Calculating eigenstates of an operator

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  • #1
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Homework Statement


Consider a two-dimensional space spanned by two orthonormal state vectors [itex] \mid \alpha \rangle [/itex] and [itex] \mid \beta \rangle [/itex]. An operator is expressed in terms of these vectors as
[tex] A = \mid \alpha \rangle \langle \alpha \mid + \lambda \mid \beta \rangle \langle \alpha \mid + \lambda^* \mid \alpha \rangle \langle \beta \mid + \mu \mid \beta \rangle \langle \beta \mid [/tex]

Determine the eigenstates of A for the case where (i) [itex] \lambda = 1, \mu = \pm 1 [/itex], (ii) [itex] \lambda = i, \mu = \pm 1 [/itex]. Do this problem also by expressing A as a 2 X 2 matrix with eigenstates as the column vectors.

Homework Equations


Just linear algebra rules.

The Attempt at a Solution


I started with [itex] \lambda = 1, \mu = 1 [/itex]. Then A is:
[tex] A = \mid \alpha \rangle \langle \alpha \mid + \mid \beta \rangle \langle \alpha \mid + \mid \alpha \rangle \langle \beta \mid + \mid \beta \rangle \langle \beta \mid [/tex]

[tex] A \mid \alpha \rangle = \mid \alpha \rangle \langle \alpha \mid \alpha \rangle + \mid \beta \rangle \langle \alpha \mid \alpha \rangle + \mid \alpha \rangle \langle \beta \mid \alpha \rangle + \mid \beta \rangle \langle \beta \mid \alpha \rangle = \mid \alpha \rangle + \mid \beta \rangle [/tex]

[tex] A \mid \beta \rangle = \mid \alpha \rangle \langle \alpha \mid \beta \rangle + \mid \beta \rangle \langle \alpha \mid \beta \rangle + \mid \alpha \rangle \langle \beta \mid \beta \rangle + \mid \beta \rangle \langle \beta \mid \beta \rangle = \mid \alpha \rangle + \mid \beta \rangle [/tex]

The eigenstate is [itex] \mid a_n \rangle [/itex] with eigenvalue [itex] a_n [/itex]. Then the following holds:
[tex] A \mid a_n \rangle = a_n \mid a_n \rangle [/tex]

The eigenstate [itex] \mid a_n \rangle [/itex] can be expressed in the basis vectors [itex] \mid \alpha \rangle [/itex] and [itex] \mid \beta \rangle [/itex]:

[tex] \mid a_n \rangle = c_1 \mid \alpha \rangle + c_2 \mid \beta \rangle [/tex]

Then the earlier equation becomes:
[tex] A \mid a_n \rangle = A( c_1 \mid \alpha \rangle + c_2 \mid \beta \rangle ) = a_n (c_1 \mid \alpha \rangle + c_2 \mid \beta \rangle [/tex]

But this is also:
[tex] A( c_1 \mid \alpha \rangle + c_2 \mid \beta \rangle ) = c_1 A \mid \alpha \rangle + c_2 A \mid \beta \rangle = c_1 (\mid \alpha \rangle + \mid \beta \rangle) + c_2 (\mid \alpha \rangle + \mid \beta \rangle) \\ = (c_1 + c_2) \mid \alpha \rangle + (c_1 + c_2) \mid \beta \rangle [/tex]

This gives the equations :
[tex] a_n c_1 = c_1 + c_2 [/tex]
[tex] a_n c_2 = c_1 + c_2 [/tex]

The only solution is if [itex] c_1 = c_2 = 0 [/itex]. Obviously I am doing something wrong but I can't see it.
 

Answers and Replies

  • #2
TSny
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This gives the equations :
[tex] a_n c_1 = c_1 + c_2 [/tex]
[tex] a_n c_2 = c_1 + c_2 [/tex]
The only solution is if [itex] c_1 = c_2 = 0 [/itex]. Obviously I am doing something wrong but I can't see it.
Are you sure that's the only solution?
 
  • #3
59
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I see, but I still get one non trivial solution out of it.
If one is 0, then both are 0. So they are either both nonzero or both 0. If they are nonzero, I can write:
[tex] a_n = \frac{c_1 + c_2}{c_1} = \frac{c_1 + c_2}{c_2} [/tex]
If [itex] c_1 + c_2 =0 [/itex] then [itex] a_n =0 [/itex] or the c's are 0. If [itex] c_1 + c_2 [/itex] is nonzero, then I can divide by it and gives me [itex] c_1 = c_2 [/itex] with [itex] a_n = 2 [/itex] Unnormalized I can take as eigenstate [itex] \mid \alpha \rangle + \mid \beta \rangle [/itex].
Now I can't find the other eigenstate. [itex] \mid \alpha \rangle - \mid \beta \rangle [/itex] is orthogonal to this eigenstate but letting A operate on it gives me 0. Or does this just mean that it is an eigenstate with eigenvalue 0?
 
  • #4
TSny
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Or does this just mean that it is an eigenstate with eigenvalue 0?
That's it.

Another approach is to subtract the two equations:
##a_nc_1 = c_1+c_2##
##a_nc_2 = c_1+c_2##
 
  • #5
59
2
Thanks again
 

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