# Schrodinger's equation problem

1. Feb 22, 2008

### neelakash

1. The problem statement, all variables and given/known data

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

(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 $$\int\ u1(x) \ (xp-px) \ u2(x) \ dx =0$$. 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.

2. Relevant equations
3. 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:

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

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

$$\ - \frac{\hbar^2}{2m}\frac{\partial^2\ u(-x)}{\partial\ x^2 }\ + \ V(x) \ u(-x) =\ E \ u(-x)$$.............(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:

$$\int\ u1(x) \ast\ u2(x) \ dx =0$$

This is OK.But what about the given one?

$$\int\ u1(x) \ (xp-px) \ u2(x) \ dx =0$$

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

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,$$\int\psi\ast\ H \psi\ dx=(\psi,\ H \psi)=(\ H \psi, \psi)=\int\ H \psi\ast\psi\ dx$$

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:

$$\frac{\partial^2\psi_1}{\partial\ x^2 }\psi_2=\frac{\partial^2\psi_2}{\partial\ x^2 }\psi_1$$

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.

Last edited: Feb 22, 2008
2. Feb 23, 2008

### Cinimod

For the last part, just divide both sides of the equation by $\psi_1 \psi_2$ and then you will end up with two separated differential equations. All you then need to do is show that they have the same general solution, and that there is just a phase factor between them. The other stuff looks right, but I'm no expert, so wait for someone else to confirm that its ok.

Last edited: Feb 23, 2008
3. Feb 23, 2008

### neelakash

OK........I got the way for the last part.It is a simple seperation of variables.But I did not get it.

I am also awaiting confirmation/improvement suggestion for the other parts.

Thank you very much.

4. Feb 23, 2008

### Cinimod

$$\frac{\partial^2 \psi_1}{\partial x^2} \psi_2 = \frac{\partial^2 \psi_2}{\partial x^2} \psi_1$$

$$\frac{1}{\psi_1}\frac{\partial^2 \psi_1}{\partial x^2} = \frac{1}{\psi_2}\frac{\partial^2 \psi_2}{\partial x^2}$$

Which can be split up into two equations...
$$\frac{1}{\psi_1}\frac{\partial^2 \psi_1}{\partial x^2} = \lambda$$

$$\frac{1}{\psi_2}\frac{\partial^2 \psi_2}{\partial x^2} = \lambda$$

Which clearly will have the same general solution. The phase difference comes about from the constant(s) of integration.

Last edited: Feb 23, 2008
5. Feb 23, 2008

### neelakash

6. Feb 23, 2008

### cepheid

Staff Emeritus
You've reversed the definitions of degenerate and non-degenerate. If spectrum of an operator is non-degenerate, it means that every distinct eigenstate has a distinct eigenvalue. If the spectrum is degenerate, it means that some distinct eigenstates share eigenvalues. As a result of this reversal, your starting points for ii and iv need revision.

7. Feb 23, 2008

### neelakash

(ii)Yes...I made a very bad mistake in saying that "degenerate solutions(=>different eigenvalues)"

Rather Hu1=Eu1 and Hu2=Eu2.For a state u,we must have from the definition of Hamiltonian:

$$\ - \frac{\ p^2 }{2m}\ u(x)=\ (E-V) \ u(x)$$

=>$$\frac{\partial^2\ u(x)}{\partial\ x^2 }\ + \frac{2\ m}{\hbar^2}\ (E-V) \ u(x)$$

Depending upon E>Vor E<V, we get two degenerate solutions u1 and u2 for u(x).For E>V,we have harmonic solution and for E<V,we have exponential solution.

The trick is that they have given you (xp-px) sandwitched between u1 and u2.Do the integrations dividing those two parts.In each case their will be an x involved.The integral overall will vanish.And you have to do for both E>V and E<V,otherwise the proof is not complete.

After playing with the problem for sometimes,I reach this conclusion.Hopefully I am correct.Let me check it.

(iii) It looks to me the problem is halfdone.I cannot be sure.Can anyone please check it?

(iv)Going to do.

Last edited: Feb 24, 2008
8. Feb 24, 2008

### neelakash

(ii)Hmmm.I just checked for harmonic solution.What I told in the earlier post is correct.

The solutions 1 and u2 are also orthogonal.Because, for exponential real solution,u1* is same as u1 so the analysis remains valid.And for harmonic solution, the constants of the solution changes.But the x dependence remains essentially the same.

Please give feedback for other solutions like this so that I may identify my mistakes and rectify myself.

Last edited: Feb 24, 2008
9. Feb 24, 2008

### neelakash

(iv)I approached like (ii) and concluded that two types of solutions are possible:harmonic solutions for E>V and real exponential solution for E<V.

For E>V: $$\psi=\ A \ e^{\ ikx}\ +\ B \ e^{\ - \ ikx}$$.We may write them as sinusoidal components.They do not vanish at infinity.So, we must discard this solution.

For E<V: $$\psi=\ A \ e^{\ kx}\ +\ B \ e^{\ - \ kx}$$.To statisfy the condition that the wave function must vanish at infinity,we must set either A or B equal to zero.Thus the degeneracy is gone.

The solutions are then degenrate,real apart from the constant A.

This solves the problem.But I wonder if this would do.Because,since we do not know the form of the potential function,a wide range of functions are already excluded from our considereation.

Last edited: Feb 24, 2008
10. Feb 24, 2008

### neelakash

(i) It looks that u(x) and u(-x) will both be the solutions of the Schrodinger's equation if they are degenerate solutions.That is both have the same eigenvalue E.

[The mistake in the previous attempt was that the Hamiltonian got changed with the replacement of x by -x]

11. Feb 24, 2008

### neelakash

(i) and this happens when u(x) and u(-x) are linearly independent.That is c1 u(x)+c2u(-x)=0 requires that c1=c2=0