# 7. More on Energy Eigenstates

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01:15:53   |   Jun 18, 2014

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• PROFESSOR: OK, so I want to start out
• by reviewing a few things and putting
• some machinery together.
• Unfortunately, this thing is sort of stuck.
• We're going to need a later, so I don't know.
• I'll put it up for now.
• So first just a bit of notation.
• This symbol, you should think of it
• like the dot product, or the inner product.
• It's just saying that bracket f g is the integral.
• It's a number that you get.
• So this is a number that you get from the function
• f and the function g by taking f, taking
• its complex conjugate, multiplying it by g,
• and then integrating overall positions.
• All right?
• So it's a way to get a number.
• And you should think about it as the analog for functions
• of the dot product for vectors.
• It's a way to get a number out of two vectors.
• And so, for example, with vectors we could do v dot w,
• and this is some number.
• And it has a nice property that v dot v,
• we can think it as v squared, it's something like a length.
• It's strictly positive, and it's something
• like the length of a vector.
• Similarly, if I take f and take its bracket with f,
• this is equal to the integral dx of f squared,
• and in particular, f could be complex, so f norm squared.
• This is strictly non-negative.
• It could vanish, but it's not negative at a point,
• hence the norm squared.
• So this will be zero if and only if what?
• f is 0, f is the 0 function, right.
• So the same way that if you take a vector,
• and you take its dot product with itself, take it the norm,
• it's 0 if an only if the vector is 0.
• So this beast satisfies a lot of the properties of a dot
• product.
• You should think about it as morally equivalent.
• We'll talk about that in more detail later.
• Second, basic postulate of quantum mechanics,
• to every observable is associated an operator,
• and it's an operator acting on the space of functions
• or on the space of wave functions.
• And to every operator corresponding
• to an observable in quantum mechanics
• are associated a special set of functions called
• the eigenfunctions, such that when the operator acts
• on that function, it gives you the same function
• back times a constant.
• What these functions mean, physically,
• is they are the wave functions describing configurations
• with a definite value of the corresponding observable.
• If I'm in an eigenfunction of position with
• eigenvalue x naught, awesome.
• Thank you, AV person, thank you.
• So if your system is described by a wave function which
• is an eigenfunction of the position
• operator with eigenvalue x naught,
• that means you can be confident that the system is
• in the configuration corresponding
• to having a definite position x naught.
• Right?
• It's not a superposition of different positions.
• It is at x naught.
• Similarly, momentum, momentum has eigenfunctions,
• and we know what these guys are.
• These are the exponentials, e to the iKX's.
• They're the eigenfunctions, and those are the wave functions
• describing states with definite value of the momentum,
• of the associated observable.
• Energy as an operator, energy is described by an operator, which
• has eigenfunctions which I'll call phi sub n, with energy
• as E sub n, those are the eigenvalues.
• And if I tell you that your wave function is the state phi
• sub 2, what that tells you is that the system has
• a definite energy, E sub 2, corresponding
• to that eigenvalue.
• Cool?
• And this is true for any physical observable.
• But these are sort of the basic ones
• that we'll keep focusing on, position, momentum, and energy,
• for the next while.
• Now a nice property about these eigenfunctions
• is that for different eigenvalues,
• the associated wave functions are different functions.
• And what I mean by saying they're different
• functions is that they're actually
• orthogonal functions in the sense of this dot product.
• If I have a state corresponding to be at x 0,
• definite position x 0, that means
• they're in eigenfunction of position with eigenvalue x 0,
• and I have another that corresponds
• to being at x1, an eigenfunction of the position operator
• or the eigenvalue x1, then these wave functions are
• orthogonal to each other.
• And we get 0 if x 0 is not equal to x1.
• Everyone cool with that?
• Now, meanwhile not only are they orthogonal
• but they're normalized in a particular way.
• The inner product gives me a delta function,
• which goes beep once, so that if I integrate against it
• I get a 1.
• Same thing with momentum.
• And you do this, this you're checking on the problem set.
• I don't remember if it was last one or this one.
• And for the energies, energy 1, if I
• know the system is in state energy 1, and let's say e sub n
• and e sub m, those are different states
• if n and m are not equal to each other.
• And this inner product is 0 if n and m are not
• equal to each other and 1 if they are.
• Their properly normalized.
• Everyone cool with that?
• Yeah.
• AUDIENCE: Is it possible that two eigenfunctions
• have the same eigenvalue?
• PROFESSOR: Absolutely.
• It is absolutely possible for two eigenfunctions
• to have the same eigenvalue.
• That is certainly possible.
• AUDIENCE: [INAUDIBLE]
• PROFESSOR: Yeah, good.
• Thank you, this is a good technicality
• that I didn't want to get into, but I'll go and get into it.
• It's a very good question.
• So the question is, is it possible for two
• different eigenfunctions to have the same eigenvalue.
• Could there be two states with the same energy ,
• different states, same energy?
• Yeah, that's absolutely possible.
• And we'll run into that.
• And there's nice physics encoded in it.
• But let's think about what that means.
• The subsequent question is well, if that's the case,
• are they really still orthogonal?
• And here's the crucial thing.
• The crucial thing is, let's say I take one function,
• I'll call the function phi 1, consider the function phi 1.
• And let it have energy E1, so that E acting on phi 1
• is equal to E1 phi 1.
• And let there be another function, phi 2,
• such that the energy operator acting on phi 2
• is also equal to E1 phi 2.
• These are said to be degenerate.
• Degenerate doesn't mean you go out and trash your car,
• degenerate that the energies are the same.
• So what does this tell me?
• This tells me a cool fact.
• If I take a wave function phi, and I will call this phi star,
• in honor of Shri Kulkarni, so I've
• got this phi star, which is a linear combination alpha
• phi 1 plus beta phi 2, a linear combination
• of them, a superposition of those two states.
• Is this also an energy eigenfunction?
• Yeah, because if I act on phi star with E, then it's linear,
• so E acting on phi star is E acting on alpha phi 1,
• alpha's a constant, doesn't care.
• Phi 1 gives me an E1.
• Similarly, E acting on phi 2 gives me an E1.
• So if I act with E on this guy, this
• is equal to, from both of these I get an overall factor of E1.
• So notice that we get the same vector back, times
• a constant, a common constant.
• So when we have degenerate eigenfunctions,
• we can take arbitrary linear combinations to them,
• get another degenerate eigenfunction.
• Cool?
• So this is like, imagine I have a vector,
• and I have another vector.
• And they share the property that they're both eigenfunctions
• of some operator.
• That means any linear combination of them
• is also, right?
• So there's a whole vector space, there's
• a whole space of possible functions
• that all have the same eigenvalue.
