22. Quantum mechanics IV: Measurement theory, states of definite energy

Uploaded by YaleCourses on 23.03.2011

Prof: So I got an interesting email this weekend
from two of your buddies, Jerry Wang and Emma Alexander.
The title of the email is this.
It says Casaplanka, so I already know these guys
are up to no good.
And the question is the following.
They said, "You've written a wave function ψ(x) and
There seems to be no reference of time, so where is time in
this," and they couldn't stop there.
That would be a good place to stop but they went on to say,
"Are you saying the ψ is just a ψ
as time goes by?"
So that's the kind of stuff that appeals to me,
so I don't care if you don't learn any quantum mechanics,
but if you can do this kind of stuff I'm not worried about you.
But now I have to give a serious answer to that serious
question, which I sort of mentioned
before, which is that everything I've done so far is for one
instant in time.
I hope I made it very clear.
So here's the analogy.
If someone comes to you and says, "Tell me all about
Newtonian mechanics.
How does it work?
What's the scheme like?"
You say, "At any instant someone has to give you the x
and p of that particle at that time," and that's all you
need to know.
That's all you can ask and that's all you need to know.
Everything about that one particle is completely given by
this x and p.
That is called a state.
That's the complete description of any system is called at
state, and a state in classical
mechanics for one particle in one dimension is just a pair of
It's a state because given that I can predict the future for
I can predict the future because if I knew the initial
momentum, namely initial velocity.
I know how fast the guy is moving, and I know where it is
at the initial x.
A tiny amount of time later it'll be at x vt,
but t is very small.
Then since I know the acceleration from Newton's Laws,
which is rate of change of velocity,
if I knew the initial velocity or initial momentum I can get
the momentum a little later.
Then in that manner I can inch forward in time.
So you need Newton's Laws, and Newton's Laws tell you,
if you like, rate of change of momentum is
the force.
So if you like you can write this mdx/dt squared is the
So this part is called dynamics and this part is called
So dynamics is how do things change with time?
Kinematics is what do you need to know about the particle?
In the quantum version of this if you say,
"What constitutes complete knowledge of the
particle?," the answer is this function
If you knew this function you know all that there is to know
about the particle.
We're not saying it's real.
It can be complex.
Then you can ask, "If you knew the ψ
at one time what's it going to be later on?"
That's the analog of this one.
So I have not come to that yet.
So this is a pretty long story.
It takes so long just to tell you what to do at one time.
Then I'm going to tell you how to go from now to later.
In Newtonian mechanics it's just a one-word answer,
x and p now is the whole story.
So anyway, since we've got four lectures including today I will,
of course, tell you before it's all over what is the analog of
Namely I'll tell you a formula for dψ/dt.
How does ψ change with time?
So let's summarize what we know.
I want to do this every class so that it gets into your
If it is true that ψ(x) tells me everything I need to
know let's ask questions of this ψ.
We can say, "Where is the particle?"
So there is some ψ.
You're told then the probability that you will find
it somewhere is the absolute value.
Probability density is proportional to ψ^(2).
So it doesn't tell you where the particle is.
It gives you the odds, and the odds are this.
Then you can say, "Okay, x is just one of
the variables of interest to me.
How about momentum?
If I measure momentum what answers will I get,
and what are the odds for the different answers?"
That question has a longer answer and it goes as follows.
Take the ψ that's given to you,
and write is as a sum with some coefficients of functions that I
call ψ_p(x), but I will tell you what they
They are e^(ipx/ℏ) by square root of L.
Then I said, "If you manage to write
the ψ that's given to you in this form,
summing over all the allowed values of p,
then the probability that you will get one of those allowed
values is equal to the absolute square of that
And then final question can be, "How do I know what
A(p) is given some function ψ?,"
and the answer is A(p) is the integral of the complex
conjugate of this ψ(x) dx.
In our case it was a ring of length L.
Then it's 0 to L, otherwise it's whatever your
allowed region is.
Student: Can you explain ________ one more time
Prof: What this guy means?
Student: Yeah.
Prof: So let me give an example.
Suppose ψ of p was cosine
4Πx over L?
Then that looks like e^(4Πix/L) e^(-4Πix/L)
divided by 2.
Let me put the 2 here, put the 2 here.
Then my ψ is not quite in the standard
form because I don't have the square root of Ls.
So let me put a square root of L here, a square root of L here
and also put one outside.
So my ψ looks like a square root of L.
Let me see.
Oh, I know what the problem is.
This ψ is not normalized because the rule is if you
don't-- first let's say it's not
normalized, so don't worry about any of
these numbers.
Can you see that it's made up of two such functions with equal
That means if you measure the momentum you'll only get the
answer corresponding to what's here and what's here.
If you compare it to e^(ipx/ℏ) you can see the p
that goes here is really 4Πâ„ /L or -4Πâ„ /L.
So a particle in this state can have only two values of momentum
if you measure it, and the probability for the two
you can see by symmetry must be 50/50 for each because they come
with equal weight.
But if you don't want to do that you can,
if you like, like this ψ, normalize it,
and then we square it, integrate it and put a number
in front so it comes out to have length 1.
Then the coefficients here directly when squared will give
you probabilities.
Oh, a lot of questions.
Student: So A_p has to be a real
number, right?
Prof: No.
A_p does not have to be a real number just like
ψ(x) does not have to be a real number A_p--
because I'm taking the absolute value of A^(2).
So even if it's a complex number the absolute value of A
will always come out to be real and positive.
Is that your question?
See A_p is not a probability.
The absolute value squared of A_p is a probability
which, of course, must be positive.
But A does not have to be positive, does not have to be
Student: You said A_p like the integral
if you're multiplying ψ* times ψ would that be in
real numbers?
Prof: You should be very careful.
Student: Really?
Prof: A particular ψ times that same ψ* is a
real positive number.
This is not the same function.
I think I remember now from previous years that students
will get confused on this.
This ψ(x) is the function that was given to me of which it
was asked, "If I measure momentum
what answer will I get?"
This ψ_p is not the function that was given to
This is another function that describes a particle whose
momentum is p.
So you've got to multiply that function with the given function
to get these coefficients.
So you can change your mind and say, "Oh,
I'm not interested in that ψ.
I want to know the answer to that ψ."
