Basics of quantum physics

[PG1995]

Hi

Could you please help me with these queries? Thank you.

Regards
PG

 

[steveB]

Q1: (i) Your answer is basically correct. The proper treatment of these interactions involves quantum field theory and is not trivial. But, physicists create simplified descriptions of these interactions. Basically, the nearby nucleus allows the obeying of all laws related to both mass/energy conservation and momentum conservation, as required.

Q1: (ii) The need for passing close to the nucleus is so that it can interact with it and balance the momentum. The nucleus is specified because it is relatively massive and increases the probability of a reaction of that type.

Q2: That's just a wording issue. It says the masses of x and y are z. This means that each individual x or y has mas z. If it said the mass of x and y is z, then the "and' would be interpreted as an addition operation and it would be saying that the total is z.

Q3: Yes, that's just a typo. I often make that mistake too. We are so used to associated electrons and protons together, that when we occasionally mention electrons and positrons, we slip up.

 

[PG1995]

Hi

It is said that a photon has a rest mass of zero but it has relativistic mass where relativistic mass is given as:

[LATEX]m=\frac{m_{o}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/LATEX]
If 'm0' is zero then how can 'm' be greater than zero because 0/anything is zero? Perhaps, I need to take a limit here. Kindly guide me. Thanks.

Regards
PG

 

[steveB]

A photon is never at rest, so it doesn't mean much in simple theory.

If v=0, then you have 0/1 which is zero, which means the mass would be zero at any speed. But, the speed is never anything but v=c, so it's not meaningful, unless you get into much more complicated theory.

However, when v=c, you do have 0/0, but in the limit 0/x where x goes to zero, the answer is still zero.

Also, remember that the photon mass is just assumed to be zero. It has been measured by many techniques and measurements can only say the mass is less than a small number. It was once thought that the neutrino is massless, but that turned out to not be true.

 

[steveB]

PG,

Something else I should mention here. The idea of a rest mass versus a relativistic mass, as you showed, is an archaic concept. Many things in physics are points of view, and this is one of them. Modern physicists prefer to not deal with this concept any more. They only deal with rest mass, and don't even call is rest mass any more. It is just mass. So modern physicists prefer to say that all particles (electrons, protons, neutrons, etc.) all have characteristics that don't change including mass and charge, and various other properties.

This doesn't mean that you can't use the concept. We are all familiar with it. Still, I think it's important to point this out. If you delve into modern treatments of special and general relativity, or even quantum field theories, you will encounter this point of view.

 

[Miles Prower]

Q1) According to relativity theory, observers won't agree on when and where an event occurred. However, all observers have to agree that the event did, indeed, occur. Photons in isolation may look like they're coming from a hard X-ray source to one observer, while another may see the same source as a radio station (blue shift/red shift). The observer who detects the low energy photons of a radio wave can't observe these forming particle pairs. However, he might observe a high energy nucleus collide with a low energy photon, and a resulting particle pair while a different observer sees a powerful X-ray photon strike a slow moving, low energy nucleus. Both observers will agree that a particle pair did, indeed, form.

Yes, indeed. A high energy photon that doesn't come close to the nucleus simply scatters the electrons surrounding that nucleus (ionizing radiation).

Q2) There is a multiplication by two there.

Q3) It's a misprint. Should have been "positron", not "proton"

 

[PG1995]

Hi

Could you please help me with these queries? The text was taken from this source. Thank you.

steveB It's really a nice picture. Your face very well reflects your goodness and kindness. No flattery intended!

Regards
PG


Helpful links:
1: Maxwell–Boltzmann distribution

 

[steveB]

Thanks for your nice complement about the picture. That is me in younger days at my physical and mental peak. My old friend and office mate from grad school just sent me that picture. I haven't seen him in 26 years, and he will be visiting from Australia in a couple of months. He took that picture while we were studying for a brutal exam in probability theory. The teacher flunked every student in the class on the midterm exam and threatened to fail every student for the course if we did not do well on the final. We were studying for days and days at that point. I was determined to do well on that exam and you may see the fire in my eyes in that picture. That final exam required every ounce of creativity and visualization I had. But, I was in the zone that day and nothing could distract me. Even my calculator battery dying just before the exam did not bother me, I just borrowed my friends calculator during the exam, which I wasn't even familiar with. I worked with perfect efficiency for 3 hours straight and calmly finished the exam one second before the Prof called the exam over. This may sound as exaggeration, but it is exactly what happened. As I walked out of the room with my friend he turned to me with a look of shock and said, "that was brutal" (no one else had finished in time), but I knew I had done well and just said, "I think I did ok". I ended up with the highest grade which gave me an A for the course. The Prof. (a well known guy in his field, who has written many text books) failed half the class for the course that year. He wasn't fooling around.

