Some thoughts on the quenching of spectroscopic factor

Spectroscopic factor plays the central role in unfolding the nuclear structure. In the simplest manner, the total Hamiltonian of the nucleus is transformed into a 1-body effective potential and the many-body residual interaction, i.e.,

The effective single-particle Hamiltonian has solution:

where is the single-particle energy. The solution for is

where is the set of basis for state from , and is the eigenenergy.

The residual interaction is minimized by adjusted the mean-field . Thus, the residual interaction can be treated as a perturbation. This perturbs the nuclear wave function

The normalization requires .

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In the Slater determinant , a single-particle wave function for a particular orbital can be pull out.

where is anti-symmetric, angular coupling operator. Thus,

The is the spectroscopic factor. There are another sum-rule for adding and removing a nucleon. so that the sum is equal to the number of particle in a particle orbital.

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I always imagine the quenching is because we did not sum-up the SFs from zero energy to infinity energy (really???), thus, we are always only observing a small fraction of the total wave function. For example,  the total wavefunction would look like this:

In experiment, we observe the overlap between ground-state to ground-state transition

for ground-state to 1st excited state transition for the same orbital is

And since we can only observed limited number of excited states, bounded by either or boht :

• experimental conditions, say incident energy
• the excited states that are beyond single-particle threshold.
• finite sensitivity of momentum

Thus, we cannot recover the full spectroscopic factor. This is what I believe for the moment.

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Experimentally, the spectroscopic factor is quenched by 40% to 50%. The “theory” is that, the short-range interaction quench ~25%, the long-range interaction quench ~20%. The long- and short-range interaction correlate the single-particle orbital and reduce the degree of “single-particle”.

The short-range interaction is mainly from the “hard-core” of nucleon, i.e., the interaction at range smaller than 1 fm. The long-range interaction is coupling with nearby vibration states of the rest of the nucleus.

For example, from the 19F(d,3He) reaction, the spectroscopic factor for 19F 1s1/2 state is ~0.4, and 0d5/2 is ~0.6.

Thus, the wavefunction of 19F is

It is worth to note that the above SFs is not re-analysised and the “quenching” is not shown. Many old data had been re-analysised using global optical model and the SF is reduced and show that the sum of SFs is ~ 0.55.

If it is the case for 19F, the wavefunction would become,

Here I use for the “correlated wavefunction” that the single-particle orbital cannot simply pull out. Nevertheless, if and are correlated,

Am I misunderstood correlation?

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My problem is, What does a correlated wave function look like?

In my naive understanding, the Slater determinant is a complete basis for N-nucleon system. A particular single-particle orbital can ALWAYS be pull out from it. If it can not, therefore, the Slater determinant is NOT complete. The consequence is that all theoretical calculation is intrinsically missed the entire CORRELATED SPACE, an opposite of Slater determinant space (of course, due to truncation of vector space, it already missed somethings).

If the theory for correlation is correct, the short-range interaction is always there. Thus, the spectroscopic factor for deuteron 0s1/2 orbital is ~0.8, assuming no long-range correlation. However, we already knew that 96% of deuteron wavefunction is from 0s1/2 and  only 4% is from 1d5/2 due to tensor force. Is it not mean the spectroscopic factor of deuteron 0s1/2 state is 0.96?  Is deuteron is a special case that no media-modification of nuclear force? But, if the short-range correlation is due to the hard core of the nucleon, the media-modification is irrelevant. Sadly, there is no good data such as d(e,e’p) experiment. Another example is 4He(d,p)5He experiment. What is the spectroscopic factor for g.s. to g.s. transition, i.e. the 0p3/2 orbital? is it ~0.6 or ~ 1.0?

Since the experimental spectroscopic factor has model dependency (i.e. the optical potential). Could the quenching is due to incomplete treatment of the short- and long-range correlation during the interaction, that the theoretical cross section is always bigger?

In the very early days, people calibrate their optical potential using elastic scattering for both incoming and out-going channel, and using this to produce the inelastic one. At that time, the spectroscopic factors are close to ~1. But since each optical potential is specialized for each experiment. It is almost impossible to compare the SF from different experiments. Thus, people switch to a global optical potential. Is something wrong with the global optical potential? How is the deviation?

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Let me summarize in here.