Well, you take that ψ and put it here.
These guys will not change.
Student: I'm sort of confused.
Right next to what you were explaining you have written
ψ of p.
Prof: Oh, I'm sorry.
No, no, no.
You have every right to be confused.
I did not mean that.
This is ψ of x.
Student: And is that just ψ of x or that
a ψ_p also?
Prof: Here?
Student: That's just the ψ of x...
Prof: Yes, this is ψ
of x.
Student: Okay.
Prof: This ψ(x)--well,
if I'm more disciplined, I can do that for you.
Maybe I should do it one more time.
It doesn't hurt.
So let's so let's say ψ(x) is cosine 4Πx/L.
Now first job is to rescale the ψ so that if you square it
and integrate it you get 1.
I know the answer to that is square root of 2/L.
You can check that.
If you take this guy and you square it and integrate it it'll
work because the square of this will be 2/L,
and the average value of cosign squared is half.
And we integrate it from 0 to L it will cancel.
You'll get 1.
If you do this particular function then all you have to do
is write this in terms of-- first write this one,
4Πix/L divided by 2 e^(-4Πix/L) divided by 2
because exponential is cosine θ is e iθ
e to the -iθ over 2.
Now I'm going to write this as 1/√2 times e^(4Πix/L)
divided by square root of L e^(-4ΠIx/L).
Divided by square root of L.
I'm just manipulating the function that was given to me.
Why do I like this form?
Because this is of the form I wanted, A of p times ψ(p)
where ψ(p) I this and that's also ψ(p).
It's a normalized function associated with momentum p.
By comparing this you can see that A is equal to 1 over square
root of 2 for p = 4Πâ„ /L, and A's also equal to
1/√2 for p = -4Πâ„ /L.
How do I know what p is?
States of definite momentum look like e^(ipx/ℏ).
So whatever multiplies ix/ℏ is the momentum that you can see
it's 4Πâ„ .
You see what I'm saying?
If you want to put an ℏ down here and put it on the top and
make the comparison.
So these are the only two values for momentum.
So this is a state which is simultaneously in the state of
momentum 4Πâ„ /L, and -4Πâ„ /L.
It has no definite momentum yet.
If you measure it then you can get this answer with probability
1 over root 2 squared, which is 1 half,
or you can get this answer 1 over root 2 squared which is 1
half and no probability for anything else.
You cannot get any other value.
See, normally when you take a ψ(x),
a generic ψ(x) that I draw, it typically has some non-zero
value in all of space, so you can find it anywhere.
But this guy has only two values of p in the sum and only
those will be possible.
Student: ________ you're drawing ℏ over there
Prof: This guy here?
Student: Yeah.
Prof: Yes, I borrowed an ℏ,
put it on top and bottom so you can compare the expression.
I'm just saying compare, if you want without these ℏs,
compare this expression to that and you can see p is equal to
Now this problem was simple in the sense that you didn't have
to do that integral, but last time before end of
class I took the following function,
ψ(x) = e^(−α|x|).
On a ring the function looked like this, falling very rapidly
and dying within a width roughly Δx = 1/α.
Beyond that it is gone.
Then if you square that and normalize it in the circle you
find this is the correct normalization.
In other words, if you took this ψ
and you square and integrate it you'll get 1.
If you're very, very careful it is not strictly
1 because this function the x goes from −L/2 to L/2
whereas in the integral I took it to go from minus to plus
That's because this function is falling so rapidly.
The function is falling so rapidly you don't care if you
cut if off at L/2 or go to infinity,
so I did that to simplify the math,
but the idea is the same.
Then if you want to know what's the probability to get some
number p you've got to take A of p a square root of α
times e^(ipx/â„ ) square root of L times e^(−α|x|)
You understand that?
This is the ψ with the root α
in it.
That's my ψ.
That's my ψ_p*(x).
If you come to me with a new function tomorrow,
I don't know, maybe a function that looks
like this, a constant in some region and 0 beyond.
We'll put that function here and do these integrals.
You'll get a different set of A of p's.
So for every function ψ(x) that's a set of A_p's
that you find by doing the integral.
Student: So if you measure the momentum and you get
one of those answers so it becomes that answer?
Prof: Right.
So I'm coming to that point.
So the second thing is another postulate.
In the end I'll give you the list of postulates.
And the postulate says that if you measure the momentum you'll
get only those values for which the corresponding A_p
is not 0, and the ones for which it is
not 0 the probability for getting the number is the
absolute value squared of that A_p.
And right after the measurement the wave function will change
from this sum over all kinds of momenta to the one term
corresponding to the one answer you got.
You understand that?
So in this simple example there are only two values for
p you can get.
No other values are possible.
They're all theoretically allowed.
For example, 6Πâ„ /L is a perfectly
allowed momentum in that ring because it corresponds to a
periodic function, but that's not contained in
this particular wave function.
This wave function is built out of only two of them,
namely 4Πâ„ /L and -4Πâ„ /L.
Once you measure it suppose you got -4ℏ/L.
The whole wave function, which I had here in the simple
example, has got only two terms.
This part will simply disappear, and the function
after the measurement will be this.
The rough logic is that if I measure momentum,
and I got an answer, that's got to mean something in
any real sense that that is the momentum of the particle,
therefore it has to be true if I immediately remeasure it.
Immediately remeasure momentum, and I want to get exactly the
same answer, that means the function after
the first measurement should contain only one momentum in its
expansion, so it'll reduce to that one
Student: Is there a difference between ψ(x) and
negative ψ(x) because it seems like everything's just...
Prof: That is correct.
I think I also explained the other day that in quantum theory
ψ(x) and say 92i times ψ(x) are treated as
physically equivalent because they contain the same relative
Of course if ψ(x) had been properly normalized to 1 this
guy would not be normalized to 1.
It's as if you live in a world where all you care about is the
direction of the vector, but not the length of the
We don't live in such a world.
If you say, "Where is Stop & Shop,"
you say, "Well, go in that
direction," and you're not told how far,
well, you're missing some information.
Imagine a world where all you need to know is which way to go.
In fact, maybe if you go to some town and ask people where
is something they'll say, "Go in that
direction," that's useful information.
It's as if the direction is the only thing that matters,
not the length of the vector.
