Finite Square Well: Constructing Initial Wave Function with E < 0

genxhis
Messages
37
Reaction score
1
I read through the derivation of bound and scattering states for a finite square well. The logic made sense to me, but I am not entirely sure how to accommodate an arbitrary initial wave function (with mean E < 0). Afterall, there are only a finite number of bound states. My guess was that the initial state might be constructed as a (unique?) mixture of both bound and scattering states. But how could this build a state with energy less than that of the lowest bound state?

Could someone clarify? thx.
 
Physics news on Phys.org
Maybe such particles cannot exist (?)

edit: okay, i think i have it figured out in terms of uncertainty principles.
 
Last edited:


Your guess is correct, in order to construct an initial wave function with an energy less than the lowest bound state, we need to consider a mixture of both bound and scattering states. This is because the bound states represent the discrete energy levels available in the finite square well, while the scattering states represent the continuous energy spectrum. By combining these two types of states, we can create an initial wave function with any energy, including those below the lowest bound state.

To understand this concept better, let's consider the time-independent Schrodinger equation for a particle in a finite square well:

$\frac{d^2\psi}{dx^2} = \frac{2m}{\hbar^2}(V_0 - E)\psi$

where $V_0$ is the depth of the well and $E$ is the energy of the particle. This equation has two types of solutions: bound states and scattering states. Bound states have discrete energy levels given by:

$E_n = \frac{\hbar^2k_n^2}{2m}$

where $k_n = \frac{n\pi}{L}$ and $n$ is a positive integer. These bound states are represented by the wave functions:

$\psi_n(x) = A_n\sin(k_nx)$

where $A_n$ is a normalization constant.

On the other hand, scattering states have a continuous energy spectrum and are represented by the wave functions:

$\psi_k(x) = B_ke^{ikx} + C_ke^{-ikx}$

where $k$ is the wave number and $B_k$ and $C_k$ are normalization constants.

Now, to construct an initial wave function with energy less than the lowest bound state, we can take a linear combination of both bound and scattering states:

$\psi(x) = \sum_{n=1}^{\infty}a_n\psi_n(x) + \int_{0}^{\infty}b_k\psi_k(x)dk$

where $a_n$ and $b_k$ are coefficients that determine the weight of each state in the linear combination. By choosing the appropriate values for these coefficients, we can create an initial wave function with any energy, including those below the lowest bound state.

In summary, the key idea is to combine both bound and scattering states to create an initial wave function with the desired energy. This approach is valid because any wave function can be represented as a
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

Similar threads

Replies
10
Views
2K
Replies
12
Views
3K
Replies
6
Views
2K
Replies
1
Views
2K
Replies
9
Views
698
Replies
1
Views
2K
Replies
2
Views
1K
Back
Top