1. The unperturbed wave function should be complete, i.e. all function can be expressed as a linear combination of them.
2. A particular single-particle orbital can be pull out from the Slater determinate .
3. The residual interaction perturbs the wave function. The short-/long-range correlation should be in the residual interaction by definition or by construction of the mean field.
4. The normalization of wave function required the sum of all SF to be 1.
5. Another sum rule of SFs equals to the number of particle.
6. By mean of the correlation, is that many excited states have to be included due to the residual interaction. No CORRELATED space, as the Slater determinant is complete. (pt. 1)
7. Above points (1) to (7) are solid mathematical statements, which are very hard to deny.
8. The logical result for the quenching of the observed SF is mainly due to not possible to sum up all SFs from all energy states for all momentum space.
9. The 2-body residual interaction can create virtual states. Are they the so called collective states?
10. But still, collective states must be able to express as the Slater determinant (pt. 1), in which a particular single-particle orbital can be pull out (pt. 2).
11. May be, even the particular single-particle orbital can be pull out, the rest cannot experimentally observed ? i.e. is not experimentally reachable. That go back to previous argument for limitation of experiments (pt. 8).
12. For some simple systems, say doubly magic +1, deuteron, halo-nucleon, very weakly bounded exited state, resonance state, the sum of SF could be close to 1. Isn’t it?
13. The theoretical cross section calculation that, the bound state wave function is obtained by pure single-particle orbital. I think it is a right thing to do.
14. The use of global optical potential may be, could be not a good thing to do. It may be the METHOD to deduce the OP has to be consistence, instead of the OP itself has to be universal. Need more reading from the past.

Simple model for 4He and NN-interaction

Starting from deuteron, the binding energy, or the p-n interaction is 2.2 MeV.

From triton, 3H, the total binding energy is 8.5 MeV, in which, there are only 3 interactions, two p-n and one n-n. Assume the p-n interaction does not change, the n-n interaction is 4.1 MeV.

The total binding energy of 3He is 7.7 MeV. The p-p interaction is 3.3 MeV.

Notices that we neglected the 3-body force in 3H and 3He. And it is strange that the n-n and p-p interaction is stronger then p-n interaction.

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In 4He, the total binding energy becomes 28.3 MeV. I try to decompose the energy in term of 2-body, 3-body, and 4-body interaction.

If we only assume 2-body interaction, the interaction strength from n-p, n-n, and p-p are insufficient. One way to look is the 1-particle separation energy.

The neutron separation energy is 20.6 MeV = 2(p-n) + (n-n).
The proton separation energy is 19.8 MeV = 2(p-n) + (p-p).
The total energy is 28.3 MeV = 4(p-n) + (n-n) + (p-p).

There is no solution for above 3 equations. Thus, only consider 2-body interaction is not enough.

The neutron separation energy is 20.6 MeV = 2(p-n) + (n-n) + 2(n-n-p) + (n-p-p)
The proton separation energy is 19.8 MeV = 2(p-n) +(p-p) + 2(n-p-p) + (n-n-p).
The total energy is 28.3 MeV = 4(p-n) + (n-n) + (p-p) + 2(n-n-p) + 2(n-p-p) + (n-n-p-p).

Assuming the 2-body terms are the same in 2H, 3H, and 3He, the (n-p-p) and (n-n-p) is 4.03 MeV, which is strange again, as the Coulomb repulsion should make the (n-p-p) interaction smaller then the (n-n-p) interaction. The (n-n-p-p) interaction is -4.03 MeV.

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Lets also add 3-body force in 3H and 3He.

The neutron separation energy of 3H is 6.3 MeV = (p-n) + (n-n) + (n-n-p)
The toal energy of 3H is 8.5 MeV = 2(p-n) + (n-n) + (n-n-p)
The toal energy of 3He is 7.7 MeV = 2(p-n) + (p-p) + (n-p-p)
The neutron separation energy of 4He is 20.6 MeV = 2(p-n) + (n-n) + 2(n-n-p) + (n-p-p)
The proton separation energy is 4He 19.8 MeV = 2(p-n) +(p-p) + 2(n-p-p) + (n-n-p).
The total energy of 4He is 28.3 MeV = 4(p-n) + (n-n) + (p-p) + 2(n-n-p) + 2(n-p-p) + (n-n-p-p).

