24. Quantum Mechanics VI: Time-dependent Schrödinger Equation
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Fundamentals of Physics II24 / 25
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Transkript: Englisch(automatisch erzeugt)
00:02
All right, today's topic is the theory of nearly everything. Okay, you wanted to know the theory of everything. You're almost there, because I'm finally ready to reveal to you the laws of quantum dynamics that tells
00:21
you how things change with time. So that's the analog of F equals ma. That's called the Schrodinger equation. And just about anything you see in this room or in this planet, anything you can see or use is really described by this equation I'm going to write down today.
00:42
It contains Newton's laws as part of it, because if you can do the quantum theory, you can always find hidden in it the classical theory. That's like saying if I can do Einstein's relativistic kinematics at low velocities, I will regain Newtonian
01:00
mechanics. So everything is contained in this one. There are some things left, of course, that we won't do, but this goes a long way. So I'll talk about it probably next time near the end, depending on how much time there is. But without further ado, I will now tell you what the
01:21
laws of motion are in quantum mechanics. So let's go back one more time to remember what we have done. The analogous statement is, in classical mechanics, for a particle moving in one dimension,
01:40
all I need to know about it right now is the position and momentum. That's it. That's the maximal information. You can say, what about other things? What about angular momentum? What about kinetic energy? What about potential energy? What about total energy? They're all functions of x and p. For example, in three dimensions,
02:01
x will be replaced by r, p will be replaced by some vector p, and there's a variable called angular momentum, but you know that once you know r and p by taking the cross product. That's it. And you can say, what happens when I measure any variable for a classical particle in this state,
02:24
xp? Well, if you know the location, it's guaranteed to be x,100 percent. Momentum is p,100 percent. Any other function of x and p, like r x p, is guaranteed to have that particular value. So everything is completely known. That's the situation at one time.
02:43
Then you want to say, okay, what can you say about the future? What's the rate of change of these things? And the answer to that is d2x over dt squared times m is the force, and in most problems you can write the force as a derivative of some potential.
03:02
So if you knew the potential, you know, ½kx squared or whatever it is at mgx, you can take the derivative on the right hand side, and the left hand side tells you the rate of change of x. I want you to note one thing. We know an equation that tells you something about the
03:21
acceleration. Once the forces are known, there's a unique acceleration. So you are free to give the particle any position you like and any velocity dx dt. That's essentially the momentum. You can pick them at random, but you cannot pick the acceleration at random, because the acceleration is not
03:41
for you to decide. The acceleration is determined by Newton's laws to be equal to essentially the force divided by mass. That comes from the fact mathematically that this is a second ordered equation in time, namely involving the second derivative, and that in such mathematics from a mathematical point of
04:00
view, if the second derivative is determined by external considerations, initial conditions are given by initial x and the first derivative. And all higher derivatives are slaved to the applied force. You don't assign them as you wish. You find out what they are from the equations of motion. So that's really all of classical mechanics.
04:23
Now you want to do quantum mechanics, and we have seen many times the story in quantum mechanics is a little more complicated. You ask a simple question and you get a very long answer. The simple question is, how do you describe the particle in quantum mechanics? It's in one dimension, and you say, I want assigned to it a function Y of x.
04:43
Y of x is any reasonable function which can be squared and integrated over the real line. Anything you write down is a possible state. That's like saying any x and any p are allowed. Likewise, Y of x is nothing special.
05:01
It can be whatever you like as long as you can square it and integrate it to get a finite answer over all of space. That's the only condition. And if your all of space goes to infinity, then Y should vanish at and minus infinity. That's the only requirement. Then you say, okay, you say it tells me everything. Why don't you tell me what the particle is doing?
05:23
And you can say, what do you want to know? Well, I want to know where it is. That's when you don't get a straight answer. You are told, well, it can be here, it can be there, it can be anywhere else, and the probability density that it's at point x is proportional to the absolute square of Y.
05:41
That means you take this Y and you square it, so it will have nothing negative in it. Everything will be real and positive. And Y itself may be complex, but this Y squared I told you over and over is defined to be Y star Y. That's real and positive. Then you can say, what if I measure momentum?
06:02
What answer will I get? That's even longer. First you're supposed to expand. I'm not going to do the whole thing too many times. You're supposed to write this Y as some coefficient times these very special functions in a world of size L. You have to write the given Y in this fashion.
06:22
And the coefficients are determined by the integral of the complex conjugate of this function times the function Y, and you gave me Yx. Now I gave some extra notes, I think.
06:40
Did people get that, called the Quantum Cookbook? That's just the recipe. Quantum mechanics is a big fat recipe, and that's all we can do, tell you this, you do this, you get these answers. That's my old goal, to simply give you the recipe. So the recipe says, what's interesting about quantum
07:01
mechanics, what makes it hard to teach, is that there is some physical principles which are summarized by these rules, which are like axioms. Then there are some purely mathematical results, which are not axioms. They are consequences of pure mathematics. You have to keep in mind what is purely a mathematical
07:21
result, therefore is deduced from the laws of mathematics, and what's a physical result that's deduced from experiment. The fact that Y describes everything is a physical result. Now, it tells you to write Y as the sum of these functions, and then the probability to obtain any momentum p is
07:45
A of p squared, where A of p is defined by this. The mathematics comes in in the following way. The first question is, who told you that I can write every function Y in this fashion? That's called the Fourier's theorem that guarantees you that in a circle of size L, every periodic function,
08:03
meaning that returns to a starting value, may be expanded in terms of these functions. That's a mathematical result. The same mathematical result also tells you how to find the coefficients. The postulates of quantum mechanics tell you two things. A of p squared is the probability that you will get
08:21
the value p when you measure momentum. Okay, that's a postulate, because you could have written this function 200 years before quantum mechanics. It would still be true, but these functions did not have a meaning at that time as states of definite momentum.
08:41
How do I know it's a state of definite momentum? If every term vanished except one term, that's all you have, then one coefficient will be something equal to 1. Everything is 0. That means the probability for getting momentum has a non-zero value only for that momentum. All other momenta are missing in that situation.
