On the collision of two projectiles on two targets in the BFKL approach

Abstract

High-energy collisions of two nucleons on two nucleons are studied in the BFKL approach in the leading approximation in α _s N _c . Diagrams with redistribution of color are considered. It is found that intermediate BKP states consisting of four reggeized gluons give a contribution which may be leading in deuteron–deuteron scattering and thus experimentally observable.

Appendices

Appendix A: Deuteron in the Glauber approach

1.1 A.1 Scattering on the deuteron

To formulate the Glauber approach to the collisions with deuteron in the diagrammatic technique we first have to relate the relativistic dpn vertex Γ with the deuteron wave function. To this end we study the electromagnetic form-factor of the deuteron, illustrated in Fig. 9, where vertices Γ are shown with blobs. In the lab. system and at zero momentum transfer it is equal to 2M where M=2m−ϵ is the deuteron mass. So we get the normalization condition

$$ \int\frac{d^4l_2}{((2\pi)^4 i}\frac{2l_{10}\varGamma^2(\mathbf {l}_2)}{ (m^2-l_2^2-i0)(m^2-l_1^2-i0)^2}=2M. $$

(81)

Here l ₁+l ₂=2l and 4l ²=M ². We neglect spins and consider all particles as scalar for simplicity. We have l ₀=m−ϵ/2, and l=0, so that putting l ₂=l+λ we find

$$m^2-l_2^2=-2m\lambda_0- \lambda_\perp^2,\qquad m^2-l_1^2=2m \lambda_0-\lambda_\perp^2, $$

where we used the orders of magnitude λ ₀∼ϵ, $|\lambda_{\perp}|\sim\sqrt{m\epsilon}$. Integration over λ ₀ transforms (81) into

$$ \int\frac{d^3l_2}{(2\pi)^3} \frac{\varGamma^2(\mathbf{l}_2)}{ 8m(m\epsilon+\mathbf{l}_2^2)^2}=2. $$

(82)

Comparing with the standard normalization of the deuteron wave function ψ _d (l) we find the desired relation

$$ \psi_d(\mathbf{l})=\frac{\varGamma(\mathbf{l})}{ 4(2\pi)^{3/2}\sqrt{m}(m\epsilon+\mathbf{l}^2)}, $$

(83)

which allows to relate the relativistic dpn vertex with the deuteron wave function in the momentum space.

Fig. 9

Electromagnetic form factor of the deuteron

In the impulse approximation, Fig. 10, with the spectator neutron the corresponding amplitude is given by

$$ \mathcal{A}^{\mathrm{imp}}=a\int\frac{d^4l_2}{(2\pi)^4 i} \frac {\varGamma^2(\mathbf{l}_2)}{ (m^2-l_2^2-i0)(m^2-l_1^2-i0)^2}, $$

(84)

where a is the forward scattering amplitude on the proton. Using (81) we find that the integral is equal to 2 and we get $\mathcal{A}^{\mathrm{imp}}=2a$. But the relativistic flux on the deuteron is twice that on the proton, so that we get σ _d =σ _p +σ _n , where the second term takes into account the diagram of Fig. 10 with the spectator proton.

Fig. 10

Scattering on the deuteron in the impulse approximation

Now consider double scattering on the deuteron illustrated in Fig. 11. The amplitude is given by

(85)

where H is the high-energy part and it is taken into account that it can only depend on the z-component of the transferred momentum, since it is the only of the spatial components which enters multiplied by the high projectile momentum.