We have 6 equations, with 6 unknown [ (p-n) , (n-n) , (p-p) , (n-n-p), (n-p-p), and (n-n-p-p)]. Notice that the equation from proton separation energy of 3He is automatically satisfied. The solution is

(p-n) = 2.2 MeV
(p-p) = – 4.7 – (n-n) MeV
(n-n-p)  = 4.1 – (n-n) MeV
(n-p-p) = 8 + (n-n) MeV
(n-n-p-p) = 0 MeV

It is interesting that there is redundant equation. But still, the (p-n) interaction is 2.2 MeV, and 4-body (n-n-p-p) becomes 0 MeV. Also the (n-p-p) is more bound than (n-n-p) by 3.9 + 2(n-n) MeV. If (n-p-p) should be more unbound, than (n-n) must be negative and smaller than -1.95 MeV.

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Since the interaction strength has to be on the s-orbit (mainly), by considering 2H, 3H, 3He, and 4He exhausted all possible equations (I think). We need other way to anchor either (n-n), (p-p), (n-p-p), and (n-n-p) interactions.

Use the Coulomb interaction, the Coulomb interaction should add -1.44 MeV on the NN pair (assuming the separation is a 1 fm). Lets assume the (n-n) – (p-p) = 1.44 MeV

(n-n) = -1.63 MeV
(p-p) = – 3.07 MeV
(n-n-p)  = 5.73 MeV
(n-p-p) = 6.37  MeV

The (p-p) is more unbound than (n-n) as expected, but the (n-p-p) is more bound than (n-n-p) by 0.64 MeV. This is surprising! We can also see that, the 3-body interaction play an important role in nuclear interaction.

According to this analysis, the main contribution of the binding energies of 3H and 3He are the 3-body force.

In 3H:  (n-n) + 2(n-p) + (n-n-p) = -1.6 + 4.4 + 5.7 = 8.5 MeV
In 3He: (p-p) + 2(n-p) + (n-p-p) = -3.1 + 4.4 + 6.4 = 7.7 MeV

Worked on the algebra, when ever the difference  (n-n) – (p-p)  > 0.8 MeV, the (n-p-p) will be more bound that (n-n-p). Thus, the average protons separation should be more than 1.8 fm. I plot the interactions energies with the change of Coulomb energy below.

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The (n-n) and (p-p) are isoscalar pair, where tensor force is zero. While the (n-p) quasi-deuteron is isovector pair. Thus, the difference between (n-n) and (n-p) reflect the tensor force in s-orbit, which is 3.8 MeV. In s-orbit, there is no spin-orbital interaction, therefore, we can regard the tensor force is 3.8 MeV for (n-p) isovector pair.

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Following this method, may be, we can explore the NN interaction in more complex system, say the p-shell nuclei. need an automatic method. I wonder the above analysis agreed present interaction theory or not. If not, why? 

Radial Density of 3D Harmonic Oscillator

From the previous post, we have the radial formula for the 3D harmonic oscillator.

For , the are plotted below.

Assume all orbits are fully filled.

The red is , the pink is , the blue is , the cyan is , and the green is .

Next, we are summing all the shells to get the nucleon density.

In here, the red is the 4He nucleon density function, the pink is the nucleus with all 1p-shell are filled, that is 16O. The blue is the sd-shell are filled, that is 40Ca. The cyan is fp-shell filled.

We can see some systematic trend there. The nuclei has shape, which is somewhat similar to the Wood-Saxon shape. But for the nuclei, the shape is like a V.

Note that the is fixed in this calculation. In fact, from Samuel Wong (2004) (Introductory Nuclear Physics),

 

Hartree-Fock for Helium excited state II

This times, we will show the energy level diagram for helium atom. We already calculated the ground state, the 1s2s singlet and triplet excited states. We will calculate higher excited states, for example, 1s3s, 1s2p or 1s3p, etc, and included in the diagram.

The most difficult part is the evaluation of the mutual interaction matrix element

The angular integral was evaluated in the last post. And the radial integral is done using Mathematica.

I will use hydrogen 1s, 2s, 2p, 3s, 3p, 4s, and 4p states for basis. Thus, the diagram will contain some of the excited states from some possible combination. The 1s4s and 1s4p state will not be calculated, because there is no room for mixing with higher excited states but only for lower excited states. This will make the eigen state be unbound.

During the calculation, one of the tricky point is the identify of the n-th s-state. It is because the mixing is always there, and the mixing can be very large. In order to determine the principle quantum number, we have to check the wave function can see how many peaks in there.

Here is the energy level diagram obtained by Hartree-Fock method using limited hydrogen wave functions as basis set.

 

Hartree method for Helium ground state

After long preparation, I am ready to do this problem.

