Linear Operator: Eigenfunctions and Boundary Conditions for Energy Calculation

[lokofer]

let be the linear operator: (Hermitian ??)

$$L = -i(x\frac{d}{dx}+1/2)$$

then the "eigenfunctions" are $$y_{n} (x)=Ax^{i\lambda _{n} -1/2$$

then my question is how would we get the energies imposing boundary conditions? (for example y(0)=Y(L)=0 wher L is a positive integer )...

 

[HallsofIvy]

Energies? That is certainly not a Hamiltonian. In any case, since it is a first order operator, you can't satisfy two boundary conditions, only one.

 

[tacman]

To calculate the energies for a given set of boundary conditions, we can use the eigenfunctions of the linear operator L. These eigenfunctions are solutions to the equation L[y(x)] = \lambda y(x), where \lambda is a constant known as the eigenvalue. In this case, the eigenfunctions are y_n(x) = A x^{i\lambda_n - 1/2}.

To impose the boundary conditions, we can plug in the values of x=0 and x=L into the eigenfunctions. This will give us two equations: y_n(0) = 0 and y_n(L) = 0. Solving these equations for \lambda_n will give us the possible eigenvalues and corresponding energies.

For example, if we have the boundary condition y(0) = y(L) = 0, this means that both ends of the system are fixed at zero. Plugging in x=0 and x=L into the eigenfunction y_n(x), we get A(0)^{i\lambda_n - 1/2} = 0 and A(L)^{i\lambda_n - 1/2} = 0. This leads to the equations 0 = 1 and 0 = e^{i\lambda_n L}. The first equation gives us no solution, while the second equation gives us \lambda_n L = 2\pi n, where n is a positive integer. Solving for \lambda_n, we get \lambda_n = \frac{2\pi n}{L}.

Substituting this back into the eigenfunction, we get y_n(x) = A x^{i\frac{2\pi n}{L} - 1/2}. This gives us a set of possible eigenfunctions and corresponding energies, which we can then use to calculate the total energy of the system.

In summary, by imposing the boundary conditions on the eigenfunctions of the linear operator L, we can determine the possible eigenvalues and energies for the system. This allows us to accurately calculate the total energy of the system and understand its behavior.