I in fact wrote the wrong matrix but I am still confused after PART A.
The actual matrix is:
H = 1 2 0
2 0 2
0 2 -1
Where the eigenvalues are E1 =0, E2=3hw, E3=-3hw and as you said the trace(H) =0 = sum of eigenvalues.
I also found the eigenvectors using Hx = Ex and they were:
for E1 =0 : |v1> = <-1, 1/2, 1>
for E2 =3: |v2> = <2, 2, 1>
for E3 =-3: |v3> = <1/2, -1, 1>
After this, I am confused. Do I need to normalize those eigenvectors?
if so then I get the normalized vectors |v1>, |v2>,|v3> to be:
|v1> = <-2/3 , 1/3, 2/3>
|v2> = <2/3 , 2/3, 1/3>
|v3> = <1/3 , -2/3, 2/3>
Now I write the initial state as a linear combination of the normalized eigenvectors:
|x> = |v1> + |v2> +|v3>
We are given that at t=0, our initial state is 1
So as a linear combination:
<1,0,0> = (-2/3)<-2/3 , 1/3, 2/3> + (2/3)<2/3 , 2/3, 1/3> + (1/3)<1/3 , -2/3, 2/3>
=(-2/3)|v1> + (2/3)|v2> +(1/3)|v3>
After this I am totally lost. The question asks me to find the state vector at time t. Can you explicitly do it so I can see. Thanks.
A = 1 0 0
0 2 0
0 0 0
We need to find its expectation at time t and also compute the probability that measuring A at time t gives 1.
My main problem is understanding how the eigenvectors and eigenvalues of the Hamiltonian matrix relate to the different probabilities and states.© BrainMass Inc. brainmass.com July 20, 2018, 8:58 am ad1c9bdddf
To be compatible with the Dirac notation you should write vectors like e.g.
|-2/3 , 1/3, 2/3>
The dual vector is written the other way around:
<-2/3 , 1/3, 2/3|
You can think of vectors as column vectors and dual vectors as row vectors. An inner product is then just a matrix product of the two. The notation <A> means the average of the operator A. The average of A in some state |x> is given by:
Now back to your problems:
You found that:
|x> = (-2/3)|v1> + (2/3)|v2> +(1/3)|v3>
You must indeed normalize the ...
A detailed solution is given.