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neelakash

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## Homework Statement

Given u(x) is a solution of Schrodinger's equation: [tex]\ - \frac{\hbar^2}{2m}\frac{\partial^2\ u(x)}{\partial\ x^2 }\ + \ V(x) \ u(x) =\ E \ u(x) [/tex]

(i)Under what condition, u(-x) will also be a solution?

(ii) If u1(x) and u2(x) be two degenerate wave functions, prove that [tex]\int\ u1(x) \ (xp-px) \ u2(x) \ dx =0[/tex]. u1(x) and u2(x) are orthogonal to each other.

(iii) Conservation of tota probability requires that Hamiltonian is hermitian operator.

(iv) u(x) must be non-degenerate and hence, real; apart from a overall phase factor.

## Homework Equations

## The Attempt at a Solution

I am posting my solution below in a few minutes.Just check it and tell me if I am correct.

(i)Replace all x by -x in the Schrodinger's equation:

[tex]\ - \frac{\hbar^2}{2m}\frac{\partial^2\ u(-x)}{\partial\ x^2 }\ + \ V(-x) \ u(-x) =\ E \ u(-x) [/tex]...(1)

Now, If u(-x) is a solution,put u(-x) into Schrodinger's equation:

[tex]\ - \frac{\hbar^2}{2m}\frac{\partial^2\ u(-x)}{\partial\ x^2 }\ + \ V(x) \ u(-x) =\ E \ u(-x) [/tex]....(2)

Comparing (1) and (2) we get the condition V(x)=V(-x).That is the potential must be an even function of x.

***Please check if I am correct.

(ii)Since hamiltonian is a hermitian operator, the degenerate solutions (=> different eigenvalues) must be orthogonal.Hence, u1 and u2 are orthogonal:

[tex]\int\ u1(x) \ast\ u2(x) \ dx =0[/tex]

This is OK.But what about the given one?

[tex]\int\ u1(x) \ (xp-px) \ u2(x) \ dx =0[/tex]

Since (xp-px)=iħ, hence the integral reduces to:[tex]\int\ u1(x) \ u2(x) \ dx =0[/tex]

If u1 and u2 are real, this is same as the orthogonality condition.Can threre be any other method to prove this?

(iii)We have Hψ=Eψ and Hψ*=Eψ* by virtue of the fact that H is real, Hermitian and eigenvalues of hermitian operator are real.

Then,[tex]\int\psi\ast\ H \psi\ dx=(\psi,\ H \psi)=(\ H \psi, \psi)=\int\ H \psi\ast\psi\ dx[/tex]

Now using the eigenvalue equations, we see that the probability is the same in both sides of the equation.And this turns out to be the consequence of the Hermiticity of H

***Please check if the argument is valid.

(iv) Non-degeneracy means to have two different states corresponding to the same value of energy.So, Let Hψ1=Eψ1 ----(3) and Hψ2=Eψ2---(4)

Writing the expression of the Hamiltonian operator,then multiplying (3) by ψ2 and (4) by ψ1 and subtracting one from another, we ultimately get:

[tex]\frac{\partial^2\psi_1}{\partial\ x^2 }\psi_2=\frac{\partial^2\psi_2}{\partial\ x^2 }\psi_1[/tex]

I think that something is to be done from here...

So still trying for it.

Please tell me if I am going along the correct way.

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