Fig. 11

Double scattering on the deuteron

Integrations over the zero components of the nuclear momenta factorize and we get

$$ \mathcal{A}=\int\frac{d^3l_2\,d^3l'_2}{(2\pi)^3} H\bigl(l_{2z}-l'_{2z} \bigr) \frac{\varGamma(\mathbf{l}_2)}{4m(m\epsilon+\mathbf {l}_2^2)} \frac{\varGamma(\mathbf{l}'_2)}{4m(m\epsilon+{\mathbf {l}'_2}^2)}, $$

(86)

or using (83)

$$ \mathcal{A}=\frac{1}{m}\int\frac{d^3l_2\,d^3l'_2}{(2\pi)^3 } H\bigl (l_{2z}-l'_{2z} \bigr)\psi_d(\mathbf{l}_2)\psi_d\bigl( \mathbf{l}'_2\bigr). $$

(87)

Integrations over the transverse components are done immediately to convert the wave functions into those with the transverse coordinates zero:

(88)

Transforming completely to the coordinate space we find

(89)

where we introduced the z-component of the transferred momentum putting $l'_{z}=l_{z}+q_{z}$. Integration over l _2z gives our final expression

$$ \mathcal{A}=\frac{1}{m}\int\frac{dq_{z}}{2\pi} \,dz H(q_z)\bigl| \psi_d(r_\perp=0,z)\bigr|^2 e^{-iq_zz}. $$

(90)

The Glauber approximation follows if H(q _z ) has a singularity at q _z =0. Typically

$$ \operatorname{Im}H(q_z)=-\hat{D}(2\pi)\delta(2kq)=-\hat{D} \frac {\pi}{k_0}\delta(q_z). $$

(91)

Here we use q ₀≪|q _z | and k ₀=k _z >0. In this case we get the Glauber approximation for the double scattering amplitude

$$ \operatorname{Im}\mathcal{A}=-\frac{1}{s}\hat{D} \int dz\bigl| \psi_d(r_\perp=0,z)\bigr|^2=-\frac{1}{ s} \hat{D}\biggl\langle\frac {1}{2\pi r^2}\biggr\rangle_d, $$

(92)

where 〈…〉 _d means the average in the deuteron. The cross-section is

$$ \sigma_d=-\frac{1}{2s^2}\hat{D}\biggl\langle\frac{1}{2\pi r^2} \biggr\rangle_d. $$

(93)

To see how this formula works consider the simplest case of the double scattering corresponding to double elastic collision shown in Fig. 12. In this case

(94)

so that $\hat{D}=-a_{p}a_{n}$. Using (92) we find

$$ \operatorname{Im}\mathcal{A}=a_pa_n \frac{1}{s} \biggl\langle \frac{1}{2\pi r^2}\biggr\rangle_d= -s\sigma_p \sigma_n\biggl\langle\frac{1}{2\pi r^2}\biggr\rangle_d. $$

(95)

From this dividing by 2s and doubling to take into account the crossed diagram find the double cross-section

$$ \sigma^{\mathrm{double}}=-\sigma_p\sigma_n\biggl\langle \frac{1}{2\pi r^2}\biggr\rangle_d. $$

(96)

Fig. 12

Double elastic collision on the deuteron

For the nuclear target instead of (93) we have at fixed impact parameter b

$$ \sigma_A(b)=-\frac{1}{2s^2}\hat{D}T_A^2(b) $$

(97)

and for double elastic collisions instead of (96)

$$ \sigma_A(b)=-\frac{1}{2}A(A-1)\sigma_N^2T_A^2(b). $$

(98)

To conclude we note that H and F are both Lorenz invariant. So (91) can be used to find F in any system.

1.2 A.2 Double scattering in d–d collisions

Now consider the case when two deuterons collide at high energies and each one experiences double collision, illustrated in Fig. 13. Our treatment is to consider subsequently the two systems in which the deuteron is well understandable, the rest systems of the target and projectile deuterons.

Fig. 13

Double scattering in deuteron–deuteron collisions

We start from the rest system of the target deuteron. We use Eq. (92) and write

$$ \operatorname{Im}\mathcal{A}=-\frac{1}{ s}\hat{D}_1\biggl\langle \frac {1}{2\pi r^2}\biggr\rangle_d, $$

(99)

where it is assumed that $\operatorname{Im} H_{1}=-\hat{D}_{1}2\pi\delta (2kq)$ and we include into H ₁ all the rest part of the diagram in Fig. 13 including the coupling to the projectile nucleons.