09:04
Another postulate of quantum mechanics is that once you measure momentum and you get one of these values, the state will go from being a sum over many such functions and collapse to the one term in the sum that corresponds to the answer you got.
09:21
Then here's another mathematical result. p is not every arbitrary real number you can imagine. We make the requirement, if you go around on a circle, the function should come back to the starting value. Therefore p is restricted to be 2Πh-bar over l times the integer n. That's a mathematical requirement, because if you think
09:43
Ψ squared is a probability, Ψ should come back to where you start. It cannot get two different values of Ψ when you go around the circle. That quantizes momentum to these values. The last thing I did was to say, if you measure energy, what answer will you get?
10:02
That's even longer. There you're supposed to solve the following equation. v to Ψ over dx squared of Ψ sub E v of x, the Ψ sub E of x is E times Ψ sub E of x.
10:21
In other words, for energy the answer is more complicated, because before I can tell you anything, I want you to solve this equation. This equation says, if in classical mechanics the particle was in some potential v of x and the particle had some mass m, you have to solve this
10:42
equation, then it's a purely mathematical problem. You try to find all solutions which behave well at infinity, that don't blow up at infinity, that vanish at infinity. That quantizes E to certain special values. And there are corresponding functions, Ψ sub E for each allowed value.
11:02
Then you're done, because then you make a similar expansion. You write the unknown Ψ, and namely some arbitrary Ψ that's given to you, and you write it as a sum of A sub E Ψ sub E of x, where A sub E is found by a similar rule. You just replace p by E and replace this function by these
11:22
functions. Then if you square A sub E, you will get the probability you will get that energy. So what makes the energy problem more complicated is that, whereas for momentum, we know once and for all, these functions describe a state of definite momentum,
11:40
where you can get only one answer. States of definite energy depend on what potential is acting on the particle. If it's a free particle, v is 0. If it's a particle that in Newtonian mechanics is a harmonic oscillator, v of x will be ½kx squared and so on. So you should know the classical potential,
12:01
and you've got to put it in. For every possible potential, you have to solve this. But that's what most people in physics departments are doing most of the time. They're solving this equation to find states of definite energy. So today I'm going to tell you why states of definite energy
12:20
are so important. What's the big deal? Why is the state of momentum not so important? Why is the state of definite position not so interesting? What is privileged about the states of definite momentum? And now you will see the role of energy. So I'm going to write down for you the equation that's analog of F equals ma.
12:42
So what are we trying to do? ψ of x is like x and p. You don't have a time label here. These are like saying at some time a particle has a position and a momentum. In quantum theory at some time it has a wave function ψ of x.
13:01
So the real question in classical mechanics is how does x vary with time and how does p vary with time? The answer is according to F equals ma. And here the question is how does ψ vary with time? First thing you've got to do is to realize that ψ itself can be a function of time, right? That's the only time you've got to ask what does it do with
13:23
time? So at t equal to 0 it may look like this. A little later it may look like that. So it's flopping and moving, just like say a string. So changing with time, and you want to know how it changes with time. So this is the great Schrodinger equation. It says ih bar partial derivative with respect to time.
13:44
It's partial because this depends on x and t. So this is the t derivative. It's equal to the following, minus h bar squared over 2m d2ψ over dx squared V of x ψ of x and t.
14:05
That's the equation, x,t. So write this down, because if you know this equation you'll be surprised how many things you can
14:20
calculate. From this follows the spectrum of the atoms. From this follows what makes a material a conductor, a semiconductor, a superconductor. Everything follows from this famous Schrodinger equation. This is an equation in which you must notice that we're
14:41
dealing for the first time with functions of time. Somebody asked me long back, where is time? Well, here is how ψ varies with time. So you're supposed to take, suppose someone says, here's my initial ψ of x and 0. Tell me, what is ψ a little later? One millisecond later?
15:01
Well, it's the rate of change of ψ with time multiplied by one millisecond. The rate of change of ψ at the initial time is obtained by taking that derivative of ψ and adding to it V times ψ you get something. That's how much ψ changes. Multiply by Δt, that's the change in ψ. That'll give you ψ at a later time. This is a first order equation in time.
15:22
What that means mathematically is the initial ψ determines the future completely. This is different from a position where you need x and dx dt at the initial time, because the equation will only tell you what d to x dt squared is. But in quantum mechanics, dψ dt itself is
15:42
determined, so you don't get to choose that. You just get to choose the initial ψ. That means an initial wave function completely determines the future according to this equation. So don't worry about this equation. I don't expect you all to see it and immediately know what to do, but I want you to know that there is an equation
16:02
that is known now. That's the analog of F equals ma. If you solve this equation, you can predict the future to the extent allowed by quantum mechanics given the present, and the present means ψ of x,0 is given. Then you go to the math department and say, this is my ψ of x,0. Please tell me by some trick what is ψ of x and
16:21
t. It turns out there is a trick by which you can predict ψ of x and t. Note also that this number i is present in the very equations of motion. So this is not like the i we used in electrical circuits, where we really meant sines and cosines,
16:40
but we took e to the iθ and e to the iωt, always hoping in the end to take the real part of the answer, because the functions of classical mechanics are always real. But in quantum theory, ψ is intrinsically complex, and it cannot get more complex than that by putting an i in the equations of motion. But that's just the way it is. You need the i to write the equations.
17:05
Therefore, our goal then is to learn different ways in which we can solve this. Now remember, everybody noticed, this looks kind of familiar here, this combination. I mean, it's up there somewhere. It looks a lot like this. You see that?
17:22
But it's not quite that. That is working on a function only of x. This is working on a function of x and t. And there are partial derivatives here, and there are total derivatives there. They are very privileged functions. They describe states of different energy. This is an arbitrary function, just evolving with time.
17:42
So you should not mix the two up. This ψ is a generic ψ changing with time. So let's ask, how can I calculate the future given the present? How do I solve this equation? So here is what you do. I'm going to do it at two levels.
