Trying to find the eigenvectors of a hamiltonian operator

  • #1

Homework Statement



I am given the Hamiltonia operator of a system in two-dimensional Hilbert space:

H = i[tex]\Delta[/tex](|w1><w2| + |w2><w1|) and am asked to find the corresponding eigenstates.


I wrote this operator as a matrix, where H11 = 0, H22 = 0, and H12= i[tex]\Delta[/tex] and H21= -i[tex]\Delta[/tex]

The eigenvalues are then, I think, the solutions to

(0-E)2 - [tex]\Delta[/tex]2= 0

so E = [tex]\Delta[/tex], -[tex]\Delta[/tex]

But when I construct eigenstates from these E values I am getting

(i , 1) and ( 1 , -i) (except with normalization factors) which seem to be orthornormal and satisfy the two eigenvalues respectively but aren't linearly independent.

Why is this? Is the matrix I'm using for the Hamiltonian correct? And if not why? I thought I can find the matrix values by <wm|H|wn>, where <wm|wn> = [tex]\delta[/tex]m,n




Edit: In addition-

What I'm ultimately trying to do is write the eigenfunctions |w1> and |w2> in terms of the eigenstates |E1> and |E2>. I have missed some class and I'm actually really lost

The homework problem states that |w1> and |w2> are eigenstates of the observable operator[tex]\Omega[/tex]. Is [tex]\Omega[/tex] just any general observable operator, so then what is the significance of its eigenstates? What exactly are |w1> and |w2> as opposed to |E1> and |E2>? Why is the Hamiltonian, an observable, written in terms of the eigenstates of [tex]\Omega[/tex]? Right now I'm trying to do this homework before the deadline purely through a shoddy understanding of linear algebra, but it would really help if I understood what is going on...
 
Last edited:

Answers and Replies

  • #2
diazona
Homework Helper
2,175
8
H = i[tex]\Delta[/tex](|w1><w2| + |w2><w1|) and am asked to find the corresponding eigenstates.

I wrote this operator as a matrix, where H11 = 0, H22 = 0, and H12= i[tex]\Delta[/tex] and H21= -i[tex]\Delta[/tex]
Where did the negative sign come from?
(i , 1) and ( 1 , -i) (except with normalization factors) which seem to be orthornormal and satisfy the two eigenvalues respectively but aren't linearly independent.
Those are not the correct eigenvectors for the matrix you wrote
Why is this? Is the matrix I'm using for the Hamiltonian correct? And if not why? I thought I can find the matrix values by <wm|H|wn>, where <wm|wn> = [tex]\delta[/tex]m,n
That is true, but you seem to have made some mistake in applying that formula.
What I'm ultimately trying to do is write the eigenfunctions |w1> and |w2> in terms of the eigenstates |E1> and |E2>. I have missed some class and I'm actually really lost

The homework problem states that |w1> and |w2> are eigenstates of the observable operator[tex]\Omega[/tex]. Is [tex]\Omega[/tex] just any general observable operator, so then what is the significance of its eigenstates? What exactly are |w1> and |w2> as opposed to |E1> and |E2>? Why is the Hamiltonian, an observable, written in terms of the eigenstates of [tex]\Omega[/tex]? Right now I'm trying to do this homework before the deadline purely through a shoddy understanding of linear algebra, but it would really help if I understood what is going on...
[itex]\vert w_1\rangle[/itex] and [itex]\vert w_2\rangle[/itex] are clearly not the eigenstates of the Hamiltonian. Eigenstates have the property that when you compute the matrix elements of an operator in the basis of that same operator's eigenstates, the matrix will be diagonal (i.e. it has nonzero elements only on the main diagonal). That is clearly not the case with the Hamiltonian in the w basis.

The E basis, consisting of [itex]\vert E_1\rangle[/itex] and [itex]\vert E_2\rangle[/itex], is the set of eigenstates of the Hamiltonian. So what you get when you find the eigenvectors is the set of coefficients you need to write the w states as linear combinations of the E states, or vice-versa. These are the eigenvectors you'll find:
[tex]\begin{pmatrix}\langle w_1 \vert E_1 \rangle \\ \langle w_2 \vert E_1 \rangle\end{pmatrix}, \begin{pmatrix}\langle w_1 \vert E_2 \rangle \\ \langle w_2 \vert E_2 \rangle\end{pmatrix}[/tex]
and then you can write
[tex]\vert w_1 \rangle = \vert E_1\rangle\langle E_1 \vert w_1 \rangle + \vert E_2\rangle\langle E_2 \vert w_1 \rangle[/tex]
just taking the coefficients from the eigenvectors you found.

