Interaction between light and matter - no room for the photon

One of the fundamental aspects of physics is the study of light and how that interacts with matter. I have been putting off this post - mainly because I am not a quantum mechanic (I am a classical mechanic). There are lots of things that could be done in this post, but I am going to try and keep it limited (and maybe come back to the interesting points later). Also, most of my posts are aimed at the intro-college level or advanced high school level. This will be a little higher. If you are in high school, there is still a lot of stuff for you here.

Let me summarize where I am going to take this post. I am going to try and briefly describe the quantum nature of matter. I am then going to show how this relates to light. In the end, I will show that the common idea that light has a dual wave-particle nature is not a necessary model. Just about everything that normal physicists (especially undergraduates) look at can be explained with light as a wave.

One final point. I am not really that knowledgeable in this area (especially compared to some). I did not "figure" this stuff out. Instead, I am repeating the arguments that David Norwood proposed after he summarized the arguments of others.

Now on with the fun.

This is the Shrödinger Equation (in one dimension):

i-8dc50ed1fb2474b9bd4d2e83da403be9-screenshot-62.jpg

  • Notice the German characters - (can't remember what it is called). That is the correct way to spell it, but due to my extreme laziness, I will use a normal "o".
  • i is the imaginary number, sqrt(-1)
  • i-d7b70595cea5a9ac64c51450bdbb4e6f-screenshot-93.gif (called h-bar) is a constant (I will talk about this later)
  • i-d5abf04b32a14f3a62a6903d4aa18e9e-screenshot-94.gif notation represents a partial derivative - or "how does this variable change as t or x changes." i-d5abf04b32a14f3a62a6903d4aa18e9e-screenshot-94.gif2 notation means "do it twice".
  • i-e5fb01d9432db6dc22da7fefdce6bf40-screenshot-95.gif is called the wavefunction. What does it mean? I will get to this in a moment.
  • V is the potential that the particle is in. It could depend on both time and x, but I will have time-independent potential.


I am not going to talk about the historical development of the Schrodinger equation (not now anyway), but let me just say that this model seems to work. But what is i-e5fb01d9432db6dc22da7fefdce6bf40-screenshot-95.gif anyway? i-e5fb01d9432db6dc22da7fefdce6bf40-screenshot-95.gif is not something that one can observe, but i-e5fb01d9432db6dc22da7fefdce6bf40-screenshot-95.gif*i-e5fb01d9432db6dc22da7fefdce6bf40-screenshot-95.gif is (where the * means "take the complex conjugate" or basically, replace i with -i). i-e5fb01d9432db6dc22da7fefdce6bf40-screenshot-95.gif*i-e5fb01d9432db6dc22da7fefdce6bf40-screenshot-95.gif(x,t) gives the probability density so that

i-1ef124d8de20edd5c142554431d2eb5c-screenshot-63.jpg

Where P is the probability of finding the particle between x1 and x2. And this is one of the main points of quantum mechanics: the Schrodinger equation basically gives us probabilities. Ok - enough about the Schrodinger equation.

Suppose I have a particle in an infinite well. Basically this means that the potential is infinite at x=0 and at x=a (the length of the well) and zero in the middle. Since the potential is infinite outside the well, there is a zero probability of finding it there.

So, how do I get a solution to the Schrodinger equation for this situation? First, I will make the assumption that I can separate the wave function into a part that depends on x and a part that depends on time:

i-58c461d828997f34d70e492d78726187-screenshot-65.jpg

If I put this into the Schrodinger Equation, it becomes:

i-d45e608c0271c5f96eafe0f7fbf7db18-screenshot-66.jpg

Here, when I take the partial with respect to time, the space portion is a constant and comes out front. The same is true for the partial with respect to x. Now, I if I multiply both sides of the equation by 1/(i-9b0b8240c9fd30c55d2183980f412aa2-screenshot-96.jpgf), I get:

i-60be5be4e2f781dfe4f539efd96f3917-screenshot-70.jpg

That looks complicated enough that there is probably a mistake in there somewhere. I am sure Uncle Al will find it if it exists. So, what now? Well, I could rearrange this just a little and get something that looks like this:

i-b76e4e813c98299d6ee5ee6d2488ed51-screenshot-69.jpg

So, here I have two pieces that add up to zero. The first piece only depends on t and the second only on x. The only way that these will add up to zero is if they are both equal to a constant. This constant turns out to be the energy (E). Now there are the following two equations (since the partial derivatives only deal with one variable, I can write it as a normal derivative)

i-ee8c060899af4de3bf7f0fbfeb27a368-screenshot-71.jpg

I don't want to go into the details, but the time equation can easily be solved (if the potential does not depend on time). This gives the time-portion of the wave function as:

i-c0b19b07619443fb74c562de4ff5f3e2-screenshot-72.jpg

Now for the x-portion. I can multiply both sides by

i-9b0b8240c9fd30c55d2183980f412aa2-screenshot-96.jpg and get:

i-03fd73a3b4700b6b2a9ae3509f09f6f8-screenshot-73.jpg

Typically, when looking at a particle in a box (infinite well) this is the starting equation. Now for the infinite well. Inside the well, V=0 and there is no solution outside the well (because the potential is infinite). This gives:

i-a6f4a78c14f7e6ad388ff96e2df78e79-screenshot-74.jpg

So, from here, I can guess a solution for i-9b0b8240c9fd30c55d2183980f412aa2-screenshot-96.jpg. This equation says that if I take the derivative twice with respect to x, I get some constant times the same thing back (with a negative sign). Two functions that meet those requirements are the sine and cosine functions. If you want, you can test for yourself that the following satisfies the above equation:

