How Do You Apply Orthonormality and Completeness in Quantum Finite Square Wells?

[v_pino]

Homework Statement

1. Mixed Spectrum
The finite square well has a mixed spectrum or a mixed set of basis functions. The set of
eigenfunctions that corresponds to the bound states are discrete (call this set {ψ_i(x)}) and
the set that corresponds to the scattering states are continuous (call this set {ψ_k(x)}). Thus
the complete set of basis functions are {ψ_i(x), ψ_k(x)}. Write down for this set of basis
functions

a) the orthonormal condition;
b) the completeness condition;
c) the expansion for an arbitrary wavefunction ψ(x) in terms of the basis functions; and
d) the expressions for the expansion coefficients in part c).

Homework Equations

An orthonormal basis may be formed from a linear combination of basis vectors.

We can write vectors as expansions of orthonormal basis so that

|a>=\sum_{n}^{i=1}a_i|i>

<a|b>=\sum_{n}^{i,j=1}a_i^*b_j<i|j>=\sum_{n}{i}a_i^*b_i

The Attempt at a Solution

I am having trouble starting this problem. In (a), do I simply normalize the wavefunctions given?

 

[Thaakisfox]

Do you know what "orthonormal" means? What is the condition for two vectors two be orthonormal? Write that down and you will know how to proceed.
Just beware that one of the basis wave functions has a continuous and the other a discrete index.

 

[v_pino]

I found the following for part (a) and (b). How do I apply them to the basis function of \left \{ \psi_i(x),\psi_k(x)\right \}? Are the basis functions that the below equations apply to the things on the left hand side of the equations?

Orthonormality:

-Discrete:
(\psi_i,\psi_j)=\int dx \psi_i^*(x) \psi_j(x)=\delta_{ij}

-Continuous:
(\psi_k,\psi_{k'})=\int_{-\infty}^{\infty} dx \psi_k^*(x) \psi_{k'}(x)=\delta(k-k')

Completeness:

-Discrete:
\sum_{i}\psi^*_i(x')\psi_i(x)=\delta(x-x')

-Continuous
\int_{-\infty}^{\infty}dk \psi^*_k(x)'\psi_k(x)=\delta(x-x')

 

[dextercioby]

The 2 othonormality conditions in post 3 are correct. For completeness, think how the unit operator acts in a (rigged) Hilbert space of functions.

 

[v_pino]

Thank you for the reply but I don't think I fully understand what a rigged Hilbert Space is. Is it just a transformation from x to exp(ikx)?

Can you please also check my attemps at part c and d?

c.) \psi(x)=\sum_{i}^{n}c_i \psi_i + \int \phi(k) \psi_k(x)dk

d.) let \hspace{1pc} f(k)=\sum_{i}^{n} c_i \psi_i(k)
\int \psi_{i'}^*f(k)dk=\int \psi_i^* \sum c_i \psi_i(k)dk=c_{i'}
c_{i'}=(\psi_{i'},\psi)=\int \psi_{i'}^*(x) \psi(x)dx
\psi(x) = \sum_{i}^{n}\int \psi_{i'}^*(x) \psi(x)dx \psi_i + \int \phi(k) \psi_k(x)dk

Is there a way of writing this in delta function form:

\psi(x)= \int \psi(x') \delta(x-x')dx