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Eigenvalues, eigenvectors, and time evolution

Dear Mitra,

I in fact wrote the wrong matrix but I am still confused after PART A.

(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>

(PART B)

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
0
0

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.

(PART C)

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

Solution Preview

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:

<x|A|x>

Now back to your problems:

You found that:

|x> = (-2/3)|v1> + (2/3)|v2> +(1/3)|v3>

You must indeed normalize the ...

Solution Summary

A detailed solution is given.

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