Fundamentals of Physics II

PHYS 201 - Lecture 25 - Quantum Mechanics VII: Summary of Postulates and Special Topics

Chapter 1: Major Postulates of Quantum Mechanics [00:00:00]

Professor Ramamurti Shankar: So the rules of quantum mechanics are going to be stated for one last time. Now we’re using the fact that you’ve heard them over and over, it’s going to be more compact than before. So it’s really the form of many postulates, because it’s a theory which is based on experiment; it’s not deduced by mathematics. You make up some postulates that condense all the experimental facts. There are new mathematical results you will need, but they are separate. They are deducible mathematically. They are not part of the physics postulates.

The first postulate — by the way, this is the postulate for Physics 201. If you know more mathematics and so on, there’s another way to write the postulate, but we’re not interested in that. We just want the rules for this class, and I don’t even care how much of this you carry in your head, but I want you to know that in principle, every problem I gave you, every question I asked, can be answered just by turning to these postulates, so you should know what they are. So the first postulate is that every particle — this is for 1 dimension, 1 particle only — is given by a certain wave function, Y(x), that contains all the possible information on that particle. This is the analog of x and p in classical mechanics. And we always choose Y so that Y²dx is normalized to 1 in all of space. That is a convention; it’s not required. If you take a Y and you multiply it by a number, it stands for the same physical state. So you pick the number so that the squared integral of Y is 1.

The second thing is, a state of definite momentum. In other words, is there a wave function that describes a particle guaranteed to have a definite momentum when I measure it? The answer is yes. That function is denoted by Y _p(x). It is not an arbitrary function. This Y is whatever you like. This is a particular Y which assures you that if you measure my momentum when I’m in this state, you will get the value p, and it looks like this: e^ipx/ℏ divided by square root of the length of that universe in which you’re living.

That’s the second way to define this function. Either I can say this is the answer, or I can give you a little way to find this function in an equivalent way. The equivalent way says states of definite momentum obey the following equation. If you solve this equation, the function you get is in fact Y_p. So Y_p can be defined either way, either by saying, solve this equation and then the function you get is Y_p, but you can easily verify that this function in fact is a solution to that equation.

Now there’s a new piece of mathematics that’s not connected with this. This is a postulate. Mathematics says, if you live on a ring, you go around in a circle, and the x goes to x + L, you’ve got to come back to the same function. That quantizes p, I have the value as (2ph/L) times m, where m is an integral. Let me call it n. m is the particle mass. That is a mathematical deduction. You don’t have to make that a postulate. That’s the postulate; that’s a deduction.

Second postulate says — or third postulate — a state of definite position, x₀. Namely, if you want to describe a particle which is guaranteed to be at x₀, what function describes that. I’m going to give the function the name Y _X0. You have to be very clear on what this x₀ is and what x is, okay? This p is a label that says something about the wave function. This is a function of x, so x varies from - to + infinity. This is a label for the state. What’s special about this state? It’s got a momentum p. What’s special about this state? It’s got a definite location x₀. What is the function? I think you should know by now what the function looks like. Anybody know? State of definite x₀? Any guess? Yes?

Student: [inaudible]

Professor Ramamurti Shankar: Yes, so I’m going to make it a little simpler. It’s the function, I’m just going to call it big spike at x₀. Now big spike at x₀ is a little vague and deliberately left so, because this spike has a little bit of non 0 value away from x₀. And you can make it narrower and narrower and taller and taller and that’s what has a technical name, but I don’t want to go there. Let’s imagine we can only measure position to one part in 10 to the 97 meters. Yes?

Student: Is it the Dirac delta function?

Professor Ramamurti Shankar: This one is not, but eventually it will become that. But I just want to keep it like this. But instead of saying it’s a spike, I can also say it obeys the following equation. x Y_x0 (x) = x₀ times same function(ie Y_x0 (x)).

So let me take a minute to discuss both these equations. They are very special mathematical forms, and this is just for your own — you don’t have to know this part of it. This will do for this course. If you want to know the whole story — see, normally, when you take a function Y(x), and you differentiate it, you will get a new function, right? Take a sine, you’ll get a cosine. If you take x², you will get 2x. So the act of taking derivatives usually gives you a brand new function, a different function. This one says, except for those constants here and here, if you take the derivative, it should look like a multiple of the same function. Obviously, that’s a very special function. You’re saying upon differentiating, it’s going to look the same. Well, not every function has a property, but we all know one function does, the exponential function.

