Calculating eigenstates of an operator

In summary, the conversation discusses determining the eigenstates of an operator expressed in terms of two orthonormal state vectors. The equations and attempts at finding the eigenstates are shown, with the final solution being that one eigenstate is \mid \alpha \rangle + \mid \beta \rangle and the other is \mid \alpha \rangle - \mid \beta \rangle with an eigenvalue of 0.
  • #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.
 
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  • #2
barefeet said:
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
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
barefeet said:
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
Thanks again
 

What is an eigenstate?

An eigenstate is a state in which a physical system can exist with a definite value for a particular observable quantity, such as energy or momentum. It is a state in which the system's wave function is a solution to the Schrödinger equation for a given operator.

What is an operator?

An operator is a mathematical object that acts on a function, transforming it into another function. In quantum mechanics, operators represent physical observables such as position, momentum, and energy.

How do you calculate eigenstates of an operator?

To calculate eigenstates of an operator, you need to solve the Schrödinger equation for the given operator. This involves finding the eigenvalues and eigenvectors of the operator, which represent the possible values and corresponding states of the observable quantity.

What is the significance of eigenstates in quantum mechanics?

Eigenstates are important in quantum mechanics because they represent the fundamental states of a system, which can be used to describe the behavior of the system and make predictions about its future evolution. They also play a key role in measuring physical observables and determining the probabilities of different outcomes.

Can eigenstates of different operators be simultaneously measured?

Yes, eigenstates of different operators can be simultaneously measured if they commute, meaning their corresponding operators can be multiplied in any order without changing the result. This is known as the principle of uncertainty, and it allows for the simultaneous measurement of compatible observables.

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