PROFESSOR: We ask, is psi of x, 0 real?
And I told you the answer is no.
And how would I know that this is not real?
Well, we can take the complex conjugate.
And at the end of the day, this will boil down
to some property of phi of k.
You see, you have an expression phi of x in terms of phi of k.
So it would not be surprising that the requirement
that psi is real means something about phi of k.
So let's just say, suppose psi is given by that,
then psi x, 0 star, the complex conjugate
would be 1 over square root of 2 pi integral phi of k star
e to the minus ikx dk.
I conjugated everything in that equation for psi of x and 0.
Now, you want to compare this with psi of x and 0
to see if it's real.
Or let's consider what is the condition that this be real.
So I want to simplify here a little more.
So what I'm going to do is going to change variables,
by changing k to minus k.
If you prefer to go a little more slowly,
you could say you're going to change k prime--
you're going to be a new k, called k prime,
equals to minus k.
But it's possible to do it this way.
Now, there's going to be a couple of changes.
Wherever you see k, you're now going to see minus k--
so 1 over 2 pi integral phi of minus k.
And I'll just put this star here, not so many parentheses--
e to the minus ikx becomes ikx.
And the dk will go to minus dk, but the order of integration,
that was from minus infinity to plus infinity, would switch.
So those two signs cancel.
So there's a sign from doing dk to minus dk,
and 1 from the limit of integration--
so at the end of the day, you have dk,
and you still have this--
minus infinity to infinity.
And you can say, well, is this equal to--
or what is the condition of for psi to be real?
Well, is this equal to 1 over 2 pi minus phi of k, e
to the ikx dk--
is that-- question mark-- is that equal to it?
That would mean that the psi of x, 0
is real, because this thing is just psi of x and 0.
So this is a question mark-- this is a condition.
So here you could say, exploring the reality condition--
condition-- when is a psi of x real?
So what must be true is that these two terms
must equal each other.
So, in fact, this requires--
reality requires that 1 over square root of 2 pi integral
from minus infinity to infinity phi of k
minus phi star of minus k, e to the ikx dx is equal to 0.
I brought the two terms to one side.
Both are of the same type--
they're integrated against an e to the ikx.
And therefore, we can combine them,
and that's what must be true in order for the function
to be real.
And now you can say, so what is it?
What's the answer?
Well, this integral should vanish.
Now, this integral should vanish--
it should vanish for all values of k.
So actually, what you want to conclude
is that this thing is identically 0.
AUDIENCE: Excuse me, shouldn't that be dk and not dx?
PROFESSOR: Yes, thank you.
Thanks very much.
So this property that this whole integral be equal to 0, you
were tempted to conclude that it means
that this thing is equal to 0.
And that is correct--
that is a perfectly legal argument.
And it basically-- if you want to express it more precisely,
you could base it on the Fourier theorem, again.
These two sets of equalities here are Fourier's theorem.
And look what this is saying--
this is saying that this quantity has a 0 Fourier
transform.
Because how do you do the Fourier transform
of a function of k?
You multiply by e to the ikx and integrate.
And therefore, this function has a 0 Fourier transform.
So, but if a function has a 0 Fourier transform,
the function must be 0.
Because already this is 0, and the integral is 0--
0.
So this is absolutely rigorous.
And therefore, you get the conclusion
that phi of minus k star must be equal to phi of k,
and that's the condition for reality.
So if a phi of k satisfies this property, that psi of x
will be real, and our phi of k doesn't satisfy this property,
what do you see in this property?
Basically, if you have phi that exists for some value of k,
it should also exist for the value of minus k.
And in fact, should be the complex conjugate
of the other value.
But here, you have some phis of k, and no phis at minus k.
So the phi that we wrote above doesn't satisfy this condition.
And therefore psi is not real, and it all makes sense.
OK, so basically, if you were plotting
not the absolute value, but the real part
and the imaginary parts of psi, you
would see some sort of funny waves.
I think if you were plotting the real part, for example,
you would see a wave like that.
And if you were plotting the imaginary part,
you'd presumably see some other wave like that.
And the absolute value, it's much nicer and simpler.