Similarly, if you took a function, any function Y (x) and multiplied it by x, you will get a new function, right? Cos x will become x times cos x. Sine x will become x times sine x. e^x will become x times e^x. It becomes a completely new function. But I am demanding that for this magical function, when I multiply by x, it is essentially a constant times the same function. You’ve got to ask yourself, how can I take a function, f(x), multiply it by x everywhere and it looks the same except for multiplicative factor. It looks impossible. But the spike is in fact such a function. Let us see why.

If we take a random function, if you multiply it by x, which looks like this, it will get taller and taller as you go to the right. But suppose the function in question is a spike at x₀. If you multiply it by x, you’re basically multiplying only by x₀, because it has a life only of x₀. It doesn’t exist anywhere else. The only place it’s even non-zero is at the point x₀. So there’s no difference between multiplying it by x, and multiplying it by x₀. Multiplying by x all over here is a waste, because the function is 0 there. That’s why for that spike function, this is actually true. You see that? It’s the second way that characterizes it. So again it says multiplying by x is generally an operation that changes the nature of the function, but there are special functions for whom the effect is to simply rescale the function by a number. Find me that function. That’s the state of definite x₀, and the only answer to that is a spike.

So we have the option of simply accepting the left hand side as postulate 2 and postulate 3, but there are equivalent ways to say this. And I will tell you in a minute why I’m saying this, because you can ask the following question: what is the state of definite energy? The state of definite energy, I’m going to write it this way in the right hand side, is also a solution to some equation. That equation is −(ℏ²/2m)(d²Y_E/dx²) + V(x)Y_E(x) = EY_E(x).

But I’m trying to tell you that I’m not going to elevate this to a postulate, because I think one of you noticed already something very familiar about this equation. Because, you see, in classical mechanics, energy is given by (p²/2m) + V(x). Do you see that on the left hand side, except for the function Y, I have what here looks like ½m x(-iℏd/dx)² + VY. That’s the left-hand side. So that once you know states of definite p and states of definite x obey these equations, states of definite anything else, any function of x and p, obeys a similar equation, but every p is replaced by −iℏd/dx, and every x is just simply left as an x. So this energy formula does not require a new axiom, but if you didn’t follow this logic, you can simply say I have one more postulate, which is that if I have states of definite energy, I must solve that equation.

But then you will need more and more postulates, because you’ve got zillions of possible variables you’re interested in. What if somebody wants to know, what is x² + p²? That’s a variable in classical mechanics. But I’m not going to write another postulate for that, because if you wanted x² + p², states of definite x² + p², it will obey the equation ((-iℏ)²d²/dx²) x Y = some constant, whatever it is, times Y. Namely, p is going to be replaced by the square of this derivative. Square of a derivative means do it twice.

So there’s a universal rule for finding states of definite anything. Let g be a general variable in classical mechanics. In quantum theory, states of definite g satisfy the equation where you first write the classical formula for g. Maybe g is x² + p². Replace x simply by x; replace p by -iℏd/dx and let that multiply as a function or differentiate some function, and that will be a state of definite g.

So I’m just giving you two ways to do this postulate. Either you can say states of definite energy are solutions to the following equation, that’s HY= E Y, and that’s what I’ve been doing so far, because I didn’t mind in this class to introduce it as one more postulate. If you’re looking further ahead, you should realize that it is really not an independent postulate, that once you know what happens to x and once you know happens to p, you can predict what the equation will be for any function of x and p. So if you see that analogy and it’s helpful to you, you’re welcome to the right hand side of the board. If you find it complicated, add on to your list of things to remember one more thing. If you want a state of definite energy, you’ve got to solve this equation.

Okay, now we come to another postulate, postulate 4. Is that right? Postulate 4. If Y(x) is written as a Σ_gA_g Y_g(x) where g could be p, g could be E, g could be position, whatever you like, then the probability that you will get a value g is simply |A_g|², and A_g is given by ∫Y_g*(x)Y (x)dx. Perhaps you can see now, if I put g = p, this is what I call |A_p|², and this will be e^ipx etc. If I put g = energy, this will be A_E, this will be Y*E. Whatever variable you want, you do these integrals and you get the probabilities.

Then there’s another postulate, postulate 5. Postulate 5 says if g is measured and you get one value, say g₁₀ — this is a postulate, you don’t have to write g₁₀. You can say one particular value — then the Σ_g A_gY_g collapses to one term, just Y_g0. g₀ is a particular value of g, namely the third value, the fourth value or the ninth value. This says the state collapses. So originally, you have a state, a general one, which is potentially capable of giving any possible answer that’s contained in this sum, with the probability proportional to the square of the coefficient. But once you made a measurement and you got a particular answer g, call it g₀, all the terms of the sum disappear except the one term, corresponding to the one answer you got. That’s a new property of quantum mechanics, that the act of measurement changes the state from being able to do many things to being able to do one thing. The one thing could be g, if you measure g. Namely, it can be momentum, it can be energy, it can be position, but you’ve got to be very careful that if you measure momentum and got a state of definite momentum, then you go and measure position and got a state of definite position, and you go back and measure momentum, you won’t get the same answer. States of definite momentum don’t know what the position is, and states of definite position don’t know what the momentum is. You will never be able to get them both. You’ve got to pick one or the other.