The two electron in the helium ground state occupy same spacial orbital but difference spin. Thus, the total wavefunction is

Since the Coulomb potential is spin-independent, the Hartree-Fock method reduce to Hartree method. The Hartree operator is

where the single-particle Hamiltonian and mutual interaction are

In the last step, we use atomic unit, such that . And the energy is in unit of Hartree, .

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We are going to use Hydrogen-like orbital as a basis set.

I like the left the , because in the integration , the can be cancelled. Also, the is a compact index of the orbital.

Using basis set expansion, we need to calculate the matrix elements of

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Now, we will concentrate on evaluate the mutual interaction integral.

Using the well-known expansion,

The angular integral

where the integral .

From this post, the triplet integral of spherical harmonic is easy to compute.

The Clebsch-Gordon coefficient imposed a restriction on .

The radial part,

The algebraic calculation of the integral is complicated, but after the restriction of from the Clebsch-Gordon coefficient, only a few terms need to be calculated.

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The general consideration is done. now, we use the first 2 even states as a basis set.

These are both s-state orbital. Thus, the Clebsch-Gordon coefficient

The radial sum only has 1 term. And the mutual interaction becomes

The angular part

Thus, the mutual interaction energy is

The radial part

We can easy to see that . Thus, if we flatten the matrix of matrixes, it is Hermitian, or symmetric.

We also notice that is a matrix of matrixes. The inner matrix runs for indexes, while the bigger matrix runs for index. This will simplify our minds.

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Now, we can start doing the Hartree method.

The general solution of the wave function is

The Hartree matrix is

In this simple 2-states example, index runs from 1 to 2. And remember the is a matrix of matrixes, using the identify the sum becomes:

,

where the represent the bigger matrix in .

The first trial wave function is the Hydrogen-like orbital,

Solve for eigen system, we have the energy after 1st trial,

After 13th trial,

Thus, the mixing of the 2s state is only 3.7%.

Since the eigen energy contains the 1-body energy and 2-body energy. So, the total energy for 2 electrons is

In which ,

So the energies for

From He to He++.  
From He+ to He++, .
From He to He+, is

The experimental 1 electron ionization energy for Helium atom is

The difference with experimental value is 2.175 eV. The following plot shows the Coulomb potential, the screening due to the existence of the other electron, the resultant mean field, the energy, and

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Usually, the Hartree method will under estimate the energy, because it neglected the correlation, for example, pairing and spin dependence. In our calculation, the energy is under estimated.

From the , we can see, the mutual interaction between 1s and 2s state is attractive. While the interaction between 1s-1s and 2s-2s states are repulsive. The repulsive can be easily understood. But I am not sure how to explain the attractive between 1s-2s state.

Since the mass correction and the fine structure correction is in order of , so the missing 0.2 eV should be due to something else, for example, the incomplete basis set.

If the basis set only contain the 1s orbit, the mutual interaction is 1.25 Hartree = 34.014 eV. Thus, the mixing reduce the interaction by 5.07 eV, just for 3.7% mixing

I included the 3s state,

The mutual energy is further reduced to 1.05415 Hartree = 28.6848 eV. The . If 4s orbital included, the . We can expect, if more orbital in included, the will approach to .

Method II (decay)

there is another way to study nuclear physics, which is by observe the decay process.

there are 3 major decays, the alpha, beta and gamma. the alpha decay is an excited nucleus go to a lower energy state by emitting an Helium ( 2 protons and 2 neutrons) nucleus. This process change the nucleus constitution and make it lighter. the beta decay is an excited nucleus go to a lower energy state by emitting an electron ( or positron, or capture an electron). This process also change the nucleus constitution but the change in mass is very small. the gamma decay is an excited nucleus go to a lower energy state by emitting a photon ( or the light particle). this process does not change the nucleus constitution.

there are many other decays, like neutron, proton, and even fission can regard as decay. in general, decay is a process that a nucleus go to a lower energy state and become more stable. But other decays has too short for lifetime, so that after the earth was formed, they are almost gone. and only alpha, beta and gamma has long left time to survive.

Via a decay process, we can know the energy level, lifetime, parity of each level of a nucleus. These informations can help us to build nuclear models and theories to understand and predict nuclear properties, like the strong and weak force.

i will present a series about decay in future posts. watch out!

Radioactivity

On 11st, March, a M9.0 earthquake strike east Japan around 3pm. this caused malfunction of the nuclear plant, Fukushima Daiichi.

i am not going to talk about the malfunction, but on the radioactivity.