18:01
One is to tell you a little bit about how you get there, and for those of you who say, look, spare me the details, I just want to know the answer. I will draw a box around the answer, and you are free to start from there. But I want to give everyone a chance to look under the hood and see what's happening. So given an equation like this, which is pretty old stuff in
18:22
mathematical physics from after Newton's time, people always ask the following question. They say, look, I don't know if I can solve it for every imaginable initial condition. It's like saying, even in the case of the oscillator, you may not be able to solve every initial condition. You say, let me find a special case where ψ of x and
18:41
t, which depends on x and on t, has the following simple form. It's a function of t alone times the function of x alone, okay? I want you to know that no one tells you that every
19:02
solution of the equation has this form. You guys have a question about this over there? Okay, good. All right, so this is an assumption. You want to see if maybe there are answers like this to the problem. The only way to do that is to take that assumed form,
19:22
put it into the equation and see if you can find a solution of this form. Not every solution looks like this. For example, you can write e to the x minus some number times t squared. That's the function of x and t. But it's not a function of x times the function of t.
19:42
You see that? You see that x and t are mixed up together. You cannot rip it out into two parts. So it's not the most general thing that can happen. It is a particular one. Right now you're eager to get any solution. You want to say, can I do anything? Can I calculate, even in the simplest case, what the future is given the present?
20:01
You're asking, can this happen? So I'm going to show you now that the equation does admit solutions of this type, okay? So are you guys with me now on what I'm trying to do? I'm trying to see if this equation admits solutions of this form.
20:24
So let's take that and put it here. Here's where you've got to do the math, okay? Take the ψ and put it here and start taking derivatives. Let's do the left hand side first. Left hand side I have ih bar. Then I bring the d by dt to act on this product.
20:45
d by dt partial means only time is to be differentiated. x is to be held constant. That's the partial derivative. That's the meaning of the partial derivative. It's like an ordinary derivative where the only variable you'd ever differentiate is the one in the derivative.
21:01
So the entire ψ of x doesn't do anything. It's like a constant. So you just put that ψ of x there. Then the derivative d by dt comes here and I claim it becomes the ordinary derivative. That's the left hand side. You understand that, why that is true?
21:22
Because on a function only of time there's no difference between partial derivative and ordinary derivative. It's got only one variable. The other potential variable this d by dt doesn't care about so it's just standing there. That's the left hand side. Now look at the right hand side. All of this stuff, and imagine putting for this
21:43
function ψ this product form. Is it clear to you in the right hand side the situation's exactly the opposite? You get all these d by dx's, partials. They are only interested in this function because it's got x dependence. f of t doesn't do anything.
22:02
All the derivatives they go right through f. So you can write it as f of t. Now you have to take the derivative with respect to ψ. That looks like d squared over dψ v of x.
22:24
If you follow this you're almost there, but take your time to understand this. The reason you write it as a product of two functions is the left hand side is only interested in differentiating the function f where it becomes the total derivative. The right hand side is only taking derivatives with respect to
22:40
x so it acts on this part of the function. It depends on x and all partial derivatives become total derivatives because if you've got only one variable there's no need to write partial derivatives. This combination I'm going to write to save some time as hψ.
23:01
Let's just say between you and me it's a shorthand. hψ is a shorthand for this entire mess here. Don't ask me why it looks like h times ψ where are the derivatives. It is a shorthand, okay? If I don't feel like writing the combination over and over I can call it hψ.
23:23
So what equation do I have now? I have ih bar ψ of x times df dt is equal to f of t times hψ, but all I want you to notice, hψ depends only on x.
23:41
It has no dependence on time. Do you see that? There's nothing here that depends on time. Okay, now this is a trick which if you learn you will be quite pleased because you'll find that as you do more and more stuff, at least in physics or economics or statistics, the trick is a very old
24:02
trick. The problem is quantum mechanics but the mathematics is very old. So what do you do next? You divide both sides by ψf. So I say divide by f of t ψ of x.
24:24
What do you think will happen if I divide by f of t ψ of x? On the left hand side I say divide by ψf. On the right hand side I say divide by ψf. Can you see that the ψ cancels here? You have a 1 over f and f cancels here and you have a
24:42
1 over ψ. The equation then says ih-bar 1 over f of t df dt is equal to 1 over ψ of x hψ. I've written this very slowly because I don't know.
25:02
You will find this in many advanced books, but you may not find it in our textbook. So if you don't follow something you should tell me there's plenty of time to do this. So I'm in no rush at all. Do you follow the? These are purely mathematical manipulations. We have not done anything involving physics. You all follow this?
25:23
Yes? Okay. Now you have to ask yourself the following. I love this argument. Even if you don't follow this I'm just going to get it off my chest. It is so clever, and here is a clever part. This is supposedly a function of time. You agree? All full of function of time.
25:42
This is a function of x. This guy doesn't know what time it is. This guy doesn't know what x is, and yet they're supposed to be equal. What can they be equal to? They cannot be equal to a function of time because then
26:00
as you vary time, suppose you think it's a function of time. Suppose it's not so. Then as time varies, this part is okay. It can vary with time to match that, but this cannot vary with time at all because there is no time here. So this cannot depend on time. And it cannot depend on x because if it was a function of x that it was equal to, as you vary x,
26:22
this can change with x to keep up with that. This has no x dependence. It cannot vary with x. So this thing that they're both equal to is not a function of time and it's not a function of space. It's a constant. That's all it can be. So the constant is going to be very cleverly given the symbol
26:41
E. You're going to call the constant E. You turn out E is connected to the energy of the problem. So now I have two equations. This equals E and that equals E, and I'm going to write it down. So one of them says Iℏ1 over F
27:03
dF dt equal to E. So let me bring the F here. The other one says Hℏ is equal to Eℏ. These two equations, if you solve them,
27:22
will give you the solution you're looking for. In other words, going back here, yes, this equation does admit solutions of this form, of the product form, provided the function F you put in the product that depends on time obeys this equation, and the function ψ that depends only on x
27:42
obeys this equation. Remember Hℏ is the shorthand for this long bunch of derivatives. We'll come to that in a moment. Well, let's solve this equation first. Now can you guys do this in your head? Iℏ dF dt equal to E times F.