i-7e7358569d8fe8455c41d110e30f6fe4-screenshot-75.jpg

Where A and B are some constants. I can find A and B by applying the boundary conditions. If the well goes from x = 0 to x = a, then i-9b0b8240c9fd30c55d2183980f412aa2-screenshot-96.jpg(0) = i-9b0b8240c9fd30c55d2183980f412aa2-screenshot-96.jpg(a) = 0. When x = 0, sin(0) = 0, so the first term is ok. The only way to make the second term go to 0 is if B = 0. Now, I have the following:

i-90987b54b9b980ae4e7dea0113848b41-screenshot-76.jpg

The way to make i-9b0b8240c9fd30c55d2183980f412aa2-screenshot-96.jpg(a) = 0 is if

i-44c6a32519a3a30ff2ddec462a2557e7-screenshot-77.jpg

Since k is related to E, and k can only have certain values, E can only have certain values. Energy is quantized.

i-0b8e9b9a8f65e24c1e0255b54b7aa6e6-screenshot-78.jpg

One more thing, what is the wavefunction? (need to add that time part back in)

i-abdba0f59dd43b21fe43ecbb40ce58a1-screenshot-79.jpg

I never solved for A. This is not too difficult to do. The probability of finding the particle between x=0 and x=a is 1 (it has to be there somewhere). So, if I set the following:

i-263e14c39990813d1dd777ad7c88f072-screenshot-80.jpg

I can solve for A, and I get:

i-7f6df1fddb4c3a2e5d41b2b9ca09bcfd-screenshot-81.jpg

Wow. This is getting to be a long post. I haven't even really done anything cool. All of the above stuff can be found in any intro modern physics or quantum textbook. What if I put the energy at E1? If I do this, and plot i-e5fb01d9432db6dc22da7fefdce6bf40-screenshot-95.gif*i-e5fb01d9432db6dc22da7fefdce6bf40-screenshot-95.gif, the time dependency drops out. Let me write this out explicitly:

i-8e6233782a7eb39fc735f4a5471bafcb-screenshot-82.jpg

Notice that the time portion goes away. If you plot this, it would look something like this:

i-8daa839f8c7bda45c9cb558db2c3a160-screenshot-831.jpg

*This is a screen shot from a java program available at Open Source Physics. Download the .jar file and you can do all sorts of cool things. As an exercise, try running the program for the wavefunction in the second energy level.

When I ran that program, I put the wavefunction only in the first energy level. What if I put a combination of two energy levels (E1 and E2)?

i-6b89c921ace20fc234a3934023a11b66-screenshot-84.jpg

Now, if I find the probability density (i-e5fb01d9432db6dc22da7fefdce6bf40-screenshot-95.gif*i-e5fb01d9432db6dc22da7fefdce6bf40-screenshot-95.gif) I get (I am skipping some of the algebra, you can redo it if you like):

i-65d744c3db229e341d331635b568a3aa-screenshot-85.jpg

Notice in this case that the time terms do not cancel. I will now return to the infinite square well simulator and let there be both and E1 and E2 state. This time, things change:

i-1a8d2540804d3a34117772b99017b761-twostate-2a.gif

(forgive me if the animated gif is too big - I tried to make it small and manageable). Here you see that there is indeed a time dependency. What is the frequency that this thing "oscillates?" I know I can't really say it oscillates, this is the probability of finding it somewhere. According to the equation above, the probability oscillates at:

i-6d6c0215fd14da146d249c87022a2f76-screenshot-92.gif

This is a meaningful relation. However, this post is getting extremely long. I think this is a great place to pause and I will post Part II.

Summary:

  • Start with the Schrödinger Equation. Where does this come from? I have not said, but if you like the Schrödinger equation, then the rest of the stuff I did follows.
  • The wavefunction is the solution to the Schrödinger equation. The "square" (not technically correct) of the wavefunction gives the probability distribution.
  • For a particle in an infinite well, there are only certain allowable energies. We (me and you both) say that energy is quantized.
  • If you put a particle in the well with the ground state energy (or any single allowed energy) the probability distribution has NO time dependence. (technically called a stationary state)
  • If you put a particle in there with a combination of energy states, the probability will oscillate back and forth with a frequency (E2 -E1)/h where h is a constant.
  • I am not a quantum mechanic and have probably made some technical errors that are worthy of a counter attack, but the general idea is ok.

Here is PART II. In the second part, I talk more about absorption and stimulated emission.

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Schrödinger or Schroedinger; o-umlaut. For synthetic organikers pummeled with theory best acquired and applied in a software package... the Screwdinger equation. Theory has its place. The rest of us are busy being empirical. (and yer gonna need relativistic correction for heavy atoms) "8^>)

Your paper is very nice. You should focus on conceptual physics. Experimental and theoretical physics are flawed. Experiments will give results but you still have to understand what the results are saying. Theoretical physics generates errors due to increased assumptions. Conceptual physics is the mathematics of physics but not like mathematics. Conceptual physics is hard because you cannot advance a theory that does not explain previous theories. It is like creating a proof from axioms. Einstein was most successful as a conceptual physicist he failed when he became a theoretical physicist. Faraday was a conceptual physicist and Maxwell quantified his concepts.

By Jamahl A. Peavey (not verified) on 17 Nov 2008 #permalink