Not trying to brag because that guy in the picture is almost like a different person to me now, and this "operation in the zone" will happen to everyone at some point in their life. I was just fortunate that it came at a critical time when I needed it. But, I think there is a lesson here. If you prepare to the fullest extent of your ability and KNOW you deserve to do well, there is a sense of calm you will have when you enter the "arena". This confidence can do nothing but improve your chances of success.

Anyway, on to the questions!

Q1: There are two ways to answer this.

First, consider the formula itself. By definition E2 is the excited state, which means it has higher energy than E1, and this means E2-E1 must be positive. If T is very large, then exp(0)=1 which means that N2 is about equal to N1, but since the -(E2-E1)/kT never goes to zero, but is a very small negative number, N2/N1 is equal to a number near to but less than one, which means N2<N1.

Also, if T is small, then the exponential argument looks like exp(negative large number) which means N2/N1 ~ 0 meaning that N2<< N1.

Second, consider the physics. Thermal excitation occurs when the thermal energy KT is comparable to or greater than the energy difference E2-E1. If kT is small, there is not enough energy to cause the excitation and N2<<N1. If KT is large, then there is plenty of energy to cause excitation. This might lead you to believe that this leads to N2>N1, but if that happened, you would need an energy input to the system to maintain the population inversion. However, an energy source causing excitation is not a case of thermal equilibrium. In other words if N2 could be greater than N1, you would be able to make a laser without an input energy source and you would have free energy. So energy conservation requires N2< N1 in thermal equilibrium.

Q2: This result follows from the fact that, at most, N2 is just slightly less than N1, If N2<N1, then you have N2<Ntotal-N2 which means 2N2<Ntotal, which means N2/Ntotal< 1/2.

 

[PG1995]

and this "operation in the zone" will happen to everyone at some point in their life. I was just fortunate that it came at a critical time when I needed it. But, I think there is a lesson here. If you prepare to the fullest extent of your ability and KNOW you deserve to do well, there is a sense of calm you will have when you enter the "arena". This confidence can do nothing but improve your chances of success.

This is why I always say that I learn from you more than you think!

Best wishes
PG

 

[PG1995]

Hi

Second, consider the physics. Thermal excitation occurs when the thermal energy KT is comparable to or greater than the energy difference E2-E1. If kT is small, there is not enough energy to cause the excitation and N2<<N1. If KT is large, then there is plenty of energy to cause excitation. This might lead you to believe that this leads to N2>N1, but if that happened, you would need an energy input to the system to maintain the population inversion. However, an energy source causing excitation is not a case of thermal equilibrium. In other words if N2 could be greater than N1, you would be able to make a laser without an input energy source and you would have free energy. So energy conservation requires N2< N1 in thermal equilibrium.

I'm using this text.

I don't really understand the significance of what you said above. I would say that in case of two-level energy state system even when the temperature T is infinite, the number of atoms in the excited state, N2, cannot exceed the number of atoms in the ground state, N1. It simply means that a material with two-level energy states cannot support population inversion, i.e. N2 > N1, even when a lot of energy is input into the system. On the other hand, three-level energy system can support population inversion due to its metastable states assuming it is supplied with constant source of energy.

I'm also confused about one other point. It is seen, using Maxwell-Boltzmann distribution, that in case of two-level energy system we needed to raise the temperature to such high level just to get some fraction of the atoms into the excited state. Perhaps, we needed such a high temperature because a lot of energy from the raised temperature goes into other processes such as heating effect. I understand that Maxwell-Boltzmann distribution is used for gases. But I was just thinking that perhaps using an electric current could be more efficient in the way that most of the energy goes only into exciting the atoms. In other words, raising temperature in order to excite the atoms is not an efficient way to do this.

I think it means that these special laser materials can support population inversion due to their metastable states and further to maintain this population inversion they require very less energy.

Please help me with the queries above. Thank you.