The analogous thing here is if you multiply the function by any
number you don't change the information.
They're all considered as the same state.
And I told you from this huge family of functions all
describing the same state I will pick that function,
that member whose square integral happens to be 1.
Student: So after the wave function breaks down how
long do you have to wait for the wave function to rematerialize,
or does it...
Prof: Oh, it's not break down.
It's another wave function because e^(ipx/ℏ),
which is ψ(p), this guy is as much a wave
function as any other ψ(x) you write down.
So these are not different creatures.
It's like saying the following.
I think the analogy may be helpful to you.
It may not be.
ψ is like some vector in three dimensions,
and you know there are these three unit vectors,
i, j and k, and you can write the
vector as some V_x times
i V_y times j
V_z times k where
V_y = V⋅j
Think of ψ as a vector and each of these
directions is corresponding to a possible momentum,
and you want to expand that vector in terms of these
This is ψ_p for p = p_1.
This is ψ_p for p = p_2 and so
So it's how much of the vector is in each direction that
determines the likelihood you'll get that answer,
that answer or that answer.
But right after the measurement, if you've caught it
in this direction, the entire vector,
only the component of the vector in the direction where
you got your answer remains.
The rest of it gets chopped out.
It's really like Polaroid glasses.
If you've got Polaroid glasses light can come in polarized this
way or polarized that way, but once it goes through the
glass it's polarized only in the one direction corresponding to
the way the polaroid works.
If it filters light with the e field going this way the light
on the other side will have only up and up.
So measurement is like a filtering process.
It filters out of the sum over many terms the one term which
corresponds to the one answer you got.
So I don't mind telling this to you any number of times,
but that is the way that quantum mechanics works.
Similarly, if you're chosen to measure position starting with
this ψ, and you found the guy here,
once you found it-- see, before you found it the
odds may vary like this.
Once you found it it's there, and if you remeasure it
infinitesimally later it'd better be there.
So the only function which has the property is the big spike at
wherever you found it.
That's called a collapsible of the wave function.
The real difference is the following.
I gave you an example of somebody tracking me to see
where I am.
I drew a probability graph, and the probability graph may
look like this.
And if you caught me here, if someone says,
"Where will I find him next?"
The answer is right there because the probability has
collapsed from being anywhere to where we just saw him.
The only difference between me and the quantum particle is if
you got me here I really was here.
You couldn't have gotten me anywhere else at that right now.
Can anybody find me anywhere else except right now here?
You cannot.
That's because my location's being constantly measured.
Photons are bouncing off to see where I am, so my measurement is
constantly measured.
A quantum particle with a function like this,
which is then found here, was not really here.
It was in this state of limbo which has no analog in which it
could have been anywhere.
It's the act of measuring it that nailed it to there,
but then right after that if you measure it we expect it to
be still there.
Now the time over which you can do that is
If you wait long enough the wave function will change;
I will tell you the laws for the dynamics that will tell you
how it will change, but we all believe that if a
measurement is immediately repeated for that same variable
we should get the same answer.
Need not be, it could have been even worse,
but at least that much is true.
So it's not going to get any more familiar.
It's a strange thing, but hopefully you will know
what the rules are.
When I said no one understands quantum mechanics what I meant
was, of course, by now you know the recipe.
It doesn't mean you like it, or it doesn't mean it looks
like anything in daily life.
Things in daily life they have a location before you measure
it, while you measure it and right after you measure it.
They're always in one place doing one thing.
Only in the quantum world they can be in many places at the
same time.
But you should be very careful if this is a wave function for
an electron.
The charge of the electron is not spread out.
I told you that.
An electron is one little guy you'll only catch in one place.
It's the wave function that's spread out.
Anyway, I think what'll happen is you will have to do a lot of
problems, and you will have to talk to a
lot of people, and you have to read a lot of
You can teach quantum mechanics for a whole semester,
sometimes I taught it for a whole year.
There's more and more stuff.
I want you at least to know how it works.
In real life when you go forward I don't think you need
all of this.
If you go into physics, or something,
or chemistry, they'll teach you quantum
mechanics again, but I wanted people doing
something else; they are the people I want to
send a message to.
Is here's a part of the world.
If you ever hear the word quantum, where does the word
quantization come from?
What's the funny business in the microscopic world?
I want you to have a felling for how that works.
And I claim that the mathematics you need is not very
You should know how to do integrals, and you should know
what e^(iθ) is.
So the postulates right now are the state has given me a
function ψ of x.
State of momentum, definite momentum looks like
this, and if you want the odds for
any particular momentum expand it in this fashion which is the
same as A of p ψ_p of x,
but A of p, again, is that integral.
I don't want to write it over.
Then if you measure p you'll get one of the answers
with the probability given by the square of the corresponding
number, and right after the measurement
the function collapses.
So it always collapses to whatever variable you measure.
Yes, another question?
Student: It looks like the opposite.
If you collapse it and you measure x then what
happens to p?
Prof: Very good.
His question was, "If you collapse it to
x what happens to p?"
I'm glad you asked.
I mean, that is the real problem.
You had the same question?
Student: No.
I was going to say well don't you not know about p?
Prof: That is correct.
So I will repeat what she said, and I'll repeat what you said.
I want to have this discussion with you guys because it's very
important because everyone's thinking the same thing.
If I first take a generic ψ, so I want everyone to
know what the answer's going to be to any of these questions.
So I take this ψ somebody prepared for me.
Let's not worry about how that person knew this is ψ,
you're given.
The electron is in this stage ψ.
If you measure x, and you got x equal to
this point the function, of course, becomes a big spike.
This spike in principle should be infinitesimally thin,
but I don't care.
Let this be the width of a proton.
If after that you say, "What momentum will I
get," well, you know what you have to do.
You've got to write spike function equal to sum over these
functions, right?
You've got to take the integral of the spike with this
exponential and do that integral,
and you'll get a bunch of numbers A of p for
all values of p.
And then if you measure momentum, and you got the one
corresponding to p = 6Πâ„ /L,
the state which contains many, many things will reduce to the
one term corresponding to p = 6Πâ„ /L.
If you plot that function it will be some oscillatory
The real part and imaginary part will both oscillate with
some wavelength given by p.
Be very careful.
