# Quantum field theory 1, lecture 21

#### 10.3 Functional integral for photons.

For photons, the field one integrates over in the functional integral is the gauge field $$A_\mu (x)$$. The field theory is described by the partition function \begin{equation*} \begin{split} Z_2[J] &= \int DA~\exp \left [iS_2[A]+i\int J^\mu A_\mu \right ]\\ &= \int DA ~ \exp \left [i \int d^4x~\left \{-\frac{1}{4} F^{\mu \nu }F_{\mu \nu } + J^\mu A_\mu \right \}\right ]. \end{split} \end{equation*} One can go to momentum space as usual \begin{equation*} A_\mu (x) = \int \frac{d^4 p}{(2\pi )^4}~e^{ipx} A_\mu (p), \end{equation*} and finds for the term in the exponential \begin{equation*} \begin{split} & \int _x\left \{-\frac{1}{4}F^{\mu \nu }F_{\mu \nu }+J^\mu A_\mu \right \} \\ & =\frac{1}{2}\int \frac{d^4 p}{(2\pi )^4}\left \{-A_\mu (-p) \left (p^2\eta ^{\mu \nu }-p^{\mu }p^{\nu } \right ) ~A_{\nu }(p)+J^{\mu }(-p) A_{\mu }(p) + A_{\mu }(-p)J^{\mu }(p)\right \}. \end{split} \end{equation*}

Attempt to invert the inverse propagator and gauge fixing. The next step would now be to perform the Gaussian integral over $$A_\mu$$ by completing the square. However, a problem arises here: The “inverse propagator” for the gauge field \begin{equation*} p^2\eta ^{\mu \nu }-p^\mu p^\nu = p^2 \mathscr{P}^{\mu \nu }(p), \end{equation*} is not invertible. We wrote it here in terms of \begin{equation*} \mathscr{P}_{\mu }^{\;\;\nu }(p) = \delta _{\mu }^{\;\;\nu }-\frac{p_\mu p^\nu }{p^2}, \end{equation*} which is in fact a projector to the space orthogonal to $$p_\nu$$ \begin{equation*} \mathscr{P}_{\mu }^{\;\;\nu }(p) \mathscr{P}_{\nu }^{\;\;\rho }(p) =\mathscr{P}_{\mu }^{\;\;\rho }(p). \end{equation*} As a projector matrix it has eigenvalues $$0$$ and $$1$$, only. However, \begin{equation*} \mathscr{P}_{\mu }^{\;\;\nu }(p) \; p_\nu = 0. \end{equation*} The field $$A_\nu (p)$$ can be decomposed into two parts, \begin{equation*} A_\nu (p) = \frac{i}{e}p_\nu \beta (p) + \hat{A}_\nu (p), \end{equation*} with \begin{equation*} \hat{A}_{\nu }(p) = \mathscr{P}_{\nu }^{\;\;\rho }(p) A_{\rho }(p), \end{equation*} such that $$p^\nu \hat{A}_\nu (p) = 0.$$ Moreover \begin{equation*} \beta (p)=\frac{e}{ip^2}p^\nu A_\nu (p). \end{equation*} When acting on $$\hat{A}_\nu (p)$$, the projector $$\mathscr{P}_{\mu }^{\;\;\nu }(p)$$ is simply the unit matrix.

Recall that gauge transformations shift the field according to \begin{equation*} A_\mu (x) \to \frac{1}{e}\partial _\mu \alpha + A_\mu (x) \end{equation*} or in momentum space \begin{equation*} A_\mu (p) \to \frac{i}{e}p_\mu \alpha (p) + A_\mu (p). \end{equation*} One can therefore always perform a gauge transformation such that $$\beta (p) =0$$ or \begin{equation*} \partial ^\mu A_\mu (x) =0. \end{equation*} This is known as Lorenz gauge or Landau gauge. We will use this gauge in the following and restrict the functional integral to field configurations that fulfil the gauge condition.