Regards
PG

 

[steveB]

I would say that in case of two-level energy state system even when the temperature T is infinite, the number of atoms in the excited state, N2, cannot exceed the number of atoms in the ground state, N1.

Yes, that's exactly what I said. And, this is for thermal equilibrium only.

It simply means that a material with two-level energy states cannot support population inversion, i.e. N2 > N1, even when a lot of energy is input into the system. On the other hand, three-level energy system can support population inversion due to its metastable states assuming it is supplied with constant source of energy.

A two or three level system can support population inversion. But neither will happen in thermal equilibrium. What is needed is an external source of energy other than thermal energy. When I mentioned energy input, I was implying energy other than thermal energy. For example, light or electricity would qualify as an energy source.

I'm also confused about one other point. It is seen, using Maxwell-Boltzmann distribution, that in case of two-level energy system we needed to raise the temperature to such high level just to get some fraction of the atoms into the excited state. Perhaps, we needed such a high temperature because a lot of energy from the raised temperature goes into other processes such as heating effect.

In a sense, the two levels are competing with each other. When the lower level absorbs a phonon (phonons are the quantum particles associated with heat and act similarly to photons in that they can be emitted and absorbed in a material) and gets excited to the higher level, that higher level can then emit the same type of phonon, either spontaneously or in a stimulated way. This is why thermal equilibrium can not give you a net energy in usable form. The more excited states there are, the more phonons are emitted. This competition can never let N2=N1 unless the temperature becomes infinite, which is not possible by known physics.

I understand that Maxwell-Boltzmann distribution is used for gases. But I was just thinking that perhaps using an electric current could be more efficient in the way that most of the energy goes only into exciting the atoms. In other words, raising temperature in order to excite the atoms is not an efficient way to do this..

Thermal equilibrium does not provide a usable energy source. If it did, we could power all our devices from ambient temperature. But, we know that we can only get energy via temperature differences, which is not thermal equilibrium. When you provide power to a system (electric power, optical power, mechanical power etc) you no longer have thermal equilibrium.

I think it means that these special laser materials can support population inversion due to their metastable states and further to maintain this population inversion they require very less energy..

Yes, the long metastable lifetime makes the threshold energy/power less, and makes it easier to achieve population inversion and lasing.

 

[PG1995]

Thanks a lot.

A two or three level system can support population inversion. But neither will happen in thermal equilibrium. What is needed is an external source of energy other than thermal energy. When I mentioned energy input, I was implying energy other than thermal energy. For example, light or electricity would qualify as an energy source.

So, population inversion is also possible in two-level energy system but it required an source of energy other than thermal energy, right? I had thought that in two-level system the population inversion is impossible. Perhaps it would be more right to say that in two-level system population inversion is achievable but overall the process is very inefficient and a lot of energy, say electrical, needs to be pumped in to maintain the inversion.

If you don't mind, I'm still confused about the thermal energy. Why population inversion is not possible with thermal energy and yet with electrical energy it's quite possible. Thanks.

Regards
PG

 

[steveB]

Yes, 2 level systems can achieve population inversion. Really, both 2 and 3-level systems are relatively inefficient because the optical transition terminates at the ground state which is difficult to deplete in population. It is 4-level lasers that are considered more efficient and easier to achieve inversion. Usually, 3- and 2-level lasers are pulsed lasers, for this reason, but an erbium doped optical fiber is able to achieve continuous lasing because the optical intensity is very high due to the small core of a fiber.

As far as thermal energy, let's make the distinction between "thermal energy" and "thermal equilibrium". Thermal energy can be useful and could be used to create an energy input to create population inversion. For example, a temperature difference is thermal potential energy. This temp difference could drive thermoelectric coolers (in reverse) to generate electricity, which could then power a laser diode, which could then pump a laser-crystal an achieve lasing. Thermal equilibrium is the state where all temperatures are equalized, and you can't extract useful energy from ambient temperature that is equalized. This is a fundamental thermodynamic law. Hence, any theory, such as the one you are asking about, will automatically have this law of nature built into it and this is why you see that your equation can't show N2>N1 in thermal equilibrium. If it did show such a thing, we would know the equation is wrong. When you describe a laser, you end up writing Einstein rate equations that allow energy input, and there you will see that population inversion is possible, but those are not cases of thermal equilibrium (they are quite the opposite).