28:03
So it's saying F is a function of time whose time derivative is the function itself. Everybody knows what such a function is. It's an exponential. And the answer is, I'm going to write it down and you can check F of t is F of 0,
28:20
E to the minus iEt over Hℏ. If you want now, take the derivative and check. F of 0 is some constant. I call it F of 0 because if t is equal to 0, this goes away and F of t is equal to F of 0. But take the time derivative and see. When you take a time derivative of this,
28:41
you get the same thing times minus iE over Hℏ, and when you multiply by iHℏ, everything cancels except Ef. So this is a very easy solution. So let's stop and understand. It says that if you are looking for solutions that are products of F of t times I of x,
29:00
F of t necessarily is this exponential function. It's the only function you can have. But now once you pick that E, you can pick E to be whatever you like, but then you must also solve this equation at the same time. But what is this equation? This says Hℏ squared over 2m d 2Ψ over dx squared V times
29:24
Ψ equals E times Ψ. And you guys know who that is, right? What is it? What can you say about the function that satisfies that equation?
29:40
Have you seen it before? Yes? What is it? It's the state of definite energy. Remember, we said functions have definite energy over that equation. So that Ψ is really just Ψ E.
30:02
So now I'll put these two pieces together, and here is where those of you who drifted off can come back, because what I'm telling you is that the Schrodinger equation in fact admits a certain solution which is a product of a function of time and a function of space. And what we found by fiddling around with it is that F
30:22
of t and Ψ are very special, and f of t must look like e to the minus iEt over Hℏ, and Ψ is just our friend Ψ's at E of x, which are functions associated with a definite energy.
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Yep, okay, the question is, are there other solutions for which this factorized form is not true? Yes, and I will put you out of your suspense very soon by
31:01
talking about that. But I want everyone to understand that you can at least solve one case of Schrodinger's equation. So what does this mean? I want you guys to think about it. This says if you start Ψ in some arbitrary configuration, that's my initial state, let it evolve with time, it obeys this rather crazy,
31:23
complicated Schrodinger equation. But if I start it at t equal to 0 in a state which is a state of definite energy, namely a state obeying that equation, then its future is very simple. All you do is attach this phase factor,
31:40
e to the minus iEt over Hℏ. Therefore, it's not a generic solution, because you may not in general start with a state which is a state of definite energy. You'll start with some random Ψ of x, and it's made up of many, many Ψ's at ease that come in the expansion of that Ψ. So it's not going to always work. But if you picked it so that there's only one such term,
32:01
where there's a sum over E, namely one such function, then the future is given by this. For example, if you have a particle in a box, you remember the wave function Ψn looks like square root of 2 over L sin nΠx over L. An arbitrary Ψ doesn't look like any of these.
32:23
These guys, remember, are nice functions that do many oscillations. But if you chose it initially to be exactly the sine function, for example, which is size of 1, then I claim as time evolves, the future state is just this initial sine function times the simple exponential.
32:44
This behavior is very special, and it's called normal modes. It's a very common idea in mathematical physics. It's the following. It's very familiar even before you did quantum mechanics.
33:00
Take a string tied at both ends, and you plug the string and you release it. Most probably, if you pluck it at one point, you'll probably pull it in the middle and let it go. That's the initial Ψ of x and t, this time for a string. Pull it in the middle, let it go.
33:21
That's an equation that determines the evolution of that string. I remind you what that equation is, is d2Ψ over dx squared minus 1 over velocity squared, d2Ψ over dt squared. That's the wave equation for a string. It's somewhat different from this problem because it's a
33:41
second derivative in time that's involved. Nonetheless, here is an amazing property of this equation derived by similar methods. If you pull a string like this and let it go, it'll go crazy when you release it. I don't even know what it'll do. It'll do all kinds of things. Stuff will go back and forth, back and forth.
34:00
But if you can deform the string at t equal to 0 to look exactly like this, sinΠx over L times the number A, that's not easy to do. Do you understand that to produce the initial profile, one hand is not enough, two hands are not enough? You've got to get an infinite number of your friends who are
34:23
infinitesimally small. You line them up along the string and tell each one, tell the person here to pull it exactly this height. The person here has to pull it exactly that height. You all lift your hands up, then I follow you. You follow this perfect sign. Then you let go.
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What do you think will happen then? What do you think will be the subsequent evolution of the string? Do you have any guess? Yep? It'll go up and down, and the future of that string will look like cosΠxvt over
35:02
L. Look at this. This is the time dependence. At t equal to 0 this goes away, this is the initial state. But look what happens at a later time. Every point x rises and falls with the same period. It goes up and down all together.
35:22
That means a little later it'll look like that. A little later it'll look like that. Then it'll look like this. Then it'll look like this. Then it'll look like that. Then it'll go back and forth. But at every instant this guy and this guy and this guy are all rescaled by the same amount from the starting value. That's called a normal mode. Now your question was, are there other solutions to
35:42
string? Of course. Typically if you don't think about it and pluck the string your initial state will be a sum of these normal modes, and that will evolve in a complicated way. But if you engineered it to begin exactly this way, or in any one of those other functions where you put an extra n here, they all have the remarkable property that they rise and fall in step.
36:02
What we have found here is in the quantum problem, if you start the system in that particular configuration, then its future has got a single time dependence common to it. That's the meaning of the factorized solution.
36:20
So we know one simple example. Take a particle in a box. If it's in the lowest energy state or ground state, the wave function looks like that. Then the future of that will be Ψ1 of x and t will be the Ψ1 of x times e to the −iE1 of t over h-bar, where E1 is h-bar
36:43
square π square 1 square over 2m l-square. That's the energy associated with that function. That's how it'll oscillate. Now you guys follow what I said now with an analogy with
37:01
the string and the quantum problem. There are slightly different equations. One is second order, one is first order. One has cosines in it, one has exponentials in it. But the common property is this is also a function of time times a function of space. Here this is a function of time and a function of space.