By the way, I want to point out to you one postulate seems to be somewhat missing, the oldest thing we ever learned in this class, right? Why didn’t I mention that? That’s actually contained here. It’s contained here because if I say what’s the probability that x has a value x₀, I will take integral Y(x₀)*Y(x) dx. Look at what this is. Y(x₀)* is a spike at x₀. Y(x) is some function. But the only place where the products survives is near x₀. Everywhere else is 0, so the only place that integral receives a contribution is when x is at x₀. So up to some constant proportional to area of this spike, the answer is proportional to the wave function at x₀. Therefore — I’m sorry, I mean to say square this, and that is what I’ve been saying is the rule for position. So again, for position you can write a separate postulate, but it’s not. The postulate that I gave you here, if you put g = position, will give you the same answer, namely this one.

Again, if you don’t want to get into too many details, I would say the following is what you should know. You want a stated definite momentum, or the probability for definite momentum, take the integral with Y_p. For energy, take Y_E. For definite position, forget all integrals. Just take Y at the point x and then square it. What I’m telling you is taking Y at the point x and squaring it is what you will do if you took the integral of Y with a spike, which will filter out, if you like, Y at only the one point x₀. And if you square it, you will get Y (x₀).

And the last postulate, all-important postulate 6, says that the state changes with time according to the Schrödinger equation. I’m just going to call it HY, but from now on, it will be -d² Y/dx² + V(x)Y(x). If I get tired of writing it, I will call it HY. That’s of course a postulate, because you can never derive Schrödinger’s equation any more than you can derive Newton’s laws. This is, as I said, one of the most powerful equations in modern physics, because it contains all of non-relativistic quantum mechanics and all of classical mechanics, and all of solid state physics and super fluids and superconductors. Everything comes from this equation. A very powerful equation.

Okay, now what did we learn from this equation? I will mention one thing that’s very important. It’s a consequence of the equation, which is, if you start the system out at time t = 0, in a state of definite energy and you see what happens if I wait some time, well, it becomes the following function at time t. It’s essentially the same function, multiplied by [eⁱ]Et/ℏ. You realize, generally a wave function, when you let it evolve, will flop and wiggle and change its shape in a complicated way. But if you release it in this initial condition, all that happens to it is that. These are called stationary states, because despite the time dependence I showed you, the probability for measuring anything is time independent. Because when you take the probability, you find the absolute value of some integral, and the absolute value of this factor just goes away. So it’s time independent. That’s why we’re very interested. If you want an atom and the atom has been sitting for a long time, it will be in one of these states of definite energy.

So that’s the case of starting in a state of definite energy, but then I said, “Maybe I don’t want to start in a state of definite energy; I want to start in an arbitrary state.” At time 0 I’m going to give you a Y (x ,0) and I’m going to ask you, “What is the fate of this state, if I wait some time p?” The simple answer is, take Y(x ,0), calculate A_E=∫Y_E* Y(x,0)dx and then Y(x,t) = A_EY_E(x) e^−iEt/ℏ. In other words, if you take your initial state and expand it — this is called expanding it — as a sum over states of definite energy with some coefficients you compute at t = 0, then at any future time, the coefficients are given simply by the initial coefficients, times this factor. You can check that this will satisfy the Schrödinger equation, because every term in it is a solution to the equation. And at t = 0, when you drop this guy, it agrees with the initial state and that’s all you want. Because in this problem, because it’s first order equation in time, if my initial state matches your initial state and satisfies Schrödinger equation, it is the answer. There are no two answers.

Chapter 2: Applications of Quantum Mechanics [00:20:32]

So I want to give you one — first let me tell you something that’s going to make you relax. Everything that I say after this moment is not in your exam. I don’t want you to worry about that, because it’s an interesting thing and I want to tell you all the things you can do with quantum mechanics. I don’t want you to worry about anything. Just try to follow this and ask questions, and get a glimpse of what you could possibly do with this theory.