• So now you say, well, look, are these two orthogonal
• to each other?
• No.
• These two?
• No.
• But here's the thing.
• If you have a vector space, if you have a the space,
• you can always find orthogonal guys and a basis
• for that space, yes?
• So while it's not true that the eigenfunctions are always
• orthogonal, it is true--
• we will not prove this, but we will discuss the proof of it
• later by pulling the mathematician out
• of the closet--
• the proof will say that it is possible to find
• a set of eigenfunctions which are orthogonal in precisely
• this fashion, even if there are degeneracies.
• OK?
• That theorem is called the spectral theorem.
• And we'll discuss it later.
• So it is always possible to do so.
• But you must be alert that there may be degeneracies.
• There aren't always degeneracies.
• In fact, degeneracies are very special.
• Why should two numbers happen to be the same?
• Something has to be forcing them to be the same.
• That's going to be an important theme for us.
• But it certainly is possible.
• Good question.
• Other questions?
• Yeah.
• AUDIENCE: [INAUDIBLE]
• PROFESSOR: Yeah, so using the triangular brackets--
• so there's another notation for the same thing, which is f g,
• but this carries some slightly different weight.
• It mean something slightly-- you'll see this in books,
• and this means something very similar to this.
• But I'm not going to use this notation.
• It's called Dirac notation.
• We'll talk about it later in the semester,
• but we're not going to talk about it just yet.
• But when you see this, effectively it
• means the same thing as this.
• This is sort of like dialect.
• You know, it's like French and Quebecois.
• Other questions?
• Other questions?
• OK.
• So given this fact, given the fact
• that we can associate observables
• to operators, operators come with special functions,
• the eigenfunctions, those eigenfunctions corresponding
• to have a definite value of the observable,
• and they're orthonormal.
• This tells us, and this is really
• the statement of the spectral theorem,
• that any function can be expanded in a basis of states
• with definite values of some observable.
• So for example, consider position.
• I claim that any wave function can
• be expanded as a superposition of states
• with definite position.
• So here's an arbitrary function, here's
• this set of states with definite position, the delta functions.
• And I can write any function as a superposition
• with some coefficients of states with definite position,
• integrating over all possible positions, x0.
• And this is also sort of trivially true,
• because what's this integral?
• Well, it's an integral, dx0 over all possible positions
• of this delta function.
• But we're evaluating at x, so this is 0
• unless x is equal to x0.
• So I can just put in x instead of x0,
• and that gives me psi of x.
• Sort of tautological We can do the same thing
• for momentum eigenfunctions.
• I claim that any function can be expanded
• in a superposition of momentum eigenfunctions, where
• I sum over all possible values in the momentum
• with some weight.
• This psi tilde of K is just telling
• me how much amplitude there is at that wave number.
• Cool?
• But this is the Fourier theorem, it's a Fourier expansion.
• So purely mathematically, we know that this is true.
• But there's also the physical statement.
• Any state can be expressed as a superposition of states
• with definite momentum.
• There's a math in here, but there's also physics in it.
• Finally, this is less obvious from a mathematical point
• of view, because I haven't even told you
• what energy is, any wave function can be expanded
• in states with definite energy.
• So this is a state, my state En, with definite energy,
• with some coefficient summed over all possible values
• of the energy.
• Given any physical observable, any physical observable,
• momentum, position, angular momentum,
• whatever, given any physical observable,
• a given wave function can be expanded
• as some superposition of having definite values of that.
• Will it in general have definite values of the observable?
• Well a general state be an energy eigenfunction?
• No.
• But any state is a superposition of energy eigenfunctions.
• Will a random state have definite position?
• Certainly not.
• You could have this wave function.
• Superposition.
• Yeah.
• AUDIENCE: Why is the energy special such
• that you can make an arbitrary state with a countable number
• of energy eigenfunctions rather than having
• to do a continuous spectrum?
• PROFESSOR: Excellent question.
• So I'm going to phrase that slightly differently.
• It's an excellent question, and we'll
• come to that at the end of today's lecture.
• So the question is, those are integrals, that
• is a sum over discrete things.
• Why?
• Why is the possible values of the position continuous,
• possible values of momentum continuous, and possible values
• of energy discrete?
• The answer to this will become apparent
• over the course of your next few problem sets.
• You have to do some problems to get
• your fingers dirty to really understand this.
• But here's the statement, and we'll
• see the first version of this at the end of today's lecture.
• Sometimes the allowed energies of a system, the energy
• eigenvalues, are discrete.
• Sometimes they are continuous.
• They will be discrete when you have bound states, states that
• are trapped in some region and aren't allowed
• to get arbitrarily far away.
• They'll be continuous when you have states that
• can get arbitrarily far away.
• Sometimes the momentum will be allowed to be discrete values,
• sometimes it will be allowed to be continuous values.
• And we'll see exactly why subsequently.
• But the thing I want to emphasize
• is that I'm writing this to emphasize that it's possible
• that each of these can be discrete or continuous.
• The important thing is that once you pick your physical system,
• you ask what are the allowed values of position, what
• are the allowed values of momentum,
• and what are the allowed values of energy.
• And then you sum over all possible values.
• Now, in the examples we looked at yesterday, or last lecture,
• the energy could have been discrete,
• as in the case of the infinite well,
• or continuous, as in the case of the free particle.
• In the case of a continuous particle
• this would have been an integral.
• In the case of the system such as a free particle, where
• the energy could take any of a continuous number
• of possible values, this would be a continuous integral.
• To deal with that, I'm often going
• to use the notation, just shorthand, integral sum.
• Which I know is a horrible bastardization
• of all that's good and just, but on the other hand,
• emphasizes the fact that in some systems
• you will get continuous, in some systems discrete,
• and sometimes you'll have both continuous and discrete.
• For example, in hydrogen, in hydrogen
• we'll find that there are bound states
• where the electron is stuck to the hydrogen nucleus,
• to the proton.
• And there are discrete allowed energy levels
• for that configuration.
• However, once you ionize the hydrogen,
• the electron can add any energy you want.
• It's no longer bound.
• It can just get arbitrarily far away.
• And there are an uncountable infinity, a continuous set
• of possible states.
• So in that situation, we'll find that we
• have both the discrete and continuous series
• of possible states.
• Yeah.
• AUDIENCE: [INAUDIBLE]
• PROFESSOR: Yeah, sure, if you work on a lattice.
• So for example, consider the following quantum system.
• And that undergraduate has been placed in 1 of 12 boxes.
• OK?
• Now, what's the state of the undergraduate?
• I don't know.
• Is it a definite position state?
• It might be.
• But probably it's a superposition, an arbitrary
• superposition, right?