The radioactivity is caused by nucleus decay into a more stable state. like Carbon-14 will decay to Carbon-13 by beta decay. there are 3 common types of radioactivities: alpha – a Helium nucleus, beta – a electron and gamma – high energy light.

the radioactivity was measured by Bq ( Becquerel ), which is 1 decay per second. for a typical radioactivity substance, there may be 10 billion decay per second. think about how many atom in 1 mole of matter.

each radiation carry some energy, for alpha decay, energy is about 5MeV, beta decay is 1 MeV, and for gamma decay, it is around 0.1 to 1 MeV. 1 MeV is about Joule. for boiling a water to increase 1 degree, it take 2400Joule. you can see how small it is.

1 more thing to consider is, not every radiation will be absorbed. some will just pass through and leave and no effect. the chance for being absorbed is proportion to the density of the material, or for simplicity, how much weight of the material.

for easy understanding, we multiple these 3 factors together, to give us how much energy absorbed per kg. and we call it Gy (Gray), the absolute does.  and we can have Gy per hour, Gy per second, depends on the time scale.

Thus, for 1Gy, an alpha source, typical 5MeV per decay and a total absorption, it takes 6241 billion hits. for 10 billion decay a second, it takes 10 mins for this does.

in other system, people use rad for radioactivity, for example in the game fallout. which is 1 Gy = 100 rad.

Here is a list of different does and the health effect

• 0.25 Gy or less  –  no effect
• 0.50 Gy  – temporary reduction of while blood cells
• 1 Gy  –  feel nausea, vomit
• 1.5 Gy – 50%  more chance to have cancer, cataracts or leukemia
• 2 Gy – 5% people will die
• 4 Gy – 50% people will die in 30 days
• 6 Gy – 90% people will die in 14 days
• 7 Gy – all die

However, different part of our body have different does limit. so, for 1Gy radiation, our eye will be blinded, but our skin has no effect. therefore, we introduce another does, called Effective does Sv ( Seivert ). it is produced by a weight factor.

different part of out body has different weight factor, for lower weight factor, the more easy the tissue will get hurt.

here is the list of weight factor

• whole body  – 1
• skin — 0.01
• lung — 0.12
• liver — 0.05
• breast — 0.05
• stomach — 0.12
• gonads – 2

we see that the weight factor for whole body is 1, therefore, in general we use Gy or Sv in equal way.

And we also observed that, when we eat some radioactive matter inside, the limit does will be decreased by 8 times in out stomach. or to say, the radioactive matter is 8 times dangerous than it is at outside. the reason is, when the matter is inside our body, we will absorb all energy.

a typical does is 3.75mSv per year. which means we received 3.75mSv in a year, 0.4uSv per hours.

for radioactive effect, there are 3 types of effect: acute ( short term ), late ( long term) and genetic effect ( permanent ) .

• short term – Erythema, hair loss, etc
• long term – Cateracts, Leukemia or cancer, etc

and the effects can be divided into 2 groups – deterministic and stochastic

• deterministic mean the chance of having those effect is increase with the radiation increased.
• stochastic is similar to deterministic, but there is a threshold. below the threshold there is no effect. – Cancer and genetic effect are belong to this group.

Differential Cross Section II

Last time, the differential cross section discussion is based on quantum mechanics. This time, i try to explain it will out any math. so, that my mum ask me, i can tell her and make her understand. :)

in a scattering experiment, think about a target, say, a proton fixed in the center, it is positive charged. if another proton coming with some energy. it will get repelled, due to the repulsive nature of Coulomb force of same charge. it should be easy to understand, if the proton coming with high energy, it will get closer to the target, or even enter inside the target.

the repelling angle of the proton is not just depend on the energy it carry, but also on the impact parameter ( we usually call it b , but i like to call it r). the impact parameter is the shortest distance between the target and the line of the moving direction of the proton at long long away.

if the impact parameter is large, the proton miss the target. it almost cannot feel the target affection. thus, it go straight and unaffected. when the r is zero, it will hit the target head on head. and due to the repulsion. it will return back.  so, we can understand. the smaller the impact parameter, the deflection will be larger. since it can feel the force stronger.

For same impact parameter, the higher energy proton will have less deflection, since it travel faster, spend less time by the force, and the deflection get less.

Thus, we have an idea that the angle of deflection is high for small impact parameter and high energy. And most important, it only depends on these 2 factors and the effect from the target.

since our detector can only detect some small angle over some small area. So, we can place out detector on some angle, get the yield, and this is the name –  differential cross section come from.

