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Quantum Mechanics

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}

where

\psi=\left( \begin{array}{c} \psi_1 \\ \psi_2 \\ \psi_3 \\ \psi_4 \\ \end{array}\right)

and

D_\mu=\partial_\mu+{i\over \hbar}eA_\mu

is the gauge covariant derivative and (e is the elementary charge, which is 1 in atomic units, i.e. the electron has a charge -e)

F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu

is the electromagnetic field tensor. It’s astonishing, that this simple Lagrangian can account for all phenomena from macroscopic scales down to something like 10^{-13}\rm\,cm. So it’s not a surprise that Feynman, Schwinger and Tomonaga received the 1965 Nobel Prize in Physics for such a fantastic achievement.

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

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

Perturbation Theory

We want to solve the equation:

(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:

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

Substituting this into (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:

(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 (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 (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 (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}

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' {\braket{{\bf r}|{\bf k'}}\braket{\bf{k'}|\bf{r'}}\over E_k-E_{k'}+i\epsilon} =\int\d^3k' {e^{i{\bf k'}\cdot({\bf r}-{\bf r'})}\over E_k-E_{k'}+i\epsilon} ={2m\over\hbar^2}\int\d^3k' {e^{i{\bf k'}\cdot({\bf r}-{\bf r'})}\over k^2-{k'}^2+i\epsilon}=

={4\pi m\over\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\hbar^2i|{\bf r}-{\bf r'}|} (2\pi i)k{e^{i k|{\bf r}-{\bf r'}|}\over 2k}=

={4\pi^2 me^{i k|{\bf r}-{\bf r'}|}\over\hbar^2|{\bf r}-{\bf r'}|}

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 ={4\pi^2 me^{-i k|{\bf r}-{\bf r'}|}\over\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)

and simplify the result even further:

G_+({\bf r}, {\bf r'}) ={4\pi^2 m\over\hbar^2}{e^{ikr}\over r} e^{-i k{\bf r'}\cdot{\bf\hat r}}

G_-({\bf r}, {\bf r'}) ={4\pi^2 m\over\hbar^2}{e^{-ikr}\over r} e^{i k{\bf r'}\cdot{\bf\hat r}}

Note: both functions may be divided by the factor (2\pi)^3 due to the momentum integration.

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}} + {4\pi^2 m\over\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}} + {4\pi^2 m\over\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}} + {4\pi^2 m\over\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}} + {4\pi^2 m\over\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)= {4\pi^2 m\over\hbar^2} \braket{k{\bf\hat r}|T|{\bf k}} = {4\pi^2 m\over\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=

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

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.

Appendix

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.

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})

Multipole expansion

{1\over |{\bf r}-{\bf r'}|} ={1\over \sqrt{({\bf r}-{\bf r'})^2}} ={1\over \sqrt{r^2-2{\bf r}\cdot {\bf r'} + r'^2}} ={1\over r\sqrt{1-2\left(r'\over r\right){\bf\hat r}\cdot {\bf\hat r'} + \left(r'\over r\right)^2}} =

={1\over r}\sum_{l=0}^\infty\left(r'\over r\right)^l P_l({\bf\hat r}\cdot {\bf\hat r'}) =

={1\over r}\left( P_0({\bf\hat r}\cdot {\bf\hat r'}) + P_1({\bf\hat r}\cdot {\bf\hat r'}){r'\over r} + P_2({\bf\hat r}\cdot {\bf\hat r'})\left(r'\over r\right)^2 + O\left(r'^3\over r^3\right) \right) =

={1\over r}\left( 1 + {\bf\hat r}\cdot {\bf\hat r'} {r'\over r} + \half\left(3({\bf\hat r}\cdot {\bf\hat r'})^2-1\right)\left(r'\over r\right)^2 + O\left(r'^3\over r^3\right) \right) =

={1\over r} +{{\bf r}\cdot {\bf r'}\over r^3} +{3({\bf r}\cdot {\bf r'})^2-r^2r'^2\over 2r^5} + O\left(r'^3\over r^4\right)

We can also use the formula:

\sum_m \braket{{\bf\hat r}|lm}\braket{lm|{\bf\hat r}'}={4\pi\over 2l+1} \braket{{\bf\hat r}\cdot{\bf\hat r'}|P_l}

and rewrite the expansion using spherical harmonics:

{1\over |{\bf r}-{\bf r'}|} ={1\over r}\sum_{l=0}^\infty\left(r'\over r\right)^l P_l({\bf\hat r}\cdot {\bf\hat r'}) =

={1\over r}\sum_{l,m}\left(r'\over r\right)^l {2l+1\over4\pi}\braket{{\bf\hat r}|lm}\braket{lm|{\bf\hat r}'} ={1\over r}\sum_{l,m}\left(r'\over r\right)^l {2l+1\over4\pi}Y_{lm}({\bf\hat r})Y_{lm}^*({\bf\hat r}')

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