\over 2m} + O\left(p^4\over m^3c^2\right)$

the following relations are also useful:

$p^2 = p_\mu p^\mu = -\hbar^2\partial_\mu \partial^\mu \equiv -\hbar^2\partial^2 = -\hbar^2\left(\partial_0\partial^0 + \partial_i\partial^i\right) = -\hbar^2\left(\partial_0\partial_0 - \partial_i\partial_i\right) = = -\hbar^2\left({1\over c^2}{\partial^2\over\partial t^2} - \nabla^2\right) = -{\hbar^2\over c^2}{\partial^2\over\partial t^2} + \hbar^2\nabla^2$

For the signature $+2$ we get:

$p^2 = p_\mu p^\mu = -\hbar^2\partial_\mu \partial^\mu \equiv -\hbar^2\partial^2 = -\hbar^2\left(\partial_0\partial^0 + \partial_i\partial^i\right) = -\hbar^2\left(-\partial_0\partial_0 + \partial_i\partial_i\right) = = -\hbar^2\left(-{1\over c^2}{\partial^2\over\partial t^2} + \nabla^2\right) = {\hbar^2\over c^2}{\partial^2\over\partial t^2} - \hbar^2\nabla^2$

So for example the Klein-Gordon equation:

$\left({\hbar^2\over c^2}{\partial^2\over\partial t^2} - \hbar^2\nabla^2 +m^2c^2\right)\psi = 0$

can be for signature $-2$ written as:

$(+\hbar^2\partial^2 + m^2 c^2)\psi = (-p^2 + m^2c^2)\psi = 0$

and for $+2$ as:

$(-\hbar^2\partial^2 + m^2 c^2)\psi = (p^2 + m^2c^2)\psi = 0$

Note: for the signature +2, we would get $p^\mu=-i\hbar\partial^\mu$ and $p_\mu=-i\hbar\partial_\mu$ .

For the minimal coupling $D_\mu = \partial_\mu + {i\over\hbar}e A_\mu$ we get:

$D^0 = \partial^0 + {i\over\hbar}e A^0$

$D^j = \partial^j + {i\over\hbar}e A^j=-{i\over\hbar}(i\hbar\partial^j-eA^j) =-{i\over\hbar}({\bf p}-e{\bf A})$

and for the lower indices:

$D_0 = \partial_0 + {i\over\hbar}e A_0$

$D_j = \partial_j + {i\over\hbar}e A_j=-{i\over\hbar}(i\hbar\partial_j-eA_j) ={i\over\hbar}(i\hbar\partial^j-eA^j) ={i\over\hbar}({\bf p}-e{\bf A})$

8.6.4. Adding Angular Momenta¶

Angular momenta are added using the Clebsch-Gordan coefficients (or equivalently $3j$ symbols):

(8.6.4.1)¶ $\ket{j_1 j_2 j_3 m_3} = \sum_{m_1 m_2} (j_1 m_1 j_2 m_2 | j_3 m_3 ) \ket{j_1 m_1 j_2 m_2} = = \sum_{m_1 m_2} (-1)^{j_1-j_2+m_3}\sqrt{2j_3+1} \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & -m_3 \end{pmatrix} \ket{j_1 m_1 j_2 m_2}$

Spin Orbit Coupling (Spin Spherical Harmonics)¶

This is just a special case of (8.6.4.1) for:

$j_1 = l m_1 = m j_2 = \half m_2 = s$

So the kets $\ket{j_1 m_1 j_2 m_2}$ can be written as:

$\ket{j_1 m_1 j_2 m_2} = \ket{l m \half s} = Y_{lm} \Phi_s$

Where:

$\Phi_{1\over2} = \begin{pmatrix}1\\0\end{pmatrix} \Phi_{-{1\over2}} = \begin{pmatrix}0\\1\end{pmatrix}$

Where $s$ is a spin, $s=\pm\half$ . Then we get:

$\ket{l \half j_3 m_3} = \sum_{m=-l}^l \sum_{s=-\half}^\half (-1)^{l-\half+m_3}\sqrt{2j_3+1} \begin{pmatrix} l & \half & j_3 \\ m & s & -m_3 \end{pmatrix} \ket{l m \half s} = = (-1)^{l-\half+m_3}\sqrt{2j_3+1} \sum_{m=-l}^l \left( \begin{pmatrix} l & \half & j_3 \\ m & -\half & -m_3 \end{pmatrix} \ket{l m \half (-\half)}\right. + +\left. \begin{pmatrix} l & \half & j_3 \\ m & \half & -m_3 \end{pmatrix} \ket{l m \half \half}\right)= =(-1)^{l-\half+m_3}\sqrt{2j_3+1} \left( \begin{pmatrix} l & \half & j_3 \\ m_3+\half & -\half & -m_3 \end{pmatrix} \ket{l (m_3+\half) \half (-\half)}\right. + +\left. \begin{pmatrix} l & \half & j_3 \\ m_3-\half & \half & -m_3 \end{pmatrix} \ket{l (m_3-\half) \half \half}\right)= =(-1)^{l-\half+m_3}\sqrt{2j_3+1} \begin{pmatrix} \begin{pmatrix} l & \half & j_3 \\ m_3-\half & \half & -m_3 \end{pmatrix} Y_{l, m_3-\half} \\ \begin{pmatrix} l & \half & j_3 \\ m_3+\half & -\half & -m_3 \end{pmatrix} Y_{l, m_3+\half} \end{pmatrix}$