• Very impressive undergraduates at MIT.
• OK, other questions.
• Yeah.
• AUDIENCE: Do these three [INAUDIBLE]
• hold even if the probability changes over time?
• PROFESSOR: Excellent question.
• We'll come back to that.
• Very good question, leading question.
• OK, so we have this.
• The next thing is that energy eigenfunctions satisfy
• some very special properties.
• And in particular, energy eigenfunctions
• have the property from the Schrodinger equation i h
• bar d t on psi of x and t is equal to the energy operator
• acting on psi of x and t.
• This tells us that if we have psi
• x 0 time t 0 is equal to phi n of x, as we saw last time,
• then the wave function, psi at x at time t
• is equal to phi n of x.
• And it only changes by an overall phase, e to the minus i
• En t over h bar.
• And this ratio En upon h bar will often
• be written omega n is equal to En over h bar.
• This is just the Dupre relations.
• Everyone cool with that?
• So are energy eigenfunctions--
• how to say.
• No wave function is more morally good than another.
• But some are particularly convenient.
• Energy eigenfunctions have the nice property
• that while they're not in a definite position
• and they don't necessarily have a definite momentum,
• they do evolve over time in a particularly simple way.
• And that and the superposition principle
• allow me to write the following.
• If I know that this is my wave function at psi at x at time 0,
• so let's say in all these cases, this is psi of x at time 0,
• how does this state evolve forward in time?
• It's kind of complicated.
• How does this description, how does psi tilde of k
• evolve forward in time?
• Again, kind of complicated.
• But when expressed in terms of the energy eigenstates,
• the answer to how it evolves forward in time
• is very simple, because I know that this
• is a superposition, a linear combination of states
• with definite energy.
• States with definite energy evolve with a phase.
• And the Schrodinger equation is linear,
• so solutions of the Schrodinger equation
• evolve to become solutions of the Schrodinger equation.
• So how does this state evolve forward in time?
• It evolves forward with a phase, e to the minus i omega n t.
• One for every different terms in this sum.
• Cool?
• So we are going to harp on energy functions,
• not because they're more moral, or more just, or more good,
• but because they're more convenient for solving the time
• evolution problem in quantum mechanics.
• and qualitative features of energy eigenfunctions.
• Cool?
• OK.
• And just to close that out, I just
• want to remind you of a couple of examples
• that we did last time, just get them on board.
• So the first is a free particle.
• well, actually let me not write that down.
• Actually, let me skip over the free particle,
• because it's so trivial.
• Let me just talk about the infinite well.
• So the potential is infinite out here,
• and it's 0 inside the well, and it
• goes from 0 to L. This is just my choice of notation.
• And the energy operator, as usual,
• is p squared upon 2m plus u of x.
• You might say, where did I derive this,
• and the answer is I didn't derive this.
• I just wrote it down.
• It's like force in Newton's equations.
• You just declare some force and you ask,
• what system does is model.
• So here's my system.
• It has what looks like a classical kind of energy,
• except these are all operators.
• And the potential here is this guy, it's 0 between 0 and L,
• and it's infinite elsewhere.
• And as we saw last time, the solutions
• to the energy eigenvalue equation
• are particularly simple.
• Phi sub n of x is equal to root properly
• normalized 2 upon L sine of Kn x,
• where kn is equal to n plus 1 pi, where
• n is an integer upon L.
• And these were chosen to satisfy our boundary conditions,
• that the wave function must vanish here, hence the sine,
• and K was chosen so that it turned over and just hit 0
• as we got to L. And that gave us that the allowed energies were
• discrete, because the En, which you can get by just plugging
• into the energy eigenvalue equation,
• was equal to h bar squared Kn squared upon 2m.
• So this tells us a nice thing.
• First off, in this system, if I take a particle
• and I throw it in here in some arbitrary state
• so that at time t equals zero the wave
• function x 0 is equal to sum over n phi n of x Cn.
• OK?
• Can I do this?
• Can I just pick some arbitrary function
• which is a superposition of energy eigenstates?
• Sure, because any function is.
• Any function can be described as a superposition
• of energy eigenfunctions.
• And if I use the energy eigenfunctions,
• it will automatically satisfy the boundary conditions.
• All good things will happen.
• So this is perfectly fine initial condition.
• What is the system at time t?
• Yeah, we just pick up the phases.
• And what phase is this guy?
• It's this, e to the minus i omega n t.
• And when I write omega n, let me be more explicit about that,
• that's En over h bar.
• So that's h bar Kn squared upon 2m t.
• Cool?
• So there is our solution for arbitrary initial conditions
• to the infinite square well problem in quantum mechanics.
• And you're going to study this in some detail on your problem
• set.
• let's look at the wave functions and the probability
• distributions for the lowest lying states.
• So for example, let's look at the wave
• function for the ground state, what I will call psi sub 0.
• And this is from 0 to L. And I put these bars here
• not because we're looking at the potential.
• I'm going to be plotting the real part of the wave function.
• But I put these walls here just to emphasize
• that that's where the walls are, at x equals 0 and x equals L.
• So what does it look like?
• Well, the first one is going to sine of Kn x.
• n is 0.
• Kn is going to be pi upon L. So that's again just this guy.
• Now, what's the probability distribution
• associated with psi 0?
• Where do you find the particle?
• So we know that it's just the norm squared of this wave
• function and the norm squared is here at 0, it's 0
• and it rises linearly, because sine
• is linear for small values.
• That makes this quadratic, and a maximum,
• So there's our probability distribution.
• Now, here's a funny thing.
• Imagine I take a particle, classical particle,
• and I put it in a box.
• And you put it in a box, and you tell it, OK, it's
• got some energy.
• So classically it's got some momentum.
• So it's sort of bouncing back and forth
• and just bounces off the arbitrarily hard walls
• and moves around.
• Where are you most likely to find that particle?
• Where does it spend most of its time?
• It spends the same amount of time at any point.
• It's moving at constant velocity.
• It goes boo, boo, boo, boo, right?
• So what's the probability distribution
• for finding it at any point inside, classically?
• Constant.
• Classically, the probability distribution is constant.
• You're just as likely to find it near the wall
• as not near the wall.
• However, quantum mechanically, for the lowest lying
• state that is clearly not true.
• You're really likely to find it near the wall.
• What's up with that?
• So that's a question that I want to put in your head
• and have you think about.
• You're going to see a similar effect arising over and over.
• And we're going to see at the very end
• that that is directly related, the fact that this goes to 0,
• is directly related, and I'm not kidding,
• to the transparency of diamond.
• OK, I think it was pretty cool.
• They're expensive.
• It's also related to the transparency
• of cubic zirconium, which I guess is less impressive.