This is not the absolute value of ψ.
It's the real or imaginary part.
They're both sines and cosines.
Absolute value will be flat.
So it'll go from a particle of known location to a particle
whose probability's completely flat on the circle.
You understand?
The wave function can look like 6Πi/L square root of
L right after the measurement.
Let's call is ψ_6.
The absolute value of ψ is a constant,
but the real and imaginary parts of ψ
oscillate with some wavelength.
So right after the measurement of momentum you don't know where
the guy is, and you say, "Let me find this
You catch it somewhere then it's a spike at that point,
but then you have no guarantee on the momentum.
So you can never produce for me something of perfectly well
defined position and momentum because once you squeeze it in
x it gets broad it in p.
Once you squeeze it in p it gets broad in x.
This is really mathematical property of Fourier analysis,
that functions which are very narrow in x when you do
the Fourier expansion have many, many wavelengths in them.
And likewise, a function with a very well
defined wavelength, because it's a complex
exponential, has a magnitude which is flat.
So now I'm going to ask the following question.
So when I did ψ(x) the probability for x was
very easy, squared the ψ.
When I said, "Okay, I want to look at
momentum," the answer was long and
complicated, namely, take these exponential
functions, write the ψ
in terms of those, find the coefficient, etcetera.
Now I can say I want some other variable I'm interested in.
I want to know what happens if I measure energy.
Energy is a very, very important variable.
It's very, very important because it turns out that if a
particle starts out in a state of definite energy,
I will show that to you later, it remains in that state.
That's the only state that will remain the way it is.
If you start in a state of definite momentum two seconds
later it can have a different momentum or it can be a mixture
of different momenta.
But if it starts in the state of definite energy it will
remain that way.
That's not obvious.
I'm going to prove that to you later.
That's why it's very important.
So most atoms are in a state of definite energy and they can
stay that way forever, but once in a while when
they're tickled by something they will either absorb light or
they will emit light.
So we draw a picture like this.
We will see that the allowed energies of the systems are some
special values.
Not every value's allowed.
And this can be called n=1, n=2,
n=3, etcetera.
And an atom, for example,
can sometimes jump from doing that to doing that,
and in that process it will emit an energy which is
E(n=3) − E(n=1).
That difference of energy will come in the form of a photon and
the energy of the photon is â„ ω,
or if you like, 2Πâ„ f where f
is what you and I call frequency.
And from the frequency you can find the wavelength.
The wavelength is just the velocity of light divided by
So an atom will have only certain allowed energies and
when it jumps form one allowed energy to another allowed energy
it will emit a photon whose frequency--
in fact, you should probably call this frequency
f_31 meaning what I get when I jump from the
level 3 to level 1.
Similarly, if you shine light on this atom it won't take any
It'll only take those frequencies that connect it from
one allowed energy to another allowed energy.
That's the fingerprint of the atom.
Both emission and absorption betray the atom.
That's how we know what atoms there are in this star,
or that star, or what the composition is.
No one's gone to any of these stars, but we know because of
the light they emit, and it's all controlled by
So the question I'm going to ask is here's the function
Someone gave it to me in some context, and I say if I measure
the energy of this particle what are the answers,
and what are the odds?
So how do you think that will play out?
You have to make a guess.
Suppose you're inventing quantum mechanics and someone
says, "What do you think is going to be the deal with
You know what the scenario might look like?
You can take a guess.
I mean, as I told you many times I don't expect you to
invent quantum mechanics on the fly, but you should be able to
What form do you think the answer will take?
You want to guess?
Student: ________ they'll be ________?
Prof: Okay, but if I want to know what
energies I can get and with what odds in analogy with momentum
what do you think will happen?
Student: Would you use the formula p^(2)/2m?
Prof: He said use p^(2)/2m.
That's a good answer.
His answer was we know that the energy is equal to 1 half m v
squared which I can write as p^(2)/2m,
So you're saying if I measure the momentum and I got a certain
answer, well, the energy's that
That's actually correct except the energy of a particle is not
always just the kinetic energy.
For a free particle this is the kinetic energy.
For a particle moving in a potential you know that you have
to add V(x).
That is the total energy.
Now that's when we have a problem.
In classical mechanics once I measure the x and
p of the particle I don't have to make another measurement
of energy.
Do you understand that?
I just plug the values I got into this formula.
For example, particle connected to a spring
V is ½kx^(2) where
k is the force constant of the spring,
and the kinetic energy's always p^(2)/2m.
Say if I measured p and I got some number,
and measured x I got some number, I can put that in
the formula and get the energy.
You don't have to do another energy measurement.
And you don't have to do an angular momentum measurement
In higher dimensions angular momentum is
r x p, and if you already measured the
position and you measured the momentum just take the cross
So in classical mechanics you only need to measure x
and p.
In quantum mechanics he made a pretty good guess that if you
measured p, the p^(2)/2m is the
That is true if the particle is not in a potential.
But if the particle is in a potential can you tell me how to
compute p^(2)/2m V(x)?
You realize you cannot really compute it because if you knew
the p exactly you have no idea where it is,
and if you knew the x exactly you don't know what the momentum
Maybe you know a little bit of both, but still,
how are you going to find the energy?
So the answer is you have to do a separate energy measurement.
You cannot infer that from x and p,
because first of all you cannot even get a pair of x and
p at a given time.
I hope I convinced you.
You measure this guy you screw up that guy.
Measure that one you mess up this one.
You can never get a state of well defined x and
p anyway.
So the way to find energy is to do a whole other calculation.
So I will tell you what the answer is, and hopefully you
will realize it's not completely different from the recipe we had
I'm going to give you a rule for the functions which
correspond to a state in which the particle has a definite
energy E.
Let's not worry about how you get it, some function.
You find all those functions, or you're given all those
functions, then can you imagine what will happen next?
If I give you all those functions what do you think the
rule is going to be?
Student: You separate it into a sum of all the
energies with coefficients.
Prof: Very good.
Let me repeat what she said.
I hope at least some of you were thinking about the same
Her answer is take the function ψ,
write it in terms of these functions ψ
E of x with some coefficient A_E
summing this over the allowed values of E,
whatever they may be.
And now that you said that what do you think
A_E is going to be in a given case?
Would you like to continue?