Quadratic partition function. Now we can easily perform the Gaussian integral, \begin{equation*} \begin{split} Z_2[J] = & \int DA\; \exp \left [\frac{i}{2}\int _p\left \{-\left (A_\mu (-p)-J_\rho (-p)\frac{\mathscr{P}^{\rho }_{\;\;\mu }}{p^2}\right ) p^2\mathscr{P}^{\mu \nu } \left (A_\nu (p)-\frac{\mathscr{P}_{\nu }^{\;\;\sigma }}{p^2}J_\sigma (p)\right )\right \}\right ] \\ & \times \exp \left [\frac{i}{2}\int _p J^\mu (-p) \frac{\mathscr{P}_{\mu \nu }(p)}{p^2} J^\nu (p)\right ]\\ &= \text{const} \times \exp \left [\frac{i}{2}\int _{x,y} J^{\mu }(x)\Delta _{\mu \nu }(x-y) J^{\nu }(y)\right ]. \end{split} \end{equation*} In the last line we used the photon propagator in position space (in Landau gauge) \begin{equation*} \Delta _{\mu \nu }(x-y) = \int \frac{d^4 p}{(2\pi )^4}\;e^{ip(x-y)}\frac{\mathscr{P}_{\mu \nu }(p)}{p^2-i\epsilon }. \end{equation*} In the last step we have inserted the $$i\epsilon$$ term as usual.

Photon propagator in position space. In the free theory one has \begin{equation*} \left \langle A_\mu (x) A_\nu (x) \right \rangle = \frac{1}{i^2}\left (\frac{1}{Z[J]} -\frac{\delta ^2}{\delta J^\mu (x)\delta J^\nu (y)} Z[J]\right )_{J=0} = \frac{1}{i}\Delta _{\mu \nu }(x-y). \end{equation*} We use the following graphical notation

$$(x,\mu )$$ $$(y,\nu )= \frac{1}{i}\Delta _{\mu \nu }(x-y),$$

or with sources $$iJ^\mu (x)$$ at the end points

$$= \frac{1}{2} \int _{x,y} iJ^{\mu }(x) \, \frac{1}{i} \Delta _{\mu \nu }(x-y) \, iJ^{\nu }(y).$$

Free solutions. To describe incoming and outgoing photons we need to discuss free solutions for the gauge field. In momentum space, and for the gauge-fixed field (Landau gauge), the linear equation of motion (Maxwell’s equation) is simply \begin{equation*} p^2 \mathscr{P}_{\mu }^{\;\;\nu }(p) \hat{A}_\nu (p) = p^2 \hat{A}_{\mu }(p) =0. \end{equation*} Non-trivial solutions satisfy $$p^2 =0$$. Without loss of generality we assume now $$p^{\mu } = (E, 0 , 0 , E)$$; all other light like momenta can be obtained from this via Lorentz-transformations.

Polarizations. Quite generally, a four-vector can be written as \begin{equation*} \hat{A}_{\nu }(p)=\left (b, \, \frac{a_1+a_2}{\sqrt{2}}, \,\frac{-ia_1+ia_2}{\sqrt{2}}, \, c\right ). \end{equation*} From the Landau gauge condition $$p^{\nu }\hat{A}_{\nu } = 0$$ it follows that $$b=-c,$$ so that one can write \begin{equation*} \hat{A}_\nu (p) = \tilde c \times (-E,0,0,E)+ a_1 \epsilon ^{(1)}_{\nu }+a_2 \epsilon ^{(2)}_{\nu }, \end{equation*} with \begin{equation*} \epsilon ^{(1)}_\nu = \left (0,\frac{1}{\sqrt{2}},\frac{-i}{\sqrt{2}},0\right ), \quad \quad \quad \epsilon ^{(2)}_\nu = \left (0,\frac{1}{\sqrt{2}},\frac{i}{\sqrt{2}},0\right ). \end{equation*} However, the term $$\sim \tilde c$$ is in fact proportional to $$p_\nu = (-E,0,0,E).$$ We can do another gauge transformation such that $$\tilde c=0$$. This does not violate the Landau gauge condition because of $$p^\nu p_\nu = 0.$$ In other words, the photon field has only two independent polarization states, chosen here as positive and negative circular polarizations, or helicities.

Mode expansion. In summary, we can expand free solutions of the photon field like \begin{equation*} A_{\mu }(x) = \sum _{\lambda = 1}^2 \int \frac{d^3 p}{(2\pi )^3} \frac{1}{\sqrt{2E_p}}\left \{a_{\vec{p},\lambda }\; \epsilon ^{(\lambda )}_{\mu }(p)\;e^{ipx} + a^{\dagger }_{\vec{p},\lambda }\; \epsilon _{\mu }^{(\lambda )*}(p) \; e^{-ipx} \right \}, \end{equation*} where $$E_p = |\vec{p}|$$ is the energy of a photon. The index $$\lambda$$ labels the two polarization states.