37:26
Okay, so I'm going to spend some time analyzing this particular function. ψ of x and t equal to e to the −iEt over h-bar ψ of E of x.
37:41
Now I'm going to do an abuse of notation and give the subscript e to this guy also. What I mean to tell you by that is this ψ, which solves Schrodinger equation, by the way, I invite you to go check it. Take the ψ and put it into Schrodinger equation. You will find it works. So in the notes I've given you, I merely tell you that this
38:01
is a solution to Schrodinger's equation. I don't go through this argument of assuming it's the product form and so on. That's optional. I don't care if you remember that or not, but this solves Schrodinger equation and I call it ψ of E because the functions on the right hand side are identified with states of definite energy.
38:26
Okay, what will happen if you measure various quantities in this state? For example, what's the position going to be? What's the probability for definite position? What's the probability for definite momentum? What's the probability for definite anything?
38:41
How will they vary with time? I will show you nothing depends on time. You can say, how can nothing depend on time? I see time in the function here, but it'll go away. Let us ask, what is the probability that the particle is at x at time t for a solution like this?
39:04
You know the answer is ψ star xt ψ of xt. And what do you get when you do that? You will get ψe star of x ψe of x.
39:21
Then you will get the absolute value of this guy squared, e to the iEt times e to the minus iEt, and that's just 1. I hope all of you know that this e to the iθ absolute value squared is 1. So it does not depend on time.
39:42
Even though ψ depends on time, ψ star ψ has no time dependence. That means the probability for finding that particle will not change with time. That means if you start the particle in the ground state, ψ, and let's say ψ squared, in fact, looks pretty much the same, it's a real function.
40:04
This probability does not change with time. That means you can make a measurement any time you want for position, and the odds don't change with time. It's very interesting. It depends on time and it doesn't depend on time. It's a lot like e to the ipx over h-bar.
40:22
It seems to depend on x, but the density does not depend on x because the exponential goes away. Similarly, it does depend on time. Without the time dependence, it won't satisfy Schrodinger equation, but the minute you take the absolute value, this goes away. That means for this particle, I can draw a little graph
40:41
that looks like this. That is the probability cloud you find in all the textbooks. Have you seen the probability cloud? They've got a little atom. There's a little fuzzy stuff all around it. They are the states of the hydrogen atom or some other atom. How do you think you get that? You solve a similar equation, except it will be in 3 dimensions instead of 1 dimension,
41:01
and for v of x, you write −zE squared over r. If you want r is x squared Y squared Z squared. zE is the nuclear charge and −E is the electron charge. You put that in, you solve the equation, and you will find a whole bunch of solutions that behave
41:23
like this. They are called stationary states, because in that stationary state, if a hydrogen atom starts out in the state, which is a state of definite energy, as time goes by, nothing happens to it essentially. Something trivial happens. It picks up the phase factor, but the probability for finding the electron never changes with time.
41:42
So if you like, you can draw a little cloud whose thickness, if you like, measures the probability for finding it at that location. So that will have all kinds of shape. It looks like dumbbells pointing to the North Pole, South Pole, maybe uniformly spherical distribution. They're all a probability of finding the electron in that
42:02
state, and it doesn't change with time. So a hydrogen atom, when you leave it alone, will be in one of these allowed states. You don't need the hydrogen atom. This particle in a box is a good enough quantum system. If you start it like that, it will stay like that. If you start it like that, it will stay like that, times that phase factor. So these are the stationary states that are important,
42:23
because that's where things will settle down. Okay, now you should also realize that that's not a typical situation. Suppose you have in one dimension, there's a particle on a hill, and at t equal to 0, it's given by some wave function that looks like this.
42:43
So it's got some average position, and if you expand it in terms of exponentials of p, it's got some range of momenta in it. What will happen to this as a function of time? Can you make a guess? Let's say it's right now got an average momentum to the left.
43:01
What do you think will happen to it? Pardon me? It'll move to the left. This is just, except for the fuzziness, you can apply your classical intuition. It's got some position, maybe not precise. It's got some momentum, maybe not precise. But when you leave something on top of a hill, it's going to slide down the hill. The average x is going to go this way,
43:21
and the average momentum will increase. So that's the situation where the average is the physical quantities change with time. That's because this state is not a function size of v of x. It's some random function you picked. Random functions you picked in some potential will in fact evolve with time in such a way that measurable quantities will
43:42
change with time. The odds for x, the odds for p, odds for everything else will change with time. Okay, so stationary states are very privileged, because if you start them that way, they stay that way, and that's why when you look at atoms, they typically stay that way. But once in a while, an atom will jump from one stationary state to another one,
44:02
and you can say that looks like a contradiction. If it's stationary, what's it doing jumping from here to there? You know the answer to that? Why does an atom ever change then if it's in a state of gravity that way forever? Why do they go up and down?
44:22
Want to guess? That's correct. So she said by absorbing photons, and what I really mean by that is this problem, v of x, involves only the Coulomb force between the electron and the proton. If that's all you have, an electron in the field of a
44:40
proton, it'll pick one of these levels. It can stay there forever. When you shine light, you're applying electromagnetic field. The electric field and magnetic field apply extra forces in the charge, and v of x should change to something else, so that this function is no longer a state of definite energy for the new problem, because you've changed the rules of the game.
45:01
You've modified the potential. Then, of course, it will move around and it will change from one state to another. But an isolated atom will remain that way forever. Well, it turns out even that's not exactly correct. You can take an isolated atom in the first excited state of hydrogen.
45:21
You come back a short time later, you'll find the fellow has come down. And you say, look, I didn't turn on any electric field. E equal to 0, B equal to 0. What made the atom come down? Do you know what the answer to that is? Any rumors? Yep. Its photon is emitted, but you need an extra
45:44
electric or magnetic field to act on it before it'll emit the photon. But where is the field? I've turned everything off. E and B are both 0. So it turns out that the state E equal to B equal to 0 is like a state, say, in a harmonic
46:01
oscillator potential, x equal to p equal to 0, sitting at the bottom of the well. We know that's not allowed in quantum mechanics. You cannot have definite x and definite p. It turns out in quantum theory, E and B are like x and definite p. That means a state of definite E is not a state of definite B.