First thing you can do is you can say, “How do I understand atoms?” You know the simplest atom in the world is a hydrogen atom. It’s got a proton and it’s got an electron. One has charge e, one has charge -e. And the proton can be taken to be so heavy that it is fixed. Just like when the earth goes round the sun, in principle, the sun is also moving, but we ignore that, so we’re going to ignore the motion of the proton. Then the electron, in classical mechanics, it has an energy which is (p_x² + p_y² + p_z²)/2m. That’s the kinetic energy, and the potential energy -ze²/r, and r is √(x² + y², z²). Do you understand that? That’s the classical formula for energy. Then our recipe tells us that in quantum theory, states of definite energy will obey the following equation: -ℏ²/2m x (d²Y/dx² +d²Y/dy²+d²Y/dz² ) - (ze²/√(x² + y² + z²))Y= EY. You have to solve that equation.

Now don’t worry about how you solve it. In fact, even Schrödinger did not know how to solve it. He had to ask a mathematician friend of his how to solve it. A mathematician owed him a favor, because at that time, the mathematician was having an affair with Schrödinger’s wife and you might say, “Who owed the favor to whom?” Actually, Schrödinger owed a favor to this guy, because meanwhile, Schrödinger was trying to seduce a pair of underage twins. These are all very interesting things you don’t know about the lives of famous people, but if you read his biography, you will find that Schrödinger’s wave function was all over the classically forbidden region.

Anyway, he’s a very interesting person. But what I want you to know is that even he couldn’t solve the equations. I don’t care if you solve it or not, because he had already done the great thing. So you go to this equation and you solve it. So this math guy helped him solve it. Nowadays, undergraduates, graduates, everybody knows how to solve it, but the first time it came, it was quite an unfamiliar equation. If you solve this thing, what you find is that the energy can take only certain values. Those values have some number in front of them a 13.6 electron volts. That 13.6 electron volts is some combination of the electric charge of Planck’s constant and mass of the electron. Forget all that. Some number, divided by n² where n is an integer of 1, 2, 3, etc. That’s it. These are the only allowed energy levels of hydrogen.

If you solve this equation in 3 dimensions and you demand the functions vanish at infinity, rather than blow up at infinity, you will find you can get solutions only at certain special energies and these are the energies you get. So let’s plot these energies. Remember this is energy 0 and this is n = 1. n = 1 is at -13.6eV. n = 2 will be -13.6/4eV. n = 3 will be that thing divided by 9 and so on. These are the allowed energies.

See, that’s a great result, because you are able to find out for the first time what energies are available to this atom, and furthermore, you also learn that if this atom were able to absorb light, it can do that only say by going from here to here or going from here to here, or going from one allowed level to another allowed level. And the difference in energy is E_f - E_i will be ℏw of the photon. That will allow it to go up. Or it can also come down, and if it comes down from some level, n = 4 to n = 2, it may, let’s see, 1, 2, 3, 4, it can go from 5 to 2, it will emit some light. In fact, this is the only way we know anything about the hydrogen atom. No one can see the hydrogen atom. You cannot actually see it because the act of observation will destroy everything. But we know it’s there; we know it’s doing its thing by shining light on it and seeing what it absorbs and what it emits.

The understanding of the quantum world is very different from classical world. For example, if Newton says, “I know the orbits or ellipses,” you can go see them. That’s what Kepler did, looked at them for 40 years, you can see they go in elliptical orbits. For hydrogen, there are no orbits. There are energy levels and there are these corresponding wave functions, Y (x, y ,z). And you can plot them and you know if you start out with any one particular state, it will stay that way forever. And the probability density, I told you, is also fixed in time. They are the clouds one finds in textbook as shape of various atomic orbits.

Okay, yes, there’s only one other thing I should mention, which is that even though these are the only allowed energies, there’s more than one state at a given energy. You may remember when I did a particle on a ring, at a given energy, there are two states of momentum. One has got momentum = square root of 2mE, other has got momentum square root of -2mE, because when you take the square and divide by 2m, you get the same energy. So a given energy can have two states. Here a given energy can have many states. There’s only one here and here you can have 4, and here you can more. There are ways to calculate them. Then it turns out that even this 1 is actually 2, because it’s another variable which does not enter our physics, called the spin of the electron. I don’t want to even go there, but it can do one of two things. So everything I do here, you’ve got to double. So really, there are two states of energy, lowest energy and there are 8 here and so on. These magic numbers come out of doing the Schrödinger equation.

All right, so that’s an example of what you can do with quantum mechanics. You can solve for the spectrum of atoms and you can see what light they will emit, what light they will absorb. You can even go beyond that. You can even ask, what’s the rate at which it will absorb light? What’s the rate at which it will emit light? Everything can be computed, because you see, you not only know the energies, you also know the corresponding functions and they are very important in calculating the rate of absorption and the rate of emission.