• So the first state, again, let's look
• at the real part of psi 1, the first excited state.
• Well, this is now a sine with one extra--
• with a 2 here, 2 pi, so it goes through 0.
• So the probability distribution associated with psi 1,
• and I should say write this as a function of x, looks
• like, well, again, it's quadratic.
• But it has a 0 again in the middle.
• So it's going to look like--
• oops, my bad art defeats me.
• OK, there we go.
• So now it's even worse.
• Not only is unlikely to be out here,
• it's also very unlikely to be found in the middle.
• In fact, there is 0 probability you'll find it in the middle.
• That's sort of surprising.
• But you can quickly guess what happens as you
• go to very high energies.
• The real part of psi n let's say 10,000, 10 to the 4,
• what is that going to look like?
• and every time you increase n by 1,
• you're just going to add one more 0 to the sign.
• That's an interesting and suggestive fact.
• So if it's size of 10,000, how many nodes
• are there going to be in the middle of the domain?
• 10,000.
• And the amplitude is going to be the same.
• I'm not to be able to do this, but you get the idea.
• And now if I construct the probability distribution,
• what's the probability distribution going to be?
• Probability of the 10,000th psi sub 10 to the 4 of x.
• Well, it's again going to be strictly positive.
• And if you are not able to make measurements on the scale of L
• upon 10,000, but just say like L over 3, because you have
• a thumb and you don't have an infinitely accurate meter, what
• do you see?
• You see effectively a constant probability distribution.
• And actually, I shouldn't draw it there.
• I should draw it through the half,
• because sine squared over 2 averages to one half,
• or, sorry, sine squared averages to one half over many periods.
• So what we see is that the classical probability
• distribution constant does arise when we look
• at very high energy states.
• Cool?
• But it is manifestly not a good description.
• The classical description is not a good description.
• Your intuition is crappy at low energies,
• near the ground state, where quantum effects are dominating,
• because indeed, classically there was no minimum energy.
• Quantum effects have to be dominating there.
• And here we see that even the probability distribution's
• radically different than our intuition.
• Yeah.
• AUDIENCE: [INAUDIBLE]
• PROFESSOR: Keep working on it.
• So I want you all to think about what--
• you're not, I promise you, unless you've already
• seen some quantum mechanics, you're
• not going to be able to answer this question now.
• But I want you to have it as an uncomfortable little piece
• of sand in the back of your oyster mind--
• no offense-- what is causing that 0?
• Why are we getting 0?
• And I'll give you a hint.
• In quantum mechanics, anytime something interesting
• happens it's because of superposition and interference.
• All right.
• So with all that said, so any questions now over this story
• in a basis, et cetera, before we get moving?
• No, OK.
• In that case, get out your clickers.
• We're going to test your knowledge.
• Channel 41, for those of you who have to adjust it.
• [CHATTER]
• Wow.
• That's kind of worrying.
• Aha.
• OK, channel 41, and here we go.
• So go ahead and start now.
• Sorry, there was a little technical glitch there.
• So psi 1 and psi 2 are eigenstates.
• They're non-degenerate, meaning the energies are different.
• Is a superposition psi 1 plus psi 2 also an eigenstate?
• All right, four more seconds.
• All right.
• I want everyone to turn to the person next to you
• and discuss this.
• You've got about 30 seconds to discuss, or a minute.
• [CHATTER]
• All right.
• I want everyone, now that you've got an answer, click again,
• put in your current best guess.
• Oh, wait, sorry.
• For some reason I have to start over again.
• OK, now click.
• This is the best.
• I'm such a convert to clickers, this is just fantastic.
• So you guys went from, so roughly you all
• went from about 30, 60, 10, to what are we now?
• 8, 82, and 10.
• So it sounds like you guys are predicting answer b.
• I like the suspense.
• There we go.
• B, good.
• So here's a quick question.
• So why?
• And the reason why is that if we have E on psi 1 plus psi 2,
• this is equal to E on psi 1 plus E on psi 2, operator, operator,
• operator, but this is equal to E 1
• psi 1 E 2 psi 2, which if E1 and E2 are not equal,
• which is not equal to E times psi 1 plus psi 2.
• Right?
• Not equal to E anything times psi 1 plus psi 2.
• And it needs to be, in order to be
• an eigenfunction, an eigenfunction of the energy
• operator.
• Yeah.
• if this was kind of a silly random case where
• one of the energies is 0.
• Does this only happen if you have something that's infinite?
• PROFESSOR: Yeah, that's a really good question.
• So first off, how do you measure an energy?
• Do you ever measure an energy?
• Do you ever measure a voltage, the actual value
• of the scalar potential, the electromagnetic scalar
• potential?
• No.
• You measure a difference.
• Do you ever measure the energy?
• No, you measure a difference in energy.
• So the absolute value of energy is sort of a silly thing.
• But we always talk about it as if it's not.
• We say, that's got energy 14.
• It's a little bit suspicious.
• nothing hallowed about the number 0,
• although we will often refer to zero energy
• with a very specific meaning.
• What we really mean in that case is
• the value of the potential energy at infinity.
• So when I say energy, usually what I mean
• is relative to the value at infinity.
• Your question is it possible to have energy 0?
• Absolutely, and we'll see that.
• And it's actually going to be really interesting what's
• true of states with energy 0 in that sense.
• Second part of your question, though,
• is how does energy being 0 fit into this?
• Well, does that save us?
• Suppose one of the energies is 0.
• Then that says E on psi 1 plus psi 2 is equal to,
• let's say E2 is 0.
• Well, that term is gone.
• So there's just the one E1.
• Are we in energy eigenstate?
• No, because it's still not of the form
• E times psi 1 plus psi 2.
• So it doesn't save us, but it's an interesting question
• for the future.
• All right.
• Next question, four parts.
• So the question says x and p commute to i h bar.
• We've shown this.
• Is p x equal to i h bar, and is ip plus cx
• the same as cx plus ip?
• If you're really unsure you can ask the person next to you,
• but you don't have to.
• OK, so this is looking good.
• Everyone have an answer in?
• No?
• Five, four, three, two, one, OK, good.
• So the answer is C, which most of you got, but not everyone.
• A bunch of you put D. So let's talk through it.
• So remember what the definition of the commutator is.
• x with p by definition is equal to xp minus px.
• If we change the order here, px is
• equal to minus this, px minus xp.
• It's just the definition of the commutator.
• So on the other hand, if you add things, does 7 plus 6
• equal 6 plus 7?
• Yeah.
• Well, of course 6 times 7 is 7 times 6.
• So that's not a terribly good analogy.
• Does the order of addition of operators matter?
• No.
• Yeah.
• Yeah, exactly.