As for the significance of [itex]\Omega[/itex], presumably you have some physical system in which the particle exists in an eigenstate of [itex]\Omega[/itex] rather than an eigenstate of the Hamiltonian. Since [itex]\Omega[/itex] and [itex]H[/itex] are non-commuting operators, you wouldn't be able to measure the eigenvalues of [itex]H[/itex] (the energies) directly (just like position and momentum). I can't think of a realistic example that matches your problem off the top of my head, but I'm sure you could look one up online somewhere (although that's probably best left for after you actually do the problem :wink:).
 
  • #3
Thank you Diazona for your reply,

Firstly, sorry- I wrote the problem wrong, it should actually be
H = i[tex]\Delta[/tex](|w1><w2| - |w2><w1|)

in which case, is my matrix correct?

I still can't manage to get linearly independent eigenvectors...


Also your explanation made quite a lot of sense to me, thank you immensely. Can you give me an example of two commuting operators where I COULD directly get the value of one from the other?
 
Last edited:
  • #4
diazona
Homework Helper
2,175
8
Firstly, sorry- I wrote the problem wrong, it should actually be
H = i[tex]\Delta[/tex](|w1><w2| - |w2><w1|)

in which case, is my matrix correct?
Yep, in that case it looks good.
I still can't manage to get linearly independent eigenvectors...
The correct eigenvectors are linearly independent. Without seeing the details of your work, I can't tell you any more.
Also your explanation made quite a lot of sense to me, thank you immensely. Can you give me an example of two commuting operators where I COULD directly get the value of one from the other?
What I had in mind there was something like angular momentum and the Hamiltonian for the hydrogen atom. Those two operators commute, so they have a set of common eigenstates: an electron can exist in a quantum state which is an eigenstate of both the Hamiltonian and the z-component of angular momentum at the same time.

By the way, nothing in the problem as given states that Ω does not commute with H. They might be commuting operators. I was just trying to come up with a situation in which this sort of calculation might be justified, and the non-commutativity of the operators was the first thing that sprang to mind, so I just put that assumption in myself.

To be perfectly honest, all this is probably more of a mathematical exercise than anything else. Real systems tend to be more complicated.
 
  • #5
Okay- I guess I was being stupid and trying to make them also orthogonal to each other. I assumed that the Hamiltonian had to be Hermitian- but the matrix I obtained isn't symmetric, (by the way, why not, I thought the Hamiltonian counts as an observable and has to be Hermitian?)

So I got |E1> = 1/(sqrt(2))( 1|w1> + i|w2>)
and |E2> = 1/(sqrt2))( i|w1> + 1|w2>)
 
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,228
1,830
Your Hamiltionian is hermitian. There's a factor of i in it, so it won't be symmetric.
 
  • #7
diazona
Homework Helper
2,175
8
Okay- I guess I was being stupid and trying to make them also orthogonal to each other. I assumed that the Hamiltonian had to be Hermitian- but the matrix I obtained isn't symmetric, (by the way, why not, I thought the Hamiltonian counts as an observable and has to be Hermitian?)

So I got |E1> = 1/(sqrt(2))( 1|w1> + i|w2>)
and |E2> = 1/(sqrt2))( i|w1> + 1|w2>)
Yep, those look right. And actually, they are orthogonal, with respect to the inner product
[tex]\langle a \vert b \rangle = a^\dagger b = (a^*)^T b[/tex]
which is the one we generally use for state vectors in quantum mechanics. I think you may be forgetting to take the complex conjugate when you transpose one of the eigenvectors.
 

Related Threads on Trying to find the eigenvectors of a hamiltonian operator

  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
18
Views
1K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
4
Views
1K
Replies
5
Views
3K
Replies
9
Views
2K
Replies
2
Views
651
  • Last Post
Replies
10
Views
5K
Replies
12
Views
1K
Top