Chapter 3: Energy-Time Uncertainty Principle [00:28:00]

Okay, that’s one thing you can do. Now I’m going to tell you about another uncertainty principle which takes the following form. It says DEDt > =ℏ. I’ve got to tell you what it means. It looks a lot like this one, DpDx > = ℏ, but it’s very different in nature, because x is the position of something. You can try to measure it. p is the momentum of something, you can try to measure it. t is not the time of something. t is just time. It’s not time of this or that, and it can be measured arbitrarily accurately. That is not the problem. So I will have to tell you the meaning of this uncertainty relationship. It is different from the others. There are many such things in quantum mechanics, but this is unique in the sense that unlike x and p which are physical variables, time is not a variable describing a particle. It’s an independent parameter along which everything happens.

But I’ll tell you what this means, but before that, I want to go back to the uncertainty principle, which you notice I never mention, because it is not a postulate, it’s a consequence. And I told you all about trying to look at something with a microscope and have you shine photons. And a lot of you tried to find ways to get around that, and say, “Maybe if you did this, maybe if you did that, you can beat the uncertainty principle.” So I’m going to put you out of your misery.

You cannot beat the uncertainty principle for the following reason. Forget about quantum mechanics. Let’s ask the following question: what is the wavelength of this signal? This is 1 meter long. You might say it’s 1 meter, but that’s not a correct answer. This signal doesn’t have a 1 meter wavelength. You know who has a 1 meter wavelength is this guy who goes on forever. That is a wave of 1 meter wavelength. This was 1 meter for a while, then completely dead on either side. So this does not have a definite wavelength, even though you think it does. A wavelength is a repetitive phenomenon in space. It’s got to repeat itself indefinitely to have a unique wavelength. So let’s take a wave train that looks like this. You somehow chop it off nicely near the end. It does some number of oscillations and then it stops. What is the wavelength associated with this?

You can associate an approximate wavelength as follows. First let me remind you that we want to think in terms of k which is 2p/l. It’s a reciprocal of a wavelength, and if you remember, functions look like kx - wt when we studied waves. k is that. What’s the meaning of k? k is the rate of change of phase of the wave, because if you go a distance x, the phase changes by kx. You watch the wave change. Every cycle is worth 2p and it changes by some amount over some distance. k is the ratio of how much phase change you had divided by the distance you travel. Do you understand that? That’s k.

So let me take this wave and the k for this will be 2p times the number of cycles here divided by the length of that wave train. But there is no unique n you can associate with it. These are all full cycles, but once you come near the end there’s a little bit of confusion as to what to do at the end. How do you close it off? So there’s an error or uncertainty of about 1 cycle coming from the 2 ends. So the uncertainty in k is 2 p/L times the uncertainty in n which is 1 or 2, but that’s all it is. You understand? A finite wave train, when you round out the edges, it may not have a full number of cycles. You taper it off in some fashion, but here you’ve got a well defined number of cycles. So it’s + or - 1 coming from the ends. So Dk is that. But this is a particle which is now confined to a region of length L because the wave function is 0 beyond that. So it’s a particle whose position is known to lie inside this interval of length L. So the uncertainty in position is L. Again, uncertainty is a technical definition, but this is our qualitative explanation. So L is really Dx. You can write Dx. Dk is 2p.

This has nothing to do with quantum mechanics. Where did I mention quantum mechanics? I’m saying something intuitively very clear. If you’re defining a wavelength, you have to let it run through many cycles. For example, if you say this guy goes to New York every week, and you observed him for only one week, that’s got no meaning. God knows what will happen next week. But if you studied a person for 50 weeks and found 50 times the person went to New York, it’s more credible when you say this person goes to New York once a week. So periodic phenomenon, either in time or in space, are defined only after many, many, many periods. Therefore a wavelength cannot be defined for an arbitrarily short interval. In fact, the longer the interval, the more wavelengths you can fit in to say that’s my wavelength. So wavelength gets more and more defined as the train gets longer and therefore the incompatibility between wavelength and the length of the train, the fact that if it’s too short in one, it’s too big in the other one, is a classical result.

Quantum mechanics comes in with a new relation that the momentum of a particle is connected to this wave number by ℏk. Multiply both sides by ℏ, ℏ there and h here, then it becomes DxDp is roughly about an ℏ. So once you concede that a particle has definite momentum only if it has a definite wavelength in its wave function, that wave function has to be very extended to have a definite wavelength and therefore a definite momentum. So there is no way you can make it — there’s no way you can take a wave arbitrarily short in its extent with an arbitrarily well defined wavelength. You’ve got to let it do its thing many times.