• Exactly.
• So it's slightly sneaky.
• OK, next question.
• OK, this one has five.
• f and g are both wave functions.
• c is a constant.
• Then if we take the inner product c times f with g,
• this is equal to what?
• Three, two, one, OK.
• so this one definitely discuss.
• Discuss with the person next to you.
• [CHATTER]
• All right.
• OK, 10 seconds.
• Wow.
• OK, fantastic.
• That works like a champ.
• Yes, complex conjugation.
• Don't screw that one up.
• It's very easy to forget, but it matters a lot.
• Cursor keeps disappearing.
• OK, next one.
• A wave function has been expressed
• as a sum of energy eigenfunctions.
• Here I'm calling them mu rather than phi, but same thing.
• Compared to the original wave function,
• the set of coefficients, given that we're using the energy
• basis, the set of coefficients contains more or less
• the same information, or it can't be determined.
• OK, five seconds.
• All right.
• And the answer is C, great.
• OK, next one.
• So right now we're normalizing.
• OK.
• All stationary states, or all energy eigenstates,
• have the form that spatial and time dependence
• is the spatial dependence, the energy eigenfunction,
• times a phase, so that the norm squared is time independent.
• Consider the sum of two non-degenerate energy
• eigenstates psi 1 and psi 2.
• Non-degenerate means they have different energy.
• Is the wave function stationary?
• Is the probability distribution time
• independent or is it time dependent?
• This one's not trivial.
• Oh, shoot.
• I forgot to get it started.
• Sorry.
• Right.
• Whoo, yeah.
• This one always kills people.
• No chatting just yet.
• Test yourself, not your neighbor.
• It's fine to look deep into your soul,
• but don't look deep into the soul of the person sitting next
• to you.
• All right.
• So at this point, chat with your neighbor.
• Let me just give you some presage.
• The parallel strategy's probably not so good, because about half
• of you got it right, and about half of you got it wrong.
• [CHATTER]
• All right.
• Let's vote again.
• And hold on, starting now.
• OK, vote again.
• You've got 10 seconds to enter a vote.
• Wow.
• OK, two seconds.
• Good.
• So the distribution on this one went from 30, 50,
• 20 initially, to now it is 10, 80, and 10.
• Amazingly, you guys got worse.
• The answer is C. And I want you to discuss with each other
• why it's C.
• [CHATTER]
• All right.
• OK.
• So let me talk you through it.
• So the wave function, we've said psi of x and t
• is equal to phi 1 at x, e to the minus i omega 1 t plus phi
• 2 of x e to the minus i omega 2 t.
• So great, we take the norm squared.
• What's the probability to find it at x at time t.
• The probability density is the norm
• squared of this guy, psi squared, which
• is equal to phi 1 complex conjugate e to the plus i omega
• 1 t plus phi 2 complex conjugate e to the plus i omega 2t times
• the thing itself phi 1 of x e to the minus i
• omega 1 t plus phi 2 of x e to the minus i omega 2t, right?
• So this has four terms.
• The first term is psi 1 norm squared.
• The phases cancel, right?
• You're going to see this happen a billion times in 804.
• The first term is going to be phi 1 norm squared.
• There's another term, which is phi 2 norm squared.
• Again the phases exactly cancel, even the minus i omega 2
• t to the plus i omega 2 t.
• Plus phi 2 squared.
• But then there are two cross terms, the interference terms.
• Plus phi 1 complex conjugate phi 2 e to the i omega 1 t
• e to the plus i omega 1 t, i omega 1 t, and e to the minus
• i omega 2t, minus omega 2.
• So we have a cross-term which depends on the difference
• in frequencies.
• Frequencies are like energies modulo on h-bar,
• so it's a difference in energies.
• And then there's another term, which
• is the complex conjugate of this guy,
• phi 2 star times phi 1 phi 2 complex conjugate phi 1
• and the phases are also the complex conjugate
• e to the minus i omega 1 minus omega 2 t of x of x of x of x.
• So is there time dependence in this, in principle?
• Absolutely, from the interference terms.
• Were we not in the superposition,
• we would not have interference terms.
• Time dependence comes from interference, when we expand
• in energy eigenfunctions.
• Cool?
• However, can these vanish?
• When?
• Sorry, say again?
• Great, so when omega 1 equals omega 2, what happens?
• Time dependence goes away.
• But omega 1 is e 1 over h bar, omega 2 is e 2 over h bar,
• and we started out by saying these are non-degenerate.
• So if they're non-degenerate, the energies are different,
• the frequencies are different, so that doesn't help us.
• How do we kill this time dependence?
• Yes.
• If the two functions aren't just orthogonal
• in a functional sense, but if we have the following.
• Suppose phi 1 is like this.
• It's 0 everywhere except for in some lump that's phi 1,
• and phi 2 is 0 everywhere except here.
• Then anywhere that phi 1 is non-zero, phi 2 is zero.
• And anywhere where phi 2 is non-zero, phi 1 is zero.
• So this can point-wise vanish.
• Do you expect this to happen generically?
• Does it happen for the energy eigenfunctions
• in the infinite square well?
• Sine waves?
• No.
• They have zero at isolated points,
• but they're non-zero generically.
• Yeah, so it doesn't work there.
• What about for the free particle?
• Well, those are just plain waves.
• Does that ever happen?
• No.
• OK, so this is an incredibly special case.
• We'll actually see it in one problem
• on a problem set later on.
• It's a very special case.
• So technically, the answer is C. And I
• want you guys to keep your minds open
• on these sorts of questions, when does a spatial dependence
• matter and when are there interference terms.
• Those are two different questions,
• and I want you to tease them apart.
• OK?
• Cool?
• Yeah?
• to think that you're fixing the initial [INAUDIBLE]
• PROFESSOR: That's a very good way to think about it.
• That's exactly right.
• That's a very, very good question.
• Let me say that subtly differently, and tell me
• if this agrees with what you were just saying.
• So I can look at this wave function,
• and I already know that the overall phase of a wave
• function doesn't matter.
• That's what it is to say a stationary state is stationary.
• It's got an overall phase that's the only thing,
• norm squared it goes away.
• So I can write this as e to the minus i omega 1 t times phi
• 1 of x plus phi 2 of x e to the minus i omega
• 2 minus omega 1 t.
• Is that what you mean?
• So that's one way to do this.
• We could also do something else.
• We could do e to the minus i omega 1 plus omega 2 upon 2 t.
• And this is more, I think, what you
• were thinking of, a sort of average frequency
• and then a relative frequency, and then
• the change in the frequencies on these two terms.
• Absolutely.
• So you can organize this in many, many ways.
• But your question gets at a very important point,
• which is that the overall phase doesn't matter.
