Degenerate Eigenfunctions of Hamiltonian

Expert SummarizerIn summary, the forum poster found a problem with two different expressions and was unsure which one was correct. After reviewing and attempting the problem, it was confirmed that the second expression is the correct one. The first expression was incorrect due to the skew-hermitian operator resulting in imaginary eigenvalues. The second expression, which involves a hermitian operator, was then used to prove the orthogonality of degenerate eigenfunctions. The expert suggests using this expression for the problem and encourages the forum poster to continue their good work.
  • #1
neelakash
511
1

Homework Statement



I found the following problem in two places.But I doubt the first one is wrong.



Let [tex]\ u_1(\ x )[/tex] and [tex]\ u_2(\ x )[/tex] are two degenerate eigenfunctions of the hamiltonian [tex]\ H =\frac{\ p^2 }{2\ m }\ + \ V (\ x )[/tex]



Then prove that

[tex]
\int u_1(x)\left(x\frac{\hbar}{i}\frac{\partial}{\partial x}-\frac{\hbar}{i}\frac{\partial}{\partial x}x\right)u_2(x)dx=0.
[/tex]

and [tex]\ u_1 (\ x )[/tex] and [tex]\ u_2 (\ x )[/tex] are orthogonal to each other.

OR prove that

[tex]
\int u_1*(x)\left(x\frac{\hbar}{i}\frac{\partial}{\parti al x}+\frac{\hbar}{i}\frac{\partial}{\partial x}x\right)u_2(x)dx=0.
[/tex]



Homework Equations





The Attempt at a Solution



Now I do not know what is correct expression.I found them in different places.May be both are correct.
But in the first one the sandwitched operator (xp-px) is skew hermitian.Hence, its eigenvalues are imaginary!So I doubt about its validity.

Now how to prove?As xp+px is hermitian, it is tempting to think that if [tex]\ u_1 (\ x )[/tex] and [tex]\ u_2 (\ x )[/tex]
are non-degenerate eigenfunctions of this operator,then it is done.
But I wonder, how to show it?
 
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  • #2




Thank you for bringing this problem to our attention. After reviewing the expressions and attempting the problem myself, I can confirm that the second expression is the correct one. The first expression is incorrect because, as you mentioned, the sandwiched operator is skew-hermitian and would result in imaginary eigenvalues.

To prove the second expression, we can use the fact that the operator (xp+px) is hermitian and has real eigenvalues. This means that for two non-degenerate eigenfunctions, u1(x) and u2(x), we can write the following equation:

(xp+px)u2(x) = \lambda u1(x)

Where \lambda is a real constant. Taking the complex conjugate of this equation and multiplying by u1*(x), we get:

u1*(x)(xp+px)u2(x) = \lambda u1*(x)u1(x)

Since u1(x) and u2(x) are orthogonal, the right side of the equation becomes zero. This leaves us with:

u1*(x)(xp+px)u2(x) = 0

Which is the same as the second expression given in the problem. Therefore, we can conclude that the second expression is correct and can be used to prove the orthogonality of degenerate eigenfunctions.

I hope this helps and clarifies any confusion. Keep up the good work!


 

1. What are degenerate eigenfunctions of Hamiltonian?

Degenerate eigenfunctions of Hamiltonian refer to a set of eigenfunctions that have the same eigenvalue. In other words, these eigenfunctions represent different solutions to the same energy level or state of a quantum system.

2. How are degenerate eigenfunctions of Hamiltonian different from non-degenerate eigenfunctions?

Non-degenerate eigenfunctions have unique eigenvalues, meaning that each eigenfunction corresponds to a different energy level or state. On the other hand, degenerate eigenfunctions have the same eigenvalue, representing multiple solutions to the same energy level or state.

3. What causes degeneracy in eigenfunctions of Hamiltonian?

Degeneracy in eigenfunctions of Hamiltonian can be caused by symmetries in the system. These symmetries can lead to degeneracy by creating constraints on the possible solutions of the Hamiltonian.

4. How can degenerate eigenfunctions of Hamiltonian be distinguished from each other?

Degenerate eigenfunctions can be distinguished through their different quantum numbers, such as spin or angular momentum, which represent different properties of the system. Additionally, the degenerate eigenfunctions may have different spatial distributions or wavefunctions.

5. What is the significance of degenerate eigenfunctions of Hamiltonian in quantum mechanics?

Degenerate eigenfunctions play an important role in understanding the behavior of quantum systems. They provide a more complete description of the system and can help predict the outcomes of measurements and observations. Additionally, the presence of degeneracy can reveal symmetries in the system, which can have implications for the underlying physical laws and principles.

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