Mathematically what happens is, there’s a mathematical theorem that says any function you write down can be built out of waves of all possible wavelengths with suitably chosen coefficients. If your wave has an almost well defined wavelength, then the coefficients, as the function of wavelength, will have a very sharp peak at the wavelength corresponding to this. And as the wave repeats more and more and more times, the coefficients will become sharper and sharper and sharper. As the wave becomes very small, if you say “What’s the wavelength of this guy?” it will be very broad. It’s nothing to do with quantum mechanics. It’s just the incompatibility between two qualities, you understand? Wavelength needs some time to express itself. You cannot do it in a tiny region. So if you want to localize the particle, how can it tell you in the tiny region what its frequency of oscillation is in space? So once you concede that, once you concede that wavelength is connected to momentum, you cannot escape the uncertainty principle.

So now I’m going to come back to energy and time, and I come back in the following way. Remember, a state of definite energy E behaves as follows in time. It looks like Y(x) times e^−iEt/ℏ. So state of definite energy, you know the particle has definite energy if you observe the wave function and it oscillates in time. And you can write this as e^-iwt, where w is E/ℏ. So your particle that has been in that state forever has a well defined energy. But if a particle was just put in that state right now, the wave function really does this only from t = 0, before that it was something else, then this function has not oscillated enough times for you to define its time period. So if you’re watching an oscillation, let’s say it’s an oscillation in time, and you say, “What’s the time period of oscillation?” again, if it extends over a certain time T, the frequency w is just the phase change per unit time. So it’s 2p times the number of cycles, but there’s an uncertainty in N of order 1 that’s 2p/T. Therefore if the system has been in existence for a time T and I call that time as Dt — that’s the meaning of Dt — Dt Dw is 2p, then quantum mechanics comes in when you put an ℏ here, you put an ℏ there and you call that the uncertainty in energy.

In other words, if someone tells me, “Go see if that clock is a regular clock. Tell me if it’s running rhythmically and periodically.” Well, if you give me enough time, I observe it for 100 cycles, I say it’s a good clock. But if you don’t give me time even to finish one cycle, how can I tell you anything about its period yet? It hasn’t had time to establish a period. Periodic phenomena have to repeat themselves, so it takes some time. Therefore for a state to have a well-defined period, that is to say, a well defined energy, it has to be in existence for some time. Then it has a well-defined energy. Okay, therefore you can say that if you’ve had time to measure energy only for a finite time, then energy will not be defined to better than this accuracy, DEDt is of order ℏ.

Let me give you an example that is from classical mechanics, nothing to do with quantum mechanics. You take a bunch of reeds, you attach them to the wall. Each one has a different length maybe. They have many frequencies of vibration, natural frequencies of vibration, depending on the length and whatnot. Now if you connect this whole thing, apparatus, to some vibrating gadget, start shaking the whole thing, suppose you shake it at a certain frequency, omega 0 that matches maybe this guy. Namely, it resonates with this guy, so your expectation is, this one will oscillate like crazy and the others will not. So let me look at this rod from the edge, okay? I look at them end on, they all look like this. They’re coming out of the blackboard. So I expect this one to be resonating wildly up and down and the others to be ignoring the signal because they are not at the resonant frequency. But what will happen in real life is that you will take the system and you start shaking it, your intention is to shake it at a very definite frequency that matches this guy, but the system does not know your plans. After quarter of a cycle, it knows you’ve exerted that force. It doesn’t know you plan to keep doing this. So at that time, what it will do is it will try to write this as a sum of many waves of definite frequency. They will contain many, many frequencies, so what you will find is, this guy also oscillates a little bit, that guy also oscillates a little bit, and as you wait longer and longer, so the system caught on to the fact that you are really serious and sending a signal of definite frequency. And when it has stabilized, you will find these guys don’t move, these guys don’t move; the guy at the resonant frequency moves.

So you’ve got to understand, it takes some time for the system to know what frequency you’re sending. If you have not sent many cycles, if you have sent only this much, it does not consider that a periodic function, because to it, the function would really do that or it could do something else. It goes with the information it has. It goes to the mathematical tables and finds the Fourier series for this and sees what frequencies are there. And anything that’s not 0 can excite all these things. And after a while, when the frequency is well defined, one reed moves.

Same thing happens with atoms. I don’t know, somewhere I drew a picture of atoms absorbing energy.So take this atom here in the ground state of hydrogen. Send light of the frequency just correct to go to the first excited state, to n = 2. You take a laser or something of that frequency and you hit the atom. Take a collection of atoms and you will think they will all jump from the ground state to the first state but nowhere else. But you will find the minute you turn on the laser, even though the laser dials say this is my frequency, atoms don’t know that. So initially, it will jump like crazy to all kinds of energy states and only after sufficient time, only after many cycles have happened, it will realize, “Hey, this is the frequency this guy’s sending me.” Then it will start going preferentially to the one state.