In the current setup, $$a_{\vec{p},\lambda }$$ and $$a^{\dagger }_{\vec{p},\lambda }$$ are simply expansion coefficients, while they become annihilation and creation operators in the operator picture. The non-trivial commutation relation becomes then \begin{equation*} \left [a_{\vec{p},\lambda }, a^{\dagger }_{\vec{p}^\prime ,\lambda ^\prime } \right ] = (2\pi )^3 \delta ^{(3)}(\vec{p}-\vec{p}^\prime ) \delta _{\lambda \lambda ^\prime }. \end{equation*}

LSZ reduction formula for photons. We also need a version of the Lehmann-Symanzik-Zimmermann reduction formula for photons. Recall that for non-relativistic bosons we could replace for the calculation of the interacting part of the S-matrix \begin{equation*} a_{\vec{q}}(\infty ) \to i\left [-q^0 + \tfrac{\vec{q}^2}{2m}+V_0\right ] \varphi (q), \end{equation*} \begin{equation*} a^{\dagger }_{\vec{q}}(-\infty ) \to i\left [-q^0 + \tfrac{\vec{q}^2}{2m}+V_0\right ] \varphi ^{*}(q). \end{equation*} For relativistic fields this is in general somewhat more complicated because of renormalization. This will be discussed in more detail in the second part of the course. In the following we will discuss only tree level diagrams where this plays no role. For photons one can replace for outgoing states \begin{equation*} \begin{split} & \sqrt{2E_p}\;a_{\vec{p},\lambda }(\infty ) \to i\epsilon ^{\nu *}_{(\lambda )}(p) \int d^4 x \; e^{-ipx}[-\partial _{\mu }\partial ^{\mu }]A_\nu (x) \\ & \sqrt{2E_p}\;a^\dagger _{\vec{p},\lambda }(-\infty ) \to i\epsilon ^{\nu }_{(\lambda )}(p) \int d^4 x \; e^{ipx}[-\partial _{\mu }\partial ^{\mu }]A_\nu (x). \end{split} \end{equation*} These formulas can be used to write S-matrix elements as correlation functions of fields. Note that $$[-\partial _{\mu }\partial ^{\nu }]$$ is essentially the inverse propagator in Landau gauge.

Mode expansion for Dirac fields. We also need a mode expansion for free Dirac fields in order to describe asymptotic (incoming and outgoing) fermion states. We write the fields as \begin{equation*} \begin{split} \psi (x) = & \sum _{s=1}^2 \frac{d^3 p}{(2\pi )^3}\frac{1}{\sqrt{2E_p}}\left \{b_{\vec{p},s}\; u_s(p)\; e^{ipx} + d^{\dagger }_{\vec{p},s}\;v_s(p)\;e^{-ipx}\right \}, \\ \bar \psi (x) = & \sum _{s=1}^2 \frac{d^3 p}{(2\pi )^3}\frac{1}{\sqrt{2E_p}}\left \{-i\;b^{\dagger }_{\vec{p},s}\; \bar{u}_s(p)\; e^{-ipx} - i d^{\dagger }_{\vec{p},s}\;\bar{v}_s(p)\;e^{ipx}\right \}. \end{split} \end{equation*} Again, $$b_{\vec{p},s}$$, $$d_{\vec{p},s}$$ etc. can be seen as expansion coefficients and become operators in the operator picture.