46:22
A state of definite B is not a state of definite E. It looks that way in the macroscopic world, because the fluctuations in E and B are very small. Therefore, just like in the lowest energy state, an oscillator has got some probability to be jiggling back and forth in x and also in p, the vacuum, which we think has no E and no B,
46:42
has small fluctuations. Because E and B both vanishing is like x and p both vanishing, not allowed. So you've got to have a little spread in both E and both B. They're called quantum fluctuations of the vacuum. So that's the theory of nothing. The vacuum, you think, is the most uninteresting thing, and yet it is not completely uninteresting because
47:02
it's got these fluctuations. It's those fluctuations that tickle the atom and make it come from an excited state to a ground state. Okay, so unless you tamper with the atom in some fashion, it will remain in a stationary state. Those states are states of definite energy. They're found by solving the Schrodinger equation without
47:22
time in it. HΨ equals EΨ is called the time independent Schrodinger equation. And that's what most of us do most of the time. The problem can be more complicated. It can involve two particles. It can involve ten particles. It may not involve this force. It may involve another force, but everybody is spending most of his time or her time solving the Schrodinger equation
47:43
to find states of definite energy, because that's where things will end up. All right, I've only shown you that the probability to find different positions doesn't change with time. I will show you the probability to find different anything doesn't change with time. Nothing will change with time, not just x probability.
48:02
So I'll do one more example. Let's ask, what's the probability to find a momentum p? What are we supposed to do? We're supposed to take e to the ipx over h-bar times the function at some time t,
48:22
and do that integral. I'm sorry, you should take that and do that integral, and then you take the absolute value of that. And that's done at every time. You take the absolute value, and that's the probability to get momentum p, right? The recipe was, if you want the probability,
48:41
take the given function, multiply it with a conjugate of ψ of p and do the integral dl and dx. ψ of xt in general has got complicated time dependence, but not our ψ. Remember our ψ? Our ψ looks like ψ of x times e to the
49:01
minus iet over h-bar. But when you take the absolute value, this has nothing to do with x. You can pull it outside the integral. Or let me put it another way. Just do the integral and see what you get. You will find A of p looks like A of p of 0 times
49:22
e to the minus iet over h-bar. Do you see that? If the only thing that happens to ψ is that you get an extra factor at later times, the only thing that happens to the A of p is it gets the extra factor later times. But the probability to find momentum p is the absolute value square of that, and in the absolute value
49:42
process, this guy is gone. You follow that? Since the wave function changes by a simple phase factor, the coefficient to have a definite momentum also changes by
50:01
the same phase factor. This is called a phase factor, exponential or modulus 1. But when you take the absolute value, that guy doesn't do anything. Now you can replace p by some other variable. It doesn't matter. The story is always the same. So a state of definite energy seems to evolve in time,
50:22
because e to the minus iet over h-bar, but none of the probabilities change with time. It's absolutely stationary with respect to anything you measure. That's why those states are very important. Okay, now I want to caution you that not every solution looks
50:40
like this. That's the question you raised. I'm going to answer that question now. Let's imagine that I found two solutions to the Schrodinger equation of this form. Solution ψ1 looks like ψ1 of x and t looks like e to the minus iet over
51:00
h-bar times ψ1, ψ sub e1 of x. That's one solution for energy E1. Then there's another solution, ψ2 of x looks like e to the minus iet over h-bar times ψ sub
51:20
e2 of x. This function has all the properties I mentioned. Namely, nothing depends on time. That has the same property. But because the Schrodinger equation is a linear equation, it is also true that this ψ, which is ψ1 ψ2,
51:40
add this one to this one, is also a solution. I think I've done it many, many times. If you take a linear equation, ψ1 obeys the Schrodinger equation, ψ2 obeys the Schrodinger equation, add the left hand side to the left hand side and the right hand side to the right hand side, you will find that if ψ1 obeys it and ψ2 does, ψ1 ψ2 also obeys it.
52:02
Not only that, it can be even more general. You can multiply this by any number, ψ1 of x and t, any constant, a2 ψ2 of x and t, but a1 and a2 don't depend on time, also obeys Schrodinger equation. Can you see that? That's superposition of solutions.
52:21
It's a property of linear equations. Nowhere does ψ squared appear in the Schrodinger equation, therefore you can add solutions. But take a solution of this form. Even though ψ1 is a product of some f and a ψ, and ψ2 is a product of some f2 and ψ2, the sum is not a product of some f and some ψ.
52:45
You cannot write it as a product. You understand? That's a product, that's a product, but the sum is not a product, because you cannot pull out a common function of time from the two of them. They're different time dependence. But that is also a solution. In fact, now you can ask yourself, what is the most
53:06
general solution I can build in this problem? Well, I think you can imagine that I can now write ψ of x and t as A of E, ψ of E of x and t, sum over all the allowed
53:23
E's, that also satisfies Schrodinger equation. Do you agree? Every term in it satisfies Schrodinger equation. You add them all up, multiply by any constant A of E, that also satisfies Schrodinger equation. So now I'm suddenly manufacturing more complicated
53:41
solutions. The original modest goal was to find a product form. But once you got the product form, you find if you add them together, you get a solution that's no longer a product of x and a product of t, a function of x and a function of t, because this guy has one time dependence. Another term is a different time dependence. You cannot pull them all out.