• But relative phases in a superposition do matter.
• So when does a phase matter in a wave function?
• It does not matter if it's an overall phase.
• But it does matter if it's a relative phase between terms
• in a superposition.
• Cool?
• Very good question.
• Other questions?
• If not, then I have some.
• So, consider a system which is in the state--
• so I should give you five--
• system is in a state which is a linear combination of n equals
• 1 and n equals 2 eigenstates.
• What's the probability that measurement
• will give us energy E1?
• And it's in this superposition.
• OK, five seconds.
• OK, fantastic.
• Yes, C, great.
• OK, everyone got that one.
• So one's a slightly more interesting question.
• Suppose I have an infinite well with width L.
• How does the energy, the ground state energy,
• compare to that of a system with a wider well?
• So L versus a larger L. OK, four seconds.
• OK, quickly discuss amongst yourselves, like 10 seconds.
• [CHATTER]
• All right.
• Now click again.
• Yeah.
• All right.
• Five seconds.
• One, two, three, four, five, great.
• OK, the answer is A. OK, great, because the energy
• of the infinite well goes like K squared.
• K goes like 1 over L. So the energy is, if we make it wider,
• the energy if we make it wider is going to be lower.
• And last couple of questions.
• OK, so t equals 0.
• Could the wave function for an electron in an infinite square
• well of width a, rather than L, be A sine squared of pi x
• upon a, where A is suitably chosen to be normalized?
• All right, you've got about five seconds left.
• And OK, we are at chance.
• We are at even odds, and the answer
• is not a superposition of A and B,
• so I encourage you to discuss with the people around you.
• [CHATTER]
• Great.
• What properties had it better satisfy in order
• to be a viable wave function?
• What properties should the wave function
• have so that it's reasonable?
• Yeah.
• Is it zero at the ends?
• Yeah.
• Good.
• Is it smooth?
• Yeah.
• Exactly.
• And so you can write it as a superposition.
• Excellent.
• Yeah.
• All right.
• Vote again.
• OK, I might have missed a few people.
• So go ahead and start.
• OK, five more seconds.
• All right.
• So we went from 50-50 to 77-23.
• That's pretty good.
• A. Why?
• Is this an energy eigenstate?
• No.
• Does that matter?
• No.
• What properties had this wave function better
• satisfy to be a reasonable wave function in this potential?
• Say again?
• It's got to vanish at the walls.
• It's got to satisfy the boundary conditions.
• What else must be true of this wave function?
• Normalizable.
• Is it normalizable?
• Yeah.
• What else?
• Continuous.
• It better not have any discontinuities.
• Is it continuous?
• Great.
• OK.
• Is there any reason that this is a stupid wave function?
• No.
• It's perfectly reasonable.
• It's not an energy eigenfunction, but--
• Yeah, cool?
• Yeah.
• AUDIENCE: This is sort of like a math question.
• So to write that at a superposition,
• you have to write it like basically a Fourier sign
• series?
• Isn't the [INAUDIBLE] function even, though?
• PROFESSOR: On this domain, that and the sines are even.
• So this is actually odd, but we're only looking at it from 0
• to L. So, I mean that half of it.
• The sines are odd, but we're only looking at the first peak.
• So you could just as well have written
• that as cosine of the midpoint plus the distance
• from the midpoint.
• Actually, let me say that again, because it's
• a much better question I just give it shrift for.
• So here's the question.
• The question is, look, so sine is an odd function,
• but sine squared is an even function.
• So how can you expand sine squared, an even function,
• in terms of sines, an odd function?
• Here's sine squared in our domain, and here's sine.
• Now what do you mean by even?
• Usually by even we mean reflection around zero.
• But I could just as well have said reflection
• around the origin.
• This potential is symmetric.
• And the energy eigenfunctions are symmetric about the origin.
• They're not symmetric about reflection around this point.
• But they are symmetric about reflection around this point.
• That's a particularly natural place to call it 0.
• So I was calling them sine because I was calling this 0,
• but I could have called it cosine
• if I called this 0, for the same Kx.
• And indeed, can we expand this sine
• squared function in terms of a basis of these sines
• on the domain 0 to L?
• Absolutely.
• Very good question.
• And lastly, last clicker question.
• Oops.
• Whatever.
• OK.
• At t equals 0, a particle is described by the wave function
• we just saw.
• Which of the following is true about the wave function
• at subsequent times?
• 5 seconds.
• Whew.
• Oh, OK.
• In the last few seconds we had an explosive burst for A, B,
• and C. So our current distribution
• is 8, 16, 10, and 67, sounds like 67 is popular.
• Discuss quickly, very quickly, with the person next to you.
• [CHATTER]
• OK, and vote again.
• OK, five seconds.
• Get your last vote in.
• All right.
• And the answer is D. Yay.
• So let's think about the logic.
• Let's go through the logic here.
• So as was pointed out by a student up here earlier,
• the wave function sine squared of pi x
• can be expanded in terms of the energy eigenfunction.
• Any reasonable function can be expanded
• in terms of a superposition of definite energy states
• of energy eigenfunctions.
• So that means we can write psi at some time
• as a superposition Cn sine of n pi x upon a e to the minus i e
• n t upon h bar, since those are, in fact, the eigenfunctions.
• So we can do that.
• Now, when we look at the time evolution,
• we know that each term in that superposition
• evolves with a phase.
• The overall wave function does not evolve with a phase.
• It is not an energy eigenstate.
• There are going to be interference terms due
• to the fact that it's a superposition.
• So its probability distribution is not time-independent.
• It is a superposition.
• And so the wave function doesn't rotate by an overall phase.
• However, we can solve the Schrodinger equation,
• as we did before.
• The wave function is expanded at time 0
• as the energy eigenfunctions times some set of coefficients.
• And the time evolution corresponds
• to adding two each independent term in the superposition
• the appropriate phase for that energy eigenstate.
• Cool?
• All right.
• So the answer is D. And that's it for the clicker questions.
• OK, so any questions on the clicker questions so far?
• OK, those are going to be posted on the web site
• so you can go over them.
• And now back to energy eigenfunctions.
• So what I want to talk about now is the qualitative behavior
• of energy eigenfunctions.
• Suppose I know I have an energy eigenfunction.
• What can I say generally about its structure?
• So let me ask the question, qualitative behavior.
• So suppose someone hands you a potential U of x.
• Someone hands you some potential, U of x,
• and says, look, I've got this potential.
• Maybe I'll draw it for you.
• It's got some wiggles, and then a big wiggle,
• and then it's got a big wiggle, and then--
• do I want to do that?
• Yeah, let's do that.
• Then a big wiggle, and something like this.