Now we boost the system into the rest one for the projectile. In this system we can repeat our treatment of the coupling to the two nucleons. If $\hat{D}_{1}=\hat{D}2\pi\delta(2l\kappa)$ where κ is the momentum transferred from the projectile, integration over the nucleon momenta will give the same factor (1/s)〈1/2πr ²〉 _d and we shall get

$$ \operatorname{Im}\mathcal{A}=- \biggl(\frac{1}{s}\biggr)^2 \hat{D}\biggl\langle\frac{1}{2\pi r^2}\biggr\rangle^2_d. $$

(100)

Note that we have

$$ \operatorname{Im}H=-\hat{D}(2\pi)^2\delta\bigl(2(l\kappa)\bigr) \delta\bigl(2(kq)\bigr). $$

(101)

This means that (100) is symmetric in projectile and target, as expected. From (100) we immediately get the cross-section (4) taking in to account the definition of D, Eq. (2)

Special attention is to be given for the case when the high-energy part H is disconnected, shown in Fig. 14. Then H contains a δ function corresponding to conservation laws for the two connected parts and is given by

$$ H=i\bigl(a_{pp}a_{nn}+a_{pn}^2 \bigr) (2\pi)^4\delta^4\bigl(k_1+l_1-k'_1-l'_1 \bigr). $$

(102)

Factor i combines (−i) from the definition of $\mathcal{A}$ and i ² from the two NN amplitudes. Note that

As we see, integrations over the transverse coordinates of the nucleons in the projectile and target become interdependent. Under the sign of integration over b we include the exponentials depending on the transverse momenta of the projectile and target in the corresponding integrals to obtain in (88) $\psi_{d}(b,l_{2z})\psi_{d}(b, l'_{2z})$ instead of $\psi_{d}(r_{\perp}=0,l_{2z})\psi_{d}(r_{\perp}=0, l'_{2z})$ and similarly for the projectile. All subsequent calculations remain unchanged and in the end we obtain in (92)

$$ \int dz\bigl|\psi_d(b,z)\bigr|^2\equiv T_d(b) $$

(103)

instead of

$$\int dz\bigl|\psi_d(r_\perp=0,z)\bigr|^2=\biggl\langle \frac{1}{2\pi r^2}\biggr\rangle_d. $$

So the net result of the connection between the transferred transverse momenta is to substitute 〈1/2πr ²〉 _d →T _d (b) both in the projectile and target and then integrate over b. The rest factors from H give $\hat{D}=2sa_{p}a_{n}$ and from (100) we conclude

$$ \operatorname{Im}\mathcal{A}=\frac{2}{s}\bigl(a_{pp}a_{nn}+a_{pn}^2 \bigr)\int d^2bT^2_d(b). $$

(104)

Dividing by 4s we find the cross-section

$$ \sigma_dd=-\frac{1}{2}\bigl(\sigma_{pp} \sigma_{nn}+\sigma_{pn}^2\bigr) \int d^2bT^2_d(b) $$

(105)

which looks very much like the standard Glauber formula.

Fig. 14

Double elastic collision in deuteron–deuteron scattering

For the collision of two heavy nuclei instead of Eq. (105) we shall get at fixed b

(106)

Appendix B: From 4 to 2 dimensions

Our amplitudes D _i are expressed as integrals in the 2-dimensional transverse momentum space. The initial scattering amplitudes are integrals in the 4-dimensional space with two additional longitudinal integrations. In this appendix we briefly discuss transition from the 4-dimensional picture to the 2-dimensional one and, in particular, Eq. (2). Note that we are going to calculate the amplitude itself rather than its imaginary part. In the simple case of a single pomeron this is practically the same. However, in our case calculation of the imaginary part requires summation of different cuts of a given diagram with their specific weights, which considerably complicates the matter.

