How to Find Ψ(x,t) for a Given Hamiltonian Matrix and Initial State?

AI Thread Summary
To find Ψ(x,t) for the given Hamiltonian matrix, the eigenvalues have been calculated as 2+√6 and 2-√6, but there are difficulties in obtaining usable eigenvectors. The eigenvector calculations yield complex results, and normalization poses additional challenges. The discussion highlights the importance of expressing the initial state Ψ(0) as a linear combination of the eigenvectors, using coefficients derived from the scalar product of the eigenvectors and the initial state. Clarification on the normalization of eigenvectors and the correct method for determining coefficients is emphasized. Ultimately, understanding time evolution and the correct formulation of Ψ(x,t) is crucial for solving the problem.
Mlisjak
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Homework Statement



I have the matrix form of the Hamiltonian:

H = ( 1 2-i
2+i 3)

If in the t=0, system is in the state a = (1 0)T, what is Ψ(x,t)?

Homework Equations



Eigenvalue equation

The Attempt at a Solution



So, I have diagonalized given matrix and got the eigenvalues: 2+√6 and 2-√6. I am suspecting that these are not good, since I can't get eigenvectors I can use. When trying to calculate eigenvectors, I get:
a = 1 and b=(-1+√6)/(2+i). This is the one I got when I used 2+√6 but after that I didn't even try with the other eigenvalue since it will be similar.
I don't know what to do with those and don't know how to normalize them. Also, even if I knew how to get correct eigenvectors, I am not sure how to proceed and get Ψ(x,t). [/B]
 
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I don't see how you got that value for b (I don't get the same signs), so check your math. If what bothers you is the (2+i) in the denominator, simply multiply by (2-i)/(2-i).
 
Do you get the same eigenvalues? I put it in the form:
( 1 2-i
2+i 3 ) * (a b)t = (2+√6) (a b)T

I'm sorry, I don't know to write it properly.

Then I got:
a+(2-i)b = (2+√6)a
That's where the expression comes from. And by using the second equation i got a=a in which cases we always put 1 in our class. Is that wrong?
 
Mlisjak said:
a+(2-i)b = (2+√6)a
That's where the expression comes from. And by using the second equation i got a=a in which cases we always put 1 in our class.
That gives
$$
b = \frac{1 + \sqrt{6}}{2 - i}
$$
which is not what you wrote above.

Try multiplying with ##(2 + i)/(2 + i)##.
 
Ok, I got the vectors but don't know how to proceed to get Ψ (x,t)
 
What have you learned about time evolution?
 
To be precise, I don't understand how to get right coefficients to write Ψ(x,0) as a linear combination of the vectors I got. When i get that, I believe that I just have to add time dependence e-iEt/†
 
Scalar product. You have found the eigenvectors φ1 and φ2 and want to write Ψ(t=0) = c1 φ1 + c2 φ2, you find the coefficients using
$$
c_n = \phi_n^\dagger \Psi
$$
 
Is that equal to finding the norm of the eigenvectors? For example, I have three eigenvectors: v1 = v2 = (1 0 0) and v3=1/√2 (0 -i 1).

Would I write Ψo= 2*(1 0 0)+1/√2 (0 -i 1)?
 
  • #10
I have used different example, since I get too complicated vectors in the exercise I posted originally
 
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