• And someone shows you this potential.
• And they say, look, what are the energy eigenfunctions?
• Well, OK, free particle was easy.
• The infinite square well was easy.
• We could solve that analytically.
• The next involved solving a differential equation.
• So what differential equation is this going to lead us to?
• Well, we know that the energy eigenvalue equation
• is minus h bar squared upon 2 m phi
• prime prime of x plus U of x phi x, so that's the energy
• operator acting on phi, is equal to,
• saying that it's an energy eigenfunction, phi sub E,
• says that it's equal to the energy operator
• acting on this eigenfunction is just a constant E phi sub
• E of x.
• And I'm going to work at moment in time,
• so we're going to drop all the t dependence for the moment.
• So this is the differential equation
• we need to solve where U of x is this god-awful function.
• Do you think it's very likely that you're
• going to be able to solve this analytically?
• Probably not.
• However, some basic ideas will help
• you get an intuition for what the wave function should
• look like.
• And I cannot overstate the importance of being able
• to eyeball a system and guess the qualitative features of its
• wave functions, because that intuition,
• that ability to estimate, is going to contain an awful lot
• of physics.
• So let's try to extract it.
• So in order to do so, I want to start
• by massaging this equation into a form which
• is particularly convenient.
• So in particular, I'm going to write this equation as phi sub
• E prime prime.
• So what I'm going to do is I'm going to take this term,
• I'm going to notice this has two derivatives,
• this has no derivatives, this has no derivatives.
• And I'm going to move this term over here and combine
• these terms into E minus U of x, and I'm
• going to divide each side by 2m upon h bar squared with a minus
• sign, giving me that phi prime prime of E
• of x upon phi E of x dividing through by this phi E
• is equal to minus 2m over h bar squared.
• And let's just get our signs right.
• We've got the minus from here, so this
• is going to be E minus U of x.
• So you might look at that and think, well,
• why is that any better than what I've just written down.
• But what is the second derivative of function?
• It's telling you not its slope, but it's telling you
• how the slope changes.
• It's telling about the curvature of the function.
• And what this is telling me is something very, very useful.
• So for example, let's look at the function.
• Let's assume that the function is real,
• although we know in general it's not.
• Let's assume that the function is real for simplicity.
• So we're going to plot the real part of phi
• in the vertical axis.
• And this is x.
• Suppose the real part of phi is positive at some point.
• Phi prime prime, if it's positive,
• tells us that not only is the slope positive,
• but it's increasing.
• Or it doesn't tell us anything about the slope,
• but it tells us that whatever the slope, it's increasing.
• If it's negative, the slope is increasing as we increase x.
• If it's positive, it's increasing as we increase x.
• So it's telling us that the wave function looks like this,
• locally, something like that.
• If phi is negative, if phi is negative,
• then if this quantity is positive,
• then phi prime prime has to be negative.
• But negative is curving down.
• So if this quantity, which I will call the curvature,
• if this quantity is positive, it curves away from the axis.
• So this is phi prime prime over phi greater than 0.
• If this quantity is positive, the function
• curves away from the axis.
• Cool?
• If this quantity is negative, phi prime prime upon
• phi less than 0, exactly the opposite.
• This has to be negative.
• If phi is positive, then phi prime prime has to be negative.
• It has to be curving down.
• And similarly, if phi is negative,
• then phi prime prime has to be positive,
• and it has to curve up.
• So if this quantity is positive, if the curvature is positive,
• it curves away from the axis.
• If the curvature is negative, if this quantity is negative,
• it curves towards the axis.
• So what does that tell you about solutions when the curvature is
• positive or negative?
• It tells you the following.
• It tells you that, imagine we have
• a function where phi prime prime over phi is constant.
• And in particular, let's let phi prime prime over phi
• be a constant, which is positive.
• And I'll call that positive constant kappa squared.
• And to emphasize that it's positive,
• I'm going to call it kappa squared.
• It's a positive thing.
• It's a real number squared.
• What does the solution look like?
• Well, this quantity is positive.
• It's always going to be curving away.
• So we have solutions that look like this or solutions
• that look like this.
• Can it ever be 0?
• Yeah, sure, it could be an inflection point.
• So for example, here the curvature is positive,
• but at this point the curvature has to switch to be like this.
• What functions are of this form?
• Let me give you another hint.
• Here's one.
• Is this curvature positive?
• Yes.
• Yup.
• Those are all positive curvature.
• And these are exponentials.
• And the solution to this differential equation
• is e to the plus kappa x or e to the minus kappa x.
• And an arbitrary solution of this equation
• is a superposition A e to the kappa x plus B
• e to the minus kappa x.
• Everyone cool with that?
• When this quantity is positive, we
• get growing and collapsing exponentials.
• Yeah?
• On the other hand, if phi prime prime over phi
• is a negative number, i.e. minus what I'll call k
• squared, then the curvature has to be negative.
• And what functions have everywhere negative curvature?
• Sinusoidals.
• Cool?
• And the general solution is A e to the i K x plus B
• e to the minus i K x.
• So that differential equation, also known as sine and cosine.
• Cool?
• So putting that together with our original function,
• let's bring this up.
• So we want to think about the wave functions here.
• But in order to think about the energy eigenstates,
• we need to decide on an energy.
• We need to pick an energy, because you
• can't find the solution without figuring the energy.
• But notice something nice here.
• So suppose the energy is e.
• And let me just draw E. This is a constant.
• The energy is this.
• So this is the value of E. Here we're drawing potential.
• But this is the value of the energy, which is a constant.
• It's just a number.
• If you had a classical particle moving in this potential,
• what would happen?
• It would roll around.
• So for example, let's say you gave it this energy
• by putting it here.
• And think of this as a gravitational potential.
• You put it here, you let go, and it falls down.
• And it'll keep rolling until it gets up here
• to the classical turning point.
• And at that point, its kinetic energy
• must be 0, because its potential energy
• is its total energy, at which point
• it will turn around and fall back.
• Yes?
• If you take your ball, and you put it here,
• and you let it roll, does it ever
• get here, to this position?
• No, because it doesn't have enough energy.
• Classically, this is a forbidden position.
• So given an energy and given a potential,
• we can break the system up into classically allowed zones
• and classically forbidden zones.
• Cool?
• Now, in a classically allowed zone,
• the energy is greater than the potential.
• And in a classically forbidden zone,
• the energy is less than the potential.
• Everyone cool with that?
• But this tells us something really nice.
• If the energy is greater than the potential,
• what do you know about the curvature?
• Yeah.
• If we're in a classically allowed zone,
• so the energy is greater than the potential,
• then this quantity is positive, there's a minus sign here,
• so this is negative.
• So the curvature is negative.