Student: It's going to be the integral or the wave
function ________________.
Prof: That's right.
That is correct.
The recipe's almost complete except you don't know what these
functions are, but if you knew these functions
you have to write the given function,
given wave function, as a sum of these functions
with some suitable coefficients.
Coefficients are found by the same rule,
and then the probability that you'll find an energy E
is, again,
A_E^(2). And everything else will be also
Once you measure energy you've got energy corresponding to
E_1, or E_2 or
Let's say you've got E_3,
the third possible value.
The whole wave function will collapse from being a sum over
many things to just this one guy, E_3.
The collapse is the same.
The probability rule is the same.
The only thing you don't know is who are these functions ψ
of E?
You understand?
So, again, the analogy's the following.
There is a vector that we call ψ.
Sometimes you want to write it in terms of
i, j, and k.
They are like the A(p).
Sometimes you may pick three other mutually perpendicular
vectors, i', j' and k'.
And if you know those coefficients you'll get the
probability for some other variable.
So you're expanding the same function over and over in many
possible ways depending on what variable is of interest to you.
If it's momentum you expand it in terms of exponential ipx of
If it's energy you expand it in terms of these functions.
So the question is what is the recipe going to be for these
functions ψ_E of x.
By what means do I find them?
Now you're getting more and more and more recipes every day,
but it's going to stop pretty soon.
This is about the last of the recipes.
Even this recipe I'll tell you how to get from a master recipe,
so it's not that many recipes, but I have to reveal that to
you one at a time.
Now you can say, "Okay, what function do
you want me to use for every energy, what is this function?
After all, when it was momentum you came right out and gave this
Why don't you do the same thing here," and that's a problem
The problem is the energy of a particle depends on what
potential it is in because it's got a kinetic and a potential
part, so I cannot give you a
universal answer for ψ_E(x).
I will have to first ask you; "Tell me the potential the
particle is in."
Once I know the potential I will give you the recipe.
Imagine the potential has been given to me, for example,
½kx^(2), or it could be a particle in
what's called a well.
You make a hole in the ground or you build a little barrier.
That's a possible potential.
You can have a potential that looks like harmonic isolator.
That's a possible potential.
You can have an electron and a hydrogen atom -1/r.
That's a possible potential.
There are many-potentials, and the answer's going to vary
on the problem.
There's no universal answer for the energy functions.
You tell me what the electron's doing, what field it is in,
what field of force it is in.
Then for each field of force, or for each potential,
there's a different answer.
And here is the master formula.
This is the great Schr�dinger equation.
So the answer looks like this.
The function ψ_E obeys the following equation,
-â„ ^(2)/2m times the second derivative of ψ
V(x) times ψ(x) is equal to
E times ψ(x).
Do not worry.
I will see you through this equation.
Everything you need to know I will tell you,
but you should not be afraid of what the equation says.
It says those functions that are allowed corresponding to
definite energy will have the property that if you took the
second derivative, multiplied it by this number,
add it to that V(x) times ψ_E you'll
get some function.
That function should be some number times the very same
If you can find me those functions then you will find out
when I will show you mathematically that there are
many solutions to the equation, but they don't occur for every
Only some energies are allowed, and the energies are usually
labeled by some integer n,
and for every n, 1,2, 3,4,
you'll get a bunch of energies, ψ_E1,
ψ _E2, ψ_E3,
3, and you write those functions down.
Then you can do everything thing I said.
But the only thing is here you have to do some hard work.
Whereas for momentum I gave you that and for a state of definite
positions x = x_0 I told you spike at
You didn't have to do much work.
For this you have to solve an equation before you an even
start, but we'll see how to do that.
So the first problem I want to solve is the problem where there
is no potential.
That is called a free particle.
A free particle is one for which V is 0.
And let me imagine it's living on this line,
the circle of length 2ΠR = L.
Oh, by the way, I should mention something
In terms of all the postulates you notice I never mentioned the
uncertainty principle today, ΔxΔp should
be bigger than ℏ.
I didn't mention it as a postulate because once you tell
me ψ is given by wave function and that states are
definite momentum have definite wavelength it follows from
mathematics that you cannot have a function of well defined
periodicity and wavelength also localized in space.
It's a mathematical consequence.
Similarly, who told me that I can expand every given ψ
as a sum over these functions?
There's a very general mathematical theorem that tells
you in what situations you can actually expand any given
function in terms of a set of functions,
namely are they like unit vector i,
j and k.?
Suppose I had only i and j and I don't have
I cannot expand ever vector in 3D using i and j.
So you've got to make sure you've got enough basis
functions and the theory tells you that if you find all the
solutions to that equation together they can expand any
Similarly the rule for expansion is also not arbitrary.
It all comes from that.
It's a very, very beautiful theory of the
mathematics behind quantum mechanics.
If you learn linear algebra one day, or if you've already
learned it, it's all linear algebra.
So every great discovery in physics is accompanied by some
mathematical stuff you need.
Like all of Newtonian mechanics requires calculus.
Without calculus you cannot do Newtonian mechanics.
Maxwell's theory for electromagnetism requires vector
Einstein's non-relativistic theory doesn't require anything;
it's algebra, but the general theory requires
what's called tensor calculus, and quantum mechanics requires
linear algebra.
And string theory we don't know what it requires.
People are still discovering new mathematics.
But it's very true that very often new mathematics is needed
to express the new laws of physics.
And if you don't know the laws you may find out you're not able
to write it down.
If you didn't know what a gradient was,
or if you did not know what a curl is,
and so on, then you cannot write down the laws of
electricity and magnetism.
So we have to solve this equation,
and I'm going to solve it for the easiest problem in the
world, a particle moving on a ring of
length L with no potential energy.
So what does that equation look like?
It says -â„ ;/2m d^(2)ψ/dx^(2) no
potential = E times ψ_E(x).
So let me rearrange the equation so it looks like this,
d^(2)ψ/dx^(2) k^(2)ψ
= 0 where k^(2) is defined to be 2mE/ℏ^(2)
= k^(2).
All I've done is just taken everything to one side and
multiplied everything by 2m/ℏ^(2) and called
that combination as k^(2).
So who is this number k?
Let's see.
The energy is ℏk^(2)/2m,
but we also know energy's p^(2)/2m.