Chapter 4: Quantum Mechanics of More Than One Particle [00:41:21]

Okay, now the final topic I want to talk about is really the beginning of a long topic, but I want to give an introduction to it because you’re going to need this. And the question is, what if I have not just 1 electron but 2, not 1 particle but 2. After all, in real life, there are lots of particles. What does the quantum mechanics of more than 1 particle look like? So if there are 2 particles, let’s say an electron and a proton, you will not be surprised that you will now have a wave function of 2 variables, and if you squared that wave function, you will get Y(x₁) Y(x₂) — Y(x₁), x₂², will give you the probability that the proton is at x₁ and the electron is at x₂. You’ve got 2 things, you’ve got to give 2 probabilities, but once again, probabilities come from squaring a function. This has to be a function of 2 variables, because each guy has his own position. So you can expect that in quantum theory of 2 particles, you’ll get a Y that depends on 2 coordinates, with 3 particles, Y that depends on 3. But let me just stop with 2, because you can learn some very profound things in about 5 minutes just by starting here.

Now let’s consider, so let’s take a concrete example. These guys are both in a box, let’s say. Let’s take a simple case, Yof x1 x2 = Ya of x1 Yb of x2, where a and b are wave functions of energy in a box with 1 particle. For example, this could be the state a and this could be the state b. Then we say particle a is in quantum state a, particle b is in quantum state b, right? You can put the electron and the proton into the same box, they can do what they like. Maybe one guys’ doing this and one guy’s doing that. That wave function is written as a product. This is doing this and that’s doing that. But now suppose you say, “What’s the probability of x₂ x₁ of finding the proton at x₂ and the electron at x₁?” That will be given by Y_a(x₂) Y_b(x₁). And that is not the same as this one.The probability to find the proton at some place and the electron at some place is not the same as the probability of finding them with exchanged positions. For example, the chancei of finding the electron here and the proton in the middle of the box is 0. The proton’s function vanishes in the middle of the box. But the probability of finding the electron in the middle of the box and the proton here is not 0. That’s perfectly okay, because there are two different outcomes of the experiment. I look for the particles. I find the electron here and the proton there. I say I found this guy here and that there, and I do it many, many times and you give me the odds.

But something very dramatic happens if the 2 particles are identical. Identical particles in quantum mechanics have very different connotations from identical particles in classical mechanics. So in classical mechanics, suppose you have identical twins, okay? They’re born and they’re separated, they’re moving around. Let’s say they look identical in every way. We can still follow them. We know this is Joe and this is Moe. Follow the two characters no matter how identical they are, because we can keep track of them continuously. So let’s do the following experiment involving these twins. There are 4 doors in this room and 1 twin comes out like this, the other twin comes out like that. Then they do one of 2 things. Either they cross over like this, or this guy goes back to that door, that guy goes back to this door. Now suppose you saw them running like this in the beginning, and you left the room when this was happening, and you saw them enter these 2 doors, you cannot tell which of the 2 happened, because you’ve just got 2 identical twins and these 2 identical exit doors. But somebody knows what happens. Somebody in that room who was watching them can clearly tell you, if this happened or that happened, because you can follow the twins at all times. So even though they’re identical, they’re distinguishable. They cannot swap roles without your knowing.

But imagine now that these are not classical particles but quantum particles, like electrons. For an electron, you don’t have a definite location or a trajectory. You only have probabilities. So you know an electron was emitted here and emitted here, and later on was absorbed here and absorbed here, and you cannot tell who really came here. Was it this guy or was it that guy? There’s no way to tell. So when you have identical particles whose trajectories you cannot follow, when you catch a particle here and a particle there, you cannot say Joe was here and Moe was there. It’s not allowed, because you’re not following their names. You can only say, “I found a particle here, I found a particle there.”

Therefore the theory cannot give different probabilities for finding particle 1 here and particle 2 there, and particle 2 there and particle 1 here, because the outcomes are indistinguishable, so the probabilities must be equal. So for 2 identical particles, p (x₁ x₂) must = p( x₂ x₁). That’s not true for these functions.