• Remember, curvature is negative means
• that we curve towards the axis.
• So in a classically allowed region,
• the wave function should be sinusoidal.
• What about in the classically forbidden regions?
• In the classically forbidden regions,
• the energy is less than the potential.
• That means in magnitude this is less than this,
• this is a negative number, minus sign,
• the curvature is going to be minus
• times a minus is a positive, so the curvature's positive.
• So the solutions are either growing exponentials
• or shrinking exponentials or superpositions of them.
• Everyone cool with that?
• So let's think about a simple example.
• Let's work through this in a simple example.
• And let me give you a little bit more board space.
• Simple example would be a potential that looks like this.
• And let's just suppose that we want
• to find an energy eigenfunction with energy that's E. Well,
• this is a classically allowed zone,
• and these are the classically forbidden regions.
• Now I want to ask, what does the wave function look like?
• And I don't want to draw it on top of the energy diagram,
• because wave function is not an energy.
• Wave function is a different quantity,
• because it's got different axes and I want
• it drawn on a different plot.
• So but as a function of x--
• so just to get the positions straight,
• these are the bounds of the classically
• allowed and forbidden regions.
• What do we expect?
• Well, we expect that it's going to be sinusoidal in here.
• We expect that it's going to be exponential growing
• or converging out here, exp.
• But one last important thing is that not only is the curvature
• negative in here in these classically allowed regions,
• but the magnitude of the curvature,
• how rapidly it's turning over, how big that second derivative
• is, depends on the difference between the energy
• and the potential.
• The greater the difference, the more rapid the curvature,
• the more rapid the turning over and fluctuation.
• If the differences between the potential and the true energy,
• the total energy, is small, then the curvature is very small.
• So the derivative changes very gradually.
• What does that tell us?
• That tells us that in here the wave function is oscillating
• rapidly, because the curvature, the difference
• between the energy and the potential is large,
• and so the wave function is oscillating rapidly.
• As we get out towards the classical turning points,
• the wave function will be oscillating less rapidly.
• The slope will be changing more gradually.
• And as a consequence, two things happen.
• Let me actually draw this slightly differently.
• So as a consequence two things happen.
• One is the wavelength gets longer,
• because the curvature is smaller.
• And the second is the amplitude gets larger,
• because it keeps on having a positive slope for longer
• and longer, and it takes longer to curve back down.
• So here we have rapid oscillations.
• And then the oscillations get longer and longer wavelength,
• until we get out to the classical turning point.
• And at this point, what happens?
• Yeah, it's got to be [INAUDIBLE]..
• Now, here we have some sine, and some superpositions
• of sine and cosines, exponentials.
• And in particular, it arrives here with some slope
• and with some value.
• We know this side we've got to get exponentials.
• And so this sum of sines and cosines at this point
• must match the sum of exponentials.
• How must it do so?
• What must be true of the wave function at this point?
• Can it be discontinuous?
• Can its derivative be discontinuous?
• No.
• So the value and the derivative must be continuous.
• So that tells us precisely which linear combination
• of positive growing and shrinking exponentials we get.
• So we'll get some linear combination,
• which may do this for awhile.
• But since it's got some contribution
• of positive exponential, it'll just grow exponentially off
• to infinity.
• And as the energy gets further and further away
• from the potential, now in their negative sine, what
• happens to the rate of growth?
• It gets more and more rapid.
• So this just diverges more and more rapidly.
• Similarly, out here we have to match the slope.
• And we know that the curvature has
• to be now positive, so it has to do this.
• So two questions.
• First off, is this sketch of the wave function
• a reasonable sketch, given what we
• know about curvature and this potential of a wave function
• with that energy?
• Are there ways in which it's a bad estimate?
• AUDIENCE: [INAUDIBLE]
• PROFESSOR: OK, excellent.
• AUDIENCE: On the right side, could it have crossed zero?
• PROFESSOR: Absolutely, it could have crossed zero.
• So I may have drawn this badly.
• It turned out it was a little subtle.
• It's not obvious.
• Maybe it actually punched all the way through zero,
• and then it diverged down negative.
• That's absolutely positive.
• So that was one of the quibbles you could have.
• Another quibble you could have is
• that it looks like I have constant wavelength in here.
• But the potential's actually changing.
• And what you should chalk this up to,
• if you'll pardon the pun, is my artistic skills are limited.
• So this is always going to be sort of inescapable
• when you qualitatively draw something.
• On a test, I'm not going to bag you points on things like that.
• That's what I want to emphasize.
• But the second thing, is there something
• Yes, you've already named it.
• It diverges off to infinity out here and out here.
• What does that tell you?
• It's not physical.
• Good.
• Sorry?
• Excellent.
• Is this an allowable energy?
• No.
• If the wave function has this energy,
• it is impossible to make it continuous,
• assuming that I drew it correctly,
• and have it converge.
• Is this wave function allowable?
• No, because it does not satisfy our boundary conditions.
• Our boundary conditions are that the wave function must vanish
• out here and it must vanish out here at infinity
• in order to be normalizable.
• Here we failed.
• Now, you can imagine that-- so let's decrease
• the energy a little bit.
• If we decrease the energy, our trial energy just a little
• tiny bit, what happens?
• Well, that's going to decrease the curvature in here.
• We decrease, we bring the energy in just a little tiny bit.
• That means this is a little bit smaller.
• The potential stays the same.
• So the curvature in the allowed region
• is just a little tiny bit smaller.
• And meanwhile, the allowed region
• has got just a little bit thinner.
• And what that will do is the curvature's a little less,
• the region's a little less, so now we have--
• Sorry, I get excited.
• And if we tweak the energy, what's going to happen?
• Well, it's going to arrive here a little bit sooner.
• And let's imagine something like this.
• And if we chose the energy just right,
• we would get it to match to a linear combination
• of collapsing and growing exponentials,
• where the contribution from the growing exponential
• in this direction vanishes.
• There's precisely one value of the energy
• that lets me do that with this number of wiggles.
• And so then it goes through and does its thing.
• And we need it to happen on both sides.
• Now if I take that solution, so that it achieves convergence
• out here, and it achieves convergence out here,
• and I take that energy and I increase it by epsilon,
• by just the tiniest little bit, what will happen to this wave
• function?
• It'll diverge.
• It will no longer be normalizable.
• When you have classically forbidden regions,
• are the allowed energies continuous or discrete?
• And that answers a question from earlier in the class.
• And it also is going to be the beginning
• of the answer to the question, why is the spectrum of hydrogen
• discrete.
• See you next time.

### Description

MIT 8.04 Quantum Physics I, Spring 2013
View the complete course: http://ocw.mit.edu/8-04S13