54:00
So we are now manufacturing solutions that don't look like their products. This is the amazing thing about solving the linear equation. You seem to have very modest goals when you start with a product form, but in the end, you find that you can make up a linear combination of products. Then the only question is, will it cover every possible
54:21
situation you give me? In other words, suppose you come to me with an arbitrary initial state. I don't know anything about it. And you say, what is its future? Can I handle that problem? The answer is, I can, and I'll tell you why that is true. Ψ of x and t looks like A sub
54:40
E. I'm going to write this more explicitly as Ψ of x e to the minus iEt over h-bar. Look at this function now at t equal to 0. At t equal to 0, I get Ψ of x and 0 to be sum over A sub E, Ψ sub E of
55:05
x. In other words, I can only handle those problems whose initial state looks like this, but my question is, should I feel limited in any
55:23
way by the restriction? Do you follow what I'm saying? Maybe I'll say it one more time. This is the most general solution I am able to manufacture that looks like this, A sub E, Ψ of E of x, e to the minus iEt over
55:41
h-bar. It's a sum over solutions of the product formed with variable, each one with a different coefficient. That's also a solution to Schrodinger equation. If I take that solution and say, what does it do at t equal to 0, I find it does the following. At t equal to 0, it looks like this.
56:04
So only for initial functions of this form I have the future. But the only is not a big only, because every function you can give me at t equal to 0 can always be written in this form. It's a mathematical result that says that,
56:21
just like sines and cosines and certain exponentials are a complete set of functions for expanding any function, the mathematical theory tells you that the solutions of hΨ equal to eΨ, if you assemble all of them, can be used to build up an arbitrary initial function. That means any initial function you give me I can write
56:41
this way, and the future of that initial state is this guy. Yes, lots of mathematical restrictions, single valued. Physicists usually don't worry about those restrictions
57:02
until, of course, they get in trouble. Then we go crawling back, the math guys to help us out. So just about anything you can write down, by the way physics works, things tend to be continuous and differentiable. That's the way natural things are. So for any function we can think of it is true. You go to the mathematicians, they'll give you a function
57:21
that is nowhere continuous, nowhere differential, nowhere defined, nowhere something. That's what makes them really happy. But we don't, they are all functions the way they've defined it, but they don't happen in real life, because whatever happens here influences what happens on either side of it, so things don't change in a discontinuous way. Unless you apply an infinite force, an infinite potential,
57:42
infinite something, everything has got what's called C infinity, can differentiate any number of times. So we don't worry about the restrictions. So in the world of physicist functions, you can write any initial function in terms of these functions. So let me tell you then the process for solving the
58:02
Schrodinger equation under any conditions. Are you with me? You come and give me Y of x and 0, and you say as a function of time where is it going to end up? That's your question. That's all you can ask. Initial state, final state. This is given. This is needed.
58:23
So I'll give you a three step solution. Step one, find A of E equals ψ of E star of x times ψ of x and 0.
58:45
And step two, ψ of x and t is equal to this A of E that you got times e to the minus iEt over h bar times ψ of E of x. So what I'm telling you is, the fate of a function
59:11
ψ with wiggles and jiggles is very complicated to explain. Some wiggle goes into some other wiggle. That goes into some other wiggle as a function of time. But there is a basic simplicity underlying that
59:21
evolution. The simplicity is the following. If at t equal to 0, you expand your ψ as such a sum where the coefficients are given by the standard rule, then as time goes away from t equal to 0, all you need to do is to multiply each coefficient by that particular term involving that particular energy.
59:42
And that gives you the ψ at later times. A state of definite energy in this jargon will be the one in which every term is absent except 1, maybe E equal to E1.
01:00:00
That is the kind we studied. That state has got only one term in the sum and its time evolution is simply given by this and all probabilities are constant. But if you mix them up with different coefficients you can then handle any initial condition. So we have now solved really for the future of any quantum mechanical problem.
01:00:23
So I'm going to give you from now until the end of class concrete examples of this, but I don't mind again answering your questions, because it's very hard for me to put myself in your place.
01:00:40
I'm trying to remember when I did not know quantum mechanics. I'm sitting in some sandbox and some kid was throwing sand in my face, so I don't know. So I've lost my innocence and I don't know how it looks to you. Yes, so let's do the following problem.
01:01:07
Let us take a world in which everything is inside the box of length L and someone has manufactured for you a certain
01:01:22
state. Let me come to that case in a minute. Let me take a simple case and I'll build up the situation you want. Let's first take a simple case where T equal to 0, ψ of x and 0 is equal to the square root of 2mL times 2 over L sine nΠx over L.
01:01:44
That is just a function with n oscillations. You agree, that's a state of definite energy. The energy of that state, E sub n, is h bar squared π squared n squared
01:02:01
over 2mL squared. We did that last time. And the reason why would we be so interested in this function? Now I can tell you why. If this is my initial state, let me take a particular n, then the state at any future time, ψ of x and T, is very simple here,
01:02:21
square root of 2 over L sine nΠx over L times e to the minus i T over h bar, but energy is n squared π squared h bar squared over 2mL squared.
01:02:46
That's it. That is the function of time. All I've done to that initial state is multiply by e to the minus iEt, but E is not some random number. E is labeled by n and E sub n is whatever you have here.
01:03:01
That's the time dependence of that state. It's very clear that if you took the absolute value of this e to the minus i, this guy has absolute value equal to 1 at all times. See, it's like saying cos T depends on time, sine T depends on time, but cos squared plus sine squared, cos squared T plus sine
01:03:21
squared T seems to depend on time, but it doesn't. So this seems to depend on time, and it does, but when you take the absolute value, it goes away. That's the simplest problem, okay? I gave you an initial state, future is very simple. Attach that factor. Now let's give you a slightly more complicated state. The more complicated state will be, I'm going to hide that
01:03:44
for now. Let us take a ψ of x and 0. That looks like 3 times square root of 2 over L sine 2Πx over L 4 times sine.
01:04:08
This is my initial state. What does it look like? It's a sum of two energy states. This guy is what I would call ψ of 2 in my notation,
01:04:23
the second highest state. This guy is ψ of 3. Everybody's properly normalized. And these are the A's. So this state, if you measure its energy, what will you get?
01:04:41
Anybody tell me what answers can I get if I measure energy now? You want to guess what are the possible energies I could get? Yes? Any of you? Either of you, yeah. Can you tell? No?
01:05:05
Yep? So her answer was, you can get in my convention E of 2 or E of 3. Just put n equal to 2 or anything.