So this number k will turn out to be just momentum
divided by ℏ.
Well, momentum has not entered the picture, but we will see.
Let's solve this equation now.
I say the solution to this equation is ψ(x) = to
any number A times e^(ikx) any number
B times e^(-ikx).
Let's see if that is true.
Take two derivatives of ψ.
What do you get?
You understand every time I take a derivative you pull down
an ik?
If you pull it twice you'll pull an ik squared which
is -k^(2)ψ, and the same thing will happen
to this term that it'll pull down a minus ik,
but if you do it twice you'll again get −k^(2).
So both of them will have the property that the second
derivative of ψ will be equal to
-k^(2)ψ, and that is the equation you
want to solve.
And A and B is whatever you like.
A and B are not fixed by the equation because
for any choice of A and any choice of B this'll
So let me write it as follows.
Ae to the i.
Let me write k, k was a shorthand for
square root of 2mE/ℏ^(2) x B to the -i
square root of 2mE/ℏ^(2) x.
I'm trying to show you that these are really functions of
definite energy ψ_E and here is
how the energy appears.
So what does it look like to you?
Have you seen these functions before?
What does it look like?
Student: One of the spring wave theories?
Prof: Pardon me?
Student: One of the spring wave theories?
Prof: The equation is like the spring equation.
That is absolutely correct, but what does this function
look like?
That function look like something to you?
Student: Isn't that a cosine function?
Prof: It's a cosine only if A is equal to B.
Forget the sines and cosines.
They are your old flames.
What's their new quantum flame?
What's the function that means a lot more in quantum mechanics
than sines and cosines?
I think I told you long back that e to the i
times dog x over ℏ is a state where the
momentum is equal to dog.
In other words, you can put anything you want
in the exponent.
If a function looks like that, that fellow there is the
So this is a state of momentum, of definite momentum.
In fact, look at this, e to the i square
root of 2mE/ℏ x B times e to the
minus that.
Do you understand?
This must be the momentum.
It is the momentum because e^(ipx/ℏ) is a state of
momentum p, but the momentum here can
either have one value, square root of 2mE,
or it can have another value minus square root of 2mE.
So what do you think the particle is doing in these
Student: It's jumping back and forth _______________
one of those states.
Prof: But in any one of them what is it doing here?
What is the sign of the momentum here?
Student: Oh, positive.
Prof: Positive, and here it's got negative
And how much momentum does it have?
The momentum it has, if you look at any of these
things, is that p^(2)/2m
= E is what is satisfied by the p that you have
In other words, what I'm telling you is in
quantum theory, in classical theory if I said,
"I've got a particle of energy E,
what is its momentum?"
You would say, "Well, E is
p^(2)/2m, therefore p is equal to
plus or minus square root of 2mE,"
because in one dimension when I give you the energy,
the Kinetic energy if you like, the particle has to have a
definite speed, but it can be to the left or it
can be to the right.
And the momentum is not arbitrary.
If the energy is E the momentum has to satisfy the
condition p^(2)/2m = E.
That's also exactly what's happening in the quantum theory.
The state of definite energy is a sum over two possible things,
one where the momentum is the positive value for root of
2mE; other is a negative value,
and these are the two allowed values even in classical
mechanics for a particle of definite energy.
But what's novel in quantum mechanics,
whereas in classical mechanics if it's got energy E it
can only be going clockwise or anti-clockwise,
but this fellow can be doing both because it's not in a state
of clockwise or anti-clockwise.
In fact, the probability for clockwise is proportional to
Probability for anti-clockwise is proportional to B^(2).
I've not normalized it, but the relative odds are
simply proportional to A^(2) and B^(2). So that's
what is bizarre about quantum mechanics that the particle has
indefinite sign of momentum.
Student: Do we have problem with m because
m is too small to ________ properly?
Prof: Which one?
Student: With the mass is there going to be some kind
of issue if that's small?
Prof: What about the m?
I'm sorry.
Student: m stands for the mass,
Prof: Yeah, m is the mass of the
particle, that's right.
Student: But is it too small in some way for our
equations to work because we had that issue in previous
Prof: When did we have the issue?
m is whatever the mass of the particle is.
It can be small.
It can be large.
I didn't understand.
Student: >
Prof: No, no, go ahead.
I want to know.
Student: >
Prof: There are no restrictions on the correctness
of this.
If the particle weighs a kilogram then you will find
that--well, we're coming to this.
So you find that these are the allowed values of p.
There are two values, but p itself is not
arbitrary, p itself is not continuous.
Maybe that's what you meant.
There's a restriction on the allowed values of p,
and therefore a restriction on the allowed values of E.
All I'm telling you now is that if you want to solve that
equation it is obviously made up of sines and cosines as you
recognize from the oscillator, or in the quantum world it's
more natural to write them in terms of exponentials,
e^(ikx) and e^(-ikx) where k is not
independent of e, k satisfies this
condition, and if you call ℏk as p,
p satisfies this condition.
This is just a classical relation between energy and
The other difference is that not every value of momentum is
Not every value of momentum is allowed for the same reason as
when I did particles of definite momentum.
In other words, if the particle is living on a
circle and the state of energy E it's given by
Ae^(ipx/â„ ) Be^(−ipx/â„ ) where
p is related to E by E = p^(2)/2m.
We have the requirement that when you go around a circle
you've got to come back to where you start.
And that condition, if you remember,
says that p times L over ℏ has to
be a multiple of some integer, or that the allowed values of
p are labeled by some index n,
which is 2Πâ„ /L times n.
I don't want to use m because m stands for
particle mass, n is the integer now.
In other words, when we studied the state of
definite momentum, namely the first function,
we realized even then that p is quantized.
Because of the single valued condition p is quantized.
And if energy functions are made up of such functions they
also have to be singlevalued.
That means the p here or the -p here both have to
satisfy the condition given by this.
Therefore, the allowed energies are also labeled by an integer
n and they are really p n squared over
2m where p n is 2Πâ„ n
over L .
You see that?
This is the quantization of energy.
So a particle in a ring has only these allowed values of
So in a way this problem is somewhat easy because it's a
free particle.
Once you've understood the particles in terms of allowed
momenta it turns out the allowed momentum states are also allowed
energy states.