In our derivation we shall use Lipatov’s effective action approach [20]. As was shown in [21] its application demonstrates that in the Regge kinematics the relation of the 4-dimensional picture to the 2-dimensional is quite simple: the former is obtained from the latter when all s-channel particles are provided with their Feynman denominators with correct positions of poles respective to the real axis. This point will be illustrated in the following. Another point essential in the derivation is that the transferred longitudinal momenta λ ₋ and κ ₊ are assumed to be much larger that the longitudinal momenta transferred by the reggeized gluons If we call it q in the simplest diagram of Fig. 3 then consider the process in, say, the rest system of the target. In this system λ ₋ is fixed but q ₋→0 as s→∞. Similarly in the rest system of the projectile κ ₊ is fixed but q ₊→0 in the high-energy limit.

It is convenient to start from the lowest order with no s-channel gluons at all. The coupling of the reggeized gluons (R) to the target quark will be given by the effective $Q\bar{Q}$RR vertex, which is a sum of direct and crossed coupling, Fig. 15

$$ U(l,q)=-2\pi\delta\bigl(2(lq)\bigr) $$

(107)

and similar for the coupling to the projectile quark. This implies that intermediate quarks are put to their mass-shell.

Fig. 15

Effective $Q\bar{Q}$RR vertex

With the help of these vertices in the lowest order the amplitude in Fig. 3 acquires the form shown in Fig. 16a, where also the momenta of external quarks are indicated. One finds a longitudinal integral

(108)

Here k ₋=l ₊=0 as well as all transverse components of k, l, κ, and λ. The rest of the diagram gives factor $[16(kl)^{2}]^{2}N_{c}^{2}D^{(0)}$, where

$$D^{(0)}=\int\frac{d^2q_\perp}{(2\pi)^2}\frac{1}{q_\perp^4} $$

gives the contribution of the 2-dimensional diagram in Fig. 3 in the lowest order. Taking into account that the amplitude H is the contribution of the diagram multiplied by −i we obtain Eq. (2) taken in the lowest order.

Fig. 16

Lowest order diagrams without s channel gluons (a) and with one s-channel gluon (b)

Now consider introduction of a single s-channel gluon (Fig. 16b). The longitudinal integral in the new loop is

(109)

We see that the s channel gluon is put on its mass shell. Expression (109) substitutes −2πδ(2(lq)) in the lowest order amplitude. Subsequent integration over q ₊ gives an extra factor

$$ -\int\frac{dq_+}{8\pi|q_+|}=-\frac{1}{4\pi}\int dy, $$

(110)

where y is the gluon rapidity. Color summation gives an extra factor −N _c .

Combined with factors 1/4π in (110), N _c and g ² the rest of the amplitude will substitute the lowest order 2-dimensional contribution D(0) by the 2-dimensional contribution containing one interaction v(q ₁,−q ₁|q ₁+p,q ₁−p). So the amplitude H will be expressed by the same Eq. (2) with D given by the 2-dimensional diagram in Fig. 16b and an extra integration over rapidity of the s-channel gluon.

This derivation is trivially generalized to any number of s-channel gluons, irrespective whether they are intermediate in the pomeron, in between the pomerons as in Figs. 4 or 5 or inside the BKP states. In all cases one additional integration over the longitudinal variable puts the gluon on its mass shell and the other is reduced to integration over its rapidity.

Appendix C: Color factors

The explicit expressions for the color wave functions of the projectile and target in which pairs (12), (34) and (13), (24), respectively, form color singlets are

$$ \bigl|(12) (34)\bigr\rangle=\frac{1}{N_c^2}\delta_{a_1a_2}\delta_{a_3a_4}, \bigl|(13) (24)\bigr\rangle=\frac{1}{N_c^2}\delta_{a_1a_3} \delta_{a_2a_4}. $$

(111)

Here we neglect terms of the relative order $1/N_{c}^{2}$. Their scalar product is

$$ \bigl\langle(13) (24)\big|(12) (34)\bigr\rangle=\frac{1}{N_c^4}\delta_{a_1a_3} \delta_{a_2a_4}\delta_{a_1a_2}\delta_{a_3a_4} = \frac{1}{N_c^2}, $$

(112)

which is the overall damping factor accompanying all diagrams with the redistribution of color like Fig. 2.

