[Blubbs]

Hello,

I am really struggeling with the calculations of point B, C and D of the attached pdf.

I understand the theory behind the variation principle and I can solve Question A and maybe B.

But from here on I have no idea how to even start calculating the expectation value.
 



If someone can help me with this assignment, that would be great - thanks alot


My first post, so sorry if there are formating problems and I translated the assignment into english, since it's original language was german - I hope there are no translation errors.

[Corribus]

Well let's start with B. Do you know how to normalize a wavefunction? If you are given a wavefunction Ψ, then you can insert a factor n so that the normalized function is nΨ. The condition for normalization is that the integral of the square of the wavefunction (assuming the wavefunction is real; if not, the product of the wave function and its complex conjugate) over "all space" is equal to 1. At that, it's a simple integration problem.

The variational principle basically says that if you don't know the exact wavefunction, but you have a good guess, you can calculate energy eigenvalues that are close to but slightly higher than the real energy eigenvalues.

For C, we would use the variational principle equation to determine the approximate energy eigenvalue E using the normalized test wavefunction you determined in B.

I.e., E(Ψ) = <Ψ> = <Ψ*|H|Ψ>. I'm assuming you understand bra-ket formalism.

D is just asking you to compare your answer in C to the energy eigenvalue for another test wavefunction, which they give you. Given that the variational principle says that the energy eigvenalue you determine with the test wavefunction is always greater than or equal to the true energy eigenvalue (i.e., the eigenvalue of the true wavefunction), you should be able to determine which test wavefunction is "better", even without knowing the true energy eigenvalue. This is because the approximate eigenvalues can only deviate from the true eigenvalue in one direction (bigger value).

[Blubbs]

Thank you for your timely response!

So with your advice I tried calculating B and I got  \sqrt(2) as the factor n, which feels right so I think I calculated that one right.

D also makes sense to me on a theory level, but what I am still missing is what you assumed I understand, so I think this is where my confusion lies.

How do I calculate <Ψ*|H|Ψ> on a very plane level?
Is it Integral of the wave-functions multiplied with the operator?
And if I do the integral what are my integral border?
Do I have to calculate 2 different values for my E(Ψ) because if you look at the assignment, there are different potential values for V = 0 or 1

I hope I framed my question understandably - if anything is not clear, feel free to ask again.

Thanks for the help.

What I also don't see, is how

[Corribus]

I had a longer response here but I removed it because I wanted to think about part of it a bit more. I'll follow up in a little while. In the meantime, however:

Your answer to B is not quite correct. My guess is that you are not using the correct integration limits. The limits in x are not from x = 0 to x = 1 but from x = 0 to x= L. (The bottom axis in the diagram is in terms of x/L, not in terms of x. Needlessly confusing but there it is. Back in the day, unitless axes were the rule.)

Bra-ket notation is shorthand for (in one dimension):

[tex]<\psi|\hat{H}|\psi>=\int \psi^* \hat{H} \psi dx[/tex]

The integration is performed over all space.

[Corribus]

Ok, here's the rest of my reply to your questions:

Do I have to calculate 2 different values for my E(Ψ) because if you look at the assignment, there are different potential values for V = 0 or 1

At this point it's just a math problem. You are essentially evaluating

[tex]\int_0^L \psi^*\hat{H}\psi dx[/tex]
As you've noted, the Hamiltonian includes a goofy discontinuous function. Nevertheless, the problem is basically set up the same as the case where the potential term V(x) was continuous over the limits x = 0 to x = L. The only difference is that here you have to integrate over a discontinuity. Integrating over a discontinuous function just requires breaking the integral up into smaller integrals, one each for the three continuous regions, each with integration limits at the points of discontinuity.

E.g. consider a discontinuous step function with point of discontinuity at x = b:

[tex] \begin{equation*}
f(x) = \left\{
        \begin{array}{ll}
            g(x) & \quad a < x < b \\
            h(x) & \quad b < x < c
        \end{array}
    \right.
\end{equation*}
[/tex]
Then the integral over the range x = a to x = c is

[tex]\int_a^c f(x) dx = \int_a^b g(x) dx + \int_b^c h(x) dx [/tex]
This works for continuous functions as well.
Anyway, hope that helps. Feel free to hit me again if you are still having trouble.