The allowed momentum states, you remember,
they either look this or they look like this.
And I can pick A and B to be arbitrary,
so one choice is to pick A equal to
1/√L and just take e^(ipx/â„ ).
Other is to pick B equal to 1/√L and pick
e^(−ipx/â„ ), but you can also mix them up.
You don't have to mix them.
If you don't mix them up you have a particle here which has a
well-defined energy and a well-defined momentum.
This guy also has a well-defined energy and a
well-defined momentum.
This guy only has a well-defined energy,
but not a well-defined momentum because there's a two-fold
ambiguity in momentum.
You understand?
Even in classical mechanics it's true.
If I give you the momentum you can find the energy.
If I give you the energy you cannot find the momentum because
there are two square roots you can take because p^(2)/2m
is E, p is plus or minus
square root of 2mE.
It's the same uncertainty even in classical mechanics.
So what happens in quantum theory is if you pick any
particle of definite momentum on the ring it'll already have
definite energy which is simply that momentum squared over
You don't need to find a new function.
What is novel is that since the energy depends only on
p^(2) you can take a function with one value of
p, and you can take a function
with the opposite value of p and add them,
there will still be a state of definite energy because whether
it's doing this or whether it's doing that the energy will
always be p^(2)/2m and the minus signs drop out of
So what is novel here is what's called degeneracy.
Degeneracy is the name, but there's more than one
solution for a given value of the variable you're interested
You saw the energy looks like this.
Therefore, you'll find there's a state E_0
which is 0 squared over 2m.
Then there are two states E_1.
One has got momentum p going clockwise and one has
momentum going anti-clockwise.
So they look like (1/√L)
That's this guy.
Then you can have (1/√L)
e^(-2Πix/L) which is the other guy.
So at every-energy, allowed energy except 0,
there'll be two solutions.
There are two quantum states with the same energy.
When you studied hydrogen atom in high school maybe you
remember there are these shells with 2, and 4,
and 8, and 10 and so on.
They are called degeneracies where the energy is not enough
to tell you what it's doing.
There at every-energy it can have a different angular
Here at every-energy it can have two different momenta,
clockwise and anti-clockwise.
Student: Going back can you explain how you got the
________ of p L over ℏ ________?
Prof: Here?
Student: To the left.
Prof: Here?
You mean this one?
Student: I don't where that came from.
Prof: It came just like in the momentum problem.
If you've got a function like this you have a right to demand
that if you go a distance L around the circle you
come back to where you start.
So if you take any x here and add to it an L it should not
make a difference.
And what you're adding is pL/ℏ and that better be
a multiple of 2Π.
It's the same single valuedness condition.
So the momentum problem pretty much does this problem for you,
all the singlevalued stuff we dealt with before.
What is novel here is that demanding energy have one value
fixes the momentum to be one of two values,
and that double valuedness is the same as in classical theory
that a particle of energy E can have two possible
momenta plus or minus square root of 2mE,
and the quantum theory, then, is the state of E
is a sum of one value of p and the other value
of p with any coefficient you like.
So if this atom makes a jump from,
or this system makes a jump from somewhere there to
somewhere there you can find the frequency of the photons it will
emit because here are my allowed energies.
E_n is this.
Suppose it jumps from n=4 to n=3.
The energy that's liberated that goes to the photon is
â„ ω will be 4Π^(2) â„ ^(2) over
2mL squared times 4 squared - 3 squared.
I'm just using this formula with n=4,
and n=3, and take the difference.
It's 16 - 9 which 7.
You plug all that in you can solve for the ω,
or if you like frequency you can write it as 2Πâ„ f
and you can find the frequencies.
This is actually true.
If you've got a charged particle moving in a ring and
you want to excite it from one state to a higher state you will
have to give it only one of these frequencies so that the
frequency you give it, â„ ω,
must match the energy difference of the particle.
You understand?
So I've drawn these levels here.
If you want the electrons, say, in a metallic ring going
in the lowest possible energy state,
if you want to jump, if you want to crank it up to
the next level you've got to have photons at that energy,
or that energy.
Well, that happens to be the same as this,
but these are the only frequencies it'll absorb from
And when it cools down it'll emit back those frequencies,
and that's something you can test.
By the way, do you know why there's only one state at
E_0 and not two?
Student: Because 0 equals -0.
Prof: That's right.
The solution plus or minus 2m square root of
2mE has only one-answer when E is 0 because 0
momentum and -0 momentum are the same.
Otherwise, any finite positive momentum has a partner which is
minus that momentum.
There's a very interesting piece of work being done
experimentally at Yale which is the claim that if you took a
metallic ring in a magnetic field it will have a current
going one way or the other way, unbalanced current,
and it's not driven by a battery.
It's not driven by anything.
Normally if you took an ordinary ring the lowest energy
state will be a field of zero current while you go one way or
the other, but if you put it in the
magnetic field one can show it likes to go one way or the other
And now measurements are being done at Yale where you can
actually measure the tiny current due to one electron,
or one net electron going one way or the other.
So this L is either a mathematical convenience if
you're talking about free space and you want to be able to
normalize your wave functions, or it really is the
circumference of a real system.
Now that we can probe nano systems very well we can vary
the L and we can find out all the energy levels.
All right, so now I'm going to do the one problem which is
really a very standard pedagogical exercise,
and that's called a particle in a box.
I remember this example the first time I remember seeing
quantization, which is more interesting than
on a ring.
A box is the following.
If you dig a hole in the ground and you are standing somewhere
here you realize you're kind of trapped unless you can scale
this wall.
Now, you can call that as a ground level and think of it as
a hole in the ground, or you can think of this as the
ground level and that's the height of your barrier.
So imagine a particle living in a barrier that looks like this.
This is the potential energy.
This is like the height above the ground,
if you like, and it has a height
V_0 t here, and is 0 here,
and it goes from some −L/2 to L/2.
If the barrier V_0 goes to
infinity it's called a box.
So a barrier that goes to infinity, I'll just show you the
part that you can see here.
Here V is infinity, and here V is infinity
and inside V is 0.
That's a particle in a box.
So if the particle goes hits against the wall it cannot,
no matter how fast it's moving, go over the top because it's
infinitely high.
And that's not realistic.