These are called spin-angular functions or spin spherical harmonics. Using the triangle selection rule of the $3j$ symbols, we can see that there are only two options for $j_3$ :

$j_3 = l+\half j_3 = l-\half$

So we get for $j_3=l+\half$ :

$\ket{(j_3-\half) \half j_3 m_3} =(-1)^{j_3-\half-\half+m_3}\sqrt{2j_3+1} \begin{pmatrix} \begin{pmatrix} j_3-\half & \half & j_3 \\ m_3-\half & \half & -m_3 \end{pmatrix} Y_{j_3-\half, m_3-\half} \\ \begin{pmatrix} j_3-\half & \half & j_3 \\ m_3+\half & -\half & -m_3 \end{pmatrix} Y_{j_3-\half, m_3+\half} \end{pmatrix} = =(-1)^{j_3+m_3-1}\sqrt{2j_3+1} \begin{pmatrix} (-1)^{j_3+m_3-1}\sqrt{j_3+m_3\over 2 j_3 (2j_3+1)} Y_{j_3-\half, m_3-\half} \\ (-1)^{2j_3} (-1)^{j_3-m_3-1}\sqrt{j_3-m_3\over 2 j_3 (2j_3+1)} Y_{j_3-\half, m_3+\half} \end{pmatrix} = = \begin{pmatrix} \sqrt{j_3+m_3\over 2 j_3} Y_{j_3-\half, m_3-\half} \\ (-1)^{4j_3}\sqrt{j_3-m_3\over 2 j_3} Y_{j_3-\half, m_3+\half} \end{pmatrix} = = {1\over \sqrt{2j_3}}\begin{pmatrix} \sqrt{j_3+m_3}\, Y_{j_3-\half, m_3-\half} \\ \sqrt{j_3-m_3}\, Y_{j_3-\half, m_3+\half} \end{pmatrix} = = {1\over\sqrt{2l+1}}\begin{pmatrix} \sqrt{l+m_3+\half}\, Y_{l, m_3-\half} \\ \sqrt{l-m_3+\half}\, Y_{l, m_3+\half} \end{pmatrix}$

The allowed values for $m_3$ are $m_3 = -l-\half, -l+\half, -l+\half +1, \dots, l-\half, l+\half$ , total of $2l+2$ values. For the case $l=\pm(l+\half)$ , the spherical harmonic is not defined ( $m > l$ ) but its coefficient (the square root $\sqrt{l \pm m_3 + \half}$ ) is zero, so the whole element is defined as zero.

For $j_3=l-\half$ :

$\ket{(j_3+\half) \half j_3 m_3} =(-1)^{j_3+\half-\half+m_3}\sqrt{2j_3+1} \begin{pmatrix} \begin{pmatrix} j_3+\half & \half & j_3 \\ m_3-\half & \half & -m_3 \end{pmatrix} Y_{j_3+\half, m_3-\half} \\ \begin{pmatrix} j_3+\half & \half & j_3 \\ m_3+\half & -\half & -m_3 \end{pmatrix} Y_{j_3+\half, m_3+\half} \end{pmatrix} = =(-1)^{j_3+m_3}\sqrt{2j_3+1} \begin{pmatrix} (-1)^{2j_3+1}(-1)^{j_3-m_3}\sqrt{j_3-m_3+1 \over (2j_3+1)(2j_3+2)} Y_{j_3+\half, m_3-\half} \\ (-1)^{j_3+m_3}\sqrt{j_3+m_3+1 \over (2j_3+1)(2j_3+2)} Y_{j_3+\half, m_3+\half} \end{pmatrix} = = \begin{pmatrix} (-1)^{4j_3+1}\sqrt{j_3-m_3+1 \over 2j_3+2} Y_{j_3+\half, m_3-\half} \\ \sqrt{j_3+m_3+1 \over 2j_3+2} Y_{j_3+\half, m_3+\half} \end{pmatrix} = = {1\over\sqrt{2j_3+2}}\begin{pmatrix} -\sqrt{j_3-m_3+1}\, Y_{j_3+\half, m_3-\half} \\ \sqrt{j_3+m_3+1}\, Y_{j_3+\half, m_3+\half} \end{pmatrix} = = {1\over\sqrt{2l+1}}\begin{pmatrix} -\sqrt{l-m_3+\half}\, Y_{l, m_3-\half} \\ \sqrt{l+m_3+\half}\, Y_{l, m_3+\half} \end{pmatrix}$

The allowed values for $m_3$ are $m_3 = -l+\half, -l+\half +1, \dots, l-\half$ , total of $2l$ values (in particular, the values $m_3=\pm (l+\half)$ are not allowed).

