00:01
In this problem you're given the hamiltonian operator in the 1, 2 basis.
00:07
1, 2 basis is an orthonormal basis.
00:09
That means the inner product of 1 with itself is 1 and 1 with 2 is 0.
00:17
That's all we need to know at this point to do the problem.
00:22
I'll give you a representation of the 1, 2 basis later, matrix representation, but we don't need it.
00:29
As long as we know it's orthonormal, good enough for our purposes now.
00:32
So let's first find h in matrix form in that basis.
00:44
And to do that we just have to calculate these quantities.
00:52
That's for the 1, 1 matrix element, 1, 2 matrix element, 2, 1 matrix element, and the 2, 2 matrix element.
01:06
So we just need to calculate those four quantities and we would have h in matrix form in that basis.
01:15
I'll show you one of them.
01:18
It's just a matter of acting on the left with the bra and acting on the right with the ket in this formula.
01:29
There's nothing different about one over the other.
01:33
So let me do 1, h, 2.
01:37
So we're going to be doing the 1, 2 matrix elements.
01:40
So a, 1, 1, 1, 2, minus 1, 2, 2, 2, plus 1, 1, 2, 2, plus 1, 2, 1, 2, and that's everything.
02:06
Now i told you those basic vectors are orthonormal.
02:14
So 1, 1 is 1.
02:17
1, 2 though, 0.
02:19
2, 2 is 1.
02:21
And if 1, 2 is 0, even though you don't have it here in this particular form, 2, 1 would be 0.
02:29
Because remember 2, 1 and 1, 2 are just complex conjugates of each other.
02:34
So this term is gone.
02:36
This term is gone.
02:37
This term will last because both are 1.
02:40
This term is gone.
02:41
So we're just left with a for a matrix element, a 1, 2 matrix element.
02:49
And like i said, you do the same thing for the others.
02:50
There's nothing fancier than that.
02:52
In the end, you get that the matrix for h, a, a, a, minus a.
03:01
That's the matrix form of h.
03:04
Now we can get our eigenvalues.
03:11
And for that, we look at the determinant of h minus lambda i.
03:19
And notice i use square brackets for my matrix, vertical lines for my determinant.
03:24
I don't use the parentheses.
03:26
I have parentheses in other places.
03:28
I like to square.
03:31
So a minus lambda, a, a, minus a, minus lambda is equal to 0.
03:38
That's how we get our eigenvalues.
03:41
So now it's just a matter of expanding that determinant.
03:44
A minus lambda, minus a, minus lambda, minus a squared.
03:49
Remember when you do this product, you get a minus sign.
03:55
Ok, so that's what we have.
03:57
And we can expand this out.
04:00
So we get minus a squared, minus lambda a, plus lambda a, plus lambda squared, minus a squared is equal to 0.
04:12
Well, these two terms are gone, bringing the a squared, minus a squared to the right.
04:17
This gives me that lambda squared is equal to 2 a squared.
04:22
So lambda, the eigenvalue, or eigenvalues in this case, square root of 2, plus or minus, square root of 2, a.
04:35
So there's our two eigenvalues.
04:38
Plus square root of 2 a, minus square root of 2 a.
04:41
So we have that.
04:44
Next up is the eigenvectors, corresponding to each of those eigenvalues.
04:51
So let's do lambda equals plus square root of 2 a.
04:55
Now remember the eigenvalue equation.
04:59
You would have h times, and the notation i'm going to use for the plus one is plus, i'll just identify the eigenvalue in the cat, plus square root of 2 a, is equal to lambda, well, which is actually going to be, well, let me actually put it in, plus square root of 2 a, plus square root of 2 a.
05:30
This is our lambda.
05:33
Now, if you were to bring, if you were to bring this right hand side over, you end up getting a zero on the left.
05:43
I mean, you get a zero on the right.
05:45
And that's what we're going to do.
05:46
We're going to rewrite the eigenvalue equation with everything on the left hand side.
05:51
So that's going to be, in matrix form, it's going to be h minus lambda i, times the eigenvector.
06:06
A minus square root of 2 a, a, a, minus a, minus square root of 2 a, times the eigenvector, which we're looking for, and the elements will be u and v.
06:19
You can use whatever you want, or just symbols to solve for.
06:24
And that will give you zero, zero for the column vector on the right.
06:30
So that, like i said, that's just bringing everything over the side, to the right, left hand side, h minus, obviously to get this form, if you're going to be doing it in matrix form, then this, then, then this becomes lambda i, and h, we have the matrix form of h.
06:49
That's how we get this.
06:51
Alright, now, this is going to, this really gives me, as you can see, it gives me a column vector on the left also.
07:01
It's got to, it's got to be the same.
07:03
But each element of that column vector is equal to zero.
07:06
That's where i get my equations to solve to get my u and v.
07:11
Or do we? well, we're going to talk about that in a minute.
07:16
So let's, let's do the, cross the top, and then down.
07:21
That gives me my first equation.
07:23
Now you can leave it in matrix form here, and have your, you know, your column vector equal to the zero column vector.
07:31
Normally it's not done.
07:32
Normally you just write the equations at this point.
07:35
But whatever you're comfortable with.
07:37
So, but i'm just going to write out the equations.
07:39
So i'm going cross and down.
07:42
So a minus square root of 2 a u plus a v is equal to zero.
07:50
Factoring out the a, this becomes 1 minus square root of 2 u plus v is equal to zero.
07:56
That's equation one.
07:58
Now for going across the bottom and down the column vector, i get here a u minus a plus square root of 2 a v is equal to zero.
08:11
Factoring out the a, i get u minus 1 plus square root of 2 v is equal to zero.
08:19
That's equation two.
08:20
You might say, oh, wonderful.
08:22
Two equations, two unknowns.
08:23
That's, that's, that's great.
08:25
Well, it's not.
08:27
Well, it is great, but it's not two equations and two unknowns.
08:30
You got two unknowns, yes, but you don't have two equations.
08:34
If you were to multiply equation two by 1 minus square root of 2 and do some algebra, you find that these are the same equation.
08:43
It's just a matter of scaling.
08:46
So what does that mean? what does that mean? i'm free to choose.
08:50
Now this physically is very important.
08:54
You have to be able to choose to be able to normalize your eigenvector.
09:02
If you had two equations, two unknowns, everything's set.
09:05
It's done.
09:05
It says that u has to be this and v must be that.
09:10
And whatever, whatever your norm is, is your norm...