Now, we have a uniform flow of particle with energy E. they will be deflected by the target and go to some angles. If we detect at the deflection angle, see how many particles ( the yield ) we can get in each angle. we can calculate back the effect of the target. For example, for a small angle, the particle get little defected, and this means the particle is from large impact parameter. for a large angle, the particles are from small impact parameter.

In some cases, the number of particle detected will be very high at some particular angle then others angles and this means, the cross section is large. and this means something interesting.

Moreover, don’t forget we can change the energy of the beam. for some suitable energy, the particle will being absorbed or resonance with the target. that given us low or high cross section on the energy spectrum.

( the graph is an unauthorized from the link: http://www.astm.org/Standards/E496.htm )

The above diagram is the differential cross section obtained from a Deuteron to a Tritium ( an isotope of Hydrogen with 2 neutrons and 1 proton) target, and the reaction change the Tritium into Helium and a neutron get out.

The reaction notation is

where a is incident particle, X is target, b is out come particle, and Y is the residual particle.

the horizontal axis is detector angle at lab-frame. and the vertical axis is energy of Tritium. we always neglect the angle 0 degree, because it means no deflection and the particle does not “see” the target. at low energy, the d.c.s. is just cause by Coulomb force. but when the energy gets higher and higher, there is a peak around 60 degree. this peak is interesting, because it penetrated into the Helium and reveal the internal structure of it. it tells us, beside of the Coulomb force, there are another force inside. that force make the particle deflects to angle 60 degree. for more detail analysis, we need mathematic. i wish someday, i can explain those mathematics in a very simple way.

Therefore, we can think that, for higher energy beam, the size we can “see” will be smaller. if we think a particle accelerator is a microscope. higher energy will have larger magnification power. That’s why we keep building large and larger machines.

density, flux & luminosity

density is a consideration factor for scattering experiment. In low density, both for target and the beam, then the probability of collision will be small and experiment will be time consuming and uncertainly increase. Remember that the size of nuclear is 1000 times less then the atom. the cross section area of it will be 1000 x 1000 times lesser. the chance for a nucleus-nucleus collision is very small. for example, if there is only 1 particle in the area on 1 atom, the chance for hitting the nucleus is = 0.000003, 3 in 1 million. it is just more than nothing. thus, in order to have a hit, we have to send more then 3 million particles for 1 atom. in some case, the beam density is small, say, 0.3 million particles per second on an area of 1 atom. then we have to wait 10 second for 1 hit.

density is measured in particle per area for target .

for beam, since particle is moving in it, time is included in the unit. there are 2 units – flux and luminosity. flux is particle per second, and luminosity is energy per second per area. since energy of the beam is solely by the number of particle. so, density of beam is particle per second per area. but in particle physics, the energy of particle was stated. thus, the luminosity is equally understood as density of beam, and their units are the same as particle per second per area.

In daily life, density is measured by mass per volume. although the unit are different, they are the same thing – ” how dense is it? ”

in solid, the density is highest compare to other state of matter. from wiki, we can check the density. and coveted it in to the unit we want. for example, copper has density . its molar mass is . thus, it has copper atom in . and , which is a huge number, so, the number of atom on is .

how about gas? the density depends on temperature and pressure, at and 1 atm pressure, helium has density   and the molar mass is 2. thus, the number of Helium atom in is . when the temperature go to -100 degree, the density will increase.

beside of the number of atom per area. we have to consider the thickness of the target. think about a target is a layer structure, each layer has certain number of atom per area. if the particle from the beams miss the 1st layer, there will be another layer and other chance for it to hit. thus. more the thickness, more chance to hit.

For a light beam, the power and the wavelength determine the flux of photon. power is energy per second. and energy of single photon is inversely proportional to its wavelength. the density of a light beam is given by :

where, is the luminosity and is the flux. for typical green class 4 laser, which has power more then 0.5 W and wavelength is about 500nm. the flux is about   photons per second per unit area. if the laser spot light is about 5 mm in diameter. thus, the density of the beam is .

for laser pointer in office, which is class 1 laser. the power is less then0.4 mW, say, 0.1 mW. for same spot size, the luminosity  is still as high as .

on LHC, the beam flux can be . by compare the the density of solid copper. it is much denser. thus, a collision in LHC is just like smashing 2 solid head to head and see what is going on.