00:01
Okay, in this problem we have an initial quantum state that looks like this and a hamiltonian operator that looks like this so the first thing we need to do is we need to find the eigenvalues and eigenvectors of the hamiltonian so you get the eigenvect values we look at this determinant and set it equal there it is and this of course has to be zero in order for this to work and so my possible energy eigenvalues are three and plus or minus five.
00:44
So there's three of them.
00:47
The eigenvectors, for e equals three, there it is, and three equals five is vector v2, that's a zero up there.
01:09
And for minus five, and all three of these eigenvectors are mutually orthogonal.
01:19
So my initial vector size zero, i can write it in terms of v1, v2, and v3 like this, just by, really just by inspection, looking at what my vector looks like, looking at my eigenvectors.
01:59
And then for each one, i can look at the coefficients out in front to figure out the probabilities.
02:06
So all we've got to do is take this coefficient and square it.
02:10
So 0 .6 squared is 0 .36 .8 over root 2 squared is 0 .32.
02:16
And this one is also 0 .32.
02:20
The minus sign, of course, when i square it, becomes a plus.
02:23
And if i add these 0 .36 plus 0 .32 equals 1 .0.
02:30
So that's like the total probability, which is 1, which is what we kind of expect.
02:39
In part b, we want to look at what happens at a later time.
02:48
So, si of t is point, so i'll write out the answer first and i'll tell you how i got it.
02:57
So there it is.
03:01
And here's why.
03:03
Because we know our time dependent solution is going to be the sum of the v sub i's with the probabilities that go with them times e to the minus ie i, e, i, t.
03:18
So there's a probability factor in here as well.
03:23
It's the same as up here at 0 .6 .8 over root 2.
03:27
That's how we get this answer for si of t.
03:39
In part c now, we want to calculate the expectation value of the hamiltonian...