Solutions of Dirac equation. The Dirac equation \begin{equation*} (\gamma ^{\mu } \partial _{\mu } + m)\psi (x) = 0, \end{equation*} becomes for the plane waves \begin{equation*} \begin{split} (i p_\mu \gamma^\mu + m)\;u_s(\vec{p}) & =0, \\ (-i p_\mu \gamma^\mu + m )\; v_s(\vec{p}) & =0. \end{split} \end{equation*} We consider this first in the frame where the spatial momentum vanishes, $$\vec{p}=0$$, such that $$p_{\mu }=(-m,0,0,0)$$, \begin{equation*} p_\mu \gamma^\mu = -\gamma ^{0}m = im \begin{pmatrix} & \mathbb{1} \\ \mathbb{1} & \end{pmatrix}. \end{equation*} The last equation holds in the chiral basis where \begin{equation*} \gamma ^{\mu } = -i \begin{pmatrix} 0 & \sigma ^{\mu } \\ \bar \sigma ^{\mu } & 0 \end{pmatrix}. \end{equation*} with $$\sigma ^{\mu } = (\mathbb{1},\vec{\sigma })$$ and $$\bar \sigma ^{\mu } = (\mathbb{1},-\vec{\sigma }).$$ For the spinor $$u_s$$ one has the equation \begin{equation*} (i p_\mu \gamma^\mu +m) u_s = m \begin{pmatrix} +\mathbb{1} & -\mathbb{1} \\ -\mathbb{1} & +\mathbb{1} \end{pmatrix} u_s = 0. \end{equation*} The two independent solutions are \begin{equation*} u_1^{(0)}=\sqrt{m}\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix},\quad \quad \quad u_2^{(0)} = \sqrt{m}\begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \end{pmatrix}. \end{equation*} The normalization has been chosen for later convenience. Similarly \begin{equation*} (-i p_\mu \gamma^\mu+m) v_s(0) = m \begin{pmatrix} \mathbb{1} & \mathbb{1} \\ \mathbb{1} & \mathbb{1} \end{pmatrix} v_s(0) = 0 \end{equation*} has the two independent solutions \begin{equation*} v_1^{(0)}=\sqrt{m}\begin{pmatrix} 0 \\ +1 \\ 0 \\ -1 \end{pmatrix},\quad \quad \quad v_2^{(0)} = \sqrt{m}\begin{pmatrix} -1 \\ 0 \\ +1 \\ 0 \end{pmatrix}. \end{equation*} We see here that the Dirac equation has two independent solutions (for spin up and and down with respect to some basis) for particles and two more for anti-particles. One can now go to an arbitrary reference frame by performing a Lorentz transformation. That gives \begin{equation*} u_s(\vec{p}) = \begin{pmatrix} \sqrt{-p_\mu \sigma ^\mu } \, \xi _s \\ \sqrt{-p_\mu \bar{\sigma }^\mu } \, \xi _s \end{pmatrix}, \quad \quad \quad v_s(\vec{p}) = \begin{pmatrix} \sqrt{-p_\mu \sigma ^\mu } \, \xi _s \\ -\sqrt{-p_\mu \bar{\sigma }^\mu } \, \xi _s \end{pmatrix}, \end{equation*} with a two-dimensional orthonormal basis $$\xi _s$$ such that \begin{equation*} \xi ^\dagger _s \xi _r = \delta _{rs},\quad \quad \quad \sum _{s=1}^2 \xi _s \xi ^\dagger _s = \mathbb{1}_2. \end{equation*} Other identities involving $$u_s(\vec{p})$$, $$v_s(\vec{p})$$ as well as \begin{equation*} \begin{split} \bar{u}_s(\vec{p}) = u^\dagger _s(\vec{p}) i\gamma ^0 =u^\dagger _s(p) \begin{pmatrix} & \mathbb{1} \\ \mathbb{1} & \end{pmatrix}, \\ \bar{v}_s(\vec{p}) = v^\dagger _s(\vec{p}) i\gamma ^0 =v^\dagger _s(p) \begin{pmatrix} & \mathbb{1} \\ \mathbb{1} & \end{pmatrix}, \end{split} \end{equation*} have been discussed in exercises. They will be mentioned here once they are needed.

LSZ reduction for Dirac fermions. Finally, let us give the LSZ reduction formulas for Dirac fermions (again neglecting renormalization effects) \begin{equation*} \begin{split} \sqrt{2E_p}b_{\vec{p},s}(\infty ) & \to i\int d^4 x \, e^{-ipx}\bar{u}_s(\vec{p}) (\gamma ^\mu \partial _\mu +m) \psi (x), \\ \sqrt{2E_p}d^\dagger _{\vec{p},s}(-\infty ) & \to -i\int d^4 x \, e^{-ipx}\bar{v}_s(\vec{p}) (\gamma ^\mu \partial _\mu +m) \psi (x), \\ \sqrt{2E_p} d_{\vec{p},s}(\infty ) & \to -i\int d^4 x \, i \bar{\psi }_s(x) (-\gamma ^\mu \overleftarrow{\partial }_\mu +m) v_s(x)\,e^{-ipx}, \\ \sqrt{2E_p} b^\dagger _{\vec{p},s}(-\infty ) & \to i\int d^4 x \, i \bar{\psi }_s(x) (-\gamma ^\mu \overleftarrow{\partial }_\mu +m) u_s(x)\,e^{ipx}. \end{split} \end{equation*} The left-pointing arrows indicate here that these derivatives act to the left (on the field $$\bar{\psi }_s(x)$$). These relations have been obtained as part of the exercises.

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