Every barrier is finite, but just to teach you the
principles we always pick the simple example.
So now I want to solve this problem.
What are the allowed wave functions, ψ,
for a particle in this potential?
So let's go back to the same equation which says
-â„ ^(2)/2m d^(2)ψ/dx^(2)
= E - V ψ.
What I've done is I've taken V to the other side.
Now the potential is a constant in the three regions.
This is region I where the potential is 0.
This is region II where the potential is infinity.
This is region III where the potential is infinity.
I'm going to first look at the region here, region III.
What's the solution going to look like in region III is what
I'm asking.
So in this region the energy's something I don't know what the
allowed values are.
Whatever the value is V is an extremely large number.
Can you see that?
In this region don't let the barrier be infinity.
Imagine it's one zillion, a very high barrier.
Then what's the solution?
The solution will look like d^(2)ψ/dx^(2)
is = 2m/â„ ^(2) times V - E ψ.
Then the solution to that is very easy,
ψ = Ae^(κx) Be^(-κx) where
κ is equal to all of this, square root of
2m/ℏ^(2) times V - E.
I'm saying this is what I'm calling κ^(2).
Look, I'm asking you give me a function whose second derivative
is some number times the function.
Well, that's obviously exponential and it's the real
exponential because it's a positive number times the
Now A and B are free parameters,
whatever they are it'll solve the equation,
but we don't want a function that's growing exponentially
when you go to infinity because that means that particle would
rather be at infinity than in your box or near your box.
So for that mathematical, for that physical reason we
junk this function.
You pick it on physical grounds as not having a part growing
exponentially when you go to infinity,
but you do admit a part that's falling exponentially when you
go to infinity.
That's okay.
But how fast is it falling?
It's going like B e to the minus some blah,
blah, blah times square root of V - E.
That's all I want you to look at.
Forget all the ℏ's and m's.
Make V larger, and larger, and larger,
and just tell me what you think it will do.
If V is larger and larger it's like this
e^(-αx) I wrote for you where α's very large.
So this function will fall faster, and faster,
and faster, and in the limit in which V goes to infinity
it will vanish.
That basically means the particle cannot be found outside
the box.
So your function ψ is 0 here and 0 here.
Because if you made the barrier height finite you will find it's
falling exponentially on either side,
but the exponential becomes narrower and narrower as the
barrier becomes higher and higher,
and in the limit in which the wall is infinitely tall there is
nothing outside.
The wave function is non-0 only inside.
So what's the solution inside?
Let me call this 0.
Let me call this L.
Inside the box there is no potential,
so this equation is d^(2)ψ/dx^(2)
(2mE/â„ ^(2))ψ = 0.
It's like a free particle in the box, but it cannot leave the
This is what I called k, remember?
So now I'm going to purposely write the solution in terms of
trigonometric functions.
You'll see in a minute why.
So I'm going to write this Ae^(ikx)
You can see if I take any of these solutions it's going to
satisfy the equation, but k better be related to E in
this form, 2mE/ℏ^(2) is equal to
Sorry that's k^(2).
Because two derivatives of this will give me
−k^(2), and if you put that particular value for
−k^(2) here these two will cancel.
But I'm somehow going for--I'm sorry, I'm going to not write it
this way, but write it as C cosine kx
D sine kx.
Do you realize I can always go back and forth between
exponentials and trigonometric functions because one is a
linear combination of the other?
If you want, write this as cos i sine
and cos - i sine and rearrange the coefficients.
It'll look like something, something cosine.
I want to call that a C, and something,
something sine which I want to call this D,
D and C may be complex.
I'm not saying anything, but you can write a solution
either in terms of the sines and cosines or E to the plus
or minus something.
So here's the function ψ.
It looks like I have an answer for every energy I want,
because pick any energy you like, find the corresponding
k, you put it here and you're done.
But that's not allowed because we have an extra condition which
is the ψ was identically 0 here, ψ was identically 0
We're going to demand that at the two ends,
it can do whatever it wants in the middle,
it must vanish at the two ends for the continuity of ψ
because if ψ had two values you're getting
two different probabilities for the same point,
so that's not allowed.
So ψ must have--must match at the two ends.
But look at this function.
It's got to vanish at the left end at x = 0 and it's got to
vanish at the right end.
At x = 0 you can see ψ of 0 is simply C
because sine vanishes.
Cosine is 1 and C has to be 0 then because this guy has
no business being non-zero on the left end.
That's good.
This guy vanishes at the left end so I allow it,
but I have another condition.
It should also vanish at the right end.
If it should vanish at the right end I demand that sine
kL should be 0 because sine kL vanishing is
Then it'll vanish at both ends.
But sine kL = 0 means kL is a multiple of Π.
That means k is nΠ/L.
This means the allowed values of k in the problem are very
special so here is n = 1, ψ looks like
sin(Πx/L), and equal to 2 looks like
sin(2Πx/L), and so on with some numbers in
front which I have not chosen yet.
If you plot them they look like this.
That's one guy.
At a higher energy I've got that guy.
Then I've got that and so on because these are exactly like
waves on a string, a violin string clamped at two
In fact, this wave equation's identical to the wave equation
on a string.
The only requirement is that the string is clamped at two
Here the ψ is clamped at the two ends
because it's got to vanish on either side outside the box.
So the allowed wavelengths are the same except here the
wavelength is connected to the momentum that's connected to
energy and E, you remember,
is ℏ^(2)k^(2)/2m
then becomes ℏ^(2)/2m times
k will be n^(2)Π
So this particle in a box can have only these particular
So let me write it for you nicely.
The allowed energies are â„ ^(2) Π^(2)/2
mL squared times an integer n squared,
and the corresponding wave functions look like this.
They're waves in which you've got half an oscillation,
or 2 half oscillations and 3 half oscillations,
but you've got to start and finish at 0.
So it's the quantization.
This is why when Schr�dinger came up with this equation
everybody embraced it right away because you suddenly understood
why energy's quantized.
You're trying to fit some number of waves into an interval
and only some multiple of half wavelengths are allowed,
but wavelength translates into momentum.
That translates into energy.
Suddenly you understand the quantization of energy.
So it's got 1 state n = 1,
and only one state n = 2,
one state n = 3.
I'm going to come back to this next time, but you should think
about this.