01:05:21
That's all you have. Your function written as a sum over ψ of b is only two terms. That means they are the only two energies you can get. So it's not a state of definite energy. You can get either this answer or this answer. But now you can sort of see it's more likely to get this guy because it has a 4 in front of it, and less likely to get this guy, and impossible to get
01:05:41
anything else. So the probability for getting n equal to 2 is proportional to 3 squared, and the probability for getting n equal to 3 is proportional to 4 squared.
01:06:00
If you want the absolute probabilities, then you can write it as 3 squared divided by 3 squared 4 squared, which is 5 squared, or you can write it as 3 squared 4 squared 5 squared. See, if you square these probabilities, you get 25.
01:06:21
3 squared 4 squared. If you want to get 1, I think you can see without too much trouble, if you rescale the whole thing by 1 fifth, now you'll find the total probabilities add up to 1. That's the way to normalize the function. That's the easy way. The hard way is to square all of this and integrate it and
01:06:41
then set it equal to 1 and see what you have to do. In the end, all you'll have to do is divide by 5. I'm just giving you a shortcut. When you expand the ψ in terms of normalized functions, then the coefficient squared should add up to 1. If they don't, you just divide them by whatever it takes. So this has got a chance 3 is to 5 of being this or that energy.
01:07:01
But as a function of time, you will find here things vary with time. This is not going to be time independent. I want to show you that. So ψ of x and t now is going to be 3 over 5 square root of 2 over L sine 2Πx
01:07:20
over L times e to the minus iE sub 2. I don't want to write the full formula for E sub n every time. I'm just going to call it E sub 2. That's 4 over 5 square root of 2 over L sine 3Πx over L times e to the minus iE
01:07:43
3t over h-bar. Now you notice that if I found the probability to be at some x, p of x and t, I have to take the absolute square of all of this.
01:08:02
And all I want you to notice is that the absolute square of all of this, you cannot drop these exponentials now. If you've got two of them, you cannot drop them, because when you take ψ1 ψ2 absolute square, ψ1 ψ2, you multiply it by ψ1 conjugate ψ2 conjugate.
01:08:21
Let's do that. So you want to multiply the whole thing by its conjugate. So first you take the absolute square of this. You will get 9 over 25 2 over L sine 2Πx over L,
01:08:41
and the absolute value of this is just 1. You see that? That is ψ1 star ψ1. Then you must take ψ2 star ψ2. That will be 16 over 25 times square root of 2 over L. I'm sorry, no square root, 2 over L,
01:09:01
sine squared 3Πx over L times 1, because the absolute value of this guy with itself is 1. But that's not the end. You've got two more terms which look like ψ1 star ψ2 ψ2 star ψ1. I'm not going to work out all the details,
01:09:21
but let me just show you that time dependence exists. So if you take ψ1 star ψ2, you will get 3 over 5 square root of 2 over L sine 2Πx over L times e to the i E2t over h-bar times, sorry, 3 over 5 times 4 over
01:09:49
5 times square root of 2 over L sine 3Πx over L times e to the i E2 minus
01:10:05
E3t over h-bar, plus one more term. I don't care about any of these things. I'm asking you to see, do things depend on time or not? This has no time dependence because the absolute value of that vanished. This has no time dependence. The absolute value of this vanished, but the cross terms,
01:10:21
when you multiply the conjugate of this by this or the conjugate of this by that, they don't cancel. That's all I want you to know. So I'll get a term like this this complex conjugate. I don't want to write that in detail. If you combine this function with the conjugate, you will find this plus the conjugate will give me a cosine. So I should probably write it in another part of the board.
01:10:55
I don't even want to write it because it's in the notes. I want you to notice the following. Look at the first term, no time dependence.
01:11:02
Second term, no time dependence. The cross term has an e to the something and I claim the other cross term will have e to the minus something. e to the i something with e to the minus i something is the cosine of something. That's all I want you to know. So there's somewhere in the time dependence, p of x and t has got a lot of stuff which is t
01:11:23
independent, plus something that looks like a whole bunch of numbers times cosine of e2 minus e3t over h-bar. That's all I want you to notice. So that means the probability of density will be oscillating.
01:11:41
The particle will not be fixed. It will be bouncing back and forth between the walls. And the rate at which it bounces is given by the difference in energy between the two states you formed in a combination. So this is how a particle in general, if you want, not the most general one, but it's a reasonably general
01:12:03
case where you added some mixture of that. Let me see, you added 2 to 3. You added that state plus that state. You added 4 times fifth of that times 3 fifth of that as the initial condition, 4 fifth of this one guy with 2
01:12:23
wiggles and 3 fifth of 3 wiggles, and you let it go in time. You'll find then if you add these time dependencies there'll be a part that varies with time, so the density will not be constant now. You'll be sloshing back and forth in the box. And that's a more typical situation. So not every state, initial state,
01:12:41
is a function of state of definite energy. It's an admixture. I've taken the simplest case where the admixture is only two parts in it. You can imagine taking a function made of three parts and four parts and ten parts, and when you square them and all that you'll get 100 cross terms. They'll all be oscillating at different rates. But the frequencies will always be given by the
01:13:00
difference in frequencies, differences in energies that went into your mixture. The last problem, which I'm not going to do now, I will do next time, but I'll tell you what it is I want to do, which is really mathematically more involved, but idea is the same. Here I gave you the initial state on a plate.
01:13:22
I just said, here it is, 3 fifth times one function of definite energy, 4 fifth times another function of definite energy. The problem you really want to be able to solve right now is when somebody gives you an arbitrary initial state and says, what happens to it? So I'm going to consider next time a function that looks
01:13:40
something like this, x times 1 minus x is my function at time 0. That's a nice function because it vanishes at 0. It vanishes at 1. It's a box of length l equal to 1. It's an allowed initial state.
01:14:00
Then I can ask, what does this do? So you should think about how do I predict its future, what's involved. So you will see that in the notes. So what I'm going to do next time is to finish this problem and give you the final set of postulates so that you can see what the rules of quantum mechanics are.