We denote C _ij =−(T _i T _j ). For interactions connecting vacuum pairs either in the projectile or in the target C _ij =N _c

(113)

For the remaining two interactions we find

(114)

Interchange (1↔2),(3↔4) gives

$$ \bigl\langle(13) (24)\bigr|C_{23}\bigl|(12) (34)\bigr\rangle =-\frac{1}{N_c}. $$

(115)

So effectively for these interactions C _ij =−N _c .

Apart from states |(12)(34)〉 and |(13)(24)〉 in the diagrams we encounter six BKP states with different ordering of the four gluons:

$$|1234\rangle,\ |1243\rangle,\ |1324\rangle,\ 1342\rangle,\ |1423\rangle,\ 1432 \rangle. $$

In the high color limit their explicit form is

$$ |1234\rangle=\frac{1}{2N_c^2}h^{a_1a_2c}h^{a_3a_4c}, $$

(116)

where h ^abc=d ^abc+if ^abc with the properties

(117)

The states |ijkl〉 are cyclic symmetric in (ijkl).

Their scalar products with the projectile and target states are

(118)

(119)

Generally if (12) or (34) are neighbors in (ijkl) then states |(12)(kl)〉 and |ijkl〉 are orthogonal. If they are not, the scalar product is the same as in (119).

We also need matrix elements of color matrices C _ij between projectile (target) states and BKP states. Obviously we need only C _ij which do not connect vacuum pairs in the projectile (target), namely for (ij)=(13),(14),(23),(24). Then we find that for (klmn)=(1234), (2134), (2143), and (1423)

$$ \bigl\langle(12) (34)\bigr|C_{ij}|\mathit{klmn}\rangle=\pm\frac {1}{2}, $$

(120)

where the sign plus is to be taken when (ij) are neighbors in (klmn) and the sign minus when they are not. Acting on the rest two states |1324〉 and |1423〉. All matrices C ₁₃,C ₂₄,C ₁₄, and C ₂₃ give N _c /2 since neighbor gluons are in the gluon color state and the matrix elements become damped by $1/N_{c}^{2}$. E.g.

$$ \bigl\langle(12) (34)\bigr|C_{13}|1423\rangle=\frac{N_c}{2}\bigl \langle(12) (34)|1423\bigr\rangle=-\frac{1}{N_c^2}. $$

(121)

(Note that the correct derivation of (120) and 121) in some cases requires taking into account subdominant terms in (116).)

As a result the matrix elements of C ₁₃ are

which implies

(122)

Note that the summed probabilities correctly give unity.

Similarly

and

Interchanging (2↔3) we get

These relations allow to study matrix elements of the product of two matrices C _ij C _kl between the projectile and target states. They are shown in Table 1 with lines (ij) and columns (kl).

Table 1 Matrix elements of the product C _ij (lines) by C _kl (columns) between states 〈(12)(34)| and |(13)(24)〉

From these results we can find the probability to find a particular BKP state between the projectile and target. States |1234〉 and |1324〉 do not appear and we find the contribution from the double interaction V _ij V _kl where V _ij =−C _ij g ² v _ij

(123)

We are also interested in the matrix elements of products of two matrices C _ij C _kl between projectile and target states without redistribution of color, that is, between states 〈(12)(34)| and |(12)(34)〉. In particular we shall be interested in separate contribution from BKP states. In this case four different BKP states appear between the projectile and target |1234〉, |1432〉, |1342〉, and |1243〉 with equal probability and, similar to (123) we find the probabilities

(124)