Finding the change in the eigenvalue without knowing the change in the

[JBrandonS]

Hello,

I am currently teaching myself quantum mechanics using MIT's OCW and am suck on the following problem from the second problem set of the 2005 7.43 class.

Homework Statement

Consider an operator O that depends on a parameter λ and consider the λ-dependent eigenvalue equation:
$$O(λ)| \psi(λ) > = o(λ) | \psi(λ) >$$

Show that one can compute $\frac{do(λ)}{dλ}$ without knowing $\frac{d|\psi(λ)>}{dλ}$. Thus one can determine the change in the eigenvalue without knowing the change in the eigenstate.

Under what conditions would O(λ) commute with O(λ')

Homework Equations

The Attempt at a Solution

I really have no idea where to begin on this one.

 

[Dick]

Hello,

I am currently teaching myself quantum mechanics using MIT's OCW and am suck on the following problem from the second problem set of the 2005 7.43 class.

Homework Statement

Consider an operator O that depends on a parameter λ and consider the λ-dependent eigenvalue equation:
$$O(λ)| \psi(λ) > = o(λ) | \psi(λ) >$$

Show that one can compute $\frac{do(λ)}{dλ}$ without knowing $\frac{d|\psi(λ)>}{dλ}$. Thus one can determine the change in the eigenvalue without knowing the change in the eigenstate.

Under what conditions would O(λ) commute with O(λ')

Homework Equations

The Attempt at a Solution

I really have no idea where to begin on this one.

The first question looks like the Hellmann-Feynman theorem. Multiply by $<\psi(λ)|$ on the left and take the derivative and I think you need O to be Hermitian. Really not sure about the second half.

 

[JBrandonS]

The first question looks like the Hellmann-Feynman theorem. Multiply by $<\psi(λ)|$ on the left and take the derivative and I think you need O to be Hermitian. Really not sure about the second half.

Thank you, like most of these problems it became really easy once I got a push. Took about 30 seconds to show.

I also believe the only case where O(λ) would commute with O(λ') would be for degenerate eigenvalues, but maybe someone else can chime in and let me know if that is right or wrong.