Schrodinger's equation problem

In summary, we discussed the solutions to Schrodinger's equation and the conditions under which u(x) and u(-x) are solutions. We also proved that two degenerate wave functions are orthogonal and that the Hamiltonian must be a Hermitian operator for conservation of total probability. Finally, we showed that u(x) must be non-degenerate and real, apart from an overall phase factor.
  • #1

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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  • #2
For the last part, just divide both sides of the equation by [itex] \psi_1 \psi_2 [/itex] 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.
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  • #3
OK...I got the way for the last part.It is a simple separation of variables.But I did not get it.

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

Thank you very much.
  • #4
[tex] \frac{\partial^2 \psi_1}{\partial x^2} \psi_2 = \frac{\partial^2 \psi_2}{\partial x^2} \psi_1 [/tex]

[tex] \frac{1}{\psi_1}\frac{\partial^2 \psi_1}{\partial x^2} = \frac{1}{\psi_2}\frac{\partial^2 \psi_2}{\partial x^2} [/tex]

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

[tex] \frac{1}{\psi_2}\frac{\partial^2 \psi_2}{\partial x^2} = \lambda [/tex]

Which clearly will have the same general solution. The phase difference comes about from the constant(s) of integration.
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  • #5
what about the other solutions?
  • #6
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.

neelakash said:
ian operator, the degenerate solutions (=> different eigenvalues) must be orthogonal.Hence, u1 and u2 are orthogonal:

neelakash said:
(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)
  • #7
(i)What about this one?Is my conclusion correct?

(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:

[tex]\ - \frac{\ p^2 }{2m}\ u(x)=\ (E-V) \ u(x) [/tex]

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

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.
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  • #8
(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.
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  • #9
(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: [tex]\psi=\ A \ e^{\ ikx}\ +\ B \ e^{\ - \ ikx}[/tex].We may write them as sinusoidal components.They do not vanish at infinity.So, we must discard this solution.

For E<V: [tex]\psi=\ A \ e^{\ kx}\ +\ B \ e^{\ - \ kx}[/tex].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.
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  • #10
(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
(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

1. What is Schrodinger's equation problem?

Schrodinger's equation problem refers to the mathematical equation developed by Austrian physicist Erwin Schrodinger in 1926. It is a fundamental equation in quantum mechanics that describes the evolution of a quantum system over time.

2. What is the significance of Schrodinger's equation problem?

Schrodinger's equation problem is significant because it provides a mathematical framework for understanding the behavior of quantum systems, which are the building blocks of all matter. It has been instrumental in the development of numerous technologies, including transistors, lasers, and computer chips.

3. How does Schrodinger's equation problem differ from other mathematical equations?

Schrodinger's equation problem is unique in that it is a partial differential equation, which means it describes how the system changes with respect to both time and space. It also introduces the concept of wave-particle duality, where particles can exhibit both wave-like and particle-like behavior.

4. What are some real-world applications of Schrodinger's equation problem?

Schrodinger's equation problem has been used in various fields, such as quantum mechanics, chemistry, and materials science, to predict the behavior of atoms, molecules, and materials. It has also been applied in the development of technologies, such as transistors, quantum computers, and medical imaging devices.

5. Is Schrodinger's equation problem still relevant today?

Yes, Schrodinger's equation problem is still a crucial equation in modern physics and continues to be used in research and technology development. It has also been expanded upon and modified for more complex systems, such as multi-particle systems and relativistic systems.

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