Question in Principles of Quantum Mechanics by Shankar

[wbphysics]

Hi,
Im studying a book "Principles of Quantum Mechanics" by Shankar.
Im in the fifth chapter, Simple Problems in One dimension.
In page 169 to 170,
the book says about the standard procedure for finding the fate of the incident wave packet.
But i don't understand why we are finding a(E)=<ψ_E/ψ_I> , ψ_E is a normalized eigenfunction of Hamiltonian and ψ_I is an incident wave packet.
More over, in page 170, Step2, Calculating a(E),
it says a part of integral vanishes since, ψ_I in k space is peaked around k=k_0 and is orthogonal to negative momentum states. (k_0=p_0/h bar)
So, please, if you have Shankar's book, help me please~T.T

 

[DrClaude]

But i don't understand why we are finding a(E)=<ψ_E/ψ_I> , ψ_E is a normalized eigenfunction of Hamiltonian and ψ_I is an incident wave packet.

For ${\phi_n}$ forming a complete complete basis, we can write and wave function $\psi$ as
$$
\psi = \sum_n a_n \phi_n
$$
where the $a_n$ are complex coefficients given by
$$
a_n = \langle \phi_n | \psi \rangle
$$
In the present case, things are slightly different because the eigenstates are continuous, instead of discrete, but the approach is the same,
$$
\psi_I = \int a(E) \psi_E \, dE
$$
and, just as in the discrete case above, the function $a(E)$, which plays the role of the coefficients $a_n$ is found using
$$
a(E) = \langle \psi_E | \psi_I \rangle
$$

More over, in page 170, Step2, Calculating a(E),
it says a part of integral vanishes since, ψ_I in k space is peaked around k=k_0 and is orthogonal to negative momentum states. (k_0=p_0/h bar)

$\psi_I$ is an incident wave packet starting far to the left of the potential step and traveling to the right. Therefore, in its representation in momentum space, the wave function is only non-zero for positive momentum (otherwise, there would be a component moving to the left) and is chosen to be peaked at $p_0$, which corresponds to the expectation value of its momentum.