The last formula is the spin spherical harmonics given in terms of $l,m_3$ , the second last formula is in terms of $j_3, m_3$ (both are used). The spin spherical harmonics is usually denoted by $\chi_\kappa^{m_3}$ or $y^{j_3, m_3}_l$ . See the next section for the definition of $\kappa$ .

Kappa¶

In order to define the state, one needs to specify both $j_3$ and $l$ (distinguishng the two cases $j_3 = l \pm \half$ ). This can be unified into just one integer $\kappa$ , where $-\hbar \kappa$ is defined as the eigenvalue of the operator:

$K = {\bsigma \cdot {\bf L}} + \hbar = \left({2\over \hbar^2} {\bf S} \cdot {\bf L} + 1 \right) \hbar = \left({1\over \hbar^2}\left({\bf J}^2 - {\bf L}^2 - {\bf S}^2\right) + 1 \right) \hbar$

Then:

$K \psi = \left({1\over \hbar^2}\left({\bf J}^2 - {\bf L}^2 - {\bf S}^2\right) + 1 \right) \hbar \psi = = \left(j_3 (j_3+1) - l(l+1)-s(s+1) + 1 \right) \hbar \psi = = \left(j_3 (j_3+1) - l(l+1) + {1\over 4} \right) \hbar \psi = = -\kappa \hbar \psi$

from which

$\kappa = -j_3 (j_3+1) + l(l+1) - {1\over 4} = = \begin{cases} -j_3 (j_3+1) + (j_3-\half)(j_3-\half+1) - {1\over 4}; & \text{for $j_3=l+\half$} \\ -j_3 (j_3+1) + (j_3+\half)(j_3+\half+1) - {1\over 4}; & \text{for $j_3=l-\half$} \\ \end{cases} = = \begin{cases} -(j_3+\half); & \text{for $j_3=l+\half$} \\ +(j_3+\half); & \text{for $j_3=l-\half$} \end{cases} = = \begin{cases} -l -1; & \text{for $j_3=l+\half$} \\ l; & \text{for $j_3=l-\half$} \end{cases}$

The opposite relation is:

$l = \begin{cases} -\kappa -1; & \text{for $\kappa < 0$, equivalently $j_3=l+\half$} \\ \kappa; & \text{for $\kappa > 0$, equivalently $j_3=l-\half$} \end{cases}$

Code:

>>> from sympy import var, S
>>> var("j l")
(j, l)
>>> k = -j*(j+1) + l*(l+1) - S(1)/4
>>> k.subs(l, j-S(1)/2).expand()
-j - 1/2
>>> k.subs(l, j+S(1)/2).expand()
j + 1/2

Some useful relations with $\kappa$ that follow from the above for both cases $j_3 = l \pm \half$ :

$l(l+1) = \kappa (\kappa+1) l = |\kappa + \half| - \half j_3 = |\kappa| - \half$

In order to enumerate all possibilities, one needs to count all integers except zero: $\kappa=-1, 1, -2, 2, -3, 3, \dots$ :

$\begin{array}{rrrrcc} \kappa & l & j_3 & j_3 - l & \mbox{degeneracy} & \mbox{label} \\ \hline -1 & 0 & 0.5 & 0.5 & 2 & s_{1/2} \\ 1 & 1 & 0.5 & -0.5 & 2 & p_{1/2} \\ -2 & 1 & 1.5 & 0.5 & 4 & p_{3/2} \\ 2 & 2 & 1.5 & -0.5 & 4 & d_{3/2} \\ -3 & 2 & 2.5 & 0.5 & 6 & d_{5/2} \\ 3 & 3 & 2.5 & -0.5 & 6 & f_{5/2} \\ -4 & 3 & 3.5 & 0.5 & 8 & f_{7/2} \\ 4 & 4 & 3.5 & -0.5 & 8 & g_{7/2} \\ -5 & 4 & 4.5 & 0.5 & 10 & g_{9/2} \\ 5 & 5 & 4.5 & -0.5 & 10 & h_{9/2} \\ -6 & 5 & 5.5 & 0.5 & 12 & h_{11/2} \\ 6 & 6 & 5.5 & -0.5 & 12 & i_{11/2} \\ \cdots & & & \\ \end{array}$

The degeneracy of the individual states for each $\kappa$ is equal to $2j_3+1=2|\kappa|$ (which is equal to $2l+2$ for $j_3=l+\half$ and $2l$ for $j_3=l-\half$ , see the previous section), that is 2, 4, 6, 8 for $j_3=0.5, 1.5, 2.5, 3.5$ (or equivalently $\kappa=\pm 1, \pm 2, \pm 3, \pm 4$ ) respectively. All states together with the given $l$ have total degeneracy $2l+2+2l=2(2l+1)$ , that is 2, 6, 10, 14 for $l=0, 1, 2, 3$ respectively.

The states are labeled by a letter corresponding to $l=0, 1, 2, 3, \dots$ (s, p, d, f, g, h, i, j, k, l, m, n, o, q, r, t, u, v, w, x, y, z, a, b, c, e, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z, A, B, C, D, E) with a subscript equal to the total angular momentum $j_3=n/2$ with $n=1, 3, 5, 7, \dots$ .