Now I’ll write a function for which it’s actually true. Can you make a guess on what it may be, what kind of function will respect the fact that if you swap the coordinates, probabilities don’t change? We know there’s particle in state a and a particle in state b and they’re identical. This says particle 1 is in and 2 is in b, but you’re not allowed to say who is who, so the correct answer is, you can also do this. Look at this function now. Here it’s a superposition, quantum mechanical superposition of a doing this and b doing that + b doing that and a doing this, or 1 doing something, 2 doing something else, + the opposite. You add them together. Now I invite you to check that if you exchange x₁ and x₂, see what happens. This becomes Y_a(x₂), that guy sitting here. This becomes Y_b(x₁), that guy sitting here. This becomes Y_b(x₂), that sitting here. So these 2 terms exchange roles when you exchange the particles. Y(x) to x₁ is in fact Y(x) — Y(x₁, x₂) is the same as Y(x₂, x₁). So that is an allowed state in quantum mechanics, where you add them and you symmetrize the product. You guys following that?

Now I’m going to try another combination. I’m going to put a - sign here. If you put a - sign, perhaps you can see if I exchange the 2 guys and this goes into that, that goes into this, I don’t come back to where I am. I come back to where the - of this guy. If this looked like x - y, that looks like y - x. So you might say, “Hey, things are not the same when I exchange the particles.” But remember, the physical quantities are not given by Y but the square of Y. The probabilities are not Y; the probabilities are Y². So even with the - sign, the probability for x₁, x₂ here is also the same as the probability for x₂ ,x₁ here, because when you find the absolute value, the - sign goes away.

So in quantum mechanics, there are 2 options for identical particles. Either you can take the product function and add to it the product with the reversed coordinates, or subtract from it. And the particles in the world, all of them decide to go with one camp or the other. Particles called bosons always choose that sign, and particles called fermions always choose the - sign. And every particle is either boson or fermion. P mesons are bosons. Photons are bosons. Electrons are fermions. Quarks are fermions. Gravitons are bosons. So everybody is one or the other. If you put 2 bosons in a box, their wave function, when you exchange them, will remain the same. If you put 2 fermions in a box, the wave function will change sign.

But now here is the beautiful result. Take this case for 2 fermions and ask yourself — this is a fermionic wave function for 2 particles. Ask yourself the following question: can they both be in the same quantum state? In other words, can the state a be the same as the state b? Remember, here a was this and b was that. I’m asking, can they both be in the same state? So I invite you to put a = b here. Let’s see what happens. That’s a. That’s also a. What do you get? Can you see something when both are a? You get 0.

So you cannot write a wave function in which the 2 fermions are in the same state, and that’s the Pauli principle. Pauli principle says, for some of the particles, they cannot be in the same state. Bosons, on the other hand, can be in the same state and like to be in the same state. I don’t have time to talk about that, but this is the Pauli principle. And you can show if you’ve got 3 fermions and 4 fermions and so on, you will find out the quantum mechanical wave functions never allow any 2 of them to be in the same state. And that is the origin of the entire periodic table of atoms, because what you do when you’ve got many electrons in an atom is, you find these energy levels that you did, the 1/n² that you got, but the nuclear charge of course may not be 1. It’s some non 0 number, depending on how many charges are in the nucleus. You take these levels, I don’t know how many levels there are here, then you start putting electrons into them. The first electron will go here. It turns out they have something called spin, so I will just say 1 way goes with down and 1 way goes with up and down. The third electron, you have to put here. It has to go to higher energy state.

If they were bosons, you can put them all in the lowest energy state. That world will look completely different from the world we live in. In the world we live in, the levels keep filling up as you put more and more and more electrons. They’ve got to go to higher and higher energy levels. You go to higher and higher energy levels, what happens once in a while is, you fill all of this right now, then that atom becomes very passive. It is not interested in either giving up electrons or taking electrons. Whereas if you had one more electron here, it’s very happy to lose this to some other atom, which may have just one electron in its lowest state. It’s a waste of it to be here; it will go sit there. Sometimes they like to give an electron to another atom, or if they do, this will become positively charged and that will become negatively charged and they have an attraction and they stay together as a molecule.

So you can understand that as you go on piling more and more electrons, a time will come when this level is completely full. That atom, whatever its place in the periodic table is, will be also very passive. So you will find things which are electrically very active with loose electrons in the upper, called valence states, and the inner shells are filled, they are not. They are inert. So the behavior of active, inert, active, inert is periodic, and that’s the periodic table that is observed. But you get that from quantum mechanics in great detail by solving for the energy levels, then using the Pauli principle to put only 1 electron there every quantum state, and you can see a lot of this behavior can be anticipated.

So anyway, these are things you will learn if you learn chemistry, if you learn physics. You will see where everything comes from. All right, so this is the end of the quantum mechanics part. I’m just going to tell you that I want to stop here and I’ll see you for the discussion section. I’m really going to miss my Mondays and Wednesdays because for me, that’s the best time of the week. So really good to be with you guys. Thank you.

[applause]

[end of transcript]

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